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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
04ne | In an isosceles triangle $ABC$ with $|AB| = |AC|$, points $M$ and $N$ are the midpoints of the sides $\overline{AB}$ and $\overline{BC}$, respectively. The circle circumscribed to the triangle $AMC$ meets the line $AN$ at point $P$ different from $A$. The line passing through $P$ parallel to the side $BC$ meets the cir... | [
"The points $A$, $M$, $P$ and $C$ lie on the same circle and $\\angle MAP = \\angle PAC$. Therefore, $|MP| = |PC|$ because the corresponding subtended angles are equal. Since $P$ lies on the bisector of $\\overline{BC}$, we conclude that $|BP| = |CP|$. Hence $|MP| = |BP|$, which means that the point $P$ lies on the... | Croatia | Croatia_2018 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry >... | English | proof only | null | |
0cmb | Problem:
A number of 17 workers stand in a row. Every contiguous group of at least 2 workers is a brigade. The chief wants to assign each brigade a leader (which is a member of the brigade) so that each worker's number of assignments is divisible by 4. Prove that the number of such ways to assign the leaders is divisi... | [
"Solution:\n\nAssume that every single worker also forms a brigade (with a unique possible leader). In this modified setting, we are interested in the number $N$ of ways to assign leadership so that each worker's number of assignments is congruent to 1 modulo 4.\n\nConsider the variables $x_{1}, x_{2}, \\ldots, x_{... | Romanian Master of Mathematics (RMM) | Romanian Master of Mathematics Competition | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof only | null | |
03dh | Let $n \ge 2$ be a natural number. The sets $A_1, \dots, A_n$ and $B_1, \dots, B_n$ of natural numbers satisfy the properties:
* $A_i \cap B_j \neq \emptyset$ for all $i, j \in \{1, 2, \dots, n\}$;
* $A_i \cap A_j = \emptyset$ and $B_i \cap B_j = \emptyset$ for all $i \neq j \in \{1, 2, \dots, n\}$.
For each of the s... | [
"We will prove that the required smallest possible value is $n$. Let $A_i$-s are the rows and $B_j$-s are the columns of a square table $n \\times n$ in which the value in row $i$ and column $j$ is $(n-1)i + j$. So all differences for the $A_i$-s are equal to $1$ and all differences for the $B_j$-s are equal to $n$... | Bulgaria | Bulgaria 2022 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n | |
0cjh | a) Prove that, if a ring $(A, +, \cdot)$ has property (P) and $a, b$ are distinct elements of $A$ such that $a$ and $a+b$ are invertible, then $b$ is not invertible, but $1+ab$ is invertible.
b) Give an example of a unitary ring possessing (P).
where property (P) is:
$$
(P) \quad \left\{ \begin{array}{l} \text{the se... | [
"Denote $U(A)$ the set of invertible elements in $A$. For $k \\in \\mathbb{N}$, $k \\ge 2$, and $x \\in A$, define $kx = \\underbrace{x+x+\\cdots+x}_{k \\text{ terms}}$. In particular $k \\cdot 1 = k$. By the given conditions $2 \\in U(A)$. Denote by (1) the equality $x+x^4 = x^2+x^3$ for all $x \\in A$.\n\nChangin... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Ring Theory",
"Algebra > Abstract Algebra > Group Theory"
] | English | proof and answer | a) b is not invertible and 1+ab is invertible. b) Example: Z/3Z × Z/3Z. | |
0axk | Problem:
A vertical pole has a cross section of a regular hexagon with a side length of $1$ foot. If a dog on a leash is tied onto one vertex, and the leash is $3$ feet long, determine the total area that the dog can walk on. | [
"Solution:\nThe total area is indicated by the lightly shaded region below.\n\n\n\n$$\n\\begin{aligned}\n& \\frac{240}{360} \\pi (3)^2 + 2\\left(\\frac{60}{360}\\right) \\pi (2)^2 + 2\\left(\\frac{60}{360}\\right) \\pi (1)^2 \\\\\n= & \\frac{2}{3}(9\\pi) + \\frac{1}{3}(4\\pi) + \\frac{1}{3}... | Philippines | Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 23π/3 | |
06ds | Let $n$ be an integer greater than $1$. In a school there are $n^2 - n + 2$ clubs and each club has exactly $n$ members. Each pair of clubs has exactly one member in common. Show that there is one student belonging to all of the clubs. | [
"Consider an arbitrary club $C_1$. Since it shares a common member with each of the other $n^2 - n + 1$ clubs, by the pigeonhole principle, there is a member in $C_1$ who is also a member of at least\n$$\n\\left\\lfloor \\frac{n^2 - n + 1}{n} \\right\\rfloor = n\n$$\nclubs. Suppose $X$ is a member of the clubs $C_1... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0k24 | Problem:
Consider a finite set of points $T \in \mathbb{R}^n$ contained in the $n$-dimensional unit ball centered at the origin, and let $X$ be the convex hull of $T$. Prove that for all positive integers $k$ and all points $x \in X$, there exist points $t_1, t_2, \ldots, t_k \in T$, not necessarily distinct, such tha... | [
"Solution:\n\nBy the definition of convex hull, we can write $x=\\sum_{i=1}^{m} \\lambda_{i} z_{i}$, where each $z_{i} \\in T$, each $\\lambda_{i} \\geq 0$ and $\\sum_{i=1}^{m} \\lambda_{i}=1$. Consider then a random variable $Z$ that takes on value $z_{i}$ with probability $\\lambda_{i}$. We have $\\mathbb{E}[Z]=x... | United States | HMIC 2018 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Linear Algebra > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
09tt | Problem:
Vind alle positieve gehele getallen $n$ waarvoor er een positief geheel getal $k$ bestaat zodat voor iedere positieve deler $d$ van $n$ geldt dat ook $d-k$ een (niet noodzakelijk positieve) deler van $n$ is. | [
"Solution:\n\nAls $n=1$ of $n$ is een priemgetal, dan zijn de enige positieve delers van $n$ gelijk aan $1$ en $n$ (die samenvallen in het geval $n=1$). Neem nu $k=n+1$, dan moeten $1-(n+1) = -n$ en $n-(n+1) = -1$ ook delers zijn van $n$. Dat klopt precies. Dus $n=1$ en $n$ is priem voldoen met $k=n+1$.\n\nAls $n=4... | Netherlands | IMO-selectietoets II | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | All prime numbers, and 1, 4, and 6. | |
0jpu | Problem:
Compute the number of sequences of integers $\left(a_{1}, \ldots, a_{200}\right)$ such that the following conditions hold.
- $0 \leq a_{1}<a_{2}<\cdots<a_{200} \leq 202$.
- There exists a positive integer $N$ with the following property: for every index $i \in\{1, \ldots, 200\}$ there exists an index $j \in\{... | [
"Solution:\n\nAnswer: $20503$\n\nLet $m := 203$ be an integer not divisible by $3$. We'll show the answer for general such $m$ is $m\\left\\lceil\\frac{m-1}{2}\\right\\rceil$.\n\nLet $x, y, z$ be the three excluded residues. Then $N$ works if and only if $\\{x, y, z\\} \\equiv \\{N-x, N-y, N-z\\}$ $\\pmod{m}$. Sinc... | United States | HMMT February 2015 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 20503 | |
0bu7 | Problem:
a) Demonstraţi că există funcţii neperiodice $f: \mathbb{R} \rightarrow \mathbb{R}$ care verifică egalitatea
$$
f(x+1)+f(x-1)=\sqrt{5} f(x)
$$
pentru orice $x \in \mathbb{R}$
b) Demonstraţi că orice funcţie $g: \mathbb{R} \rightarrow \mathbb{R}$ care verifică egalitatea
$$
g(x+1)+g(x-1)=\sqrt{3} g(x)
$$
pent... | [
"Solution:\n\na) Căutăm soluţii printre funcţiile de forma $f(x)=a^{x}$, unde $a>0$. Obţinem egalitatea $a+a^{-1}=\\sqrt{5}$, de unde $a=\\frac{\\sqrt{5} \\pm 1}{2}$. Se verifică faptul că funcţiile $f_{1}: \\mathbb{R} \\rightarrow \\mathbb{R}, f_{1}(x)=\\left(\\frac{\\sqrt{5}-1}{2}\\right)^{x}$ şi $f_{2}: \\mathbb... | Romania | Olimpiada Naţională de Matematică | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0i6i | Problem:
A sequence is defined by $a_{0}=1$ and $a_{n}=2^{a_{n-1}}$ for $n \geq 1$. What is the last digit (in base 10) of $a_{15}$? | [
"Solution:\n\nCertainly $a_{13} \\geq 2$, so $a_{14}$ is divisible by $2^{2}=4$. Writing $a_{14}=4k$, we have $a_{15}=2^{4k}=16^{k}$. But every power of $16$ ends in $6$, so this is the answer."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic"
] | null | final answer only | 6 | |
0c2t | The product of divisors of a natural number equals the square of that number. Find it, knowing that it is with 10 less than the sum of its divisors. | [] | Romania | Shortlisted problems for the 69th NMO | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 14 | |
02qa | Problem:
Uma calculadora esquisita tem apenas as teclas numéricas de 0 a 9 e duas teclas especiais $A$ e $B$. Quando a tecla $A$ é apertada, o número que aparece no visor é elevado ao quadrado; quando a tecla $B$ é apertada, soma-se 3 ao número que aparece no visor. Nessa calculadora é possível obter 22 a partir do 1 ... | [
"Solution:\n\na) A seguir vemos o que acontece quando começamos com o número 3 no visor e apertamos as teclas na ordem $B B A B$ :\n$$\n3 \\xrightarrow{B} 3+3=6 \\xrightarrow{B} 6+3=9 \\xrightarrow{A} 9^{2}=81 \\xrightarrow{B} 81+3=84\n$$\nLogo, o número que vai aparecer no visor é 84.\n\nb) Uma maneira é apertar a... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | null | proof and answer | a) 84; b) One example: B B A B B; c) Impossible to reach 54 from 2 | |
05y1 | Problem:
Soit $x, y, z$ trois réels positifs, tel que $x \leqslant 1$. Démontrer que:
$$
x y + y + 2 z \geqslant 4 \sqrt{x y z}
$$ | [
"Solution:\n\nNotons que comme $x \\leqslant 1$, $x y + y \\geqslant x y + x y = 2 x y$. En particulier, par inégalité arithmético-géométrique\n$$\nx y + y + 2 z \\geqslant 2 x y + 2 z = 2(x y + z) \\geqslant 4 \\sqrt{x y z}\n$$\nce qui donne l'inégalité voulue."
] | France | Préparation Olympique Française de Mathématiques - ENVOI 5 : Pot-POURRI | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
00ma | Man bestimme alle Paare $(a, b)$ nichtnegativer ganzer Zahlen, die
$$
2017^a = b^6 - 32b + 1
$$
erfüllen. | [
"Antwort: Die zwei Lösungspaare sind $(0, 0)$ und $(0, 2)$.\nWeil $2017^a$ ungerade ist, muss $b$ gerade sein, also ist $b = 2c$, $c$ ganz. Folglich ist $2017^a = 64(c^6 - c) + 1$, also gilt $2017^a \\equiv 1 \\pmod{64}$. Es sind aber $2017 \\equiv 33 \\pmod{64}$ und $2017^2 \\equiv (1+32)^2 = 1+2\\cdot32+32^2 \\eq... | Austria | 48. Österreichische Mathematik-Olympiade Bundeswettbewerb für Fortgeschrittene, Teil 1 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic"
] | German | proof and answer | [(0, 0), (0, 2)] | |
01nw | Given a right-angled triangle $ABC$ with $\angle C = 90^\circ$.
Triangle $AMN$ equal to $ABC$ is constructed on the hypotenuse of $ABC$, $\angle ANM = 90^\circ$, $AN = BC$ (see the fig.). The incircle $\Gamma_1$ of triangle $AMN$ touches the hypotenuse $AM$ at point $P$, and the incircle $\Gam... | [
"Note that $\\angle MAN = 90^\\circ - \\angle BAC = \\angle ABC$. So $MN = AC$,\n\n\nFig. 1\n\n\nFig. 2\n\n$AN = BC$ and, moreover, $AP = BQ$. Let $R$ be the midpoint of $AB$. Then $\\triangle APR = \\triangle BQR$ ($AP = BQ$, $AR = BR$ and $\\angle PAR = \\angle QB... | Belarus | Belorusija 2012 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | English | proof only | null | |
0g40 | Problem:
For an integer $a \geq 2$, denote by $\delta(a)$ the second largest divisor of $a$. Let $(a_{n})_{n \geq 1}$ be a sequence of integers such that $a_{1} \geq 2$ and
$$
a_{n+1}=a_{n}+\delta\left(a_{n}\right)
$$
for all $n \geq 1$. Prove that there exists a positive integer $k$ such that $a_{k}$ is divisible by ... | [
"Solution:\n\nLet $p_{i}=a_{i} / \\delta\\left(a_{i}\\right)$ be the smallest divisor of $a_{i}$ not equal to $1$. We then have\n$$\na_{i+1}=a_{i}+\\frac{a_{i}}{p_{i}}=\\left(p_{i}+1\\right) \\frac{a_{i}}{p_{i}}\n$$\nSo, if $p_{1}$ is odd, then $a_{i+1}$ will be even. Note that the sequence $a_{n}$ fulfills the con... | Switzerland | Final round | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0f4w | Problem:
$N$ is a sum of $n$ powers of $2$. If $N$ is divisible by $2^{m} - 1$, prove that $n \geq m$. Does there exist a number divisible by $11\ldots1$ (that is, $m$ $1$s) which has the sum of its digits less than $m$? | [] | Soviet Union | 16th ASU | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | n ≥ m; No | |
0kqk | Problem:
Herbert rolls $6$ fair standard dice and computes the product of all of his rolls. If the probability that the product is prime can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$, compute $100a + b$. | [
"Solution:\n\nThe only way this can happen is if $5$ of the dice roll $1$ and the last die rolls a prime number $(2, 3, \\text{ or } 5)$. There are $6$ ways to choose the die that rolls the prime, and $3$ ways to choose the prime. Thus, the probability is $\\frac{3 \\cdot 6}{6^{6}} = \\frac{1}{2592}$."
] | United States | HMMT February | [
"Statistics > Probability > Counting Methods > Combinations",
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 2692 | |
0ibz | Problem:
$$
\binom{2003}{1} + \binom{2003}{4} + \binom{2003}{7} + \cdots + \binom{2003}{2002}
$$ | [
"Solution:\n\nLet $\\omega = -1/2 + i \\sqrt{3}/2$ be a complex cube root of unity. Then, by the binomial theorem, we have\n$$\n\\begin{aligned}\n\\omega^{2}(\\omega+1)^{2003} &= \\binom{2003}{0} \\omega^{2} + \\binom{2003}{1} \\omega^{3} + \\binom{2003}{2} \\omega^{4} + \\cdots + \\binom{2003}{2003} \\omega^{2005}... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | final answer only | (2^2003 - 2)/3 | |
003f | Se consideran $n$ números reales $a_1, a_2, ..., a_n$, no necesariamente distintos. Sea $d$ la diferencia entre el mayor y el menor de ellos y sea $s = \sum_{i<j} |a_i - a_j|$.
Demuestre que
$$
(n-1)d \le s \le \frac{n^2 d}{4}
$$
y determine las condiciones que deben cumplir estos $n$ números para que se verifique cada... | [] | Argentina | 21° OLIMPIADA IBEROAMERICANA DE MATEMÁTICA | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | Español | proof and answer | (n−1)d ≤ s ≤ (n^2/4)d.
Equality cases:
- Lower bound s = (n−1)d holds if and only if, after ordering the numbers, all interior elements are equal to a common value. Equivalently, there exist values m ≤ c ≤ M such that exactly one number equals the minimum m, exactly one equals the maximum M, and all remaining are equa... | |
07k9 | Find all prime numbers $p$ and $q$ such that $p$ divides $q + 6$ and $q$ divides $p + 7$. | [
"$p = 2$ and $p$ divides $q + 6$ together imply $2$ divides $q$. Hence $q = 2$. But then $q$ divides $p + 7$ is not satisfied. Thus $p$ is odd.\n$q = 2$ and $p$ divides $q + 6$ together imply $p = 2$ but $p = 2 = q$ contradicts $q$ divides $p + 7$. Thus $p$ and $q$ are both odd.\n$p + 7$ is even and so $q \\le (p +... | Ireland | Irish Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | p = 19, q = 13 | |
070a | Problem:
Find the smallest positive integer $n$ so that a cube with side $n$ can be divided into 1996 cubes each with side a positive integer. | [
"Solution:\nDivide all the cubes into unit cubes. Then the 1996 cubes must each contain at least one unit cube, so the large cube contains at least 1996 unit cubes. But $12^3 = 1728 < 1996 < 2197 = 13^3$, so it is certainly not possible for $n < 13$.\n\nIt can be achieved with 13 by $1 \\cdot 5^3 + 11 \\cdot 2^3 + ... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | 13 | |
081o | Problem:
Qual è il minimo valore dell'espressione $x^{2}-8 x y+19 y^{2}-6 y+14$ al variare di $x$ e $y$ fra i numeri reali? | [
"Solution:\n\nLa risposta è 11. Possiamo scrivere\n$$\n\\begin{aligned}\n& x^{2}-8 x y+19 y^{2}-6 y+14= \\\\\n= & x^{2}-8 x y+16 y^{2}+3 y^{2}-6 y+3+11= \\\\\n= & (x-4 y)^{2}+3(y-1)^{2}+11 \\geq 11 .\n\\end{aligned}\n$$\nIl valore 11 si ottiene per $y=1, x=4$."
] | Italy | Progetto Olimpiadi di Matematica | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 11 | |
0hiu | Problem:
If $a$ and $b$ are two positive numbers not greater than $1$ prove that
$$
\frac{a+b}{1+a b} \leq \frac{1}{1+a}+\frac{1}{1+b}
$$
When does the equality hold? | [
"Solution:\nIf we multiply both sides of the inequality with $(1+a b)(1+a)(1+b)$ the required inequality becomes equivalent to\n$$\n\\begin{gathered}\n(a+b)(1+a)(1+b) \\leq (1+a)(1+a b)+(1+b)(1+a b) \\Leftrightarrow \\\\\n(a+b)(1+a+b+a b) \\leq 1+a b+a+a^{2} b+1+a b+b+a b^{2} \\Leftrightarrow \\\\\na+a^{2}+a b+a^{2... | United States | Berkeley Math Circle | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Equality holds if and only if a = b = 1. | |
0lct | Given a regular 103-sided polygon with 79 vertices are colored red and the remaining vertices are colored blue. Denote $A$ to be the number of pairs of adjacent red vertices and $B$ to be the number of pairs of adjacent blue vertices.
a) Find all possible values of $(A, B)$.
b) Determine the number of pairwise non-si... | [
"a) Clearly, there are 24 blue vertices. If all red vertices form a single block then $A = 78$, if they form two blocks then $A = 77$, and similarly, if they form $k$ blocks then $A = 79 - k$.\n\nBesides, the number of red blocks is equal to the number of blue blocks, hence $B = 24 - k$. This implies that all possi... | Vietnam | VMO | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | a) All pairs (A, B) are (79 − k, 24 − k) for k = 1, 2, …, 24.
b) The number of pairwise non-similar colorings with B = 14 is (C(23, 9) · C(78, 9)) / 10. | |
0iym | $$
x^3(y^2 + z^2)^2 + y^3(z^2 + x^2)^2 + z^3(x^2 + y^2)^2 \ge xyz[xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2].
$$ | [
"We multiply both sides of the desired inequality by $\\sum_{\\text{cyc}} x = (x + y + z)$; that is, we are going to show that\n$$\n\\left(\\sum_{\\text{cyc}} x\\right) \\cdot L = (x + y + z) \\cdot L \\geq (x + y + z) \\cdot R = \\left(\\sum_{\\text{cyc}} x\\right) \\cdot R\n$$\nor\n$$\n\\left(\\sum_{\\text{cyc}} ... | United States | Team Selection Test 2009 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
06ny | Let $ABCD$ be a cyclic quadrilateral. Assume that the points $Q$, $A$, $B$, $P$ are collinear in this order, in such a way that the line $AC$ is tangent to the circle $ADQ$, and the line $BD$ is tangent to the circle $BCP$. Let $M$ and $N$ be the midpoints of $BC$ and $AD$, respectively. Prove that the following three ... | [] | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane ... | null | proof only | null | |
0143 | Problem:
Find all positive integers $n = p_{1} p_{2} \cdots p_{k}$ which divide $\left(p_{1}+1\right)\left(p_{2}+1\right) \cdots\left(p_{k}+1\right)$, where $p_{1} p_{2} \cdots p_{k}$ is the factorization of $n$ into prime factors (not necessarily distinct). | [
"Solution:\n\nLet $m = \\left(p_{1}+1\\right)\\left(p_{2}+1\\right) \\cdots\\left(p_{k}+1\\right)$. We may assume that $p_{k}$ is the largest prime factor. If $p_{k} > 3$ then $p_{k}$ cannot divide $m$, because if $p_{k}$ divides $m$ it is a prime factor of $p_{i}+1$ for some $i$, but if $p_{i} = 2$ then $p_{i}+1 <... | Baltic Way | Baltic Way 2005 | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | All n of the form 2^r 3^s with nonnegative integers r,s satisfying s ≤ r ≤ 2s (including n = 1 when r = s = 0). | |
0ktj | Problem:
Let $A_{1} B_{1} C_{1}$, $A_{2} B_{2} C_{2}$, and $A_{3} B_{3} C_{3}$ be three triangles in the plane. For $1 \leq i \leq 3$, let $D_{i}$, $E_{i}$, and $F_{i}$ be the midpoints of $B_{i} C_{i}$, $A_{i} C_{i}$, and $A_{i} B_{i}$, respectively. Furthermore, for $1 \leq i \leq 3$ let $G_{i}$ be the centroid of $... | [
"Solution:\n\nLet $P_{i}(x, y, z)$ be the point with barycentric coordinates $(x, y, z)$ in triangle $A_{i} B_{i} C_{i}$. Note that since this is linear in $x, y$, and $z$, the signed area of triangle $P_{1}(x, y, z) P_{2}(x, y, z) P_{3}(x, y, z)$ is a homogenous quadratic polynomial in $x, y$, and $z$; call it $f(... | United States | HMMT February 2022 | [
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimizat... | null | proof and answer | 917 | |
08ik | Problem:
A number $A$ is written with $2n$ digits, each of which is $4$, and a number $B$ is written with $n$ digits, each of which is $8$. Prove that for each $n$, $A + 2B + 4$ is a perfect square.
Problem:
Un număr $A$ este scris cu $2n$ cifre, fiecare dintre acestea fiind $4$; un număr $B$ este scris cu $n$ cifre, ... | [
"Solution:\n\n$$\n\\begin{aligned}\nA &= \\underbrace{44 \\ldots 44}_{2n} = \\underbrace{44 \\ldots 4}_{n} \\underbrace{44 \\ldots 4}_{n} = \\underbrace{44 \\ldots 4}_{n} \\underbrace{400 \\ldots 0}_{n} - \\underbrace{44 \\ldots 4}_{n} + \\underbrace{88 \\ldots 8}_{n} = \\underbrace{44 \\ldots 4}_{n} \\cdot (10^n -... | JBMO | 7th JBMO | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof only | null | |
06x9 | Let $ABC$ and $A'B'C'$ be two triangles having the same circumcircle $\omega$, and the same orthocentre $H$. Let $\Omega$ be the circumcircle of the triangle determined by the lines $AA'$, $BB'$, and $CC'$. Prove that $H$, the centre of $\omega$, and the centre of $\Omega$ are collinear.
(Denmark) | [
"In what follows, $\\Varangle(p, q)$ will denote the directed angle between lines $p$ and $q$, taken modulo $180^\\circ$. Denote by $O$ the centre of $\\omega$. In any triangle, the homothety with ratio $-\\frac{1}{2}$ centred at the centroid of the triangle takes the vertices to the midpoints of the opposite sides... | IMO | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane ... | English | proof only | null | |
098a | Problem:
Fie $m$ și $n$ două numere întregi, astfel încât $\frac{n^{3}}{m+n}$ este un număr întreg. Arătați că $\frac{2023 \cdot m^{4}}{m+n}$ este un număr întreg. | [
"Solution:\n\nConsiderăm numărul $\\frac{m^{3}}{m+n}$.\nCalculăm suma $\\frac{m^{3}}{m+n}+\\frac{n^{3}}{m+n}=\\frac{(m+n)\\left(m^{2}-m n+n^{2}\\right)}{m+n}=m^{2}-m n+n^{2}$.\nObservăm că $m^{2}-m n+n^{2} \\in \\mathbb{Z}$, știm că $\\frac{n^{3}}{m+n} \\in \\mathbb{Z}$, atunci $\\frac{m^{3}}{m+n} \\in \\mathbb{Z}$... | Moldova | Olimpiada Republicană la Matematică | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
08mb | Problem:
Determine all prime numbers $p_{1}, p_{2}, \ldots, p_{12}, p_{13}$, $p_{1} \leq p_{2} \leq \ldots \leq p_{12} \leq p_{13}$, such that
$$
p_{1}^{2}+p_{2}^{2}+\ldots+p_{12}^{2}=p_{13}^{2}
$$
and one of them is equal to $2 p_{1}+p_{9}$. | [
"Solution:\nObviously, $p_{13} \\neq 2$, because sum of squares of 12 prime numbers is greater or equal to $12 \\times 2^{2}=48$. Thus, $p_{13}$ is odd number and $p_{13} \\geq 7$.\nWe have that $n^{2} \\equiv 1 \\pmod{8}$, when $n$ is odd. Let $k$ be the number of prime numbers equal to $2$. Looking at equation mo... | JBMO | 2009 Shortlist JBMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (2,2,2,2,2,2,2,3,3,5,7,7,13); (2,2,2,2,2,2,2,3,3,5,7,13,17); (2,2,2,2,2,2,2,3,3,5,7,29,31) | |
0fe2 | Problem:
Se consideran las funciones reales de variable real $f(x)$ de la forma: $f(x)=a x+b$, siendo $a$ y $b$ números reales.
¿Para qué valores de $a$ y $b$ se verifica $f^{2000}(x)=x$ para todo número real $x$?
[Nota: Se define $f^{2}(x)=f(f(x))$, $f^{3}(x)=f(f(f(x)))$, y en general, $f^{n}(x)=f(f^{n-1}(x))=f(f(\ldo... | [
"Solution:\nEn primer lugar, observemos que si componemos dos funciones (lineales) del tipo $a x+b$, obtenemos una función de este tipo, cuyo coeficiente en la variable $x$ es el producto de los respectivos coeficientes de las dos funciones.\nPor lo tanto si $f(x)=a x+b$, entonces $f^{2000}(x)$ es una función del t... | Spain | null | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | Either a = 1 and b = 0, or a = −1 and b is any real number. | |
01ln | Non-zero real numbers $a$, $b$, $c$ satisfy the equality
$$
\frac{ab}{b-c} + \frac{bc}{c-a} + \frac{ca}{a-b} = \frac{ab}{b+c} + \frac{bc}{c+a} + \frac{ca}{a+b} + 6abc.
$$ | [] | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Intermediate Algebra > Other"
] | null | proof only | null | |
0ize | Problem:
Justine has a coin which will come up the same as the last flip $\frac{2}{3}$ of the time and the other side $\frac{1}{3}$ of the time. She flips it and it comes up heads. She then flips it 2010 more times. What is the probability that the last flip is heads? | [
"Solution:\n\nLet the \"value\" of a flip be 1 if the flip is different from the previous flip and let it be 0 if the flip is the same as the previous flip. The last flip will be heads if the sum of the values of all 2010 flips is even. The probability that this will happen is\n$$\n\\sum_{i=0}^{1005} \\binom{2010}{... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | (3^2010 + 1) / (2 * 3^2010) | |
04yf | Let $\alpha \neq 0$ be a real number. Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$$f(x^2 + y^2) = f(x - y)f(x + y) + \alpha y f(y)$$
holds for all $x, y \in \mathbb{R}$. | [
"Answer: For every $\\alpha \\neq 0$, the zero function and the function with value 1 at 0, but 0 elsewhere are solutions. For $\\alpha = 2$, the identity function $x \\mapsto x$ is another solution.\n\n*Solution-check.* The linear function clearly works. Consider the function $f$ such that $f(0) = 1$ and $f(x) = 0... | Czech-Polish-Slovak Mathematical Match | CAPS Match 2024 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | For any nonzero parameter: the zero function and the function that equals one at zero and zero elsewhere. Additionally, when the parameter equals two: the identity function. | |
06j8 | Let $a_n = \underbrace{677\cdots7}_{n}$. Is it possible to find infinitely many multiples of $a_{2014}$ in the sequence $\{a_n\}$? | [
"Yes. By the pigeonhole principle, two of the numbers of the form $11\\cdots1$ leave the same remainder when divided by $a_{2014}$. Their difference, which is of the form $11\\cdots100\\cdots0$, is a multiple of $a_{2014}$. Since $(a_{2014}, 10) = 1$, we know that $a_{2014}$ divides $\\underbrace{11\\cdots1}_{m \\t... | Hong Kong | 1997-2023 IMO HK TST | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | Yes | |
01ku | Prove that
$$
\frac{a^3(a+c)(a+b)}{(a-c)(a-b)} + \frac{b^3(b+a)(b+c)}{(b-a)(b-c)} + \frac{c^3(c+b)(c+a)}{(c-b)(c-a)} = abc
$$
for all admissible $a$, $b$, $c$ such that $a+b+c=0$. | [
"In view of $a+b+c=0$, we rewrite the left hand side of the required inequality in the form\n$$\n\\begin{aligned}\nM &= \\frac{a^3(a+c)(a+b)}{(a-c)(a-b)} + \\frac{b^3(b+a)(b+c)}{(b-a)(b-c)} + \\frac{c^3(c+b)(c+a)}{(c-b)(c-a)} = \\\\\n&= \\frac{a^3(-b)(-c)}{(a-c)(a-b)} + \\frac{b^3(-c)(-a)}{(b-a)(b-c)} + \\frac{c^3(... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange"
] | English | proof only | null | |
0l83 | There are $8! = 40320$ eight-digit positive integers that use each of the digits $1, 2, 3, 4, 5, 6, 7, 8$ exactly once. Let $N$ be the number of these integers that are divisible by $22$. Find the difference between $N$ and $2025$. | [] | United States | 2025 AIME I | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 279 | |
01le | Given a triangle $ABC$ with $\angle ABC = 120^\circ$, $BC = 2AB$. Find the angle between its medians $AM$ and $BK$.
(S. Mazanik) | [
"Let $P$ be the point of intersection of the medians $AM$ and $BK$. Draw the line $l$ through $A$ parallel to $BC$. Let $L$ be the point of intersection of $l$ and the line $BK$. The triangles $AKL$ and $BKC$ are equal ($AK = KC$, $\\angle AKL = \\angle BKC$, $\\angle KAL = \\angle BCK$), so $AL = BC = 2AB$. Since ... | Belarus | 60th Belarusian 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 > Miscellaneous > Constructions and loci"
] | English | proof and answer | 60° | |
0bso | Prove that, for every positive integer $n$, there exists a unique $c_n \in (0, 1)$ such that
$$
\int_0^1 \frac{1}{1+x^n} dx = \frac{1}{1+(c_n)^n},
$$
and evaluate the limit $\lim_{n \to \infty} n(c_n)^n$. | [
"Existence of $c_n$ follows from the (first) mean-value theorem (each integrand is continuous), and uniqueness follows from injectivity of each integrand.\n\nTo evaluate the required limit, let $I_n = \\int_0^1 (1+x^n)^{-1} dx$, and notice that $|I_n - 1| = \\left| \\int_0^1 \\frac{-x^n}{1+x^n} dx \\right| \\le \\i... | Romania | 67th Romanian Mathematical Olympiad | [
"Calculus > Integral Calculus > Applications",
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Limits"
] | English | proof and answer | log 2 | |
00w2 | Problem:
Integers $1, 2, \ldots, n$ are written (in some order) on the circumference of a circle. What is the smallest possible sum of moduli of the differences of neighbouring numbers? | [
"Solution:\n\nLet $a_{1} = 1, a_{2}, \\ldots, a_{k} = n, a_{k+1}, \\ldots, a_{n}$ be the order in which the numbers $1, 2, \\ldots, n$ are written around the circle. Then the sum of moduli of the differences of neighbouring numbers is\n$$\n\\begin{aligned}\n& \\left|1 - a_{2}\\right| + \\left|a_{2} - a_{3}\\right| ... | Baltic Way | Baltic Way | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2n - 2 | |
09jt | Consider an $8 \times 8$ chessboard and a game in which a knight can move from any corner of a $3 \times 4$ rectangle to the diagonally opposite corner (as shown in the figure). What is the minimum number of moves required for a knight to travel from the lower left corner of the chessboard to the upper right corner?
![... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 4 | |
011f | Problem:
Let $t \geqslant \frac{1}{2}$ be a real number and $n$ a positive integer. Prove that
$$
t^{2 n} \geqslant (t-1)^{2 n} + (2 t-1)^{n}
$$ | [
"Solution:\n\nUse induction. For $n=1$ the inequality reads $t^{2} \\geqslant (t-1)^{2} + (2 t-1)$ which is obviously true. To prove the induction step it suffices to show that\n$$\nt^{2}(t-1)^{2 n} + t^{2}(2 t-1)^{n} \\geqslant (t-1)^{2 n+2} + (2 t-1)^{n+1}\n$$\nThis easily follows from $t^{2} \\geqslant (t-1)^{2}... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0aak | Determine all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that for any $x, y > 0$ it holds that
$$
f(xy + f(x)) = y f(x) + x.
$$ | [] | North Macedonia | Fourth Memorial Mathematical Contest | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x for all positive real x | |
0jh1 | Problem:
Consider a rectangular array of single digits $d_{i, j}$ with 10 rows and 7 columns, such that $d_{i+1, j}-d_{i, j}$ is always 1 or -9 for all $1 \leq i \leq 9$ and all $1 \leq j \leq 7$, as in the example below. For $1 \leq i \leq 10$, let $m_{i}$ be the median of $d_{i, 1}, \ldots, d_{i, 7}$. Determine the ... | [
"Solution:\n\nNote that rearranging the columns does not change the medians, hence we may sort the first row, so that $d_{1,1} \\leq d_{1,2} \\leq \\ldots \\leq d_{1,7}$. The calculations are much simplified if we subtract $i-1$ from each row. In other words, we put $D_{i, j}=d_{i, j}-(i-1)$. This subtracts $i-1$ f... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | least = greatest = 9/2 | |
0fw7 | Problem:
Wie viele siebenstellige Zahlen gibt es, für die das Produkt der Ziffern gleich $45^{3}$ ist? | [
"Solution:\n\nWir untersuchen zuerst, welche Ziffern wie oft vorkommen können. Die Primfaktorzerlegung von $45^{3}$ ist $3^{6} \\cdot 5^{3}$. Aus dieser ist ersichtlich, dass $1, 3, 5$ und $9$ die einzigen natürlichen Zahlen sind, die kleiner als $10$ sind und $45^{3}$ teilen. Es können also nur diese vier Ziffern ... | Switzerland | Vorrundenprüfung | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 350 | |
06b0 | Determine all values of the integer $v$ for which the number $v^2 + 10v + 160$ is a perfect square. | [
"Let $v^2 + 10v + 160 = \\kappa^2$, where $\\kappa$ is a positive integer. Then we have:\n$$\n\\begin{align*}\nv^2 + 10v + 160 = \\kappa^2 &\\Leftrightarrow (v+5)^2 + 135 = \\kappa^2 \\Leftrightarrow \\kappa^2 - (v+5)^2 = 135 \\\\\n&\\Leftrightarrow (\\kappa - |v+5|)(\\kappa + |v+5|) = 135 = 1 \\cdot 3^3 \\cdot 5, ... | Greece | Selection Examination | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | [-72, -26, -16, -8, -2, 6, 16, 62] | |
0ag2 | Let $O$ be the center of the incircle of triangle $ABC$. The points $K$ and $L$ are the intersection points of the circumcircles of triangles $BOC$ and $AOC$, respectively with the bisectors of the angles at $A$ and $B$, $P$ is the middle point of $\overline{KL}$, $M$ is symmetrical to $O$ with respect to $P$ and $N$ i... | [
"The angles $LCA$ and $LOA$ are equal as inscribed angles upon the same arc. The angle $LOA$ is equal to the sum of the angles $OAB$ and $OBA$, as an external angle to the triangle $ABO$, therefore:\n$$\n\\begin{aligned}\n\\angle LCO &= \\angle LCA + \\angle OCA = \\angle LOA + \\angle OCA = \\angle OAB + \\angle O... | North Macedonia | null | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0kru | Problem:
An apartment building consists of 20 rooms numbered $1, 2, \ldots, 20$ arranged clockwise in a circle. To move from one room to another, one can either walk to the next room clockwise (i.e. from room $i$ to room $(i+1) \bmod 20$) or walk across the center to the opposite room (i.e. from room $i$ to room $(i+1... | [
"Solution:\n\nOne way is to walk directly from room 10 to 20. Else, divide the rooms into 10 pairs $A_{0} = (10, 20), A_{1} = (1, 11), A_{2} = (2, 12), \\ldots, A_{9} = (9, 19)$. Notice that\n- each move is either between rooms in $A_{i}$ and $A_{(i+1) \\bmod 10}$ for some $i \\in \\{0, 1, \\ldots, 9\\}$, or betwee... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | final answer only | 257 | |
00ru | Let $M = \{(a,b,c) \in \mathbb{R}^3: 0 < a, b, c < \frac{1}{2} \text{ with } a+b+c=1\}$ and $f: M \to \mathbb{R}$ given as
$$
f(a, b, c) = 4\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) - \frac{1}{abc}
$$
Find the best (real) bounds $\alpha$ and $\beta$ such that
$$
f(M) = \{f(a, b, c) : (a, b, c) \in M\} \subse... | [
"Let $\\forall (a,b,c) \\in M$, $\\alpha \\le f(a,b,c) \\le \\beta$ and suppose that there are no better bounds, i.e. $\\alpha$ is the largest possible and $\\beta$ is the smallest possible. Now,\n$$\n\\begin{align*}\n\\alpha \\le f(a, b, c) \\le \\beta &\\Leftrightarrow \\alpha abc \\le 4(ab + bc + ca) - 1 \\le \\... | Balkan Mathematical Olympiad | BMO 2017 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | proof and answer | alpha = 8 (not achievable), beta = 9 (achievable) | |
0c28 | Let $n$ be a natural number with $n \ge 2$. For any real numbers $a_1, a_2, \dots, a_n$, we denote $S_0 = 1$ and define $S_k$ by
$$
S_k = \sum_{1 \le i_1 < i_2 < \dots < i_k \le n} a_{i_1} a_{i_2} \dots a_{i_k},
$$
that is, the sum of all products of any $k$ numbers chosen among $a_1, a_2, \dots, a_n$, $k \in \{1, 2, \... | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 2^{n-1} | |
0iqb | Problem:
Find the smallest positive integer $n$ such that $107 n$ has the same last two digits as $n$. | [
"Solution:\nThe two numbers have the same last two digits if and only if $100$ divides their difference $106 n$, which happens if and only if $50$ divides $n$.\n\nAnswer: $50$"
] | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 50 | |
099g | $n$ is a positive integer and relatively prime with $6$. $a_1, \dots, a_n, b_1, \dots, b_n$ are positive integers such that $a_1 < a_2 < \dots < a_n$ and $b_1 < b_2 < \dots < b_n$. If for arbitrary natural number $t$ such that $a_i + a_j + a_k = t$ ($i < j < k$) triple's number is equal to $b_i + b_j + b_k = t$ ($i < j... | [
"Consider the following polynomials:\n$$\na(x) = x^{a_1} + x^{a_2} + \\dots + x^{a_n}\n$$\n$$\nb(x) = x^{b_1} + x^{b_2} + \\dots + x^{b_n}\n$$\nNow assume the contrary, in other words $a(x) \\neq b(x)$.\n$$\n\\begin{aligned}\n[a(x)]^3 &= \\left(\\sum_{i=1}^{n} x^{a_i}\\right) \\cdot \\left(\\sum_{i \\neq j} x^{a_i+... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
087y | Problem:
Quanto vale la somma delle seste potenze delle soluzioni dell'equazione $x^{6}-16 x^{4}+16 x^{2}-1=0$?
(A) 6375
(B) 6482
(C) 6515
(D) 6660
(E) 6662 | [
"Solution:\n\nLa risposta è $\\mathbf{(E)}$. Si ha $x^{6}-16 x^{4}+16 x^{2}-1=\\left(x^{2}-1\\right)\\left(x^{4}-15 x^{2}+1\\right)$ (tutte le radici sono reali), se dunque $z$ è una radice diversa da $\\pm 1$, si ha $z^{4}=15 z^{2}-1$, $z^{6}=15 z^{4}-z^{2}=224 z^{2}-15$. Dette $z_{1}, \\ldots, z_{4}$ queste ultim... | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | E | |
04wn | We distribute $n \ge 1$ labelled balls among nine persons $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, $I$. Determine in how many ways it is possible to distribute the balls under the condition that $A$ gets the same number of balls as the persons $B$, $C$, $D$ and $E$ together. | [
"Consider the polynomial\n$$\n(x+2)^{2n} = (x^2 + 4x + 4)^n = \\\\ = (x^2 + x + x + x + x + 1 + 1 + 1 + 1)(x^2 + x + x + x + x + 1 + 1 + 1 + 1) \\dots \\\\ (x^2 + x + x + x + x + 1 + 1 + 1 + 1)\n$$\nand suppose that we multiply out the brackets, obtaining $9^n$ summands. We show the one-to-one correspondence betwee... | Czech-Polish-Slovak Mathematical Match | Czech-Polish-Slovak Match | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | binom(2n, n) * 2^n | |
0bci | Problem:
Determinaţi toate funcţiile $f: \mathbb{R} \rightarrow \mathbb{R}$ pentru care $f(y f(x+y)+f(x))=4 x+2 y f(x+y)$ pentru toţi $x, y \in \mathbb{R}$. | [] | Romania | Olimpiada europeana de matematica a fetelor | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 2x for all real x | |
0drh | Find all functions $f: \mathbb{R} \to \mathbb{R}$, where $\mathbb{R}$ is the set of real numbers, such that
$$
f(x)f(yf(x)-1) = x^2f(y) - f(x) \quad \text{for all } x, y \in \mathbb{R}.
$$ | [
"The constant function $f(x) = 0$ is a solution.\nLet $f$ be a solution that is not identically $0$. We shall show that $f(x) = x$ for all $x$. Letting $x = 0$ in the given equation, we get\n$$\nf(0)[f(yf(0) - 1) + 1] = 0.\n$$\nSuppose $f(0) \\neq 0$. Let $x = yf(0) - 1$. As $y$ ranges over all real numbers, so doe... | Singapore | Singapur 2015 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 0 for all real x; f(x) = x for all real x | |
09h8 | For a positive integer $n$, let $\sigma(n)$ be the sum of all divisors of $n$. Show that there exist infinitely many positive integers $n$ such that $n$ divides $2^{\sigma(n)}-1$. | [
"Let $k$ be a positive integer. We choose a prime divisor $p_j$ of $2^{2^j} + 1$ for each $0 \\le j < k$. Then the product $n_k = p_0p_1\\dots p_{k-1}$ is a divisor of $2^{2^k} - 1 = \\prod_{j=0}^{k-1}(2^{2^j} + 1)$.\n\nSince $n_k$ is odd, $\\sigma(n_k) = (p_0+1)\\dots(p_{k-1}+1)$ is divisible by $2^k$. Hence $2^{2... | Mongolia | Mongolian National Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
04z5 | Let $a$ and $b$ be the lengths of the legs of a given right triangle. Prove that angle $\varphi$, where $0 < \varphi < 90^\circ$, is an acute angle of this triangle if and only if $(a \cos \varphi + b \sin \varphi)(a \sin \varphi + b \cos \varphi) = 2ab$. (Seniors.) | [
"The equality given in the problem is equivalent to\n$$\n(a^2 + b^2) \\sin \\varphi \\cos \\varphi + ab(\\sin^2 \\varphi + \\cos^2 \\varphi) = 2ab\n$$\nand hence also to\n$$\n(a^2 + b^2) \\sin \\varphi \\cos \\varphi = ab. \\qquad (1)\n$$\n\nLet $\\alpha$ and $\\beta$ be the angles opposite to legs with length $a$ ... | Estonia | Estonija 2010 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0c19 | Let $a, b, c, d$ be natural numbers such that $a + b + c + d = 2018$. Find the minimum value of the expression:
$$
E = (a - b)^2 + 2(a - c)^2 + 3(a - d)^2 + 4(b - c)^2 + 5(b - d)^2 + 6(c - d)^2.
$$ | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Linear Algebra > Vectors",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 14 | |
0hx4 | Problem:
Let $a > 2$ be given, and define a sequence $a_{0}, a_{1}, a_{2}, \ldots$ by
$$
a_{0} = 1, \quad a_{1} = a, \quad a_{n+1} = \left(\frac{a_{n}^{2}}{a_{n-1}^{2}} - 2\right) \cdot a_{n} .
$$
Show that for all integers $k \geq 0$, we have
$$
\sum_{i=0}^{k} \frac{1}{a_{i}} < \frac{a+2-\sqrt{a^{2}-4}}{2} .
$$ | [
"Solution:\nWrite the recursion as\n$$\n\\frac{a_{n+1}}{a_{n}} = \\left(\\frac{a_{n}}{a_{n-1}}\\right)^{2} - 2\n$$\nLet $b_{n} = a_{n+1} / a_{n}$; then $b_{n} = b_{n-1}^{2} - 2$ and $b_{0} = a$. Let\n$$\nt = \\frac{a + \\sqrt{a^{2} - 4}}{2}\n$$\nThen $a = t + 1 / t$ and it is a simple induction to show that $b_{n} ... | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials"
] | null | proof only | null | |
0199 | Determine all pairs $(p, q)$ of primes for which both $p^2 + q^3$ and $q^2 + p^3$ are perfect squares. | [
"**Answer.** There is only one such pair, namely $(p, q) = (3, 3)$.\n\n**Proof.** Let the pair $(p, q)$ be as described in the statement of the problem.\n\n1.) First we show that $p \\neq 2$. Otherwise, there would exist a prime $q$ for which $q^2 + 8$ and $q^3 + 4$ are perfect squares. Because of $q^2 < q^2 + 8$, ... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (3, 3) | |
028j | Problem:
Seja $\triangle ABC$ um triângulo tal que a altura relativa ao lado $BC$ não é menor do que o lado $BC$ e a altura relativa ao lado $AB$ não é menor do que o lado $AB$. Determine as medidas dos ângulos deste triângulo. | [
"Solution:\n\nSejam $h_{a}$ e $h_{c}$ as alturas relativas aos lados $BC = a$ e $AB = c$, respectivamente. Por hipótese temos que $h_{a} \\geq a$ e $h_{c} \\geq c$. Como $h_{a}$ e $h_{c}$ são os comprimentos das alturas, então $h_{a} \\leq c$ e $h_{c} \\leq a$.\n\nUm dos lados considerados é maior ou igual ao outro... | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof and answer | 45°, 45°, 90° | |
0azu | Problem:
Find the smallest positive real numbers $x$ and $y$ such that $x^{2} - 3x + 2.5 = \sin y - 0.75$. | [
"Solution:\n\nThe equation above can be rewritten as\n$$\n\\left(x - \\frac{3}{2}\\right)^{2} + 1 = \\sin y.\n$$\nSince the left side is always greater than or equal to $1$ and the right side is always less than or equal to $1$, then both sides must be equal to $1$. This means $x = \\frac{3}{2}$ and $y = (4k + 1) \... | Philippines | 20th Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | x = 3/2, y = pi/2 | |
0kez | Problem:
An infinite castle has rooms labeled $1, 2, 3, \ldots$ If room $n$ is on the same hall as rooms $2n+1$ and $3n+1$ for every $n$, what is the maximum possible number of different halls on the castle? | [
"Solution:\n\nWe claim that all rooms must be on the same hall. Suppose for contradiction there is a room other than room $1$ that is the first room on its hall. We will show by induction that its room number must be congruent to $2^{2k-1} \\pmod{3 \\cdot 2^{2k-1}}$ for every positive integer $k$, and therefore mus... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 1 | |
07ul | Show that for all $x, y \in \mathbb{R}$:
$$
x^4 + y^4 + 18 \ge 12xy.
$$
When do we get equality? | [
"If $x, y \\ge 0$, we can use a 4-term AM-GM to obtain\n$$\nx^4 + y^4 + 18 = x^4 + y^4 + 9 + 9 \\geq 4 (x^4 \\cdot y^4 \\cdot 9 \\cdot 9)^{1/4} = 12xy.\n$$\nWe get equality if and only if all four terms are equal, i.e. if and only if $x = y = \\sqrt{3}$.\nIf one of the two numbers $x, y$ is non-positive and the oth... | Ireland | IRL_ABooklet | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Equality holds if and only if x = y = √3 or x = y = −√3. | |
0jiw | Problem:
Let $ABC$ be a triangle with $CA = CB = 5$ and $AB = 8$. A circle $\omega$ is drawn such that the interior of triangle $ABC$ is completely contained in the interior of $\omega$. Find the smallest possible area of $\omega$. | [
"Solution:\n$16\\pi$\n\nWe need to contain the interior of $\\overline{AB}$, so the diameter is at least $8$. This bound is sharp because the circle with diameter $\\overline{AB}$ contains all of $ABC$. Hence the minimal area is $16\\pi$."
] | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 16π | |
092g | Problem:
Let $ABC$ be an acute triangle with $AB > AC$. Prove that there exists a point $D$ with the following property: whenever two distinct points $X$ and $Y$ lie in the interior of $ABC$ such that the points $B$, $C$, $X$, and $Y$ lie on a circle and
$$
\angle AXB - \angle ACB = \angle CYA - \angle CBA
$$
holds, th... | [
"Solution:\nLet $D$ be the point on $BC$ for which $AD$ is a tangent to the circumcircle of $ABC$. As we will show in the sequel, the point $D$ is as desired.\n\n\n\nSolution 1. We assume $B$, $C$, $X$, $Y$ lie on a circle in that order, the other case being similar. We compute the followin... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians"
] | null | proof only | null | |
09nz | Let $a_0 = 2^{2025}$. Two players take turns extending a sequence $a_0, a_1, a_2, \dots$ by the following rules:
On the $n$-th move, the player chooses $a_n = a_{n-1} + 1$ or $a_n = S(a_{n-1})$, where $S(x)$ denotes the sum of the digits of $x$.
The game ends when either three identical numbers appear in the sequence, ... | [
"Answer: Player 1 has a winning strategy for all initial values, including $a_0 = 2^{2025}$.\nLet us consider small values of $a_0$.\n\n**Case 1:** $a_0 \\le 8$. To avoid repeating a number three times, each player must choose $a_n = a_{n-1} + 1$. Then player 1 can increment three times in a row:\n$$\na_0, a_0 + 1,... | Mongolia | MMO2025 Round 4 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Player 1 | |
059u | Find all pairs of integers $(x, y)$ that satisfy the equation $y^4 = x(2x^2 + y)^3$. | [
"*Answer:* $(\\frac{z^3(z-1)}{2}, \\frac{z^6(z-1)}{2})$ where $z$ is arbitrary integer.\n\nSolution:\nIf $x = 0$ then $y = 0$. We now assume that $x \\neq 0$.\nLet $d = \\gcd(x, y) > 0$ and $x = da, y = db$. Dividing the sides of the equation by $d^4$, we get $b^4 = a(2a^2d + b)^3$. Thus $a \\mid b^4$. Since $a$ an... | Estonia | Estonian Math Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | (x, y) = (z^3(z-1)/2, z^6(z-1)/2) for all integers z | |
0j9v | Problem:
Given that $P$ is a real polynomial of degree at most $2012$ such that $P(n) = 2^{n}$ for $n = 1, 2, \ldots, 2012$, what choice(s) of $P(0)$ produce the minimal possible value of $P(0)^{2} + P(2013)^{2}$? | [
"Solution:\nAnswer: $1 - 2^{2012}$\n\nDefine $\\Delta^{1}(n) = P(n+1) - P(n)$ and $\\Delta^{i}(n) = \\Delta^{i-1}(n+1) - \\Delta^{i-1}(n)$ for $i > 1$. Since $P(n)$ has degree at most $2012$, we know that $\\Delta^{2012}(n)$ is constant. Computing, we obtain $\\Delta^{1}(0) = 2 - P(0)$ and $\\Delta^{i}(0) = 2^{i-1}... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 1 - 2^{2012} | |
0e9g | Problem:
Naj bosta $x_{1}$ in $x_{2}$ različni ničli polinoma $p(x)=x^{2}+a x+b$, $x_{1}^{2}-\frac{1}{2}$ in $x_{2}^{2}-\frac{1}{2}$ pa naj bosta ničli polinoma $q(x)=x^{2}+(a^{2}-\frac{1}{2}) x+b^{2}-\frac{1}{2}$. Določi $a$ in $b$. | [
"Solution:\n\nNičli polinoma $p(x)=x^{2}+a x+b$ sta $x_{1}=\\frac{-a+\\sqrt{a^{2}-4 b}}{2}$ in $x_{2}=\\frac{-a-\\sqrt{a^{2}-4 b}}{2}$. Ker morata biti različni, je $a^{2}-4 b \\neq 0$. Od tod izračunamo\n$$\nx_{1}^{2}-\\frac{1}{2}=\\frac{\\left(a^{2}-2 b-1\\right)-a \\sqrt{a^{2}-4 b}}{2}\n$$\nin\n$$\nx_{2}^{2}-\\f... | Slovenia | 58. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | a = 0, b = -3/4 | |
0359 | Problem:
Find all values of the real parameter $a$ such that the image of the function
$$
f(x)=\frac{\sin^{2} x-a}{\sin^{3} x-\left(a^{2}+2\right) \sin x+2}
$$
contains the interval $\left[\frac{1}{2}, 2\right]$. | [
"Solution:\nSet $t=\\sin x$ and $g(t)=\\frac{t^{2}-a}{t^{3}-\\left(a^{2}+2\\right) t+2}$.\nIf the numerator and the denominator of $g$ have a common root then $a \\geq 0$ and $t= \\pm \\sqrt{a}$. If $t=-\\sqrt{a}$ we obtain $\\sqrt{a}\\left(a^{2}-a+2\\right)=-2$, which is impossible since $a^{2}-a+2>0$ for every $a... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | null | proof and answer | (-∞, -1/2] ∪ (1, ∞) | |
0aon | Problem:
How many ways can you place $10$ identical balls in $3$ baskets of different colors if it is possible for a basket to be empty? | [
"Solution:\n\nThis is a problem of distributing $10$ identical balls into $3$ distinct baskets, where baskets may be empty.\n\nThe number of ways is given by the stars and bars formula:\n$$\n\\binom{10 + 3 - 1}{3 - 1} = \\binom{12}{2} = 66.\n$$\n\nSo, there are $66$ ways."
] | Philippines | 18th PMO Area Stage | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 66 | |
08xn | Let $\Gamma$ be the circumcircle of a triangle $ABC$, and $\ell$ be the line tangent to $\Gamma$ at point $A$. Let $D, E$ be interior points of the sides $AB, AC$, respectively, satisfying the condition $BD : DA = AE : EC$. Let $F, G$ be the two points of intersection of line $DE$ and circle $\Gamma$, $H$ be the point ... | [
"Let $X$ be the point of intersection of line $BC$ and the line going through point $D$ and parallel to line $AC$. Then, we see that $X$ lies on the line going through point $E$ and parallel to line $AB$, because we have $BX : XC = BD : DA = AE : EC$.\nSince line $IA$ (which is the same as $\\ell$) is tangent to ci... | Japan | Japan Mathematical Olympiad Final Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0224 | Problem:
Maria está planejando participar do Triatlon-Brasil que começa às 24 horas de domingo e consta de $800~\mathrm{m}$ a nado, seguido de $20~\mathrm{km}$ de bicicleta e finalmente $4~\mathrm{km}$ de corrida. Maria corre a uma velocidade constante e que é o triplo da velocidade que nada, e pedala 2,5 vezes mais r... | [
"Solution:\n\nSeja $x$ a velocidade em metros por minuto com que Maria nada. Logo, a sua velocidade na corrida é $3x$ e na bicicleta $2,5 \\times 3x = 7,5x$. Logo, o tempo total que ela gastará nas 3 etapas é:\n$$\n\\underbrace{\\frac{800}{x}}_{\\text{nadando}} + \\underbrace{\\frac{4000}{3x}}_{\\text{correndo}} + ... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | Swimming: 40/3 minutes; Running: 200/9 minutes; Cycling: 400/9 minutes. | |
0744 | Let $P$ be a path on the vertex set $V = \{1, 2, \dots, n\}$, where $j$ is joined to $j+1$, $1 \le j \le n-1$. For each subset $A \subset V$ and the induced subgraph $G(A)$ of $P$, define $\mu(A) = |A| + O(G(A))$, where $O(G(A))$ is the number of components of $G(A)$, each with an odd number of vertices; $\mu(\emptyset... | [
"If $p = r = 0$, we have $A = \\emptyset$ so that $\\mu(A) = 0$. In this case $T(0,0) = \\{\\emptyset\\}$ and $|T(0,0)| = 1$, which agrees with the formula.\n\nAssume $p \\ge r \\ge 1$. Add two two dummy vertices $0$, $n+1$ to $P$ such that $0$ is joined to $1$ and $n$ is joined to $n+1$. Let\n$$\n\\Pi = (l_1, m_1,... | India | Indija TS 2009 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
01zp | Inside an isosceles triangle $ABC$ ($AB = BC$), a point $D$ is chosen so that $\angle ADC = 150^\circ$. On the segment $CD$, a point $E$ is chosen so that $AE = AB$.
Prove that if $\angle BAE + \angle CBE = 60^\circ$, then $\angle BDC + \angle EAC = 90^\circ$. | [
"Let us rotate $\\triangle AED$ around the point $A$ so that $E \\to B$, $D \\to T$, and reflect $\\triangle ABT$ symmetrically with respect to $BT$ ($A \\to F$). Since $\\angle ADE = 150^\\circ$, then the triangle $ATF$ is equilateral. Let us prove that $FD = DC$. Denote $\\angle EAB = \\alpha$ and $\\angle AED = ... | Belarus | SELECTION TESTS OF THE BELARUSIAN TEAM TO THE IMO | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
002m | Sean $a$, $b$, $c$, $d$ cuatro elementos distintos del conjunto $\{1, 2, 3, \ldots, 2005\}$, de modo que la suma de cada tres de ellos sea múltiplo del cuarto. Determinar el mayor valor que puede tomar $a + b + c + d$. | [] | Argentina | XIV Olimpiada Matemática Rioplatense | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Español | proof and answer | 4008 | |
02ru | Problem:
Júlio Daniel tem um quadrado de papel com vértices $A$, $B$, $C$ e $D$. Ele primeiro dobra este quadrado de papel $ABCD$ levando os vértices $B$ e $D$ até a diagonal, como mostra a figura a seguir:
Em seguida, Júlio Daniel leva o vértice $C$ até o vértice $A$, obtendo assim um pent... | [
"Solution:\n\na) Abrindo o quadrado de papel dobrado, pode-se notar que o ângulo $a$ é o mesmo ângulo do vértice $A$ do quadrado $ABCD$. Logo, como todos os ângulos internos de um quadrado são iguais a $90^\\circ$, concluímos que $a = 90^\\circ$.\n\nb) Desdobrando o papel (apenas a última dobra), obtemos:\n\n![](at... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | a = 90°, b = 112.5° | |
04ay | Let $a$, $b > 1$ be relatively prime positive integers. Define a sequence
$$
x_1 = a, \quad x_2 = b, \quad x_n = \frac{x_{n-1}^2 + x_{n-2}^2}{x_{n-1} + x_{n-2}} \quad \text{for } n \ge 3.
$$
Prove that $x_n$ is not an integer for $n \ge 3$. (Tonći Kokan) | [
"Notice that $x_n > 1$, for all $n \\in \\mathbb{N}$. We also notice that all $x_n$ are rational so we can write $x_n = \\frac{p_n}{q_n}$, where $p_n$ and $q_n$ are positive integers and $M(p_n, q_n) = 1$.\n\nFirst let us prove that $p_n$ and $p_{n+1}$ are relatively prime for every $n \\in \\mathbb{N}$. We will pr... | Croatia | CroatianCompetitions2011 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0b8y | Let $n \ge 5$ be an integer. Consider $n$ distinct points in the plane, each coloured either white or black. For each positive integer $1 \le k < \frac{n}{2}$, a $k$-move consists in selecting $k$ points and reversing their colours. Find all values of $n$ for which, for any eligible $k$ and for any initial colouring, t... | [
"The problem holds for and only for $n$ odd.\n\nTo this end, suppose $n$ even. Colour exactly one point in white and the rest in black. Choose $k=2$. After performing a 2-move, the number of points of a same colour remains odd, so a monochromatic configuration is not achievable.\n\nSuppose $n$ odd. Assume $k$ odd. ... | Romania | NMO Selection Tests for the Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | All odd integers n | |
08fl | Problem:
Sia $ABC$ un triangolo non equilatero, e sia $R$ il raggio della sua circonferenza circoscritta. La circonferenza inscritta ad $ABC$ ha centro in $I$, ed è tangente al lato $CA$ nel punto $D$, ed al lato $CB$ nel punto $E$.
Sia $A_{1}$ il punto della retta $EI$ tale che $A_{1}I=R$, con $I$ che sta tra $A_{1}... | [
"Solution:\n\na. Indichiamo con $O$ il circocentro di $ABC$, ed indichiamo con $\\alpha, \\beta, \\gamma$ gli angoli di $ABC$ nei vertici $A, B, C$, rispettivamente.\n\nIl punto fondamentale è che i triangoli $IAA_{1}$ e $AIO$ sono congruenti. Per dimostrarlo, osserviamo che $IA=AI$ in quanto si tratta dello stesso... | Italy | XXXVIII Olimpiade Italiana di Matematica | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof and answer | All real values strictly between 2 and 3 | |
09ia | Let $ABCD$ be an isosceles trapezoid with $AB = CD$ and $BC < AD$. The bisector of the angle $\angle ABC$ meets the side $AD$ at point $E$ and the line passing through $E$ parallel to the bisector of the angle $\angle CDA$ meets the side $BC$ at point $F$. Prove that $\angle AFE = \angle EFB$. | [] | Mongolia | Mongolian Mathematical Olympiad Round 2 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
09ml | Let $\{P_n(x)\}_{n \ge 0}$ be a sequence of polynomials given by $P_0(x) = 0$ and
$$
P_{n+1}(x) = P_n(x) + \frac{x - P_n(x)^2}{2}
$$
for $n \ge 0$. Prove that
$$
|P_m(x) - P_n(x)| < \frac{1}{n+1}
$$
for all $m \ge n \ge 0$ and all $0 \le x \le 1$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0ifg | Problem:
Let $n > 0$ be an integer. Each face of a regular tetrahedron is painted in one of $n$ colors (the faces are not necessarily painted different colors). Suppose there are $n^{3}$ possible colorings, where rotations, but not reflections, of the same coloring are considered the same. Find all possible values of $... | [
"Solution:\nWe count the possible number of colorings. If four colors are used, there are two different colorings that are mirror images of each other, for a total of $2\\binom{n}{4}$ colorings. If three colors are used, we choose one color to use twice (which determines the coloring), for a total of $3\\binom{n}{3... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 1 and 11 | |
0gvw | Prove that for any rational numbers $a$ and $b$ the graph of the function
$$
f(x) = x^3 - 6abx - 2a^3 - 4b^3, \quad x \in \mathbb{R},
$$
has exactly one common point with the $x$-axis. | [
"Оскільки $f'(x) = 3x^2 - 6ab$, то при $ab \\le 0$ функція $f$ строго зростає на всій числовій прямій, набуває, очевидно, і від’ємних, і додатних значень, а тому твердження задачі справджується.\n\nЯкщо $a > 0$ і $b > 0$, то точками локального екстремуму будуть $x = \\pm\\sqrt{2ab}$, причому $f(\\pm\\sqrt{2ab}) = -... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
07xf | Consider three points $A$, $B$, $C$ on a circle $\Gamma$, with $\angle BAC > 90^\circ$. Let $d$ denote the line tangent to $\Gamma$ at $A$. Points $M$ and $N$ are chosen on $d$ such that $\angle MBA = \angle ABC$ and $\angle NCA = \angle ACB$.
Prove that $A$ is the midpoint of segment $MN$. | [
"Let $U$ be the intersection point of $BM$ and $CN$. Then $A$ is the incentre of triangle $UBC$, because $AB$ and $AC$ are angle bisectors by definition of $M$ and $N$. In particular, $AU$ is the angle bisector of $\\angle MUN$.\n\n\n\nBecause $MN$ is tangent to $\\Gamma$, the Alternate Seg... | Ireland | IRL_ABooklet_2024 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00tc | In an exotic country, the National Bank issues coins that can take any value in the interval $[0, 1]$. Find the smallest constant $c > 0$ such that the following holds, no matter the situation in that country:
*Any citizen of the exotic country that has a finite number of coins, with a total value of no more than 1000,... | [
"The answer is $c = \\frac{1000}{91} = 11 - \\frac{11}{1001}$. Clearly, if $c'$ works, so does any $c > c'$. First we prove that $c = 11 - \\frac{11}{1001}$ is good.\n\nWe start with 100 empty boxes. First, we consider only the coins that individually value more than $\\frac{1000}{1001}$. As their sum cannot overpa... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 1000/91 | |
0bvw | Let $ABC$ be an acute triangle in which $AB < AC$. Let $M$ be the midpoint of the side $BC$ and consider $D$ an arbitrary point of the line segment $AM$. Let $E$ be a point of the line segment $BD$ and consider the point $F$ of the line $AB$ such that lines $EF$ and $BC$ are parallel. If the orthocenter, $H$, of the tr... | [
"Let $D'$ be the orthogonal projection of $H$ onto $AM$, $E'$ the intersection point of the lines $BD'$ and $AH$, and let $F'$ be the intersection point of the lines $D'H$ and $AB$. We prove that $D' = D$, $E' = E$, $F' = F$, hence $HD \\perp AM$.\nLet $H_0$ and $D'_0$ be the reflections across the point $M$, of th... | Romania | Eleventh STARS OF MATHEMATICS Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Triangles > Tri... | null | proof only | null | |
0iyi | Problem:
The angles of a convex $n$-sided polygon form an arithmetic progression whose common difference (in degrees) is a non-zero integer. Find the largest possible value of $n$ for which this is possible. (A polygon is convex if its interior angles are all less than $180^{\circ}$.) | [
"Solution:\n\nAnswer: $27$\n\nThe exterior angles form an arithmetic sequence too (since they are each $180^{\\circ}$ minus the corresponding interior angle). The sum of this sequence must be $360^{\\circ}$. Let the smallest exterior angle be $x$ and the common difference be $d$. The sum of the exterior angles is t... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | 27 | |
0dc5 | Let $AD$ be the altitude of the right angled triangle $ABC$ with $\angle A = 90^{\circ}$. Let $DE$ be the altitude of the triangle $ADB$ and $DZ$ be the altitude of the triangle $ADC$ respectively. Let $N$ be chosen on the line $AB$ such that $CN$ is parallel to $EZ$. Let $A'$ be the symmetric of $A$ with respect to th... | [
"Suppose that the line $AA'$ intersects the lines $EZ$, $BC$ and $CN$ at the points $L$, $M$, $F$ respectively. The line $IK$ being diagonal of the rectangle $KA'IA$ passes through $L$, which by construction of $A'$ is the middle of the other diagonal $AA'$. The triangles $ZAL$ and $ALE$ are similar, so $\\angle ZA... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0gkw | Determine all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying
$$
(x^2 + y^2)f(xy) = f(x)f(y)f(x^2 + y^2)
$$
for all real numbers $x$ and $y$. | [
"Putting $x = y = 0$ in the functional equation, we get $f(0) = 0$. Putting $y = 1$ into the functional equation, we get\n$$\n(x^2 + 1)f(x) = f(x)f(1)f(x^2 + 1) \\quad (1)\n$$\nwhich shows that $f(x) \\equiv 0$ is a solution. Henceforth, we consider $f(x) \\neq 0$. We claim that if $f(x) = 0$, then $x = 0$. If ther... | Thailand | The 10th Thailand Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) ≡ 0; f(x) = x; f(x) = −x; f(x) = |x|; f(x) = −|x| | |
04pq | The product of the second and fourth term of an arithmetic sequence with the common difference $d$ is $-d^2$. Find the product of the third and fifth term of that sequence. (Matko Ljulj) | [
"Let the first term of the sequence be $a$.\n\nThe second term is $a + d$.\nThe fourth term is $a + 3d$.\n\nTheir product is $(a + d)(a + 3d) = -d^2$.\n\nWe are to find the product of the third and fifth terms:\n\nThird term: $a + 2d$\nFifth term: $a + 4d$\n\nTheir product is $(a + 2d)(a + 4d)$.\n\nLet us expand $(... | Croatia | Croatian Mathematical Society Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | 0 | |
04uh | Is it possible to fill an $n \times n$ table with numbers $1$ and $2$ such that the sum of the numbers in each column is a multiple of $5$ and the sum of the numbers in each row is a multiple of $7$? Solve this:
a) for $n = 9$
b) for $n = 12$
(Tomáš Bárta) | [] | Czech Republic | First Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof and answer | a) No; b) Yes | |
08dc | Problem:
Sia $ABC$ un triangolo isoscele su base $BC$ e siano $D, E$ punti sui lati $AB, BC$ rispettivamente, tali che che le rette $DE$ e $AC$ risultino parallele. Si consideri inoltre il punto $F$ sulla retta $DE$ che si trova dalla parte opposta di $D$ rispetto ad $E$ ed è tale che $FE$ sia congruente ad $AD$. Dett... | [
"Solution:\n\nNotiamo che il triangolo $BDE$ è isoscele su base $BE$. Infatti, considerando i segmenti paralleli $ED, CA$ e il segmento $CB$, abbiamo $\\widehat{ACB} = \\widehat{DEB}$. Quindi, dato che $\\widehat{ACB} = \\widehat{CBA}$, otteniamo $\\widehat{DEB} = \\widehat{CBA}$.\n\nConsideriamo ora i triangoli $O... | Italy | Progetto Olimpiadi della Matematica | [
"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 | |
047l | Given an integer $n \ge 2$, let $x_1, x_2, \dots, x_n$ be non-negative real numbers satisfying $x_1 + x_2 + \dots + x_n = n$. Find the minimum and maximum values of
$$
\sum_{k=1}^{n} \frac{1 + x_k^2 + x_k^4}{1 + x_{k+1} + x_{k+1}^2 + x_{k+1}^3 + x_{k+1}^4},
$$
where $x_{n+1} = x_1$. | [
"*Proof.* The minimum value is $\\frac{3}{5}n$.\nOn one hand, taking $x_1 = x_2 = \\dots = x_n = 1$, we have\n$$\n\\sum_{k=1}^{n} \\frac{1 + x_k^2 + x_k^4}{1 + x_{k+1} + x_{k+1}^2 + x_{k+1}^3 + x_{k+1}^4} = \\frac{3}{5}n.\n$$\nOn the other hand, by the AM-GM inequality, for any $1 \\le k \\le n$, we have $x_k^2+x_k... | China | 2024 CGMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | English | proof and answer | minimum = (3/5) n; maximum = n^4 + n^2 + n - 1 + (n - 1)/(n^5 - 1) |
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