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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0bmu | Let $ABC$ be a triangle, and let $r$ denote its inradius. Let $R_A$ denote the radius of the circle internally tangent at $A$ to the circle $ABC$ and tangent to the line $BC$; the radii $R_B$ and $R_C$ are defined similarly. Show that $1/R_A + 1/R_B + 1/R_C \le 2/r$. | [
"Let $a$, $h_A$ and $\\Delta$ denote the length of the side $BC$, the length of the altitude from $A$ in the triangle $ABC$, and the area of the triangle $ABC$, respectively. Consider the circle internally tangent at $A$ to the circle $ABC$ and tangent at $T$ to the line $BC$. Then $2R_A \\ge AT \\ge h_A = 2\\Delta... | Romania | 66th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line,... | null | proof only | null | |
0je2 | Problem:
The lines $y = x$, $y = 2x$, and $y = 3x$ are the three medians of a triangle with perimeter $1$. Find the length of the longest side of the triangle. | [
"Solution:\nThe three medians of a triangle contain its vertices, so the three vertices of the triangle are $(a, a)$, $(b, 2b)$, and $(c, 3c)$ for some $a$, $b$, and $c$. Then, the midpoint of $(a, a)$ and $(b, 2b)$, which is $\\left(\\frac{a+b}{2}, \\frac{a+2b}{2}\\right)$, must lie along the line $y = 3x$. Theref... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | sqrt(58) / (2 + sqrt(34) + sqrt(58)) | |
0jij | Problem:
Let $\omega$ be a circle, and let $A$ and $B$ be two points in its interior. Prove that there exists a circle passing through $A$ and $B$ that is contained in the interior of $\omega$. | [
"Solution:\nWLOG, suppose $OA \\geq OB$. Let $\\omega'$ be the circle of radius $OA$ centered at $O$. We have that $B$ lies inside $\\omega'$. Thus, it is possible to scale $\\omega'$ down about the point $A$ to get a circle $\\omega''$ passing through both $A$ and $B$. Since $\\omega''$ lies inside $\\omega'$ and ... | United States | HMMT 2014 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0hhg | Nonzero real numbers $a, b, c$ satisfy the equation $ab + bc + ac = 0$. Prove that the numbers $a + b + c$ and $\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a}$ have the same sign. | [
"If we assume that $b + c = 0$, then we have\n$$\nab + bc + ac = a(b + c) + ac = ac = 0,\n$$\nwhich contradicts the condition. Thus $(a+b)(b+c)(c+a) \\neq 0$.\nSince\n$$\n(a + b + c)(a + b) = a^2 + ab + b^2 + ab + bc + ac = a^2 + ab + b^2,\n$$\nwe have $\\frac{1}{a+b} = (a+b+c) \\cdot \\frac{1}{a^2+ab+b^2}$. Then, ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
00ue | Are there polynomials $P$ and $Q$ with real coefficients such that $P(P(x)) \cdot Q(Q(x))$ has exactly $2023$ distinct real roots and $P(Q(x)) \cdot Q(P(x))$ has exactly $2024$ distinct real roots? | [
"Answer: There exist such $P$ and $Q$.\nLet $M = 2023!$ and $0 < a_1 < a_2 < \\dots < a_{2024} < 1$. We will show that the polynomials $P(x) = (x - a_1) \\cdot (x - a_2) \\cdots (x - a_{2024}) + M$ and $Q(x) = (x - (M+1)) \\cdot (x - (M+2)) \\cdots (x - (M + 2023)) + M$ fulfill the stated conditions.\n\nFor $x \\in... | Balkan Mathematical Olympiad | BMO 2023 Short List | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
053f | Call a tuple $(b_m, b_{m+1}, \dots, b_n)$ of integers *perfect* if both following conditions are fulfilled:
1. There exists an integer $a > 1$ such that $b_k = a^k + 1$ for all $k = m, m+1, \dots, n$;
2. For all $k = m, m+1, \dots, n$, there exists a prime number $q$ and a non-negative integer $t$ such that $b_k = q^t$... | [
"*Answer:* $(2^0 + 1, 2^1 + 1, 2^2 + 1, 2^3 + 1, 2^4 + 1)$.\n\nClearly $(2^0 + 1, 2^1 + 1, 2^2 + 1, 2^3 + 1, 2^4 + 1)$ is a perfect tuple with length 5. Show in the rest that there are no other perfect tuples with length 5 or larger.\n\nFor that, let $(a^m + 1, a^{m+1} + 1, \\dots, a^n + 1)$ be an arbitrary perfect... | Estonia | IMO Team Selection Contest | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | (2^0 + 1, 2^1 + 1, 2^2 + 1, 2^3 + 1, 2^4 + 1) | |
0jsd | Problem:
A positive integer $\overline{A B C}$, where $A, B, C$ are digits, satisfies
$$
\overline{A B C} = B^{C} - A
$$
Find $\overline{A B C}$. | [
"Solution:\nThe equation is equivalent to $100A + 10B + C = B^{C} - A$.\n\nSuppose $A = 0$, so that we get $10B + C = B^{C}$. Reducing mod $B$, we find that $C$ must be divisible by $B$. $C \\neq 0$, since otherwise $10B = 1$, contradiction, so $C \\geq B$. Thus $10B + C \\geq B^{B}$ for digits $B, C$. For $B \\geq... | United States | HMMT November | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 127 | |
06bi | Let $n$ be a positive integer larger than or equal to $2$. On an $n \times n$ chessboard, $r$ chess pieces have been placed such that
(i) there is at least one chess piece in each row;
(ii) there is at least one chess piece in each column; and
(iii) there is at least one chess piece in each diagonal (here a 'diagonal' ... | [
"The smallest possible value of $r$ is $2n$ if $n$ is even, and is $2n + 1$ if $n$ is odd.\n\n**Case 1.** $n = 2k$ is even\nLet $(i, j)$ be the cell in row $i$ and column $j$. Consider the following $2n$ diagonals:\n$$\n\\begin{aligned}\n\\{(i, j) : i + j = m\\} & \\text{ for } m = 2, 3, \\dots, k + 1, \\\\\n\\{(i,... | Hong Kong | 1997-2023 IMO HK TST | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | r = 2n if n is even; r = 2n + 1 if n is odd | |
08pp | Problem:
The cells of a $8 \times 8$ table are initially white. Alice and Bob play a game. First Alice paints $n$ of the fields in red. Then Bob chooses 4 rows and 4 columns from the table and paints all fields in them in black. Alice wins if there is at least one red field left. Find the least value of $n$ such that ... | [
"Solution:\n\nWe will show that the least value of $n$ is $n=13$.\nIf $n \\leq 12$, Bob wins by painting black the 4 rows containing the highest numbers of red cells. Indeed, if at least 5 red cells remain, then one of the rows not blackened contains at least 2 red cells. Thus, each one of the rows blackened contai... | JBMO | Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 13 | |
01dq | 1. Each sequence of at least two consecutive integers contains a number that is divisible by no prime number less than the amount of members in the sequence.
2. Each sequence of at least two consecutive integers contains a number that is relatively prime to all other members of the sequence. | [
"**Answer:** Neither hypothesis is true.\n\n1. The sequence $(2, 3, 4, 5, 6, 7, 8, 9)$ contains 8 consecutive integers which all are divisible by some prime less than 8.\n\n2. By Chinese Remainder Theorem, there exists an integer $x$ that satisfies the following conditions:\n* $x$ is divisible by $2$, $5$ and $11$;... | Baltic Way | Baltic Way 2016 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0j51 | Problem:
Danny has a set of $15$ pool balls, numbered $1,2, \ldots, 15$. In how many ways can he put the balls in $8$ indistinguishable bins such that the sum of the numbers of the balls in each bin is $14$, $15$, or $16$? | [
"Solution:\n\nAnswer: $122$\n\nClearly, the balls numbered $15,14, \\ldots, 9,8$ must be placed in separate bins, so we number the bins $15,14, \\ldots, 9,8$. Note that bins $15$ and $14$ may contain only one ball while all other bins must contain at least two balls. We have two cases to examine.\n\nCase 1: Only on... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | final answer only | 122 | |
0hdf | Find all positive integers $n$ such that $n^n + 1$ is divisible by $n+1$. | [
"For odd integer $n$ it is enough to decompose:\n$$\nn^n + 1 = (n+1)(n^{n-1} - n^{n-2} + n^{n-3} - \\dots + 1).\n$$\nLet's assume that there exists an even number $n = 2k$ such that $K = (2k)^{2k} + 1$ is divisible by $2k + 1$. Then $2k + 1$ is also a divisor of $2kK = (2k)^{2k+1} + 2k = ((2k)^{2k+1} + 1) + (2k-1)$... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All odd positive integers | |
0cm6 | Problem:
Given any positive real number $\varepsilon$, prove that, for all but finitely many positive integers $v$, any graph on $v$ vertices with at least $(1+\varepsilon) v$ edges has two distinct simple cycles of equal lengths.
(Recall that the notion of a simple cycle does not allow repetition of vertices in a cycl... | [
"Solution:\nFix a positive real number $\\varepsilon$, and let $G$ be a graph on $v$ vertices with at least $(1+\\varepsilon) v$ edges, all of whose simple cycles have pairwise distinct lengths.\nAssuming $\\varepsilon^{2} v \\geq 1$, we exhibit an upper bound linear in $v$ and a lower bound quadratic in $v$ for th... | Romanian Master of Mathematics (RMM) | Romanian Master of Mathematics Competition Day 1 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Other"
] | null | proof only | null | |
0iac | Problem:
a$ and $b$ are integers such that $a + \sqrt{b} = \sqrt{15 + \sqrt{216}}$. Compute $a / b$. | [
"Solution:\n\n$\\frac{1}{2}$\n\nSquaring both sides gives $a^{2} + b + 2a \\sqrt{b} = 15 + \\sqrt{216}$; separating rational from irrational parts, we get $a^{2} + b = 15$, $4a^{2}b = 216$, so $a^{2}$ and $b$ equal $9$ and $6$. $a$ is an integer, so $a^{2} = 9$, $b = 6 \\Rightarrow a / b = 3 / 6 = 1 / 2$. (We canno... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Intermediate Algebra > Other"
] | null | final answer only | 1/2 | |
0aor | Problem:
Solve the system of equations:
$$
\left\{\begin{array}{l}
x + y + \sqrt{x y} = 28 \\
x^{2} + y^{2} + x y = 336
\end{array}\right.
$$ | [
"Solution:\n$(4,16)$ and $(16,4)$\n\nAfter rewriting the first equation into $x + y = 28 - \\sqrt{x y}$, we square to get\n$$\nx^{2} + x y + y^{2} = 784 - 56 \\sqrt{x y}\n$$\nUsing the second given equation, the last equation becomes\n$$\n336 = 784 - 56 \\sqrt{x y} \\quad \\text{or} \\quad \\sqrt{x y} = 8\n$$\nSubs... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (4, 16) and (16, 4) | |
06dx | Let $n \ge 3$ be an integer and $x_1, x_2, \dots, x_{n-1}$ be nonnegative integers such that
(i) $x_1 + x_2 + \dots + x_{n-1} = n$,
(ii) $x_1 + 2x_2 + \dots + (n-1)x_{n-1} = 2n - 2$.
Find the minimum of the sum $\sum_{k=1}^{n-1} kx_k(2n-k)$. Justify your answer. | [
"The minimum value is $3n^2 - 3n$.\nWe have\n$$\n\\begin{align*}\n\\sum_{k=1}^{n-1} kx_k(2n-k) &= 2n(2n-2) - \\sum_{k=1}^{n-1} k^2 x_k \\\\\n&= 2n(2n-2) - \\sum_{k=1}^{n-1} x_k - \\sum_{k=1}^{n-1} (k-1)(k+1)x_k \\\\\n&\\ge 2n(2n-2) - n - \\sum_{k=1}^{n-1} (k-1)nx_k \\\\\n&= 2n(2n-2) - n - n \\sum_{k=1}^{n-1} kx_k +... | Hong Kong | IMO HK TST | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 3n^2 - 3n | |
0783 | Let $ABC$ be a triangle, and let $D$ be the foot of the $A$-altitude. Points $P, Q$ are chosen on $BC$ such that $DP = DQ = DA$. Suppose $AP$ and $AQ$ intersect the circumcircle of $ABC$ again at $X$ and $Y$. Prove that the perpendicular bisectors of the lines $PX$, $QY$, and $BC$ intersect. | [] | India | IMOTC Practice Test 1 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0k6t | Problem:
Let $\mathbb{N}=\{1,2,3, \ldots\}$ be the set of all positive integers, and let $f$ be a bijection from $\mathbb{N}$ to $\mathbb{N}$. Must there exist some positive integer $n$ such that $(f(1), f(2), \ldots, f(n))$ is a permutation of $(1,2, \ldots, n)$ ? | [
"Solution:\nAnswer: No\nConsider the bijection $f$ defined by\n$$\n(f(1), f(2), f(3), f(4), \\ldots)=(2,4,6,1,8,3,10,5,12, \\ldots)\n$$\nwhich alternates between even and odd numbers after the second entry. (More formally, we define $f(n)=2n$ for $n=1,2$, $f(n)=n+3$ for odd $n \\geq 3$ and $f(n)=n-3$ for even $n \\... | United States | HMMT February 2019 Team Round | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | No | |
05pd | Problem:
Peut-on arranger les nombres $1,1,2,2, \ldots, 50,50$ tels que, pour tout $1 \leqslant k \leqslant 50$, entre les deux nombres $k$ il y ait exactement $k$ éléments. | [
"Solution:\n\nLa réponse est non. Supposons, par l'absurde, qu'on le peut. Pour tout $1 \\leqslant k \\leqslant 50$, soit $a_{k}$ un entier tel que $a_{k}$ et $a_{k}+k+1$ sont les positions des deux $k$. Évidemment, on a\n$$\n\\sum_{k=1}^{50}\\left(a_{k}+a_{k}+k+1\\right)=1+2+\\ldots+100\n$$\nEn regardant l'égalité... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | No | |
06xy | Let $\mathbb{Z}_{>0}$ be the set of positive integers. Determine all functions $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that
$$
f^{b f(a)}(a+1) = (a+1) f(b)
$$
holds for all $a, b \in \mathbb{Z}_{>0}$, where $f^{k}(n) = f(f(\cdots f(n) \cdots))$ denotes the composition of $f$ with itself $k$ times. | [
"Answer: The only function satisfying the condition is $f(n) = n+1$ for all $n \\in \\mathbb{Z}_{>0}$.\nLet $P(a, b)$ be the equality in the statement.\n\nSolution 1. We divide the solution into 5 steps.\n\nStep 1. ($f$ is injective)\nClaim 1. For any $a \\geqslant 2$, the set $\\{f^{n}(a) \\mid n \\in \\mathbb{Z}_... | IMO | International Mathematical Olympiad Shortlist | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
... | null | proof and answer | f(n) = n + 1 for all positive integers n | |
09s3 | Problem:
Op een $2014 \times 2014$-bord staat op elk van de $2014^{2}$ vakjes een lamp. Lampen kunnen aan of uit staan. In de beginsituatie is een aantal van de lampen aan. In een zet kies je een rij of kolom waarin minstens 1007 lampen aan staan en verander je van alle 2014 lampen in die rij of kolom de status (van a... | [
"Solution:\n\nNummer de rijen van 1 tot en met 2014 en de kolommen ook. Bekijk de volgende beginsituatie: in rij $i$ zijn de lampen in de kolommen $i, i+1, \\ldots, i+1005$ aan en de rest uit, waarbij we de kolomnummers modulo 2014 rekenen. Nu zijn in elke rij en in elke kolom precies 1006 lampen aan. Er is dus gee... | Netherlands | IMO-selectietoets | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2026084 | |
09ut | We want to colour the 36 squares of a $6 \times 6$ board. Every square must be coloured white, grey, or black, and the following requirement must be met:
Three adjacent squares in the same row or column, must always have three different colours.
We say that two colourings are *truly different* if you cannot get one fro... | [
"3."
] | Netherlands | Junior Mathematical Olympiad, September 2019 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 3 | |
0ale | Problem:
Let $w^{3}=1$. What is a value of $\left(1+w-w^{2}\right)^{3}+\left(1-w+w^{2}\right)^{3}$?
(a) -16
(b) -21
(c) 18
(d) 15 | [] | Philippines | Qualifying Round | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | MCQ | a | |
0h3y | Two players play the following game. They start with the polynomial $2013x^2 + 2012x + 2011$ and take moves by turns. During each move a player subtracts from the polynomial one of the following polynomials: $x^2$, $x$, $x^2 - x + 1$ or $x^2 + x - 1$, at his choice. If after a player's move the polynomial has an intege... | [
"Доведемо, що виграє перший гравець. Він може ходити так, щоб після його ходу вільний член одержаного многочлена був непарним, а коефіцієнти при $x^2$ та $x$ мали однакову парність (для цього він може спочатку відняти від многочлена $x^2$, а далі повторювати ходи другого гравця). Тоді після ходів першого гравця буд... | Ukraine | Ukrainian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | first player | |
0eef | Determine all pairs of real numbers $a$ and $b$, which satisfy the inequality
$$
a^{2}(2a - b) + b^{2}(2b - a) \geq 0.
$$ | [
"Expanding the left side of the inequality we get $2a^3 - a^2b - ab^2 + 2b^3$ and factoring it back up again results in\n$$\n\\begin{aligned}\n2a^3 - a^2b - ab^2 + 2b^3 &= 2(a^3 + b^3) - ab(a + b) = 2(a + b)(a^2 - ab + b^2) - ab(a + b) = \\\\\n&= (a + b)(2a^2 - 3ab + 2b^2).\n\\end{aligned}\n$$\nThe second factor is... | Slovenia | Slovenija 2016 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a + b ≥ 0 | |
0bgz | Solve in positive integers the equation $2^n = 3^m + 23$. | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | (n, m) = (5, 2) | |
003h | Dos personas, $A$ y $B$, juegan quitando monedas de una pila que contiene inicialmente $2006$ monedas. Los jugadores juegan por turnos quitando en cada turno de $1$ a $7$ monedas; cada jugador conserva consigo las monedas que ha quitado. Si un jugador lo desea, puede pasar (no quitar monedas en su turno) pero para ello... | [] | Argentina | XVII OLIMPIADA MATEMÁTICA DEL CONO SUR | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Español | proof and answer | A | |
0fel | Problem:
Encontrar todas las soluciones enteras positivas de
$$
\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b+c-2}=1
$$ | [
"Solution:\n\nSi $a, b, c \\geq 2$, entonces la expresión es $\\leq 1$, con igualdad si y solo si $a = b = c = 2$. Así que suponemos ahora que uno de ellos es $1$, pongamos sin pérdida de generalidad $a = 1$. Si $b, c \\geq 3$ se llega a que la expresión es nuevamente menor que $1$, así que únicamente hay que anali... | Spain | null | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (a,b,c) = (2,2,2) | |
0h3j | For a triangle *ABC*, $I_A$ is the center of excircle opposite vertex $A$. Prove that points $B$, $C$ and the circumcenters of triangles $ABI_A$ and $ACI_A$ are cyclic. | [
"Нехай точка $P$ — центр описаного кола трикутника $ABI_A$. Кут $ABI_A$ тупий, а тому точки $P$ і $B$ лежать по різні боки від прямої $AI_A$. Нехай $M$ і $N$ — точки дотику даного кола зі стороною $BC$ і продовженням сторони $AC$ відповідно, а $K$ — точка перетину цього продовження з описаним колом трикутника $ABI_... | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
01r8 | Determine the largest possible number of three-element sets that can be formed so that any two of these sets have exactly one common element, but there is not an element that belongs to all these sets. | [
"Answer: 7 sets.\nLet $N$ be the required number of 3-element sets satisfying the problem condition. Suppose that $N \\ge 8$. Let $M = \\{x, y, z\\}$ be one of these sets. Since the number of the remaining sets is greater than or equal to 7 and each such set has exactly one common element with $M$, we see that ther... | Belarus | Final Round | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 7 | |
0bmh | Let $I$ be the center of the inscribed circle of triangle $ABC$, with $AB \neq AC$. Let $M$ be the midpoint of side $BC$, and $D$ the projection of $I$ on $BC$. The circle with center $M$ and radius $MD$ intersects line $AI$ at $P$ and $Q$. Show that $\angle BAC + \angle PMQ = 180^\circ$.
Laurențiu Ploscaru | [
"Without loss of generality, we may assume that $AB < AC$ and $AP < AQ$. We shall prove that angles $BAC$ and $PMQ$ have parallel sides, which trivially implies the conclusion.\n\nFirst approach. Consider the tangency points $E \\in AC$ and $F \\in AB$ on the inscribed circle. Let $P'$ and $Q'$ be the intersection ... | Romania | 66th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | ∠BAC + ∠PMQ = 180° | |
03uo | As shown in Fig. 1, $ABCD$ is a convex quadrilateral with $\angle B + \angle D < 180^\circ$, and $P$ is a moving point on the plane. Let
$$
f(P) = PA \times BC + PD \times CA + PC \times AB.
$$
(1) Prove that $P$, $A$, $B$, $C$ are concyclic when $f(P)$ reaches the minimum.
(2) Suppose that point $E$ is on the arc $\w... | [
"(1) As shown in Fig. 1, by the Ptolemy inequality we have\n$$\nPA \\times BC + PC \\times AB \\ge PB \\times AC.\n$$\nTherefore,\n$$\n\\begin{aligned}\nf(P) &= PA \\times BC + PC \\times AB + PD \\times CA \\\\\n&\\ge PB \\times CA + PD \\times CA \\\\\n&= (PB + PD) \\times CA.\n\\end{aligned}\n$$\nThe equality ho... | China | China Mathematical Competition (Complementary Test) | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geo... | English | proof and answer | sqrt(10) | |
0ih0 | Problem:
Five people of different heights are standing in line from shortest to tallest. As it happens, the tops of their heads are all collinear; also, for any two successive people, the horizontal distance between them equals the height of the shorter person. If the shortest person is 3 feet tall and the tallest per... | [
"Solution:\n\nIf $A$, $B$, and $C$ are the tops of the heads of three successive people and $D$, $E$, and $F$ are their respective feet, let $P$ be the foot of the perpendicular from $A$ to $BE$ and let $Q$ be the foot of the perpendicular from $B$ to $CF$. Then, by equal angles, $\\triangle ABP \\sim \\triangle BC... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | sqrt(21) | |
033y | Problem:
Consider the system
$$
\begin{cases}
x^{2} + y^{2} = a^{2} + 2 \\
\dfrac{1}{x} + \dfrac{1}{y} = a
\end{cases}
$$
where $a$ is a real number.
a) Solve the system for $a=0$.
b) Find all $a$, for which the system has exactly two solutions. | [
"Solution:\n\na) If $a=0$, then $x=-y$ and hence $2x^{2}=2$. It follows that $(x, y) = (1, -1)$ or $(x, y) = (-1, 1)$.\n\nb) We know from a) that $a=0$ is one of the desired numbers. Let $a \\neq 0$. Setting $x+y=p$, $xy=q$, we have $p=a q$ and $p^{2}-2q=a^{2}+2$. Then $a^{2} q^{2}-2q-a^{2}-2=0$ and hence $(p, q)=(... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | a) (1, −1) and (−1, 1). b) a ∈ (−√2, √2). | |
0f2b | Problem:
$n$ points lie in the plane, not all on a single line. A real number is assigned to each point. The sum of the numbers is zero for all the points lying on any line. Show that all the assigned numbers must be zero. | [
"Solution:\n\nSuppose the points are $P_1, P_2, \\ldots, P_n$, and the real numbers assigned are $a_1, a_2, \\ldots, a_n$.\n\nLet $L$ be any line containing some of the points. By hypothesis, the sum of the $a_i$ for all $P_i$ on $L$ is zero.\n\nLet us show that all $a_i = 0$.\n\nPick three non-collinear points $P_... | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Functional equations",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
06t2 | Let $\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $A B C$ with $A B > B C$. The angle bisector of $\angle A B C$ intersects $\Omega$ at $M \neq B$. Let $\Gamma$ be the circle with diameter $B M$. The angle bisectors of $\angle A O B$ and $\angle B O C$ intersect $\Gamma$ at point... | [
"Let $K$ be the midpoint of $B M$, i.e., the centre of $\\Gamma$. Notice that $A B \\neq B C$ implies $K \\neq O$. Clearly, the lines $O M$ and $O K$ are the perpendicular bisectors of $A C$ and $B M$, respectively. Therefore, $R$ is the intersection point of $P Q$ and $O K$.\nLet $N$ be the second point of interse... | IMO | 55th International Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point ci... | null | proof only | null | |
0bgp | Problem:
Fie $ABC$ un triunghi ascuţitunghic cu ortocentrul $H$ şi fie $W$ un punct situat în interiorul laturii $BC$. Punctele $M$ şi $N$ sunt picioarele înălţimilor din $B$, respectiv $C$. Notăm cu $\omega_{1}$ cercul circumscris triunghiului $BWN$ şi fie $X$ punctul diametral opus lui $W$ în cercul $\omega_{1}$. An... | [] | Romania | Romania 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"
] | null | proof only | null | |
0e7p | Let $A$, $B$, $C$ and $D$ be points on a circle $\Gamma$ such that the lines $AB$ and $CD$ intersect at the point $T$, where $A$ lies between $B$ and $T$ and $D$ lies between $C$ and $T$. Let the line through $D$ that is parallel to the line $AB$ intersect the circle $\Gamma$ again at the point $E$, and let the line $E... | [
"The points $C$, $D$, $E$ and $F$ are concyclic and the lines $AB$ and $DE$ are parallel, so we have $\\angle GTF = \\angle DEF = \\angle DCF$. Hence, triangles $\\Delta TGF$ and $\\Delta CGT$ are similar and they have an equal side ratio. We notice that\n$$\n\\frac{|GF|}{|GT|} = \\frac{|GT|}{|GC|} \\text{ or } |GT... | Slovenia | Selection Examinations for the IMO 2013 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03tr | **Prove the following statements:**
(1) If $2n-1$ is a prime number, then for any group of distinct positive integers $a_1, a_2, \dots, a_n$ there exist $i, j \in \{1, 2, \dots, n\}$ such that $\frac{a_i+a_j}{(a_i, a_j)} \ge 2n-1$.
(2) If $2n-1$ is a composite number, then there exists a group of distinct positive in... | [
"(1) Let $p = 2n-1$ be a prime. Without loss of generality, we assume that $(a_1, a_2, \\dots, a_n) = 1$. If there exists $i$ ($1 \\le i \\le n$) such that $p \\nmid a_i$, then there exists $j (\\ne i)$ such that $p \\nmid a_j$. Therefore $p \\nmid (a_i, a_j)$. Then we have\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} \\ge \... | China | China Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
04vu | Determine the number of quadratic polynomials $P(x)$ with integral coefficients such that the following inequalities hold for all $x \in \mathbb{R}$:
$$
x^2 + 2x - 2023 < P(x) < 2x^2.
$$ | [
"Denote the coefficients of $P$ as $P(x) = a x^2 + b x + c$. By looking what happens for $|x|$ large, we conclude that $a \\in \\{1, 2\\}$. These two cases shall be treated separately:\n\n* Suppose that $a = 1$. The lower bound on $P$ then tells us that the inequality $2x - 2023 < b x + c$ must hold for all $x$. Cl... | Czech Republic | First Round of the 73rd Czech and Slovak Mathematical Olympiad (December 12th, 2023) | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 4042 | |
07ol | Suppose $\triangle ABC$ is a triangle. Let $D$ be the reflection of $A$ in the perpendicular bisector $L$ of $BC$. Show that $|AD| = |a - 2c \cos \angle ABC|$. | [
"Let $M$, $N$ be the feet of the perpendiculars from $A$, $D$ on $BC$. Then, $N$ is the reflection of $M$ and the right triangles $AMB$ and $DNC$ are congruent. Moreover, $AD$ is parallel to $BC$ and $|AD| = |MN|$. There are various possibilities for the position of $M$, some are shown below.\n\n![](attached_image_... | Ireland | Irska 2014 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Transformations",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
09dk | **ДБ-А2** $n$-ийн сондгой нэмэгдүүнүүд нь л тэнцүү байж болох хуваалтуудын тоо нь $n$-ийн нэмэгдүүн бүр $3$-аас илүүгүй давтагдаж болох хуваалтын тоотой тэнцүү болохыг батал. | [
"$n$-сондгой нэмэгдэхүүнүүд нь л тэнцүү байж болох хуваалтын тоог $a_n$ гээ. Түүний үүсгэгч функцийг бичвэл:\n$$\nf(x) = 1 + \\sum_{k=0}^{\\infty} a_k x^k = (1+x^2)(1+x^4)(1+x^6)\\dots(1+x+x^2+x^3+\\dots) \\\\\n(1+x^3+x^6+x^9+\\dots)(1+x^5+x^{10}+\\dots) \\\\\n\\frac{1-x^4}{1-x^3} \\cdot \\frac{1-x^8}{1-x^4} \\cdot... | Mongolia | ММО-48 | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | Mongolian | proof only | null | |
0hmb | Problem:
If $a, b, c, d$ are positive real numbers such that $\frac{5 a+b}{5 c+d}=\frac{6 a+b}{6 c+d}$ and $\frac{7 a+b}{7 c+d}=9$, calculate $\frac{9 a+b}{9 c+d}$. | [
"Solution:\nLet $\\frac{5 a+b}{5 c+d}=\\frac{6 a+b}{6 c+d}=k$. Then $5 a+b=k(5 c+d)$ and $6 a+b=k(6 c+d)$. Subtracting these two equations gives $a=k c$. Now we can easily get that $b=k d$. From $\\frac{7 a+b}{7 c+d}=\\frac{7 k c+k d}{7 c+d}=k=9$ we get $\\frac{9 a+b}{9 c+d}=9$ as well."
] | United States | Berkeley Math Circle Monthly Contest 3 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 9 | |
03uf | Given real numbers $a$, $b$, $c$ such that $abc = 1$, prove that for all the integers $k \ge 2$,
$$
\frac{a^k}{a+b} + \frac{b^k}{b+c} + \frac{c^k}{c+a} \ge \frac{3}{2}.
$$ | [
"Since\n$$\n\\frac{a^k}{a+b} + \\frac{1}{4}(a+b) + \\underbrace{\\frac{1}{2} + \\frac{1}{2} + \\dots + \\frac{1}{2}}_{k-2} \\ge k \\cdot \\sqrt[k]{\\frac{a^k}{2^k}} = \\frac{k}{2}a,\n$$\nthen\n$$\n\\frac{a^k}{a+b} \\ge \\frac{k}{2}a - \\frac{1}{4}(a+b) - \\frac{k-2}{2}.\n$$\nSimilarly,\n$$\n\\frac{b^k}{b+c} \\ge \\... | China | China Southeastern Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
05xm | Problem:
Un quadrilatère convexe $ABCD$ admet un cercle inscrit de centre $I$. Soit $I_a$, $I_b$, $I_c$ et $I_d$ les centres des cercles inscrits des triangles $DAB$, $ABC$, $BCD$ et $CDA$ respectivement. On suppose que les tangentes communes aux cercles circonscrits de $AI_bI_d$ et $CI_bI_d$ se rencontrent en $X$ et ... | [
"Solution:\n\nOn note $\\omega_{i}$ le cercle inscrit de centre $I_{i}$ et de rayon $r_{i}$ pour $i \\in\\{A, B, C, D\\}$, on note également $O_{A}, O_{B}$, $O_{C}$ et $O_{D}$ les centres des cercles $A I_{b} I_{d}, B I_{a} I_{c}, C I_{b} I_{d}$ et $D I_{a} I_{c}$. On commence par démontrer le lemme suivant:\n\nLem... | France | Préparation Olympique Française de Mathématiques - ENVOI 5 : Pot-POURRI | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"... | null | proof only | null | |
0fmc | Problem:
Sea $ABC$ un triángulo arbitrario, $P$ un punto interior y $H_{A}, H_{B}$ y $H_{C}$, respectivamente, los ortocentros de los triángulos $PBC$, $PAC$ y $PAB$. Demuestra que los triángulos $H_{A} H_{B} H_{C}$ y $ABC$ tienen la misma área. | [
"Solution:\n\nCalculemos la distancia de un vértice $A$ al ortocentro del triángulo $ABC$.\n\nEn las figuras siguientes podemos observar que los triángulos $BCC'$ y $AHC'$ son semejantes. (Recordemos que los ángulos de lados perpendiculares son iguales o suplementarios). De esta semejanza resulta\n$$\n\\frac{AH}{CB... | Spain | Spain | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Mi... | null | proof only | null | |
06tg | In a company of people some pairs are enemies. A group of people is called unsociable if the number of members in the group is odd and at least $3$, and it is possible to arrange all its members around a round table so that every two neighbors are enemies. Given that there are at most $2015$ unsociable groups, prove th... | [
"We will prove the following more general statement.\n\n**Claim.** Let $G$ be a graph with chromatic number $k \\geqslant 3$. Then $G$ contains at least $2^{k-1}-k$ unsociable groups.\n\nRecall that the chromatic number of $G$ is the least $k$ such that a proper coloring\n$$\nV=V_{1} \\sqcup \\cdots \\sqcup V_{k}\n... | IMO | 56th International Mathematical Olympiad Shortlisted Problems | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
007r | On an infinite sheet of grid paper 999 grid squares are colored black. Call *special* a rectangle with sides on grid lines if it has two black opposite corner cells (rectangles with side 1 are included). Let $N$ be the maximum number of black cells in a special rectangle for a given configuration. Find the minimum of $... | [
"The minimum of $N$ is 201.\n\nIn an arbitrary configuration take the minimal rectangle $R$ that contains all black cells. Choose a black cell on every side (there is at least one by minimality) and label these $A, B, C, D$ as in the first figure. Consider the special rectangles $[AB], [BC], [CD], [DA]$ and $[AC]$.... | Argentina | Mathematical Olympiad Rioplatense | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 201 | |
0cc5 | Let $A$ and $B$ be two matrices from $M_n(\mathbb{R})$. Show that $\text{rank}(A) = \text{rank}(B)$ if and only if there are three invertible matrices $X, Y, Z \in M_n(\mathbb{R})$ such that $AX + YB = AZB$. | [
"1. Assume that there are the invertible matrices $X, Y, Z \\in M_n(\\mathbb{R})$ such that $AX + YB = AZB$. Then $\\text{rank}(A) = \\text{rank}(AX) = \\text{rank}((AZ - Y)B) \\le \\text{rank}(B)$. Similarly, $\\text{rank}(B) \\le \\text{rank}(A)$. Therefore, $\\text{rank}(A) = \\text{rank}(B)$.\n\n2. Assume $\\te... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants",
"Algebra > Linear Algebra > Linear transformations"
] | null | proof only | null | |
0830 | Problem:
Un punto interno ad un triangolo equilatero dista $5$, $7$, $8$ dai tre lati. Quanto vale il lato del triangolo?
(A) I dati non sono sufficienti
(B) la configurazione data non può esistere
(C) il lato vale $20$
(D) il lato vale $14 \sqrt{3}$
(E) il lato vale $\frac{40}{3} \sqrt{3}$. | [
"Solution:\n\nLa risposta è (E). Con riferimento alla figura a fianco, si ha infatti che l'area del triangolo equilatero di lato $l$ incognito può essere determinata in due modi:\n\n1) come somma di tre aree di tre triangoli di base $l$ e di altezza $5, 7, 8$\n\n2) con la consueta formula dell'area, tenendo present... | Italy | Progetto Olimpiadi di Matematica 2003 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | E | |
00xf | Problem:
Given that $a^{2}+b^{2}+(a+b)^{2}=c^{2}+d^{2}+(c+d)^{2}$, prove that $a^{4}+b^{4}+(a+b)^{4}=c^{4}+d^{4}+(c+d)^{4}$. | [
"Solution:\n\nUse the identity $\\left(a^{2}+b^{2}+(a+b)^{2}\\right)^{2}=2\\left(a^{4}+b^{4}+(a+b)^{4}\\right)$."
] | Baltic Way | Baltic Way 1992 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
049f | Given $\log_a x = 3$, $\log_{ab} x = 2$, determine $\log_b x$. | [] | Croatia | Hrvatska 2011 | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | final answer only | 6 | |
093f | Problem:
Let $a_{1}, a_{2}, a_{3}, \ldots$ be the sequence of positive integers such that
$$
a_{1}=1 \quad \text{ and } \quad a_{k+1}=a_{k}^{3}+1, \text{ for all positive integers } k
$$
Prove that for every prime number $p$ of the form $3 \ell+2$, where $\ell$ is a non-negative integer, there exists a positive integer... | [
"Solution:\nLet $f(x)=x^{3}+1$. In what follows all congruences are considered modulo $p$.\nWe will prove that if $f(a) \\equiv f(b)$ for some integers $a$ and $b$, then $a \\equiv b$. This is clear when $f(a) \\equiv 1$, because then $a \\equiv b \\equiv 0$. Otherwise $a \\not \\equiv 0 \\not \\equiv b$. By Fermat... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof only | null | |
0jd8 | Problem:
Prove that every power of $3$, from $27$ onward, has an even tens digit. | [
"Solution:\n\nBy repeatedly multiplying by $3$, we see that the units digits of powers of $3$ are either $3$, $9$, $7$, or $1$. Suppose that $N$ is a power of $3$ with an even tens digit; we will prove that the tens digit of $3N$ is also even, from which it will follow inductively that every power of $3$ from $27$ ... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
029y | Problem:
Operação em uma fração - Que número se deve somar aos dois termos de uma fração para se obter o inverso dessa mesma fração? | [
"Solution:\n\nSeja $\\frac{a}{b}$ a fração procurada e seja $c$ um número tal que $\\frac{a+c}{b+c}=\\frac{b}{a}$. Esta igualdade é equivalente a $(a+c) a=(b+c) b$. Assim temos:\n$$\n(a+c) a=(b+c) b \\Longrightarrow a^{2}+a c-b^{2}-b c=0 \\Longrightarrow\\left(a^{2}-b^{2}\\right)+c(a-b)=0\n$$\nDonde\n$$\n0=\\left(a... | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | c = -(a + b), with the special case a = b (fraction equal to 1) where any c works | |
0b8r | Two rectangles of unit area overlap to form a convex octagon. Show that the area of the octagon is at least $\frac{1}{2}$. | [
"Each side of one rectangle meets the contour of the other rectangle at exactly two points situated on consecutive sides. Let $A_0A_1A_2A_3$ and $B_0B_1B_2B_3$ be circular labellings of the two rectangles such that the segments $A_iA_{i+1}$ and $B_iB_{i+1}$ have equal lengths and meet at a point labelled $C_i$. The... | Romania | NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
01mw | Several chess players took part in a chess tournament. Each participant played exactly one game with any other participant. A participant received $1$ point for a win, $0.5$ point for a draw, and $0$ point for a lose. Any two players received different numbers of points and the participant taking the last place receive... | [] | Belarus | 61st Belarusian Mathematical Olympiad | [
"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 | 5 | |
05h2 | Problem:
Soit $\omega_{1}$ et $\omega_{2}$ deux cercles qui se coupent en $P$ et $Q$. On suppose que la tangente commune à $\omega_{1}$ et $\omega_{2}$, la plus proche de $P$, touche les cercles $\omega_{1}$ et $\omega_{2}$ en $A$ et $B$ respectivement. Les tangentes en $P$ aux cercles $\omega_{1}$ et $\omega_{2}$ reco... | [
"Solution:\nDémonstration. On conjecture sur une bonne figure que les points $H, P$ et $B$ sont alignés, de même pour les points $K, P$ et $A$.\n\nLemme. Les points $H, P$ et $B$ sont alignés, ainsi que les points $K, P$ et $A$.\n\nOn note $x = \\widehat{PAB}$, $y = \\widehat{ABP}$ et $z = \\widehat{CPD}$. Comme la... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jii | Problem:
Consider a regular $n$-gon with $n>3$, and call a line acceptable if it passes through the interior of this $n$-gon. Draw $m$ different acceptable lines, so that the $n$-gon is divided into several smaller polygons.
a. Prove that there exists an $m$, depending only on $n$, such that any collection of $m$ acc... | [
"Solution:\n\nWe will prove that if $m \\geq n-4$, then there is guaranteed to be a smaller polygon with 3 or 4 sides, while if $m \\leq n-5$, there might not be a polygon with 3 or 4 sides. This will solve both parts of the problem.\n\nGiven a configuration of lines, let $P_{1}, \\ldots, P_{k}$ be all of the resul... | United States | HMMT 2014 HMIC | [
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof and answer | n - 4 | |
09eu | If $A = \{a_1, a_2, \dots, a_n\} \subset \mathbb{N}$ then let denote $A^* = \{a_i + a_j \mid 1 \le i < j \le n\}$ (where $A^*$ is a multiset). For instance: if $A = \{1, 2, 3, 4\}$ then $A^* = \{3, 4, 5, 5, 6, 7\}$. Prove that if $A \neq B$ and $A^* = B^*$ then $|A| = |B| = 2^k$. | [
"Setting $|A| = n$, $|B| = m$ we get $|A^*| = C_n^2$, $|B^*| = C_m^2$ and since $A^* = B^*$, $C_n^2 = C_m^2 \\Rightarrow n = m$. Let $A = \\{a_1, a_2, ..., a_n\\}$; $B = \\{b_1, b_2, ..., b_n\\}$.\nConsider polynomials $f(x) = x^{a_1} + x^{a_2} + ... + x^{a_n}$; $g(x) = x^{b_1} + x^{b_2} + ... + x^{b_n}$.\nFrom $A^... | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
07c2 | A special computer can store algebraic expressions in its memory, which is unlimited and initially only expression $x$ is stored in it. This computer can perform the following operations:
* If algebraic expression $f$ is stored in the computer's memory, $\frac{1}{f}$ can be stored in it too (assuming that $f$ is not eq... | [
"For simplicity, $S$ is used to denote the set of all algebraic expressions that can be stored in the computer's memory. It is claimed that $x^n \\in S$ if and only if $n$ is odd.\nTo prove one part of the assertion made above, it is claimed that all of the expressions in $S$ are odd, i.e. $f(-x) = -f(x)$ for all e... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | All odd natural numbers n | |
01ph | $N$ segments are arranged inside a unit circle $\Gamma$. The sum of the lengths of all these segments is equal to $2\sqrt{N}$.
Prove that there exists a concentric with $\Gamma$ circumference intersecting at least two of these segments. | [
"Consider $360^\\circ$ rotation of $\\Gamma$ (together with all segments) about its center. Under this rotation each of the segments covers some ring with the center at the center of $\\Gamma$. If we show that the sum of the areas of all these rings is not less than $\\pi$, then the statement will be proved.\n\nWe ... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
05gn | Problem:
Soient $\omega_{1}$ et $\omega_{2}$ deux cercles disjoints. Une tangente commune extérieure à ces deux cercles coupe $\omega_{1}$ en $A$ et $\omega_{2}$ en $B$. La deuxième tangente commune extérieure à ces deux cercles coupe $\omega_{1}$ en $D$ et $\omega_{2}$ en $C$. La droite $(BD)$ coupe $\omega_{1}$ en l... | [
"Solution:\n\n\n\nUne symétrie axiale par rapport à la droite joignant les centres des deux cercles échange $A$ et $D$ ainsi que $B$ et $C$. Donc, $AB = DC$. On pense alors à appliquer la puissance d'un point, on a alors :\n$$\nDQ \\cdot DB = DC^{2} = AB^{2} = BP \\cdot BD\n$$\nou la premiè... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0dgo | Given is an acute triangle $ABC$ with $BC < CA < AB$. Points $K$ and $L$ lie on segments $AC$ and $AB$ and satisfy $AK = AL = BC$. Perpendicular bisectors of segments $CK$ and $BL$ intersect line $BC$ at points $P$ and $Q$, respectively. Segments $KP$ and $LQ$ intersect at $M$. Prove that $CK + KM = BL + LM$. | [
"Let $D$ and $E$ be points of rays $ML \\to$ and $MK \\to$, respectively, such that $LD = AB$ and $KE = AC$. As $\\angle DLA = \\angle BLM = \\angle LBC$, $LD = AB$ and $AL = BC$, triangles $DLA$ and $ABC$ are congruent. Analogously, $EAK$ and $ABC$ are congruent. From\n$$\n\\angle DAL + \\angle LAK + \\angle KAE =... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
08ed | Problem:
Quanti sono i numeri di 6 cifre divisibili per 33 che siano palindromi, cioè che rimangano uguali se letti da destra verso sinistra?
(A) 30
(B) 33
(C) 300
(D) 333
(E) Nessuna delle precedenti. | [
"Solution:\n\nLa risposta è $\\mathbf{( C )}$. Dato che il numero è palindromo, possiamo scriverlo come $a b c c b a$, dove $a, b, c$ sono cifre e $a \\neq 0$. Il criterio di divisibilità per $11$ è allora automaticamente soddisfatto, essendo $a-b+c-c+b-a=0$.\n\nPer il criterio di divisibilità per $3$ occorre che $... | Italy | Italian Mathematical Olympiad - February Round | [
"Number Theory > Divisibility / Factorization"
] | null | MCQ | C | |
0jk3 | Problem:
Let $A=\{a_{1}, a_{2}, \ldots, a_{7}\}$ be a set of distinct positive integers such that the mean of the elements of any nonempty subset of $A$ is an integer. Find the smallest possible value of the sum of the elements in $A$. | [
"Solution:\nAnswer: $1267$\n\nFor $2 \\leq i \\leq 6$, we claim that $a_{1} \\equiv \\ldots \\equiv a_{7} \\pmod{i}$. This is because if we consider any $i-1$ of the $7$ numbers, the other $7-(i-1)=8-i$ of them must all be equal modulo $i$, because we want the sum of all subsets of size $i$ to be a multiple of $i$.... | United States | HMMT 2014 | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | 1267 | |
0enb | Show that for every natural number $n$ the product
$$
\left(4 - \frac{2}{1}\right) \left(4 - \frac{2}{2}\right) \left(4 - \frac{2}{3}\right) \dots \left(4 - \frac{2}{n}\right)
$$
is an integer. | [
"$$\n\\begin{align*}\n\\left(4 - \\frac{2}{1}\\right) \\left(4 - \\frac{2}{2}\\right) \\dots \\left(4 - \\frac{2}{n}\\right) &= \\prod_{k=1}^{n} \\left(\\frac{4k-2}{k}\\right) \\\\\n&= \\prod_{k=1}^{n} 2 \\left(\\frac{2k-1}{k}\\right) \\\\\n&= \\prod_{k=1}^{n} \\left(\\frac{2k-1}{k} \\cdot \\frac{2k}{k}\\right) \\\... | South Africa | South-Afrika 2011-2013 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0aik | A convex quadrilateral $ABCD$ is given. Let $E$ be the intersection of $AB$ and $CD$, $F$ be the intersection of $AD$ and $BC$, and $G$ be the intersection of $AC$ and $EF$. Prove that the following two statements are equivalent:
(i) $BD$ and $EF$ are parallel
(ii) $G$ is the midpoint of the segment $\overline{EF}$
Да... | [
"We draw a line $l$ through $E$ which is parallel to $BC$. Let $H$ be the intersection of $l$ and $AG$. Now we have that $G$ is the intersection of the diagonals in the trapezoid $EHFC$.\n\n(i) $\\Rightarrow$ (ii): Let the lines $BD$ and $EF$ be parallel. Then, from Thales' theorem for parallel segments we have the... | North Macedonia | Macedonian Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0793 | Prove that for every natural number $m$, there exists a natural number $N$ such that for every natural number $b$ that $2 \le b \le 1389$, the sum of digits of $N$ in base $b$, is more than $m$. | [
"We claim that $N = (1389!)^{m+1} - 1$ has the desired properties. Let $2 \\le b \\le 1389$, we know that there are at least $m+1$ zeros in the rightmost of $(1389!)^{m+1}$ in base $b$, because $(1389!)^{m+1}$ is divisible by $b^{m+1}$. Hence, the $m+1$ rightmost digits of $(1389!)^{m+1} - 1$ is $b-1$, then sum of ... | Iran | 27th Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
07mm | Let $n$ be an integer not less than $3$ and let $a_1, a_2, \dots, a_n$ be a finite sequence of positive integers, such that, for $k = 2, 3, \dots, n$
$$
n(a_k + 1) - (n-1)a_{k-1} = 1.
$$
Prove that $a_n$ is not divisible by $(n-1)^2$. | [
"The given recursion can be rewritten as $a_{k-1} = \\frac{n}{n-1}a_k + 1$ for $2 \\le k \\le n$. Let us write, for simplicity, $q = \\frac{n}{n-1}$, so that we have $a_{k-1} = q a_k + 1$. We obtain\n$$\na_{n-1} = q a_n + 1\n$$\n$$\na_{n-2} = q a_{n-1} + 1 = q^2 a_n + q + 1\n$$\nand rewrite this, using $q = \\frac{... | Ireland | Irish Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0jnz | Problem:
Find the smallest positive integer $n$ such that the polynomial $(x+1)^{n}-1$ is "divisible by $x^{2}+1$ modulo 3", or more precisely, either of the following equivalent conditions holds:
- there exist polynomials $P, Q$ with integer coefficients such that $(x+1)^{n}-1=\left(x^{2}+1\right) P(x)+3 Q(x)$;
- or m... | [
"Solution:\nAnswer: 8\n\nSolution 1. We have $(x+1)^{2}=x^{2}+2x+1 \\equiv 2x$, $(x+1)^{4} \\equiv (2x)^{2} \\equiv -4 \\equiv -1$, and $(x+1)^{8} \\equiv (-1)^{2}=1$. So the order $n$ divides 8, as $x+1$ and $x^{2}+1$ are relatively prime polynomials modulo 3 (or more conceptually, in $\\mathbb{F}_{3}[x]$), but ca... | United States | HMMT February | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 8 | |
00g8 | Given two positive integers $m$ and $n$, find the smallest positive integer $k$ such that among any $k$ people, either there are $2m$ of them who form $m$ pairs of mutually acquainted people or there are $2n$ of them forming $n$ pairs of mutually unacquainted people. | [
"Let the smallest positive integer $k$ satisfying the condition of the problem be denoted $r(m, n)$. We shall show that\n$$\nr(m, n) = 2(m + n) - \\min\\{m, n\\} - 1\n$$\nObserve that, by symmetry, $r(m, n) = r(n, m)$. Therefore it suffices to consider the case where $m \\geq n$, and to prove that\n$$\nr(m, n) = 2m... | Asia Pacific Mathematics Olympiad (APMO) | XV APMO | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | r(m, n) = 2(m + n) - \min\{m, n\} - 1 | |
05op | Problem:
Montrer que pour tout entier $n \geq 1$, il existe $m$ un multiple de $n$ tel que la somme des chiffres de $m$ fasse $n$. | [
"Solution:\n\nLa suite des puissances de $10$, donc $1, 10, 100, \\cdots$ contient une infinité de termes, qui peuvent prendre un nombre fini de résidus modulo $n$. Il existe donc un résidu $k$ tel qu'une infinité de puissances de $10$ soient congrues à $k$ modulo $n$. Soient donc $0 \\leqslant a_{1} < a_{2} < \\cd... | France | OCympiades Françaises de Mathématiques | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0ksp | Problem:
For points $P$, $Q$, $R$, let $E_{P, Q}(R)$ denote the ellipse with foci $P$ and $Q$ through $R$.
Points $T$, $A$, $B$, $C$ are points on a line in that order. A ray from $A$ intersects $E_{A, B}(T)$ at $Q$ and $E_{A, C}(T)$ at $P$. Show that $E_{B, P}(A)$ intersects $E_{C, Q}(A)$ at two points. | [
"Solution:\n\nFirst of all, we claim that $T$ is the tangency point of the $B$-excircle of $ABQ$ to line $AB$. This is because\n$$\n\\frac{AQ + BQ - AB}{2} = \\frac{AT + BT - AB}{2} = \\frac{AT + AT}{2} = AT.\n$$\nSimilarly, $T$ is also the tangency point of the $C$-excircle of $ACP$. Thus, as the center of the exc... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0kdo | Problem:
The numbers $1, 2, \ldots, 10$ are written in a circle. There are four people, and each person randomly selects five consecutive integers (e.g. $1, 2, 3, 4, 5$, or $8, 9, 10, 1, 2$). If the probability that there exists some number that was not selected by any of the four people is $p$, compute $10000 p$. | [
"Solution:\n\nThe unselected numbers must be consecutive. Suppose that $\\{1, 2, \\ldots, k\\}$ are the unselected numbers for some $k$.\n\nIn this case, $1$ cannot be selected, so there are $5$ possible sets of consecutive numbers the people could have chosen. This leads to $5^{4}$ possibilities. Moreover, $10$ mu... | United States | HMMO 2020 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 3690 | |
03q5 | Let the sides of a scalene triangle $\triangle ABC$ be $AB = c$, $BC = a$, $CA = b$, $D$, $E$, $F$ be points on $BC$, $CA$, $AB$, such that $AD$, $BE$, $CF$ are angle bisectors of the triangle, respectively. Assume that $DE = DF$. Prove that
$$
(1) \frac{a}{b+c} = \frac{b}{c+a} + \frac{c}{a+b};
$$
$$
(2) \angle BAC > 9... | [
"**Solution** Using the sine rule, we have that\n$$\n\\frac{\\sin \\angle AFD}{\\sin \\angle FAD} = \\frac{AD}{FD} = \\frac{AD}{ED} = \\frac{\\sin \\angle AED}{\\sin \\angle FAD},\n$$\nthen $\\sin \\angle AFD = \\sin \\angle AED$. So, either $\\angle AFD = \\angle AED$ or $\\angle AFD + \\angle AED = 180^{\\circ}$.... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry... | English | proof only | null | |
0i0t | Problem:
Do there exist three different prime numbers such that the sum of any two of them is a square? | [
"Solution:\n\nThe answer is no. We work modulo $4$ (i.e., we look at remainders on division by $4$). Note that any square is $\\equiv 0$ or $1 \\bmod 4$. Indeed, if $n=2k$ is even, $n^{2}=4k^{2} \\equiv 0$, while if $n=2k+1$ is odd, $n^{2}=4k^{2}+4k+1 \\equiv 1$.\n\nNow suppose some three such primes exist. If they... | United States | Berkeley Math Circle | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | No | |
0idq | Problem:
If $x$ and $y$ are real numbers with $(x+y)^{4} = x - y$, what is the maximum possible value of $y$? | [
"Solution:\n$$\n3 \\sqrt[3]{2} / 16\n$$\nBy drawing the graph of the curve (as shown), which is just a $135^{\\circ}$ clockwise rotation and scaling of $y = x^{4}$, we see that the maximum is achieved at the unique point where $d y / d x = 0$. Implicit differentiation gives $4(d y / d x + 1)(x + y)^{3} = 1 - d y / ... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | null | proof and answer | 3 * cube_root(2) / 16 | |
0g05 | Problem:
Sei $ABC$ ein Dreieck. Die Punkte $K$, $L$ und $M$ liegen auf den Seiten $BC$, $CA$ und $AB$, sodass sich die Geraden $AK$, $BL$ und $CM$ in einem Punkt schneiden. Zeige, dass man von den Dreiecken $AML$, $BKM$ und $CLK$ zwei wählen kann, sodass die Summe ihrer Inkreisradien mindestens so groß ist wie der Ink... | [
"Solution:\n\nNach Ceva gilt $\\frac{AM}{MB} \\cdot \\frac{BK}{KC} \\cdot \\frac{CL}{LA} = 1$. Wir können oBdA annehmen, dass $\\frac{BK}{KC} \\geq 1$ gilt. Wir unterscheiden nun zwei Fälle. Entweder ist $\\frac{CL}{LA} \\leq 1$, oder es gilt $\\frac{CL}{LA} > 1$.\n\n1. Fall:\nSei $X$ der Schnittpunkt der Parallele... | Switzerland | IMO-Selektion | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometr... | null | proof only | null | |
0if7 | Problem:
If $n$ is a positive integer, let $s(n)$ denote the sum of the digits of $n$. We say that $n$ is zesty if there exist positive integers $x$ and $y$ greater than 1 such that $x y = n$ and $s(x) s(y) = s(n)$. How many zesty two-digit numbers are there? | [
"Solution:\n\nLet $n$ be a zesty two-digit number, and let $x$ and $y$ be as in the problem statement. Clearly if both $x$ and $y$ are one-digit numbers, then $s(x) s(y) = n \\neq s(n)$. Thus either $x$ is a two-digit number or $y$ is. Assume without loss of generality that it is $x$. If $x = 10a + b$, $1 \\leq a \... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 34 | |
0knj | Rectangles $BCC_1B_2$, $CAA_1C_2$, and $ABB_1A_2$ are erected outside an acute triangle $ABC$. Suppose that
$$
\angle BC_1C + \angle CA_1A + \angle AB_1B = 180^\circ.
$$
Prove that lines $B_1C_2$, $C_1A_2$, and $A_1B_2$ are concurrent. | [] | United States | USAJMO | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
039i | Prove that $n = 11$ is the least positive integer such that for any coloring of the edges of a complete graph of $n$ vertexes in three colors there exists a monochromatic cycle of length 4. | [] | Bulgaria | Second selection test for IMO 2007, Vietnam | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
02md | Problem:
Duas tabelas - As linhas da primeira tabela dada são todas progressões aritméticas de uma mesma razão e as colunas dessa tabela são todas progressões aritméticas de uma mesma razão. Na segunda tabela dada foi utilizada a mesma lei de formação, mas alguém apagou alguns números deixando apenas três. Qual é o nú... | [] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 102 | |
0939 | Problem:
Let $ABC$ be an acute-angled triangle with $AB < AC$, and let $D$ be the foot of its altitude from $A$. Let $R$ and $Q$ be the centroids of the triangles $ABD$ and $ACD$, respectively. Let $P$ be a point on the line segment $BC$ such that $P \neq D$ and the points $P, Q, R$ and $D$ are concyclic. Prove that t... | [
"Solution:\n\nWithout loss of generality, we may assume that $P$ lies on the line segment $CD$. Let $K, L$ be the midpoints of the sides $AB, AC$, respectively. Obviously\n$$\n\\angle CBA = \\angle KDB = \\angle RQP \\quad \\text{and} \\quad \\angle ACB = \\angle CDL = \\angle PRQ.\n$$\nFurthermore, from the fact t... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous ... | null | proof only | null | |
0i01 | Problem:
Evaluate $\sum_{n=1}^{\infty} \frac{n^{5}}{n!}$. | [
"Solution:\nWe start by noticing that $\\sum_{n=1}^{\\infty} \\frac{n}{n!}=\\sum_{n=1}^{\\infty} \\frac{1}{(n-1)!}=\\sum_{n=0}^{\\infty} \\frac{1}{n!}=e$.\n\nNext we see that $\\sum_{n=1}^{\\infty} \\frac{n^{2}}{n!}=\\sum_{n=1}^{\\infty} \\frac{n}{(n-1)!}=\\sum_{n=0}^{\\infty} \\frac{1+n}{n!}=\\sum_{n=0}^{\\infty} ... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 52e | |
0gfk | 設三角形 $ABC$ 外接圓為 $\omega$。點 $K$, $L$, $M$ 分別位於邊 $AB$, $BC$, $CA$ 上, 且滿足 $CM \cdot CL = AM \cdot BL$。射線 $LK$ 交直線 $AC$ 於點 $P$。設圓 $\omega$ 與三角形 $KMP$ 外接圓的共同弦交線段 $AM$ 於點 $S$。證明 $SK \parallel BC$。 | [] | Taiwan | 2022 數學奧林匹亞競賽第五階段培訓營, 國際競賽實作(一) | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Chinese; English | proof only | null | |
0ikq | Problem:
Find all real numbers $x$ such that
$$
x^{2} + \left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x}{3} \right\rfloor = 10
$$ | [
"Solution:\nEvidently $x^{2}$ must be an integer. Well, there aren't that many things to check, are there? Among positive $x$, $\\sqrt{8}$ is too small and $\\sqrt{9}$ is too big; among negative $x$, $-\\sqrt{15}$ is too small and $-\\sqrt{13}$ is too big."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | -sqrt(14) | |
061f | Problem:
Für jede positive ganze Zahl $n$ bezeichne $d(n)$ die Anzahl aller positiver Teiler von $n$. (Beispiele: $d(2)=2, d(6)=4, d(9)=3$.)
Man bestimme alle positiven ganzen Zahlen $n$ mit der Eigenschaft $(d(n))^3 = 4n$. | [] | Germany | Auswahlwettbewerb zur IMO 2001 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | n = 2 and n = 128 | |
04ar | Let $n \ge 3$ be an integer. Determine the minimum number of points one has to mark inside a convex $n$-gon in order for the interior of any triangle with the vertices at vertices of the $n$-gon to contain at least one of the marked points. | [
"Since all diagonals from one vertex divide an $n$-gon into $n-2$ disjoint triangles, at least $n-2$ points are necessary.\n\nWe claim that it is possible to mark $n-2$ points so that the given condition is satisfied. Denote the vertices of the given $n$-gon with $A_1, A_2, \\dots, A_n$. Draw all the diagonals of t... | Croatia | CroatianCompetitions2011 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | n - 2 | |
0gh4 | 令 $Q_{>1}$ 為所有大於 1 的有理數所成的集合。假設函數 $f: Q_{>1} \to \mathbb{Z}$ 滿足
$$
f(q) = \begin{cases} q - 3 & \text{若 } q \text{ 為整數,} \\ \lceil q \rceil - 3 + f\left(\frac{1}{\lceil q \rceil - q}\right) & \text{若 } q \text{ 不為整數.} \end{cases}
$$
證明：對於任意滿足 $\frac{1}{a} + \frac{1}{b} = 1$ 的 $a, b \in Q_{>1}$, 必有 $f(a) + f(b) = -2$.
... | [
"Suppose that $a = m/n$ and $b = m/(m-n)$ where $\\gcd(m,n) = 1$. If $n = \\frac{1}{2}m$ then $(m,n) = (2,1)$ and the statement clearly holds in this case as $f(2) = -1$. Therefore from now on we assume that $n \\neq \\frac{1}{2}m$, and without loss of generality assume $n < \\frac{1}{2}m$. We will induct on $m$.\n... | Taiwan | 2023 數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | Chinese (Traditional) | proof only | null | |
0bgb | Problem:
Fie $A$ o matrice neinversabilă de ordin $n$ cu elemente reale, $n \geq 2$, şi fie $A^{*}$ adjuncta matricei $A$. Arătaţi că $\operatorname{tr}\left(A^{*}\right) \neq -1$ dacă şi numai dacă matricea $I_{n} + A^{*}$ este inversabilă.
Problem:
Az $n$-edrendú, nem invertálható $A$ négyzetes mátrix elemei valós s... | [
"Solution:\nDeoarece matricea $A$ este neinversabilă, avem $\\operatorname{rang}(A) \\leq n-1$. Distingem cazurile:\n\ni. $\\operatorname{rang}(A) \\leq n-2$. Atunci $A^{*} = O_{n}$ şi echivalenţa este evidentă.\n\n\nii. $\\operatorname{rang}(A) = n-1$. Atunci $A A^{*} = O_{n}$ şi, din inegalitatea lui Sylvester, $... | Romania | Olimpiada Naţională de Matematică Etapa Naţională | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | null | proof only | null | |
0cxq | a) Prove that for each positive integer $n$ there is a unique positive integer $a_{n}$ such that
$$
(1+\sqrt{5})^{n}=\sqrt{a_{n}}+\sqrt{a_{n}+4^{n}}
$$
b) Prove that $a_{2010}$ is divisible by $5 \cdot 4^{2009}$ and find the quotient. | [
"(a) Let $(1+\\sqrt{5})^{n}=x_{n}+y_{n} \\sqrt{5}$, where $x_{n}, y_{n}$ are positive integers, $n=1,2, \\ldots$ Then\n$$\n(1-\\sqrt{5})^{n}=x_{n}-y_{n} \\sqrt{5}, \\quad n=1,2, \\ldots\n$$\nhence\n$$\n\\begin{equation*}\nx_{n}^{2}-5 y_{n}^{2}=(-4)^{n}, \\quad n=1,2, \\ldots \\tag{1}\n\\end{equation*}\n$$\nIf $n$ i... | Saudi Arabia | SAMC | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | F_{2010}^2 | |
0l7z | Let $f: \mathbb{R} \to \mathbb{Z}$ be a function satisfying
$$
f(x - y) - 2f(x) + f(x + y) \geq -1
$$
for all $x, y \in \mathbb{R}$. Find all possible values of the set $\{f(x) \mid x \in \mathbb{R}\}$. | [] | United States | USA TST Selection Test for 67th IMO and 15th EGMO | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | All sets of the form {k} or {k, k+1} for some integer k. | |
030x | Problem:
Se consideră o rețea formată din 49 de puncte, ce reprezintă vârfurile a 36 de pătrate de latură 1 în care este descompus un pătrat de latură 6.
Spunem că un pătrat cu vârfurile în punctele rețelei este bun, dacă laturile și diagonalele sale nu sunt pe laturile pătratelor rețelei.
a) Aflați numărul de pătrate ... | [
"Solution:\na) Spunem că un pătrat este normal, dacă are vârfurile în punctele rețelei și laturile sale se află pe drepte ale rețelei paralele cu laturile pătratului $6 \\times 6$, sau pe laturile pătratului $6 \\times 6$. Orice pătrat bun are vârfurile pe laturile unui pătrat normal, a cărui lungime a laturii poat... | Brazil | Al doilea baraj de selecție pentru OBMJ | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | a) 70; b) sqrt(5)/5 | |
05ln | Problem:
Soit $\omega$ le cercle circonscrit à un triangle $A B C$. On désigne par $M$ et $N$ les milieux des côtés $[A B]$ et $[A C]$ respectivement, et par $T$ le milieu de l'arc $B C$ de $\omega$ ne contenant pas $A$. Les cercles circonscrits aux triangles $A M T$ et $A N T$ rencontrent les médiatrices de $[A C]$ e... | [
"Solution:\n\n\n\nNotons $O$ le centre du cercle $\\omega$. On a $O=(M Y) \\cap(N X)$. Soit $\\ell$ la médiatrice de $[A T]$. Elle passe par $O$.\n\nNotons $s$ la symétrie par rapport à $\\ell$. Comme $(A T)$ est la bissectrice de $\\widehat{B A C}$, la droite $s(A B)$ est parallèle à $(A C... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0f39 | Problem:
What is the smallest number of pieces that can be placed on an $8 \times 8$ chessboard so that every row, column and diagonal has at least one piece? [A diagonal is any line of squares parallel to one of the two main diagonals, so there are 30 diagonals in all.] What is the smallest number for an $n \times n$... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 16; in general 2n (for n ≥ 2), and 1 for n = 1 | |
00o6 | Let $ABCD$ be a rhombus with $\angle BAD < 90^\circ$. The circle passing through $D$ with center $A$ intersects the line $CD$ a second time in point $E$. Let $S$ be the intersection of the lines $BE$ and $AC$.
Prove that the points $A, S, D$ and $E$ lie on a circle.
 | [
"By the inscribed angle theorem, it is enough to show that $\\angle SED = \\angle SAD$.\nSince $ABCD$ is a rhombus, we have\n$$\n\\angle SAD = \\frac{1}{2} \\angle BAD.\n$$\nSince $ABCE$ is an isosceles trapezoid, we have by symmetry that\n$$\n\\angle SED = \\angle ECS = \\frac{1}{2} \\angle DCB = \\frac{1}{2} \\an... | Austria | AUT_ABooklet_2023 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
02qw | Problem:
A figura mostra um polígono regular de dez lados com centro $O$. Qual é a medida do ângulo $a$ ?
A) $15^{\circ}$
B) $18^{\circ}$
C) $20^{\circ}$
D) $30^{\circ}$
E) $36^{\circ}$ | [
"Solution:\nO triângulo $AOB$ é isósceles pois os lados $OA$ e $OB$ são iguais. Logo, os ângulos $OAB$ e $OBA$ também são iguais, ou seja, ambos têm medida $a$.\n\nNotamos agora que o ângulo central $AOB$ mede $\\frac{4}{10} \\times 360^{\\circ} = 144^{\\circ}$.\n\nComo a soma dos ângulos internos de um triângulo v... | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | MCQ | B | |
0ag3 | Let $p_1, p_2, \dots, p_{2009} \in \mathbb{R}^+$. Prove the inequality
$$
\sum_{i=1}^{2009} \frac{p_i}{p_{i+1} + p_{i+2} + \dots + p_{i+1004}} \ge \frac{2009}{1004},
$$
where the indexes are taken mod $2009$. | [
"Denote\n$$\nL = \\frac{p_1}{p_2 + p_3 + \\dots + p_{1005}} + \\frac{p_2}{p_3 + p_4 + \\dots + p_{1006}} + \\dots + \\frac{p_{2009}}{p_1 + p_2 + \\dots + p_{1004}}\n$$\nWe will use the Cauchy-Schwarz inequality in the form:\nFor every $a_1, a_2, \\dots, a_n \\in \\mathbb{R}$, $b_1, b_2, \\dots, b_n \\in \\mathbb{R}... | North Macedonia | null | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
03ju | Problem:
A set of $\frac{1}{2} n(n+1)$ distinct numbers is arranged at random in a triangular array:
Let $M_{k}$ be the largest number in the $k$-th row from the top. Find the probability that
$$
M_{1}<M_{2}<M_{3}<\cdots<M_{n}
$$ | [] | Canada | Canadian Mathematical Olympiad | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 2^n / (n+1)! |
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