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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
04ek | Let $ABC$ be an acute triangle with orthocentre $H$. The line through the point $A$ perpendicular to $AC$ and the line through the point $B$ perpendicular to $BC$ intersect at $D$. The circle with centre $C$ which contains $H$ intersects the circumcircle of the triangle $ABC$ at $E$ and $F$. Prove that $|DE| = |DF| = |... | [
"Let $\\alpha$, $\\beta$, $\\gamma$ be the angles of the triangle $ABC$ and $R$ the radius of its circumcircle. Without loss of generality, let $E$ lie on the arc $BC$ and $F$ on the arc $AC$.\nSince $\\angle CAD = \\angle CBD = 90^\\circ$, the point $D$ lies on the circumcircle of $ABC$ and $CD$ is the diameter.\n... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0j61 | Problem:
In a game of Fish, $R2$ and $R3$ are each holding a positive number of cards so that they are collectively holding a total of $24$ cards. Each player gives an integer estimate for the number of cards he is holding, such that each estimate is an integer between $80\%$ of his actual number of cards and $120\%$ ... | [
"Solution:\n\nAnswer: $20$\n\nTo minimize the sum, we want each player to say an estimate as small as possible—i.e., an estimate as close to $80\\%$ of his actual number of cards as possible. We claim that the minimum possible sum is $20$.\n\nFirst, this is achievable when $R2$ has $10$ cards and estimates $8$, and... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 20 | |
0hol | Problem:
Let $M$ be an interior point of a parallelogram $ABCD$. Prove that $MA + MB + MC + MD$ is strictly less than the length of the perimeter of $ABCD$. | [
"Solution:\n\nDenote by $X$ and $Y$ the points of intersection of the segments $AB$ and $CD$ with the line through $M$ parallel to $BC$. Similarly, let $U$ and $V$ denote the points of intersection of the segments $AD$ and $BC$ with the line through $M$ parallel to $AB$.\n\nThen $MA < AU + UM = XM + UM$, $MB < MX +... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0keh | Problem:
Given a quadrilateral $A B C D$ extend $A D$ and $B C$ to meet at $E$ and $A B$ and $D C$ to meet at $F$. Draw the circumcircles of triangle $A B E$, $A D F$, $D C E$, and $B C F$. Prove that all four of these circles pass through a single point. | [
"Solution:\n\nLet circumcircles of $B C F$ and $D C E$ intersect at point $M$. We find $\\angle D M F$. As $\\angle D M C = \\angle D E C$, and $\\angle M C F = 180 - \\angle C B F$, we have $\\angle D M F = 180 - (\\angle C B F - \\angle D E C) = 180 - \\angle E A B$, so $A D M F$ are concyclic. Similarly, $A B M ... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
0kcd | Problem:
Let $a$ be a positive integer such that $2a$ has units digit $4$. What is the sum of the possible units digits of $3a$? | [
"Solution:\n\nIf $2a$ has last digit $4$, then the last digit of $a$ is either $2$ or $7$. In the former case, $3a$ has last digit $6$, and in the latter case, $3a$ has last digit $1$. This gives a final answer of $6+1=7$."
] | United States | HMMO 2020 | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 7 | |
02jr | Problem:
| Seleção | Jogos | V | E | D | GM | GS | P |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Dinamarca | 3 | 2 | 1 | 0 | 5 | 2 | 7 |
| Senegal | 3 | 1 | 2 | 0 | 5 | 4 | $?$ |
| Uruguai | 3 | 0 | 2 | 1 | 4 | $\boldsymbol\{?\}$ | 2 |
| França | 3 | 0 | 1 | 2 | 0 | 3 | 1 |
Utilize as informaç... | [
"Solution:\n\n5. (C) Segundo as condições da copa, uma vitória vale 3 pontos, um empate vale 1 ponto e quem sofre uma derrota não pontua. Como Senegal teve uma vitória e dois empates, ele somou: $1 \\times 3 + 2 \\times 1 = 5$ pontos."
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | C | |
0kz5 | Problem:
Compute the number of quadruples $(a, b, c, d)$ of positive integers satisfying
$$
12 a + 21 b + 28 c + 84 d = 2024
$$ | [
"Solution:\nLooking at the equation mod $7$ gives $a \\equiv 3 \\pmod{7}$, so let $a = 7 a' + 3$. Then mod $4$ gives $b \\equiv 0 \\pmod{4}$, so let $b = 4 b'$. Finally, mod $3$ gives $c \\equiv 2 \\pmod{3}$, so let $c = 3 c' + 2$.\nNow our equation yields\n$$\n84 a' + 84 b' + 84 c' + 84 d = 2024 - 3 \\cdot 12 - 2 ... | United States | HMMT February 2024 Guts Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | final answer only | 2024 | |
0bfv | Find all integers $n \ge 2$ for which there exist $x_1, x_2, \dots, x_n \in \mathbb{R}^*$ such that
$$
x_1 + x_2 + \dots + x_n = \frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_n} = 0.
$$ | [
"Let $M$ be the set of the numbers $n$ that satisfy the conditions above. We shall show that $M = \\mathbb{N} \\setminus \\{0, 1, 3\\}$.\n\nWe can observe that $2p \\in M$, for any $p \\ge 1$, the relations being satisfied e.g. for $x_1 = x_2 = \\dots = x_p = 1$ and $x_{p+1} = x_{p+2} = \\dots = x_{2p} = -1$.\n\nMo... | Romania | The Danube Mathematical Competition | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | All integers n ≥ 2 with n ≠ 3 | |
06ya | Let $n$ and $T$ be positive integers. James has $4 n$ marbles with weights $1,2, \ldots, 4 n$. He places them on a balance scale, so that both sides have equal weight. Andrew may move a marble from one side of the scale to the other, so that the absolute difference in weights of the two sides remains at most $T$.
Find,... | [
"Consider partitioning the weights into pairs $(t, 4 n+1-t)$. Suppose that each side of the balance contains $n$ of those pairs. If one side of the balance contains the pair $(t, 4 n+1-t)$ for $1 \\leqslant t < 2 n$ and the other side contains $(2 n, 2 n+1)$, then the following sequence of moves swaps those pairs b... | IMO | IMO2024 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | 4n | |
0cyv | For any positive integer $n$ let $a_{n}$ be the number of pairs $(x, y)$ of integers satisfying $|x^{2}-y^{2}|=n$.
a. Find $a_{1432}$ and $a_{1433}$.
b. Find $a_{n}$. | [
"Let us consider the following cases.\n\nCase 1: $n$ is odd. The equation $|x^{2}-y^{2}|=n$ is equivalent to\n$$\n|x-y||x+y|=n,\n$$\nhence $|x-y|=d$ and $|x+y|=\\frac{n}{d}$, where $d$ is a divisor of $n$. The system\n$$\n\\left\\{\\begin{array}{l}\n|x-y|=d \\\\\n|x+y|=\\frac{n}{d}\n\\end{array}\\right.\n$$\nhas f... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | a_1432 = 16, a_1433 = 8; and in general
- a_n = 4·tau(n) if n is odd,
- a_n = 0 if n ≡ 2 mod 4,
- a_n = 4·tau(n/4) if n is divisible by 4. | |
0gh6 | Let $n \ge 3$ be a positive integer. Show that for any real numbers $0 \le x_1, x_2, \dots, x_n \le 1$ satisfying $x_1 + x_2 + \dots + x_n = 3$, there always exist $x_i$, $x_j$ such that $x_ix_j > 2^{-|i-j|}$.
令 $n \ge 3$ 為正整數。證明：滿足 $x_1 + x_2 + \cdots + x_n = 3$ 以及 $0 \le x_1, x_2, \ldots, x_n \le 1$ 的實數 $x_1, x_2, \... | [
"Let $1 \\le a < b \\le n$ be such that $2^{b-a}x_a x_b$ is maximal. This choice of $a$ and $b$ implies that $x_{a+t} \\le 2^t x_a$ for all $1-a \\le t \\le b-a-1$, and similarly $x_{b-t} \\le 2^t x_b$ for all $b-n \\le t \\le b-a+1$. Now, suppose that $x_a \\in (\\frac{1}{2^{u+1}}, \\frac{1}{2^u}]$ and $x_b \\in (... | Taiwan | 2023 數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | Chinese (Traditional) | proof only | null | |
0lci | Let be given a sequence $(u_n)$ defined by
$$
u_1 = 1, u_2 = 11, u_{n+2} = u_{n+1} + 5u_n, \forall n \in \mathbb{Z}^+.$$
Prove that $u_n$ is not a perfect square for all $n > 3$. | [] | Vietnam | Vietnamese Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
02ve | Problem:
Um número natural $n$ é interessante se a soma dos dígitos de $n$ é igual à soma dos dígitos de $3n+11$. Verifique que existem infinitos números interessantes. | [
"Solution:\n\nDado um número natural $n$, vamos denotar por $s(n)$ a soma dos dígitos de $n$. Podemos fazer uma tabela com os primeiros inteiros positivos para encontrar algum exemplo de número interessante.\n\n| $n$ | $3n+11$ | $s(n)$ | $s(3n+11)$ | $n$ | $3n+11$ | $s(n)$ | $s(3n+11)$ |\n| :---: | :---: | :---: | ... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0e5y | Problem:
Reši enačbo $\left(1-\left(1+x^{-2}\right)^{-1}\right)^{-1}=3,25$. | [
"Solution:\n\nUredimo notranji oklepaj $\\left(1-\\left(1+\\frac{1}{x^{2}}\\right)^{-1}\\right)^{-1}=3,25$, razširimo na skupni imenovalec $\\left(1-\\left(\\frac{x^{2}+1}{x^{2}}\\right)^{-1}\\right)^{-1}=3,25$. Upoštevamo negativni eksponent $\\left(1-\\frac{x^{2}}{x^{2}+1}\\right)^{-1}=3,25$ in znova razširimo na... | Slovenia | Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | -1.5, 1.5 | |
03m6 | Consider 70-digit numbers $n$, with the property that each of the digits $1, 2, 3, \dots, 7$ appears in the decimal expansion of $n$ ten times (and $8$, $9$, and $0$ do not appear). Show that no number of this form can divide another number of this form. | [
"Assume the contrary: there exist $a$ and $b$ of the prescribed form, such that $b \\ge a$ and $a$ divides $b$. Then $a$ divides $b-a$.\n\nClaim: $a$ is not divisible by $3$ but $b-a$ is divisible by $9$. Indeed, the sum of the digits is $10(1 + \\cdots + 7) = 280$, for both $a$ and $b$. [Here one needs to know or ... | Canada | Kanada 2011 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
01wa | Let $ABC$ be a triangle with $AB = AC$, and let $M$ be the midpoint of $BC$. Let $P$ be a point such that $PB < PC$ and $PA$ is parallel to $BC$. Let $X$ and $Y$ be points on the lines $PB$ and $PC$, respectively, so that $B$ lies on the segment $PX$, $C$ lies on the segment $PY$ and $\angle PXM = \angle PYM$.
Prove th... | [
"1. See IMO 2018 Shortlist, Problem G2."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Spiral similarity"
] | English | proof only | null | |
0kvk | Problem:
There is a unique quadruple of positive integers $(a, b, c, k)$ such that $c$ is not a perfect square and $a+\sqrt{b+\sqrt{c}}$ is a root of the polynomial $x^{4}-20 x^{3}+108 x^{2}-k x+9$. Compute $c$. | [
"Solution:\n\nThere are many ways to do this, including bashing it out directly.\nThe four roots are $a \\pm \\sqrt{b \\pm \\sqrt{c}}$, so the sum of roots is $20$, so $a=5$. Next, we compute the sum of squares of roots:\n$$\n(a+\\sqrt{b \\pm \\sqrt{c}})^2+(a-\\sqrt{b \\pm \\sqrt{c}})^2=2 a^2+2 b \\pm 2 \\sqrt{c}\n... | United States | HMMT November 2023 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 7 | |
0i44 | Problem:
Determine the number of palindromes that are less than $1000$. | [
"Solution:\n\nEvery one-digit number (there are nine) is a palindrome.\n\nThe two-digit palindromes have the form $\\underline{a}\\,\\underline{a}$ for a nonzero digit $a$, so there are nine of them.\n\nA three-digit palindrome is $\\underline{a}\\,\\underline{b}\\,\\underline{a}$ with $a$ a nonzero digit and $b$ a... | United States | HMMT 2002 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 108 | |
00db | Hallar todos los enteros $n>1$ para los que es posible escribir en las casillas de un tablero de $n \times n$ los números enteros desde $1$ hasta $n^2$, sin repeticiones, de modo que en cada fila y en cada columna el promedio de los $n$ números escritos sea un número entero. | [
"La propuesta es hacer que en cada fila y en cada columna la suma de los números escritos sea un múltiplo de $n$. Por lo tanto, reducimos el problema a completar el tablero con los números $0, 1, 2, \\ldots, n-1$, usando a cada uno de ellos exactamente $n$ veces, de modo que se verifique la condición.\n\nSi $n$ es ... | Argentina | Nacional OMA | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | Spanish | proof and answer | all integers n ≥ 3 | |
0kc4 | Problem:
Three players play tic-tac-toe together. In other words, the three players take turns placing an "A", "B", and "C", respectively, in one of the free spots of a $3 \times 3$ grid, and the first player to have three of their label in a row, column, or diagonal wins. How many possible final boards are there wher... | [
"Solution:\n\nIn all winning cases for the third player, every spot in the grid must be filled. There are two ways that player $C$ wins along a diagonal, and six ways that player $C$ wins along a row or column. In the former case, any arrangement of the $A$s and $B$s is a valid board, since every other row, column,... | United States | HMMO 2020 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 148 | |
01pc | Find all polynomials $P(x)$ such that the equality
$$
(x - 1)P(x + 1) - (x + 1)P(x - 1) = 4P(x)
$$
holds for all real $x$. | [
"Answer: $P(x) = a x(x - 1)(x + 1)$, where $a \\in \\mathbb{R}$.\nSet $x = 1$ and $x = -1$ in the initial identity\n$$\n(x - 1)P(x + 1) - (x + 1)P(x - 1) = 4P(x). \\quad (1)\n$$\nThus we obtain $-2P(0) = 4P(1)$ and $-2P(0) = 4P(-1)$ respectively. Setting $x = 0$ in (1), we obtain $-P(1) - P(-1) = 4P(0)$, so, taking... | Belarus | BelarusMO 2013_s | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | P(x) = a x(x - 1)(x + 1), where a ∈ ℝ | |
01mn | Prove that if positive numbers $a, b, x, y$ satisfy the inequality $ab \ge xa + yb$, then they satisfy the inequality $\sqrt{a+b} \ge \sqrt{x} + \sqrt{y}$. | [] | Belarus | 61st Belarusian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0kbz | Problem:
A point $(x, y)$ is selected uniformly at random from the unit square $S=\{(x, y) \mid 0 \leq x \leq 1, 0 \leq y \leq 1\}$. If the probability that $(3x+2y, x+4y)$ is in $S$ is $\frac{a}{b}$, where $a, b$ are relatively prime positive integers, compute $100a+b$. | [
"Solution:\n\nUnder the transformation $(x, y) \\mapsto (3x+2y, x+4y)$, $S$ is mapped to a parallelogram with vertices $(0,0)$, $(3,1)$, $(5,5)$, and $(2,4)$. Using the shoelace formula, the area of this parallelogram is $10$.\n\nThe intersection of the image parallelogram and $S$ is the qu... | United States | HMMO 2020 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Linear Algebra > Determinants"
] | null | proof and answer | 820 | |
0kau | Let $ABC$ be a triangle and let $M$ and $N$ denote the midpoints of $\overline{AB}$ and $\overline{AC}$, respectively. Let $X$ be a point such that $\overline{AX}$ is tangent to the circumcircle of triangle $ABC$. Denote by $\omega_B$ the circle through $M$ and $B$ tangent to $\overline{MX}$, and by $\omega_C$ the circ... | [
"**First solution using symmedians (Merlijn Staps)** Let $\\overline{XY}$ be the other tangent from $X$ to $(AMN)$.\n**Claim.** Line $\\overline{XM}$ is tangent to $(BMY)$; hence $Y$ lies on $\\omega_B$.\n\n*Proof.* Let $Z$ be the midpoint of $\\overline{AY}$. Then $\\overline{MX}$ is the $... | United States | USA IMO TST | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, ... | null | proof only | null | |
09y1 | Around a round table $n \ge 3$ players are sitting. The game leader divides $n$ coins among the players, in such a way that not everyone gets exactly one coin. Any player can see the number of coins of each other player. Every 10 seconds, the game leader rings a bell. At that moment, each player looks how many coins th... | [
"(a) Consider the situation where the first player has 2 coins, the second player has 0 coins and all other players have 1 coin. This situation looks as follows:\n$$\n\\underbrace{2011\\cdots11}_{n-2 \\text{ ones}}\n$$\nFor example, for $n=3$ the starting distribution is 201. We see that the first and the third pla... | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | a) For any n ≥ 3: starting distribution 2,0,1,1,...,1 (one player has two, the next has zero, all others have one). This pattern shifts each round and never reaches all ones.
b) For any n ≥ 4: starting distribution 2,0,0,2,1,1,...,1 (first has two, second and third have zero, fourth has two, all others have one). After... | |
01q9 | Let $\Gamma_B$ and $\Gamma_C$ be excircles of an acute-angled triangle $ABC$ opposite to its vertices $B$ and $C$, respectively. Let $C_1$ and $L$ be the tangent points of $\Gamma_C$ and the side $AB$ and the line $BC$ respectively. Let $B_1$ and $M$ be the tangent points of $\Gamma_B$ and the side $AC$ and the line $B... | [
"Let $\\angle BAC = \\alpha$, $\\angle CBA = \\beta$, $\\angle ACB = \\gamma$. Let $AH$ be the altitude of the triangle $ABC$ and $X'$ be the point of intersection of $AH$ and the line $MB_1$. Then $\\angle X'AB_1 = 90^\\circ - \\gamma$. Since $\\gamma = \\angle ACB = \\angle CB_1M + \\angle B_1MC$ and $CB_1 = CM$ ... | Belarus | Selection and Training Session | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | English | proof only | null | |
0i38 | Problem:
A right triangle has a hypotenuse of length $2$, and one of its legs has length $1$. The altitude to its hypotenuse is drawn. What is the area of the rectangle whose diagonal is this altitude? | [
"Solution:\n\nCall the triangle $ABC$, with $AC = 2$ and $BC = 1$. By the Pythagorean theorem, $AB = \\sqrt{3}$. Call the point at which the altitude intersects the hypotenuse $D$. Let $E \\neq B$ be the vertex of the rectangle on $AB$ and $F \\neq B$ be the vertex of the rectangle on $BC$. Triangle $BDC$ is simila... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 3√3/16 | |
0hji | Problem:
If $k$ is an integer, prove that the number $k^{2}+k+1$ is not divisible by $2006$. | [
"Solution:\n\nThe number $k^{2}+k+1 = k(k+1)+1$ is odd, because $k(k+1)$ is even. Hence it can't be divisible by $2006$."
] | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0hkt | Problem:
There are 10 bags full of coins. All coins look the same and all weigh 10 grams, except the coins from one bag that are fake and all weigh 9 grams. Given a scale, how could you tell which bag has the wrong coins in just one measurement? Explain your answer! | [
"Solution:\n\nTake one coin from the first bag, 2 from the second, 3 from the third, ..., 10 from the 10th and place them on the scale. The total mass shown by the scale should be $$(1+2+\\cdots+10) \\cdot 10 - x$$ grams, where $x$ is the number of fake coins (each coin weighs 10 grams, except for the fake ones tha... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Discrete Mathematics > Algorithms"
] | null | proof only | null | |
03lz | Problem:
Two circles of different radii are cut out of cardboard. Each circle is subdivided into 200 equal sectors. On each circle 100 sectors are painted white and the other 100 are painted black. The smaller circle is then placed on top of the larger circle, so that their centers coincide. Show that one can rotate t... | [
"Solution:\n\nLet $x_{0}, \\ldots, x_{199}$ be variables. Assign the value of $+1$ or $-1$ to $x_{i}$ depending on whether the $(i+1)$st segment of the larger circle (counting counterclockwise) is black or white, respectively. Similarly, assign the value of $+1$ or $-1$ to the variable $y_{i}$ depending on whether ... | Canada | CANADIAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0f0u | Problem:
In the triangle $ABC$, $\angle C = 90^\circ$ and $AC = BC$. Take points $D$ on $CA$ and $E$ on $CB$ such that $CD = CE$. Let the perpendiculars from $D$ and $C$ to $AE$ meet $AB$ at $K$ and $L$ respectively. Show that $KL = LB$. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0gcj | 平面上, 設點 $A, B, C, D, E, F$ 滿足 $\triangle BCD \stackrel{+}{\sim} \triangle ECA \stackrel{+}{\sim} \triangle BFA$ (其中 $\stackrel{+}{\sim}$ 表示正向相似), 且 $I$ 為 $\triangle ABC$ 的內心。
試證: 三角形 $AID, BIE, CIF$ 的三個外接圓的圓心共線。
註: 所謂 $\triangle ABC \stackrel{+}{\sim} \triangle DEF$ 兩三角形正向相似, 除了該兩三角形相似以外, 並要求 $A$ 到 $B$ 到 $C$ 的旋轉方向, 與 $... | [
"解: 先來證明一個引理:\n**Lemma 1.** 給定 $\\triangle ABC$. 設 D, E, F 滿足 $\\{AE, AF\\}, \\{BF, BD\\}, \\{CD, CE\\}$ 分別為 $\\angle BAC, \\angle CBA, \\angle ACB$ 的等角線, 那麼 AD, BE, CF 共點。如圖:\n\n**Proof of Lemma 1.** 設 AD, BE, CF 分別與 $\\odot(BDC), \\odot(CEA), \\odot(AFB)$ 再交於 X, Y, Z。由\n$$\n\\angle BYC = ... | Taiwan | 二〇一八數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane ... | null | proof only | null | |
0jn5 | Problem:
Compute the prime factorization of $25^{3}-27^{2}$. | [
"Solution:\n\nNoticing that $25=5^{2}$ and $27=3^{3}$, we have this is\n$$\n\\begin{aligned}\n5^{6}-3^{6} &=\\left(5^{3}-3^{3}\\right)\\left(5^{3}+3^{3}\\right) \\\\\n& =(5-3)\\left(5^{2}+5 \\cdot 3+3^{2}\\right)(5+3)\\left(5^{2}-5 \\cdot 3+3^{2}\\right) \\\\\n& =2 \\cdot 49 \\cdot 8 \\cdot 19 \\\\\n& =2^{4} \\cdot... | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 2^4 * 7^2 * 19 | |
0jwz | Problem:
A repunit is a positive integer, all of whose digits are $1$'s. Let $a_{1} < a_{2} < a_{3} < \ldots$ be a list of all the positive integers that can be expressed as the sum of distinct repunits. Compute $a_{111}$. | [
"Solution:\nLet $\\{r_{n}\\}_{n \\geq 0}$ be the repunits (so $r_{0} = 1$, $r_{1} = 11$, and so on). We see that for any $n$, there is\n$$\nr_{n-1} + r_{n-2} + \\cdots + r_{0} < \\frac{r_{n}}{10} + \\frac{r_{n}}{100} + \\cdots < \\frac{r_{n}}{9} < r_{n}\n$$\nso $r_{n}$ is only needed when all possible combinations ... | United States | HMMT November | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 1223456 | |
06lj | Find all real-valued functions $f$ defined on the set of real numbers such that
$$
f(f(x) + y) + f(x + f(y)) = 2f(xf(y))
$$
for any real numbers $x$ and $y$. | [
"The answer is any constant function $f$.\nClearly, constant functions are solutions. In the following, we show that there is no solution if $f$ is not a constant function.\nLabel the equation\n$$\nf(f(x) + y) + f(x + f(y)) = 2f(xf(y)). \\qquad (1)\n$$\nSwapping $x$ and $y$, we get $f(f(y) + x) + f(y + f(x)) = 2f(y... | Hong Kong | IMO HK TST | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | All constant functions: f(x) = c for any real constant c. | |
0gak | 設凸五邊形 $AXYZB$ 內接於一個以 $AB$ 為直徑的半圓。令 $K$ 為 $Y$ 對 $AB$ 的垂足，且令 $O$ 為 $AB$ 的中點。令 $L$ 為 $XZ$ 與 $YO$ 的交點。在直線 $KL$ 上取一點 $M$ 使得 $MA = MB$，及設 $I$ 為 $O$ 對直線 $XZ$ 的對稱點。
證明：若四邊形 $XKOZ$ 內接於一圓，則四邊形 $YOMI$ 也內接於一圓。
Let $AXYZB$ be a convex pentagon inscribed in a semicircle with diameter $AB$, and let $K$ be the foot of the altitude fr... | [
"將半圓延伸成為圓 $\\Gamma$。設直線 $LK$ 與 $\\Gamma$ 交於 $P, Q$ 兩點。在射線 $OM$ 上找一點 $W$ 滿足 $OW \\cdot OM = OA \\cdot OB$。由此知 $P, Q, W, O$ 四點共圓，令此圓為 $\\gamma$。\n\n三個圓：$\\Gamma, \\gamma$，以及 $XKOZ$ 的外接圓的根心 (radical center) 為 $L$ 點，因為直線 $XZ$ 與 $PQ$ 為根軸。所以直線 $YO$ 是 $\\Gamma$ 與 $\\gamma$ 的根軸。\n令 $XZ$ 與 $AB$ 的交點為... | Taiwan | 二〇一六數學奧林匹亞競賽第二階段選訓營 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry... | null | proof only | null | |
05bh | Call a number *interesting* if it can be represented as the sum of squares of three distinct non-negative integers. For example, the number $5$ is interesting, because $5 = 0^2 + 1^2 + 2^2$. Call a number *special* if it is not interesting, but can be represented as the product of two distinct interesting numbers.
a. ... | [
"The factorisation $(2k+1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1$, where one of the numbers $k$ and $k+1$ is always even, shows that the square of any odd number gives a remainder of $1$ upon division by $8$. The square of an even number not divisible by $4$ gives a remainder of $4$ and the square of an even number divis... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic"
] | English | proof and answer | One example is 175. There are infinitely many special numbers. | |
02c6 | Problem:
Eliane quer escolher o seu horário para a natação. Ela quer ir a duas aulas por semana, uma de manhã e a outra de tarde, não sendo no mesmo dia nem em dias seguidos. De manhã, há aulas de natação de segunda-feira a sábado, às $9 h$, às $10 h$ e às $11 h$ e de tarde, de segunda-feira a sexta-feira, às $17 h$ e... | [
"Solution:\n\nSe a aula da manhã é segunda ou sexta (em qualquer dos três horários), então o dia da aula de tarde pode ser escolhida de 3 formas diferentes (em qualquer dos dois horários), assim temos $2 \\times 3 \\times 3 \\times 2 = 36$ formas diferentes de escolher o horário.\n\nNo caso em que a aula de manhã s... | Brazil | Lista 2 | [
"Discrete Mathematics > Combinatorics"
] | null | proof and answer | 96 | |
0igj | Problem:
Calculate
$$
\lim_{x \rightarrow 0^{+}} \left(x^{x^{x}} - x^{x}\right).
$$ | [
"Solution:\nWe first calculate $\\lim_{x \\rightarrow 0^{+}} x^{x}$: it is just $\\exp\\left(\\lim_{x \\rightarrow 0^{+}} x \\ln x\\right)$. But\n$$\n\\lim_{x \\rightarrow 0^{+}} x \\ln x = \\lim_{x \\rightarrow 0^{+}} \\frac{\\ln x}{1/x} = \\lim_{x \\rightarrow 0^{+}} \\frac{1/x}{-1/x^{2}} = \\lim_{x \\rightarrow ... | United States | Harvard-MIT Mathematics Tournament | [
"Precalculus > Limits",
"Calculus > Differential Calculus > Derivatives"
] | null | proof and answer | -1 | |
06jc | $ABC$ is a triangle with integral sides. $M$ is the midpoint of $BC$. The in-circle with centre $I$ touches $AB$ and $AC$ at $E$ and $F$ respectively and $D$ is the projection of $M$ on $EF$. Suppose that $ADMI$ is a parallelogram and $AB + BC + CA = 65$. Find $AB \times BC \times CA$. | [
"The product is $9360$.\nLet $a$, $b$, $c$, $r$, $s$ be the lengths of $BC$, $CA$, $AB$, the inradius and the semiperimeter of $\\triangle ABC$ respectively. Let $N$ be the midpoint of $AC$, and let $P$ be the intersection point of $EF$ and $MN$. It is well-known that $\\angle BPC = 90^\\circ$. Therefore, we have $... | Hong Kong | 1997-2023 IMO HK TST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | null | proof and answer | 9360 | |
0h7g | Andriy, Bogdan and Olesia were walking by the same road from home to the school. Andriy was walking with velocity equal to $a$ km/h for $(2-b)$ hours, Bogdan was walking with velocity equal to $b$ km/h for $(2-c)$ hours, Olesia was walking with velocity equal to $c$ km/h for $(2-a)$ hours, where $a, b, c$ are some real... | [
"Analyzing the problem we get:\n$$\nS = a(2-b), \\quad S = b(2-c), \\quad S = c(2-a),\n$$\nwhere $S$ -- positive integer which is equal to the distance.\n\nWe may assume that $a \\ge b$. If $a > b$, then $2-b < 2-c$ or $b > c$. Analogously, $2-c < 2-a$ or $c > a$. Contradiction. So $a = b = c$. Then we have that $S... | Ukraine | UkraineMO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 1 | |
0k28 | Problem:
You are the first lucky player to play in a slightly modified episode of Deal or No Deal! Initially, there are sixteen cases marked $1$ through $16$. The dollar amounts in the cases are the powers of $2$ from $2^{1} = 2$ to $2^{16} = 65536$, in some random order. The game has eight turns. In each turn, you cho... | [
"Solution:\nFirstly, note that it is always optimal for you to take the case with the largest amount of money. To prove this rigorously, consider a strategy where you don't - then change the first move where you deviate to taking the maximal case. This can only increase your return.\n\nWe calculate the probability ... | United States | HMMT February 2018 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | (14 * 2^17 + 4) / 15 | |
09xu | In triangle *ABC*, a point $D$ lies on side $BC$ and a point $E$ lies on side $AC$ such that the line segments $BD$, $DE$, and $AE$ have the same length. The point $F$ is the intersection between the line segments $AD$ and $BE$. Angle $C$ is $68^\circ$.
What is the size of angle $F$ in triangle $AFB$?
![](attached_ima... | [
"E) $124^\\circ$"
] | Netherlands | Dutch Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation"
] | English | MCQ | E | |
078a | Prove that for all integers $k > 2$, there exists $k$ distinct positive integers $a_1, \dots, a_k$ such that
$$
\sum_{1 \le i < j \le k} \frac{1}{a_i a_j} = 1.
$$ | [
"Let us first introduce some notation: given positive integers $n$ and $k$, $e_{n,k}(x_1, \\dots, x_n)$ will denote the $k$th elementary symmetric polynomial in the $n$-variables $x_1, \\dots, x_n$.\n\nWe define a sequence of sets inductively as follows: $S_3 = \\{1, 2, 3\\}$. Now for $n > 2$, suppose $S_n = \\{a_1... | India | IMO_TSTs_India | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
048b | If
$$
\frac{x_1}{x_1 + 1} = \frac{x_2}{x_2 + 3} = \frac{x_3}{x_3 + 5} = \dots = \frac{x_{1006}}{x_{1006} + 2011}, \\
x_1 + x_2 + \dots + x_{1006} = 503^2,
$$
determine $x_{1006}$. | [
"Let us denote\n$$\n\\frac{x_1}{x_1 + 1} = \\frac{x_2}{x_2 + 3} = \\frac{x_3}{x_3 + 5} = \\dots = \\frac{x_{1006}}{x_{1006} + 2011} = a.\n$$\nFrom $\\frac{x_k}{x_k + (2k-1)} = a$ follows $x_k = \\frac{a}{1-a} \\cdot (2k-1)$ for $k = 1, 2, \\dots, 1006$. By including this in the last given equality we get\n$$\n\\fra... | Croatia | CroatianCompetitions2011 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | 2011/4 | |
0fm0 | A sequence $(a_n)_{n \ge 1}$ is defined recursively as
$$
a_1 = 1, \ a_2 = 5, \text{ and } a_n = \frac{a_{n-1}^2 + 4}{a_{n-2}}, \text{ for } n \ge 3.
$$
Prove that every term in the sequence is an integer. Find an explicit formula for $a_n$. | [
"Let us compute the first few terms:\n\n$a_1 = 1$\n\n$a_2 = 5$\n\n$a_3 = \\frac{5^2 + 4}{1} = \\frac{25 + 4}{1} = 29$\n\n$a_4 = \\frac{29^2 + 4}{5} = \\frac{841 + 4}{5} = \\frac{845}{5} = 169$\n\n$a_5 = \\frac{169^2 + 4}{29} = \\frac{28561 + 4}{29} = \\frac{28565}{29} = 985$\n\n$a_6 = \\frac{985^2 + 4}{169} = \\fra... | Spain | Spanija 2012 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | a_n = [ (sqrt(2) + 1)(3 + 2 sqrt(2))^{n-1} + (sqrt(2) - 1)(3 - 2 sqrt(2))^{n-1} ] / (2 sqrt(2)) | |
0b4x | Problem:
How many 9-term sequences $a_{1}, \ldots, a_{9}$ of nonnegative integers are there such that
- $0 \leq a_{i} < i$ for all $i = 1, \ldots, 9$; and
- there are no ordered triples $(i, j, k)$ with $1 \leq i < j < k \leq 9$, such that $a_{i}, a_{j}, a_{k}$ are all distinct? | [] | Philippines | 25th Philippine Mathematical Olympiad Area Stage | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 503 | |
0fin | Problem:
Una empresa produce semanalmente 300 bicicletas de montaña que vende íntegramente al precio de 600 euros cada una. Tras un análisis de mercados observa que si varía el precio, también varían sus ventas (de forma continua) según la siguiente proporción: por cada 7 euros que aumente o disminuya el precio de sus... | [
"Solution:\n\nAl precio actual, los ingresos semanales son $600 \\times 300 = 180.000$ euros.\n\na) Si incrementa el precio en 7 euros, entonces vende 297 bicicletas, obteniendo en este caso $607 \\times 297 = 180.279$ euros. Luego la respuesta a la primera pregunta es: Sí.\n\nb) Llamamos $x$ a la cantidad de euros... | Spain | XXXV Olimpiada Matemática Española | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | a) Yes. b) 650 euros | |
0ieb | Problem:
You are given a set of cards labeled from $1$ to $100$. You wish to make piles of three cards such that in any pile, the number on one of the cards is the product of the numbers on the other two cards. However, no card can be in more than one pile. What is the maximum number of piles you can form at once? | [
"Solution: $8$\nCertainly, the two factors in any pile cannot both be at least $10$, since then the product would be at least $10 \\times 11 > 100$. Also, the number $1$ can not appear in any pile, since then the other two cards in the pile would have to be the same. So each pile must use one of the numbers $2, 3, ... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 8 | |
030d | Problem:
Em uma competição, os competidores ocupam todos os lugares de um salão retangular onde os assentos estão organizados em filas e colunas de tal modo que há mais de duas filas e em cada fila há mais de dois assentos. Em um dado momento, esses competidores recebem a ordem de cumprimentarem com um aperto de mão a... | [
"Solution:\n\na) Um competidor que sentou em um dos 4 cantos do salão deu 3 apertos de mão. Se um competidor sentou em algum dos 4 bordos, mas sem estar posicionado nos cantos, ele deu 5 apertos de mão. Finalmente, se um competidor sentou no interior do salão, ele deu 8 apertos de mão.\nA figura a seguir ilustra es... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a) Possible handshakes per person: 3, 5, or 8.
b) Counts: 4 people have 3; 2(n−2)+2(m−2) people have 5; (n−2)(m−2) people have 8.
c) Total competitors: 280, with (n, m) = (14, 20) or (20, 14). | |
0ayv | Problem:
Suppose $a_{1}, a_{2}, \ldots$ is a sequence of integers, and $d$ is some integer. For all natural numbers $n$,
(i) $\left|a_{n}\right|$ is prime;
(ii) $a_{n+2}=a_{n+1}+a_{n}+d$.
Show that the sequence is constant. | [
"Solution:\n\nConsider the sequence $\\{b_{n}\\}$ defined by $b_{n}=a_{n}+d$ for all $n$, so that $b_{n+2}=b_{n+1}+b_{n}$ for all $n$. This sequence is determined by its first two terms $b_{1}$ and $b_{2}$, and the same holds true if we reduce the sequence $\\bmod a_{1}$. Taking remainders $\\bmod a_{1}$, pairs of ... | Philippines | 20th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0l4w | Sixteen chairs are arranged in a row. Eight people each select a chair in which to sit so that no person sits next to two other people. Let $N$ be the number of subsets of the 16 chairs that could be selected. Find the remainder when $N$ is divided by 1000. | [] | United States | AIME II | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 907 | |
00ch | Sean $A, B, C$ los colores de las tres cajas y asignamos a cada número el color de la caja que lo contiene. Ahora el problema es hallar el máximo valor de $n$ tal que es posible colorear los números $1, 2, 3, \ldots, n$ con los colores $A, B, C$ de modo que ningún par de números de un mismo color difieran en el cuadrad... | [
"La respuesta es $29$. Supongamos que los números $1, 2, 3, \\ldots, 29$ se pueden colorear con los colores $A, B, C$ de modo que ningún par de números de un mismo color difieren en el cuadrado de un entero. Sea $f(i)$ el color del número $1 \\le i \\le 29$. Como $9, 16$ y $25$ son cuadrados, a los números $1, 10$ ... | Argentina | Nacional 2019 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 28 | |
07b3 | The sequence $\{a_n\}_{n=1}^{\infty}$ of natural numbers satisfies the following relation:
$$
a_{n+2} = \left\lfloor \frac{2a_{n+1}}{a_n} \right\rfloor + \left\lfloor \frac{2a_n}{a_{n+1}} \right\rfloor,
$$
for which by $\lfloor x \rfloor$ we mean the integer part of $x$. Prove that there exists natural number $m$ such... | [
"First we prove some lemmas.\n\n**Lemma 1.** $a_n \\ge 3$ for all natural numbers $n \\ge 3$.\n*Proof.* Suppose that $a_n < 3$ for some natural number $n \\ge 3$. If $a_{n-1} \\ge a_{n-2}$, then\n$$\n\\frac{2a_{n-1}}{a_{n-2}} \\ge 2 \\Rightarrow \\left[ \\frac{2a_{n-1}}{a_{n-2}} \\right] \\ge 2 \\Rightarrow \\left[... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
05jl | Problem:
Prouver que si $n$ est un entier strictement positif, l'expression
$$
\frac{\sqrt{n+\sqrt{0}}+\sqrt{n+\sqrt{1}}+\sqrt{n+\sqrt{2}}+\cdots+\sqrt{n+\sqrt{n^{2}-1}}+\sqrt{n+\sqrt{n^{2}}}}{\sqrt{n-\sqrt{0}}+\sqrt{n-\sqrt{1}}+\sqrt{n-\sqrt{2}}+\cdots+\sqrt{n-\sqrt{n^{2}-1}}+\sqrt{n-\sqrt{n^{2}}}}
$$
est indépendant... | [
"Solution:\n\nEn calculant le carré de chacun des deux membres, on déduit que, pour tous réels $a, b$ tels que $0 \\leqslant b \\leqslant a$, on a\n$$\n\\sqrt{a+\\sqrt{a^{2}-b^{2}}}=\\sqrt{\\frac{a+b}{2}}+\\sqrt{\\frac{a-b}{2}}\n$$\nEn particulier, pour tous entiers naturels $n$ et $m$, avec $m \\leqslant n^{2}$, o... | France | Olympiades Françaises de Mathématiques, Envoi Numéro 3 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Other"
] | null | proof only | null | |
05ef | Problem:
In each cell of a $2025 \times 2025$ board, a nonnegative real number is written in such a way that the sum of the numbers in each row is equal to $1$, and the sum of the numbers in each column is equal to $1$. Define $r_{i}$ to be the largest value in row $i$, and let $R = r_{1} + r_{2} + \dots + r_{2025}$. ... | [
"Solution:\n\nAnswer: $\\frac{2025}{89}$.\nIn general, if the table is $m^{2} \\times m^{2}$, the answer is $\\frac{m^{2}}{2m - 1}$.\n\nThe example is as follows: label rows and columns from $1$ to $m^{2}$, from top to bottom and left to right. For the first $m$ columns, write $\\frac{1}{m}$ in all squares whose co... | European Girls' Mathematical Olympiad (EGMO) | EGMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2025/89 | |
0gzo | Each girl: Oksana, Olesya, Olya and Olexandra has a rectangle with sides $2010$ and $10$. They were given a task: to cut this rectangle into two pieces from which one can make a triangle without overlaps. All of them succeeded in this task. Can they make pairwise distinct triangles? | [
"**Answer:** Yes. The required example on the fig.01.\n\n\nFig.01"
] | Ukraine | 50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010) | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | Yes | |
02a9 | Problem:
Qual é o menor número, $5^{2002}$ ou $3^{2002}+4^{2002}$? | [
"Solution:\nComo $5^{2}=3^{2}+4^{2}$, temos $5^{2002}=(3^{2}+4^{2})^{1001}$. Sabemos que, para $a>0$ e $b>0$,\n$$\n(a+b)^{1001}>a^{1001}+b^{1001}\n$$\nAssim, $5^{2002}>3^{2002}+4^{2002}$."
] | Brazil | Desafios | [
"Algebra > Equations and Inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 3^{2002}+4^{2002} | |
0g1q | Problem:
Quel est le nombre maximal de Skew-Tetrominos que l'on peut placer sur un rectangle $8 \times 9$ sans recouvrement?
Remarque: Les tetrominos peuvent être tournés et réfléchis. | [
"Solution:\n\nLe nombre maximal de Skew-Tetrominos est 16.\n\nMontrons déjà qu'on peut en placer 16. On sépare le rectangle en 4 plus petits rectangles de 2 lignes et 9 colonnes. Sur chacun de ces petits rectangles on peut placer 4 Skew-Tetrominos (en les mettant tous dans l'orientation sur la feuille d'examen), do... | Switzerland | SMO - Vorrunde | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 16 | |
0d6p | Let $n \geq 4$ be a positive integer and there exist $n$ positive integers that are arranged on a circle such that:
- The product of each pair of two non-adjacent numbers is divisible by $2015 \cdot 2016$.
- The product of each pair of two adjacent numbers is not divisible by $2015 \cdot 2016$.
Find the maximum value o... | [
"See the solution in the test of level 4. $\\square$"
] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | 6 | |
0ad2 | The book that Marko reads has $242$ pages. The first day he has read $22$ pages. During the second day he has read $4$ pages more than the first and the same during the third day. How many days will Marko need to finish the book if he reads two pages more than the third day each day from the fourth day onward? | [
"Let $x$ be the number of days from the fourth day onward that Marko needs to finish the book. Then we have the equation\n$$\n22 + 2 \\cdot (22 + 4) + x \\cdot (22 + 4 + 2) = 242,\n$$\nfrom where we obtain $x = 6$. So Marko will read the book for $3 + x = 3 + 6 = 9$ days."
] | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 9 | |
0ipg | Problem:
Find the sum of all positive integers $n$ such that $n$ divides $n^{2}+n+2$. | [
"Solution:\nSince $n$ always divides $n^{2}+n$, the only $n$ that work are divisors of $2$, because if $n$ divides $a$ and $n$ divides $b$, then $n$ divides $a+b$. So the solutions are $1$ and $2$ which sum to $3$."
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 3 | |
0k13 | Problem:
Abbot writes the letter $A$ on the board. Every minute, he replaces every occurrence of $A$ with $A B$ and every occurrence of $B$ with $B A$, hence creating a string that is twice as long. After 10 minutes, there are $2^{10}=1024$ letters on the board. How many adjacent pairs are the same letter? | [
"Solution:\n\nAnswer: 341\nLet $a_{n}$ denote the number of adjacent pairs of letters that are the same after $n$ minutes, and $b_{n}$ the number of adjacent pairs that are different.\n\nLemma 1. $a_{n}=b_{n-1}$ for all $n \\geq 0$.\n\nProof. Any adjacent pair of identical letters $X X$ at stage $n$ either came fro... | United States | HMMT November 2018 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 341 | |
0gv1 | Let $ABCD$ be a cyclic quadrilateral and let the midpoints of $AB$, $BC$, $CD$ and $DA$ be $K$, $L$, $M$ and $N$, respectively. Let the reflections of the point $M$ with respect to the lines $AD$ and $BC$ be $P$ and $Q$, respectively. Finally the circumcenter of the triangle $KPQ$ be $R$. Prove that $RN = RL$. | [
"We start by noting that $KLMN$ is a parallelogram, and by the symmetry we have $NP = NM = KL$ and $LQ = LM = KN$. Moreover, we have $\\angle KNM = \\angle KLM$.\n\nSince $ABCD$ is cyclic, $MN \\parallel AC$ and $LM \\parallel BD$ we get that $\\angle DNM = \\angle DAC = \\angle DBC = \\angle MLC$ and from the symm... | Turkey | Team Selection Test for JBMO 2024 | [
"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",
"Geometry > Plane Geometry > Miscellaneou... | English | proof only | null | |
0gil | Find all functions $f$ from real numbers to real numbers such that
$$
2f((x+y)^2) = f(x+y) + (f(x))^2 + (4y-1)f(x) - 2y + 4y^2
$$
holds for all real numbers $x$ and $y$.
試求所有從實數映至實數的函數 $f$, 滿足：
$$
2f((x+y)^2) = f(x+y) + (f(x))^2 + (4y-1)f(x) - 2y + 4y^2
$$
對於所有實數 $x$ 和 $y$ 皆成立。 | [
"唯一解為 $f(x) = 2x$。\n\n解法一：\n1. 原式代入 $y=0$，得\n$$\n2f(x^2) = f(x) + f(x)^2 - f(x) = f(x)^2, \\qquad (1)\n$$\n從而有\n$$\nf(x)^2 = 2f(x^2) = 2f((-x)^2) = f(-x)^2. \\qquad (2)\n$$\n2. 原式代入 $y=-x$，得\n$$\nf(0) = f(x)^2 - (4x+1)f(x) + 2x + 4x^2. \\qquad (3)\n$$\n這表示\n$$\nf(x)^2 - (4x+1)f(x) + 2x + 4x^2 = f(0) = f(-0)^2 - (-4... | Taiwan | Taiwan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | Chinese; English | proof and answer | f(x) = 2x | |
01d0 | Set $A$ consists of 2016 natural numbers. All prime divisors of these numbers are smaller than 29. Prove that there are four distinct numbers $a$, $b$, $c$ and $d$ in $A$ such that $abcd$ is a square. | [
"There are nine prime numbers smaller than 29. Let us denote them as $p_1$, $p_2$, \\dots, $p_9$. To each number $n$ from $A$ we can assign a 9-element sequence $(n_1, n_2, \\dots, n_9)$ such that $n_i = 1$ when in the factorization of $n$ $p_i$ has odd exponent, and $n_i = 0$ otherwise. There are only 512 differen... | Baltic Way | Baltic Way 2016 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0dhq | Given is an odd integer $n \ge 1$. Let $S$ be the set of all points in the three dimensional space, whose all coordinates belong to the set $\{0, 1, ..., n\}$. Determine the maximum size of a subset $A \subset S$ with the following property: For every two distinct points $(x_1, x_2, x_3)$, $(y_1, y_2, y_3) \in A$ among... | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 3*((n+1)/2)^2 | |
05dm | Problem:
Determine all real constants $t$ such that whenever $a, b, c$ are the lengths of the sides of a triangle, then so are $a^{2}+b c t$, $b^{2}+c a t$, $c^{2}+a b t$. | [
"Solution:\n\nIf $t < 2/3$, take a triangle with sides $c = b = 1$ and $a = 2 - \\epsilon$. Then\n$$\nb^{2} + c a t + c^{2} + a b t - a^{2} - b c t = 3 t - 2 + \\epsilon (4 - 2 t - \\epsilon) \\leq 0\n$$\nfor small positive $\\epsilon$; for instance, for any $0 < \\epsilon < (2 - 3 t)/(4 - 2 t)$.\n\nOn the other ha... | European Girls' Mathematical Olympiad (EGMO) | European Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | t in [2/3, 2] | |
043g | Given function $f(x) = |2 - \log_3 x|$, positive real numbers $a, b, c$ satisfy $a < b < c$ and $f(a) = 2f(b) = 2f(c)$. Find the minimum of $\frac{ac}{b}$. | [
"Notice that $f(x) = |\\log_3(\\frac{x}{9})|$ is monotonically decreasing on $(0, 9]$ and monotonically increasing on $[9, +\\infty)$.\n\nBy the conditions satisfied by $a, b, c$, we have $0 < a < b < 9 < c$ and\n$$\n\\log_3\\left(\\frac{9}{a}\\right) = 2\\log_3\\left(\\frac{9}{b}\\right) = 2\\log_3\\left(\\frac{c}... | China | China Mathematical Competition | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | 9 | |
0bn5 | Consider a triangular pyramidal frustum $ABCA'B'C'$. Points $D \in (AA')$, $E \in (BB')$ and $F \in (CC')$ are such that the planes $(AEF)$ and $(DB'C')$ are parallel. Prove that the planes $(A'EF)$ and $(DBC)$ are also parallel. | [
"Denote by $V$ the common point of the supporting lines of the lateral edges of the frustum. As planes $(AEF)$ and $(DB'C')$ are parallel, we have $EF \\parallel B'C'$ and $DB' \\parallel AE$. Thales Theorem gives from $DB' \\parallel AE$ and $A'B' \\parallel AB$:\n$$\n\\frac{VD}{VA} = \\frac{VB'}{VE}, \\quad \\fra... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof only | null | |
033l | Problem:
Let $M$ be a point on a circle $k$. A circle $k_{1}$ with center $M$ meets $k$ at points $C$ and $D$. A chord $AB$ of $k$ is tangent to $k_{1}$ at point $H$. Prove that the line $CD$ bisects the segment $MH$ if and only if $AB$ is a diameter of $k$. | [
"Solution:\nLet $AB$ be a diameter of $k$. Since $\\Varangle D H A = \\Varangle D C H = \\alpha$, $\\Varangle C H B = \\Varangle C D H = \\beta$, $\\Varangle D M H = 2\\alpha$, $\\Varangle C M H = 2\\beta$ and $\\Varangle M D C = 90^{\\circ} - \\alpha - \\beta$, the Sine theorem for $\\triangle D M O$ gives\n$$\nMO... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0dnj | Problem:
Нека је $k$ природан број. За $n \in \mathbb{N}$ означимо са $f_{k}(n)$ најмањи природан број већи од $k n$ такав да је $n f_{k}(n)$ потпун квадрат природног броја. Ако је испуњено $f_{k}(m)=f_{k}(n)$, доказати да важи $m=n$. | [
"Solution:\n\nПретпоставимо да је $f_{k}(m)=f_{k}(n)=q$. Напишимо број $q$ у облику $q=a u^{2}$, где су $a, u \\in \\mathbb{N}$ и $a$ није дељиво ниједним потпуним квадратом већим од 1. Како је $m q=a m u^{2}$ потпун квадрат, то је и $a m$, па следи да је $m=a v^{2}$ за неко $v \\in \\mathbb{N}$. Слично је $n=a w^{... | Serbia | 9. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
01mx | Several chess players took part in a chess tournament. Each player played against every other player. How many games were played? | [] | Belarus | Belarusian Mathematical Olympiad | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | n(n-1)/2 | |
0j4j | Problem:
Rosencrantz and Guildenstern play a game in which they repeatedly flip a fair coin. Let $a_{1}=4$, $a_{2}=3$, and $a_{n}=a_{n-1}+a_{n-2}$ for all $n \geq 3$. On the $n$th flip, if the coin is heads, Rosencrantz pays Guildenstern $a_{n}$ dollars, and, if the coin is tails, Guildenstern pays Rosencrantz $a_{n}$... | [
"Solution:\n\nAnswer: $\\frac{1}{2}-\\frac{1}{2^{1341}}$\n\nSince Rosencrantz and Guildenstern have an equal chance of winning each toss, both have the same probability of ending up with a positive amount of money. Let $x$ denote the probability that they both end up with zero dollars. We wish to find $\\frac{1-x}{... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 1/2 - 1/2^1341 | |
08kd | Problem:
Let $ABCD$ be an isosceles trapezoid with $AB = AD = BC$, $AB \parallel DC$, $AB > DC$. Let $E$ be the point of intersection of the diagonals $AC$ and $BD$ and $N$ be the symmetric point of $B$ with respect to the line $AC$. Prove that quadrilateral $ANDE$ is cyclic. | [
"Solution:\nLet $\\omega$ be a circle passing through the points $A$, $N$, $D$ and let $M$ be the point where $\\omega$ intersects $BD$ for the second time. The quadrilateral $ANDM$ is cyclic and it follows that\n$$\n\\angle NDM + \\angle NAM = \\angle NDM + \\angle BDC = 180^\\circ\n$$\nand\n![](attached_image_1.p... | JBMO | OJBM | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fxm | Problem:
Finde alle Paare $(a, b)$ natürlicher Zahlen, sodass die folgende Gleichung erfüllt ist:
$$
a^{6 a}=b^{b}
$$ | [
"Solution:\nEs gilt $6 a > b$, denn sonst wäre $b^{b} \\geq (6 a)^{6 a} > a^{6 a}$. Division durch $a^{b}$ ergibt die neue Gleichung\n$$\na^{6 a-b} = \\left(\\frac{b}{a}\\right)^{b}\n$$\nin der die linke Seite nach dem eben Gesagten ganz ist, also auch die rechte. Somit ist $a$ ein Teiler von $b = k a$ und die Ungl... | Switzerland | Vorrundenprüfung | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | (1, 1), (3, 9), (16, 64), (3125, 15625) | |
0at1 | Problem:
If $0 < \theta < \pi / 2$ and $1 + \sin \theta = 2 \cos \theta$, determine the numerical value of $\sin \theta$. | [
"Solution:\n$\\frac{3}{5}$"
] | Philippines | Philippines Mathematical Olympiad | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | 3/5 | |
07z4 | Problem:
Si consideri il quadrante infinito in figura, dove tutti i quadratini hanno lato $1$. È possibile colorare di nero alcuni dei quadratini in modo che siano soddisfatte entrambe le seguenti proprietà?
- Per ogni numero naturale $n$, il quadrato con vertice in $O$ e di lato $n$ (con i lati paralleli agli assi) h... | [
"Solution:\n\nSì, è possibile. Per esempio, si colorino di nero:\n- il primo quadratino avente vertice in $O$;\n- i primi $2$ quadratini sulle due diagonali adiacenti alla diagonale centrale;\n- i primi $3$ quadratini sulle due diagonali adiacenti alle precedenti, e così via.\n\nÈ chiaro che ogni diagonale a $45^{\... | Italy | XIII GARA NAZIONALE di MATEMATICA | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
00wj | Problem:
Let $P$ be a point on the circumcircle of a triangle $ABC$. It is known that the base points of the perpendiculars drawn from $P$ onto the lines $AB$, $BC$ and $CA$ lie on one straight line (called a Simson line). Prove that the Simson lines of two diametrically opposite points $P_{1}$ and $P_{2}$ are perpend... | [
"Solution:\n\nLet $O$ be the circumcentre of the triangle $ABC$ and $\\angle B$ be its maximal angle (so that $\\angle A$ and $\\angle C$ are necessarily acute). Further, let $B_{1}$ and $C_{1}$ be the base points of the perpendiculars drawn from the point $P$ to the sides $AC$ and $AB$ respectively and let $\\alph... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Misc... | null | proof only | null | |
02o3 | a. Show a positive integer not greater than $1000$ with at least $20$ positive divisors.
b. Does there exist a positive integer not greater than $11000$ with at least $200$ positive divisors? | [
"a. For example, $900 = 2^2 \\cdot 3^2 \\cdot 5^2$, which has $(2+1) \\cdot (2+1) \\cdot (2+1) = 27$ positive divisors.\n\nb. No, there doesn't. Let $n$ be a number with at least $200$ divisors. If the $i$-th divisor is $d$, then the $i$-th to last divisor is $\\frac{n}{d}$. Let $m$ be the $100$th divisor. So $m \\... | Brazil | Brazilian Math Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | a: 900; b: No | |
043m | Let $a, b$ be real numbers and function $f(x) = x^3 + a x^2 + b x$. If there exist three real numbers $x_1, x_2, x_3$ satisfying $x_1 + 1 \le x_2 \le x_3 - 1$ and $f(x_1) = f(x_2) = f(x_3)$. Find the minimum of $|a| + 2|b|$. | [
"For function $f(x)$ and real numbers $x_1, x_2, x_3$ satisfying the conditions, let $f(x_1) = f(x_2) = f(x_3) = c$, and then $x_1, x_2, x_3$ are the three real roots of cubic equation $x^3 + a x^2 + b x - c = 0$. By Vieta's formulas, we know that\n$$\na = -(x_1 + x_2 + x_3), \\quad b = x_1 x_2 + x_2 x_3 + x_3 x_1.... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | √3 | |
0diq | For $a, b, c > 0$, denote $m = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}$, $n = \frac{a}{c} + \frac{c}{b} + \frac{b}{a}$. Find all possible values of $k$ such that the following inequality is true for all $m, n$ defined as above:
$$
(m - n)k^2 + (2m - n - 3)k - 8m + 6n + 6 \le 0.
$$ | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | [-4, -3] ∪ {2} | |
0i45 | Problem:
Nine nonnegative numbers have average $10$. What is the greatest possible value for their median? | [
"Solution:\n\nIf the median is $m$, then the five highest numbers are all at least $m$, so the sum of all the numbers is at least $5m$. Thus $90 \\geq 5m \\Rightarrow m \\leq 18$. Conversely, we can achieve $m=18$ by taking four $0$'s and five $18$'s."
] | United States | Harvard-MIT Math Tournament | [
"Statistics > Mathematical Statistics"
] | null | proof and answer | 18 | |
0he8 | It is given that there exists such prime $p$, for which $10^{17} \le p \le 10^{17} + 10$. Find $p$. | [
"We should find the last digit of this number. It cannot be even or equal to $5$.\n\nNumber $10^{17} + 1$ is divisible by $11$.\n\nConsider number $10^{17} + 9$ modulo $7$. Then, $10 \\equiv 3 \\pmod{7} \\Rightarrow 10^2 \\equiv 2 \\pmod{7} \\Rightarrow 10^3 \\equiv -1 \\pmod{7} \\Rightarrow 10^{15} \\equiv -1 \\pm... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 10^17 + 3 | |
062i | Problem:
Man zeige, dass es eine ganze Zahl $a$ gibt, für die $a^{3}-36 a^{2}+51 a-97$ ein Vielfaches von $3^{2008}$ ist. | [
"Solution:\n\n1. Lösung: Definiere das Polynom $P(x)=x^{3}-36 x^{2}+51 x-97$. Ausgehend von $P_{0}(x)=\\frac{1}{81} P(9 x+1)=9 x^{3}-33 x^{2}-2 x-1$, definiere rekursiv Polynome $P_{k}(x)=3 a_{k} x^{3}+3 b_{k} x^{2}+c_{k} x+d_{k}$ mit ganzen Zahlen $a_{k}, b_{k}, c_{k}, d_{k}$, wobei $c_{k}$ nicht durch 3 teilbar i... | Germany | IMO-Auswahlklausur | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
001k | Sea $ABC$ un triángulo isósceles con $AC = BC$. Se consideran puntos $D$, $E$, $F$ en $BC$, $CA$, $AB$, respectivamente, tales que $AF > BF$ y que el cuadrilátero $CEFD$ sea un paralelogramo. La recta perpendicular a $BC$ trazada por $B$ intersecta a la mediatriz de $AB$ en $G$. Demostrar que la recta $DE$ es perpendic... | [] | Argentina | XIX Olimpíada Matemática Argentina | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | español | proof only | null | |
08cv | Problem:
Dati due numeri reali positivi $a, b$ definiamo
$$
a \star b=\frac{a b+1}{a+b}
$$
Quanto vale $1 \star(2 \star(3 \star(\cdots(2017 \star 2018))))$ ?
(A) $1 / 2018$
(B) 1
(C) $2018 / 2017$
(D) 1009
(E) 2018 | [
"Solution:\n\nLa risposta è $\\mathbf{( B )}$. Si può osservare che per qualunque numero reale positivo $x$ si ha $1 \\star x= \\frac{1 \\cdot x+1}{1+x}=1$; prendendo come $x$ il valore $2 \\star(3 \\star(\\cdots(2017 \\star 2018)))$ si ottiene che la risposta al problema è $1 \\star x=1$."
] | Italy | GARA di FEBBRAIO | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | MCQ | B | |
092l | Problem:
There are $n \geqslant 3$ positive integers written on a blackboard. A move consists of choosing three numbers $a, b, c$ on the blackboard such that they are the sides of a non-degenerate non-equilateral triangle and replacing them by $a+b-c$, $b+c-a$ and $c+a-b$.
Show that an infinite sequence of moves cann... | [
"Solution:\n\nWe will show that the product of all the numbers on the blackboard can never increase. Indeed, for the three numbers $a, b$ and $c$ we have the inequalities\n$$\n\\begin{aligned}\n& a^{2} \\geqslant a^{2}-(b-c)^{2}=(a+b-c)(a+c-b) \\\\\n& b^{2} \\geqslant b^{2}-(a-c)^{2}=(b+a-c)(b+c-a) \\\\\n& c^{2} \\... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0hwf | Problem:
Define a sequence $a_{0}, a_{1}, a_{2}, \ldots$ in the following way: $a_{0}=0$, and for $n \geq 0$,
$$
a_{n+1}=a_{n}+5^{a_{n}} .
$$
Let $k$ be any positive integer. Prove that the remainders when $a_{0}, a_{1}, \ldots, a_{2^{k-1}}$ are divided by $2^{k}$ are all different. | [
"Solution:\n\nWe begin with a simple numerical lemma.\n\nLemma 1. For all $k \\geq 0, 5^{2^{k}}-1$ is divisible by $2^{k+2}$.\n\nProof. By induction. For $k=0$, the statement may be checked directly. To step from $k$ to $k+1$, we write\n$$\n5^{2^{k+1}}-1=\\left(5^{2^{k}}\\right)^{2}-1=\\left(5^{2^{k}}-1\\right)\\le... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
05v8 | Problem:
Une suite réelle $a_{1}, \ldots, a_{k}$ est casable dans l'intervalle $[b, c]$ si il existe des réels $x_{0}, \ldots, x_{k}$ dans $[b, c]$ tels que $\left|x_{i}-x_{i-1}\right|=a_{i}$ pour $k \geqslant i \geqslant 1$. La suite est normalisée si ses termes sont tous inférieurs ou égaux à 1.
1) Montrer que pour... | [
"Solution:\n\n1) Dans la suite, on dira qu'une suite de $(x_{i})$ satisfaisant la propriété de l'énoncé case la suite $(a_{i})$. On commence par se donner une idée du problème en essayant de montrer l'énoncé pour des petites valeurs de $n$. Si $n=0$, il faut montrer qu'existent $x_{0}, x_{1} \\in[0,1]$ tels que $\\... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
08nu | Problem:
Find the largest number of distinct integers that can be chosen from the set $\{1,2, \ldots, 2013\}$ so that the difference of no two of them is equal to $17$. | [
"Solution:\nConsider the sets $A_{mn} = \\{34m + n - 34, 34m + n - 17\\}$ for $1 \\leq m \\leq 59$ and $1 \\leq n \\leq 17$, and $B_k = \\{2006 + k\\}$ for $1 \\leq k \\leq 7$. As we cannot choose more than one number from each of these sets, we can choose at most $59 \\cdot 17 + 7 = 1010$ numbers. On the other han... | JBMO | 17th Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1010 | |
0jxd | Problem:
Regular octagon $CHILDREN$ has area $1$. Find the area of pentagon $CHILD$. | [
"Solution:\n\nThe pentagon $CHILD$ is congruent to the pentagon $NERDC$, as their corresponding angles and sides are congruent. Moreover, the two pentagons together compose the entire octagon, so each pentagon must have area one-half of the area of the octagon, or $\\frac{1}{2}$."
] | United States | HMMT November | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 1/2 | |
0do2 | Problem:
Доказати да постоји тачно један полином $P(x)$ с реалним коефицијентима за који је полином
$$
(x+y)^{1000}-P(x)-P(y)
$$
дељив полиномом $x y-x-y$. | [
"Solution:\n\nОзначимо $n=1000$. Сменом $x=u+1$ и $y=v+1$ добијамо да полином $u v-1$ дели полином $P(u+1)+P(v+1)-(u+v+2)^{n}$. Еквивалентан услов је да је $P(u+1)+P(v+1)-(u+v+2)^{n}=0$ кад год је $u v-1=0$ (видети напомену). Тако за $u \\neq 0$ и $v=\\frac{1}{u}$ имамо $P(u+1)+P\\left(\\frac{1}{u}+1\\right)=\\left... | Serbia | 12. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
0jh4 | Let $ABC$ be a triangle. Find all points $P$ on segment $BC$ satisfying the following property: If $X$ and $Y$ are the intersections of line $PA$ with the common external tangent lines of the circumcircles of triangles $PAB$ and $PAC$, then
$$
\left( \frac{PA}{XY} \right)^2 + \frac{PB \cdot PC}{AB \cdot AC} = 1.
$$
(Th... | [
"We consider the configuration shown on the left below. Let $O_B$ and $\\omega_B$ ($O_C$ and $\\omega_C$) denote the circumcenter and circumcircle of triangle $ABP$ ($ACP$) respectively. Line $ST$, with $S$ on $\\omega_B$ and $T$ on $\\omega_C$, is one of the common tangent lines of the two circumcircles. Point $X$... | United States | USAMO | [
"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",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | null | proof and answer | Exactly two points on BC: the intersection of BC with the internal angle bisector of angle A, and its reflection across the midpoint of BC (equivalently, the points dividing BC in the ratios AB:AC and AC:AB). | |
00c2 | Let $ABC$ be an acute-angled triangle with $\angle BAC = 60°$, incenter $I$ and circumcenter $O$. Let $O'$ be the point diametrically opposed to $O$ on the circumcircle of the triangle $BOC$. Prove that
$$
IO' = BI + IC.
$$ | [
"By considering the inscribed angle $\\angle BAC$ in the circumcircle of the triangle $ABC$, we have that $\\angle BOC = 2 \\cdot \\angle BAC = 120^\\circ$; then, $\\angle BO'C = 60^\\circ$. Since $O$ is a point of the perpendicular bisector of $BC$, then $O'$ is also on this line. Therefore, the triangle $BO'C$ is... | Argentina | 29° Olimpiada Matemática del Cono Sur | [
"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 | |
0hzc | Problem:
As part of his effort to take over the world, Edward starts producing his own currency. As part of an effort to stop Edward, Alex works in the mint and produces 1 counterfeit coin for every 99 real ones. Alex isn't very good at this, so none of the counterfeit coins are the right weight. Since the mint is not... | [
"Solution:\n\n$5\\%$ of the coins are sent to the lab, and only $95\\%$ of the coins are sent to the lab and counterfeit, so there is a $19\\%$ chance that a coin sent to the lab is counterfeit and an $81\\%$ chance that it is real. The lab could correctly detect a counterfeit coin or falsely accuse a real one of b... | United States | Harvard-MIT Math Tournament | [
"Statistics > Probability > Counting Methods > Other",
"Math Word Problems"
] | null | final answer only | 19/28 | |
06yy | Problem:
Show that the roots $r, s, t$ of the equation $x(x-2)(3x-7)=2$ are real and positive. Find $\tan^{-1} r + \tan^{-1} s + \tan^{-1} t$. | [
"Solution:\n\nPut $f(x) = x(x-2)(3x-7) - 2 = 3x^{3} - 13x^{2} + 14x - 2$. Then $f(0) = -2$, $f(1) = 2$, so there is a root between $0$ and $1$. $f(2) = -2$, so there is another root between $1$ and $2$. $f(3) = 4$, so the third root is between $2$ and $3$. $f(x) = 0$ has three roots, so they are all real and positi... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | null | proof and answer | 3π/4 | |
0l45 | Problem:
For all positive integers $r$ and $s$, let $\operatorname{Top}(r, s)$ denote the top number (i.e., numerator) when $\frac{r}{s}$ is written in simplified form. For instance, $\operatorname{Top}(20,24)=5$. Compute the number of ordered pairs of positive integers $(a, z)$ such that $200 \leq a \leq 300$ and $\o... | [
"Solution:\n\nIn general, $\\operatorname{Top}(r, s)=\\frac{r}{\\operatorname{gcd}(r, s)}$. We characterize all possible $(a, z)$ as follows.\n\nClaim 1. For any positive integers $a$ and $z$, we have $\\operatorname{Top}(a, z)=\\operatorname{Top}(z, a-1)$ if and only if there exists positive integers $d$ and $e$ s... | United States | HMMT November 2024 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 38 | |
0fu0 | Problem:
Sei $ABC$ ein gleichschenkliges Dreieck mit $|AC| = |BC|$ und Inkreismittelpunkt $I$. Sei $P$ ein Punkt auf dem Umkreis des Dreiecks $AIB$, der im Dreieck $ABC$ liegt. Die Geraden durch $P$, parallel zu $CA$ und $CB$, schneiden $AB$ in $D$ und $E$. Die zu $AB$ parallele Gerade durch $P$ schneidet $CA$ und $CB... | [
"Solution:\n\nDie entsprechenden Seiten der Dreiecke $PDE$ und $CFG$ sind parallel. Falls $DF$ und $EG$ nicht parallel sind, gehen diese beiden Dreiecke durch eine Streckung auseinander hervor und $DF$, $EG$ und $CP$ schneiden sich im Streckzentrum. Dies führt zu folgendem\n\nSatz 1. Sei $Q$ der Schnittpunkt von $C... | Switzerland | IMO Selektion | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid,... | null | proof only | null | |
0c14 | Determine the positive integers $a, b, c$ which satisfy the equality:
$$
\frac{a+b}{2} + \frac{a^2+b^2}{2} = \frac{7c+1}{c+1}.
$$ | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (a,b,c) = (1,2,1), (2,1,1), (2,2,5) |
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