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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
08dw | Problem:
Quanti sono i polinomi $p(x)$ a coefficienti reali, di grado compreso fra 1 e 2020 (estremi inclusi), per cui esiste un numero reale $\alpha$ tale che l'equazione $p(x)^2 = p\left(x^2\right) + \alpha p(x)$ sia verificata per ogni numero reale $x$? | [
"Solution:\n\nLa risposta è 4040. Detto $a$ il coefficiente di testa di $p(x)$, confrontando i coefficienti di testa di $p(x)^2$ e $p\\left(x^2\\right)+k p(x)$ si ottiene $a^2=a$, quindi $a=1$, cioè $p(x)$ è monico. Ora, se $p(x)$ è un monomio si ha sempre $p(x)^2=p\\left(x^2\\right)$, cioè l'uguaglianza voluta con... | Italy | Olimpiadi della Matematica | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | 4040 | |
0in3 | Problem:
There are thirteen broken computers situated at the following set $S$ of thirteen points in the plane:
$$
\begin{array}{lll}
A=(1,10) & B=(976,9) & C=(666,87) \\
D=(377,422) & E=(535,488) & F=(775,488) \\
G=(941,500) & H=(225,583) & I=(388,696) \\
J=(3,713) & K=(504,872) & L=(560,934) \\
& M=(22,997) &
\end{a... | [
"Solution:\n\nAnswer: ADHIKLEFGBCJM. This is an instance of the minimum-latency problem, which is at least NP-hard. There is an easy $O(n!)$ algorithm, but this is unavailable to teams on computational grounds (100 MHz calculators used to seem fast...) The best strategy may be drawing an accurate picture and exerci... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Algorithms",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | final answer only | ADHIKLEFGBCJM | |
0gbo | 已知正整數 $a_1, a_2, \dots, a_n$ ($a_1 < a_2 < \dots < a_n$), $k$ 為正實數且 $k \ge 1$.
試證：
$$
\sum_{i=1}^{n} a_i^{2k+1} \ge \left( \sum_{i=1}^{n} a_i^k \right)^2 .
$$ | [
"我們分兩步驟證明題設：\n\n(1) 用數學歸納法證明：\n$$\n2 \\sum_{i=1}^{n} a_i^k \\le (a_n + 1)^k a_n. \\qquad (1)\n$$\n證明：當 $n=1$ 時，易知 Eq. (1) 成立。\n假設當 $n=m$ 時，Eq. (1) 成立，即\n$$\n2 \\sum_{i=1}^{m} a_i^k \\le (a_m + 1)^k a_m .\n$$\n則當 $n=m+1$ 時，\n$$\n\\begin{aligned}\n2 \\sum_{i=1}^{m+1} a_i^k &= 2 \\sum_{i=1}^{m} a_i^k + 2a_{m+1}^k \\\\... | Taiwan | 2018 數學奧林匹亞競賽第二階段選訓營, 獨立研究(一) | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0119 | Problem:
Let $x_{1}, x_{2}, \ldots, x_{n}$ be positive integers such that no one of them is an initial fragment of any other (for example, $12$ is an initial fragment of $12$, $125$ and $12405$). Prove that
$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}<3
$$ | [
"Solution:\nLet $\\{y_{1}, \\ldots, y_{k}\\} \\subset \\{x_{1}, \\ldots, x_{n}\\}$ be a subset of numbers with the maximal number of digits, and differing from one another only by their last digits: $y_{1}=\\overline{y \\alpha_{1}}, y_{2}=\\overline{y \\alpha_{2}}, \\ldots, y_{k}=\\overline{y \\alpha_{k}}$ (here $\... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0fd1 | Problem:
Sean $a_{0}, a_{1}, a_{2}, a_{3}, a_{4}$ cinco números positivos en progresión aritmética de razón $d$. Probar que
$$
a_{2}^{3} \leq \frac{1}{10}\left(a_{0}^{3}+4 a_{1}^{3}+4 a_{3}^{3}+a_{4}^{3}\right)
$$ | [
"Solution:\nLa desigualdad dada puede escribirse como\n$$\n10 a_{2}^{3} \\leq a_{0}^{3}+4 a_{1}^{3}+4 a_{3}^{3}+a_{4}^{3}\n$$\ny sumando $6 a_{2}^{3}$ a ambos miembros se convierte en\n$$\na_{2}^{3} \\leq \\frac{1}{16}\\left(a_{0}^{3}+4 a_{1}^{3}+6 a_{2}^{3}+4 a_{3}^{3}+a_{4}^{3}\\right)\n$$\nPor otro lado como $a_... | Spain | null | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
02uc | Problem:
Um hexágono é chamado equiângulo quando possui os seis ângulos internos iguais. Considere o hexágono equiângulo $A B C D E F$ com lados $3, y, 5, 4, 1$ e $x$, da figura a seguir. Determine os comprimentos $x$ e $y$ desconhecidos.
 | [
"Solution:\n\nComo um hexágono pode ser dividido em 4 triângulos por meio de suas diagonais, a soma de seus ângulos internos é $180^{\\circ}(6-2)=720^{\\circ}$. Dado que ele é equiângulo, cada um dos ângulos internos medirá $\\frac{720^{\\circ}}{6}=120^{\\circ}$. Sabendo disso, ao prolongarmos os lados formaremos, ... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | x = 6, y = 2 | |
05qq | Problem:
On dit que deux permutations $a_{1}, \ldots, a_{4035}$ et $b_{1}, \ldots, b_{4035}$ des entiers $1, \ldots, 4035$ s'intersectent s'il existe un entier $k \leqslant 4035$ tel que $a_{k}=b_{k}$. On dit qu'un ensemble $E$ de permutations est inévitable si chaque permutation des entiers $1, \ldots, 4035$ intersec... | [
"Solution:\n\na) Pour tout $i \\in \\{1, \\ldots, 2018\\}$, on note $\\sigma^{(i)}$ la permutation\n$$\ni+1, i+2, \\ldots, 2018, 1, 2, \\ldots, i, 2019, 2020, \\ldots, 4035\n$$\nEn d'autres termes, $\\sigma^{(i)}$ est la permutation telle que:\n$\\triangleright$ pour tout $j \\leqslant 2018$, $\\sigma_{j}^{(i)}$ es... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof and answer | Yes: there exists an unavoidable set of size 2018 (e.g., the 2018 cyclic shifts of the first 2018 positions while fixing the rest). No: there is no unavoidable set of size 2017. | |
02r8 | Elaine uses each of the digits $1$ to $8$ and writes down two $4$-digit numbers.
a. If the sum of these numbers is the largest possible, what is their sum?
b. If the sum of these numbers is the least possible, what is their minimum value? | [
"a. The largest sum is obtained when the largest digits are assigned to the leftmost positions, so it is equal to $(8+7) \\cdot 1000 + (6+5) \\cdot 100 + (4+3) \\cdot 10 + 2 + 1 = 16373$.\n\nb. The smallest sum is obtained when the smallest digits are assigned to the leftmost positions. So $1$ and $2$ are in the th... | Brazil | Brazilian Math Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | a) 16173; b) 3825 | |
0k4t | Problem:
Let $a_{0} = a_{1} = 1$ and $a_{n+1} = 7 a_{n} - a_{n-1} - 2$ for all positive integers $n$. Prove that $a_{n}$ is a perfect square for all $n$. | [
"Solution:\nWe claim that $a_{n} = F_{2n-1}^{2}$, where $F_{n}$ is the $n$\\text{th} Fibonacci number. For the base case, we compute the first four values:\n$$\n\\begin{gathered}\na_{1} = 1^{2} = F_{1}^{2}, \\quad a_{2} = 7 \\cdot 1 - 1 - 2 = 2^{2} = F_{3}^{2} \\\\\na_{3} = 7 \\cdot 4 - 1 - 2 = 5^{2} = F_{5}^{2}, \... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0l02 | In the following table, each question mark is to be replaced by “Possible” or “Not Possible” to indicate whether a nonvertical line with the given slope can contain the given number of lattice points (points both of whose coordinates are integers). How many of the 12 entries will be “Possible”?
| ... | [
"If the slope is $0$, then the line is horizontal and its equation is $y = b$ for some real number $b$. If $b$ is an integer, then the line will contain infinitely many lattice points, and if $b$ is not an integer, then it will contain no lattice points. Therefore exactly two of the entries in that row of the table... | United States | AMC 10 B | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 6 | |
07je | Kimia has a strange clock. Its minute hand doesn't work correctly. At any moment, instead of moving one second, it randomly jumps $34$ or $47$ seconds. For example, if at one moment the clock shows $12{:}23{:}05$, in the next moments it might show
$12{:}23{:}39$, $12{:}24{:}13$, $12{:}25{:}00$, $12{:}25{:}34$, $12{:}26... | [
"First, notice that if at some moment the minute hand is on $36$ or $49$, our goal is achieved. In the next step, if the hand is on $2$, our goal is achieved since in the next step it must land on one of $2 + 34 = 36$ or $2 + 47 = 49$.\n\nWe prove by induction that if the clock hand is on a number of the form $X_k ... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0arm | Problem:
How many roots has the equation $\sin x - \log_{10} x = 0$? | [
"Solution:\n\n(ans. 3 .\n$\\sin x = \\log x \\Rightarrow x \\leq 10.10 < 2 \\cdot 2\\pi$ means that in $[0, 2\\pi]$ there is a complete period of $\\sin$ and part of a second period. There is an intersection in the first period, and after the first period, near $\\frac{5\\pi}{2}$ to the left and to the right, $\\si... | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | 3 | |
0bqn | A square with side length $2n + 1$, $n \in \mathbb{N}$, $n \geq 2$, is divided by parallels to its sides into $(2n + 1)^2$ squares and exactly $2n(n + 1)$ of them are colored.
We will call an horizontal line of the large square *nice* if it has more than half of its $2n + 1$ squares colored. Denote $f(n)$ the maximum ... | [] | Romania | 67th NMO Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) 4; b) 1009 | |
08vk | 4 points lie on a plane in such a way that no 3 among them lie on a same straight line. Consider 4 triangles formed by 3 of the 4 given points. If the radii of the 4 inscribed circles to these 4 triangles have the same length, prove that all of these triangles are congruent. | [
"First, let us show that the following lemma holds.\n**Lemma:** Suppose a triangle $T$ contains a triangle $T'$, and that the lengths of the radii of $T$ and $T'$ are the same. Then, we must have $T = T'$.\n**Proof:** It is clear that the incircle of a triangle is uniquely determined by the fact that it is the circ... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof only | null | |
08cs | Problem:
Siano $x_{1}, x_{2}, \ldots, x_{n}$ interi positivi. Supponiamo che, nella loro scrittura decimale, nessuno degli $x_{i}$ sia un "prolungamento" di un altro $x_{j}$. Per esempio, $123$ è un prolungamento di $12$, e $459$ è un prolungamento di $4$, ma $134$ non è un prolungamento di $123$.
Dimostrare che
$$
\f... | [] | Italy | Olimpiade Italiana di Matematica | [
"Discrete Mathematics > Other",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0hcb | How many different ways to cover the $4 \times 4$ square with five $3 \times 1$ rectangles are there, so that exactly one $1 \times 1$ cell is left uncovered?
 | [
"We will cover the $1 \\times 1$ cells two ways as shown in Fig. 13. Since any $3 \\times 1$ rectangle covers exactly one cell of each color, only white cell can be left uncovered (since there are 6 white and 5 of grey and black cells). Same positions of white cells are only on the edges of $4 \\times 4$ square. Th... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 16 | |
0bf6 | a) Prove that $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{2^m} < m$, for all $m \in \mathbb{N}^*$.
b) Let $p_1, p_2, \dots, p_n$ be the sequence of the primes less than $2^{100}$. Prove that
$$
\frac{1}{p_1} + \frac{1}{p_2} + \dots + \frac{1}{p_n} < 10.
$$ | [
"a) It is a well-known inequality that can be immediately proved by induction.\n\nb) The numbers $p_i p_j p_k p_l$, with $1 \\le i \\le j \\le k \\le l \\le n$, are distinct and all less than $2^{400}$, so\n$$ \\left(\\frac{1}{p_1} + \\frac{1}{p_2} + \\dots + \\frac{1}{p_n}\\right)^4 \\le 4! \\sum_{1 \\le i \\le j ... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof only | null | |
0jr2 | Problem:
Let $\mathcal{C}$ be a cube of side length $2$. We color each of the faces of $\mathcal{C}$ blue, then subdivide it into $2^{3}=8$ unit cubes. We then randomly rearrange these cubes (possibly with rotation) to form a new 3-dimensional cube.
What is the probability that its exterior is still completely blue? | [
"Solution:\nAnswer: $\\frac{1}{2^{24}}$ or $\\frac{1}{8^{8}}$ or $\\frac{1}{16777216}$\n\nEach vertex of the original cube must end up as a vertex of the new cube in order for all the old blue faces to show. There are $8$ such vertices, each corresponding to one unit cube, and each has a probability $\\frac{1}{8}$ ... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Geometry > Solid Geometry > 3D Shapes"
] | null | final answer only | 1/16777216 | |
08vu | Suppose $12$ boxes are lined up from left to right. We want to put a ball in each of these $12$ boxes. Balls are colored red, blue or yellow. How many distinct ways of putting balls into the $12$ boxes are there if the following condition is to be satisfied?
* Condition: For each ball placed in a box, at least $1$ of ... | [
"For an integer $n \\ge 2$, let us consider the number of ways of filling the $n$ boxes lined up from left to right by putting $n$ balls one-by-one into the boxes starting from the left-most one, assuming that any of the balls can have any of the three colors. Denote by $a_n$ the possible number of ways of filling ... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2049 | |
083u | Problem:
Quest'anno Alberto ha provato a imparare francese, inglese e tedesco. Sapendo che
(i) se sa il tedesco, allora sa anche francese e inglese;
(ii) se sa il francese, allora sa anche un'altra lingua tra inglese e tedesco;
(iii) se sa l'inglese, allora sa il tedesco ma non il francese; quante di tali lingue sa Al... | [] | Italy | UNIONE MATEMATICA ITALIANA Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO BIENNIO | [
"Discrete Mathematics > Logic"
] | null | MCQ | A | |
0a1p | Problem:
Zij $n$ een positief geheel getal. Bewijs dat de getallen
$$
1^{1}, 3^{3}, 5^{5}, \ldots,\left(2^{n}-1\right)^{2^{n}-1}
$$
in verschillende restklassen zitten modulo $2^{n}$. | [
"Solution:\nWe bewijzen het gevraagde met inductie. Voor $n=1$ kijken we enkel naar het getal $1^{1}$, dus is het gevraagde triviaal waar.\n\nStel nu als inductiehypothese dat $1^{1}, 3^{3}, 5^{5}, \\ldots,\\left(2^{n}-1\\right)^{2^{n}-1}$ in verschillende restklassen zitten modulo $2^{n}$. Ten eerste zitten deze g... | Netherlands | IMO-selectietoets II | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof only | null | |
03bz | There are $2k$ citizens in a town every two of which are either friends or enemies. For some positive integer $t$ each citizen has at most $t$ enemies and there exists a citizen having exactly $t$ enemies. A group is called *friendly* if any two members of the group are friends. It is known that a friendly group with m... | [
"Let $A = \\{a_1, a_2, \\dots, a_k\\}$ and $B = \\{b_1, b_2, \\dots, b_k\\}$ be the two friendly groups. For arbitrary group $C$ from $A$ denote by $S_C$ the group of all people from $B$ each of which is an enemy of at least one person from $C$. If $|C| > |S_C|$ then $C \\cup (B \\setminus S_C)$ is a friendly group... | Bulgaria | 55th IMO Team Selection Test | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof only | null | |
07lx | The country of Harpland has three types of coin: green, white and orange.
The unit of currency in Harpland is the shilling. Any coin is worth a positive integer number of shillings, but coins of the same colour may be worth different amounts. A set of coins is stacked in the form of an equilateral triangle of side $n$ ... | [
"Without loss of generality the coins form the pattern shown below.\n\n\n\nLet $G$, $W$ and $O$ denote the total worth in shillings of the green, white and orange coins in the triangle, respectively. The problem reduces to showing that $G$, $W$ and $O$ are all divisible by three. Now fix so... | Ireland | Irish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Other"
] | null | proof only | null | |
0bv7 | Problem:
a) Calculați
$$
\lim_{n \rightarrow \infty} \frac{(n^{2}+3) \cdot (n^{2}+5) \cdot \ldots \cdot [(n+1)^{2}+2016]}{(n^{2}+2) \cdot (n^{2}+4) \cdots [(n+1)^{2}+2015]}
$$
b) Calculați
$$
\lim_{n \rightarrow \infty}\left(\frac{\sqrt[n]{4}+\sqrt[n]{504}}{2}\right)^{2n}
$$ | [] | Romania | OLIMPIADA DE MATEMATICĂ ETAPA LOCALĂ | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | a) 1; b) 2016 | |
0gk2 | Prove that
$$
\sqrt{a^2 + b^2 - \sqrt{2} ab} + \sqrt{b^2 + c^2 - \sqrt{2} bc} \geq \sqrt{a^2 + c^2}
$$
for all positive real numbers $a$, $b$ and $c$. | [
"The inequality results from the triangle inequality,\n$PQ + PR \\ge QR$, as shown in the figure.\n\n"
] | Thailand | Thai Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | proof only | null | |
0cxb | Find all triples $(a, b, c)$ of positive integers for which
$$
a + b c = 2010 \text{ and } b + c a = 250.
$$ | [] | Saudi Arabia | SAMC | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | (a, b, c) = (3, 223, 9) | |
0aku | For any integer $n > 1$, let $s(n)$ be its smallest prime divisor and $d(n)$ be the number of its positive divisors. Is it possible to choose $2022$ positive integers $a_1, a_2, \dots, a_{2022}$ with $a_1 < a_2 - 1 < \dots < a_{2022} - 2021$ such that for all $k = 1, \dots, 2021$ it holds that
$$d(a_{k+1} - a_k - 1) > ... | [] | North Macedonia | Team Selection Test for IMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | English | proof and answer | Yes | |
04w9 | Real numbers $a, b, c, d$ are such that
$$
a + b + c + d = 0 \quad \text{and} \quad \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} = 0.
$$
How many of the equalities
$$ab = cd, \quad ac = bd, \quad ad = bc$$
can hold simultaneously? | [] | Czech Republic | National Round | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | 3 | |
0ifv | Problem:
Let $p=2^{24036583}-1$, the largest prime currently known. For how many positive integers $c$ do the quadratics $\pm x^{2} \pm p x \pm c$ all have rational roots? | [
"Solution: 0\nThis is equivalent to both discriminants $p^{2} \\pm 4 c$ being squares. In other words, $p^{2}$ must be the average of two squares $a^{2}$ and $b^{2}$. Note that $a$ and $b$ must have the same parity, and that $\\left(\\frac{a+b}{2}\\right)^{2}+\\left(\\frac{a-b}{2}\\right)^{2}=\\frac{a^{2}+b^{2}}{2}... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 0 | |
0ejr | Problem:
Nataša je zlepila kvadrat in enakostranični trikotnik v petkotnik z idejo, da lahko z njimi oblikuje neskončen vzorec, ki ravnine ne pokrije v celoti. Ali Natašin vzorec pokrije več kot $75\%$ površine ravnine?
 | [
"Solution:\nOznačimo z $a$ dolžino stranice Natašinega petkotnika. Potem je njegova ploščina enaka\n$$\na^2\\left(1+\\frac{\\sqrt{3}}{4}\\right).\n$$\nNajvečji notranji kot petkotnika je enak $90^{\\circ}+60^{\\circ}=150^{\\circ}$. Luknja v Natašinem vzorcu je torej romb z manjšim notranjim kotom enakim $360^{\\cir... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Yes, the pattern covers more than 75 percent of the plane. | |
05iu | Problem:
Un digicode s'ouvre dès qu'on fait l'unique combinaison correcte de 4 chiffres (qui peut éventuellement contenir des répétitions). Par exemple, si l'on tape la suite des chiffres 000125 le digicode s'ouvrira si le code est soit 0001, soit 0012, soit 0125. Petit Pierre ne connaît pas le code. Combien de chiffr... | [
"Solution:\n\nIl est clair qu'il faut au moins $10003$ chiffres, car il faut essayer toutes les $10000$ combinaisons. Montrons que ce nombre est aussi suffisant.\n\nDessinons $1000$ points correspondant à toutes les suites de $3$ chiffres. Dessinons une flèche entre deux points si les deux derniers chiffres de la p... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | 10003 | |
032x | Problem:
Let $a$, $b$ and $c$ be positive integers such that one of them is coprime with any of the other two. Prove that there are positive integers $x$, $y$ and $z$ such that $x^{a} = y^{b} + z^{c}$. | [
"Solution:\nWe consider two cases.\n\nCase 1. Let $(a, b) = (a, c) = 1$. Then $(a, b c) = 1$ and hence there are integers $u$ and $v$ such that $u a + v b c = 1$. This means that $a$ divides $-v b c + 1$. If $k \\geq 1$ is a positive integer such that $a$ divides $-v - k$, then $a$ divides $k b c + 1$, i.e. $k b c ... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0125 | Problem:
Given a parallelogram $A B C D$. A circle passing through $A$ meets the line segments $A B$, $A C$ and $A D$ at inner points $M$, $K$, $N$, respectively. Prove that
$$
|A B| \cdot|A M|+|A D| \cdot|A N|=|A K| \cdot|A C| .
$$ | [
"Solution:\n\nLet $X$ be the point on segment $A C$ such that $\\angle A D X=\\angle A K N$, then\n$$\n\\angle A X D=\\angle A N K=180^\\circ-\\angle A M K\n$$\n(see Figure 2).\n\nTriangles $N A K$ and $X A D$ are similar, having two pairs of equal angles, hence $|A X|=\\frac{|A N| \\cdot|A D|}{|A K|}$. Since trian... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0god | The points $A$ and $B$ lie on a circle with diameter $CD$ and on different sides of the line $CD$. A circle $\Gamma$ passing through the points $C$ and $D$ intersects the line segment $AC$ at a point $E$ different from its endpoints, and the line $BC$ at a point $F$. $P$ is the point of intersection of the tangent line... | [
"As $AB$ is the Simson line for the point $D$ and the triangle $FCE$, $DR$ is perpendicular to $EF$. We have $\\angle QEP = \\angle EQP = \\angle ECF = \\angle XEF$, where $X$ is a point on the ray $PE$ beyond $E$. Therefore $Q$, $E$, $F$ are collinear. As $PS$ is perpendicular to $EQ$, the result follows."
] | Turkey | 18th Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0gij | 令 $N$ 為一正整數。考慮張 $N \times N$ 的方格紙。一條右下行路徑是一系列的方格，其中每一個方格都在前一個方格的右方一格或下方一格。一條右上行路徑是一系列的方格，其中每一個方格都在前一個方格的右方一格或上方一格。
證明：我們無法將 $N \times N$ 的方格紙拆分成少於 $N$ 條的右下行和/或右上行路徑。下圖為一個將 $5 \times 5$ 方格拆分成 5 條路徑的範例。
Let $N$ be a positive integer, and consider a $N \times N$ grid. A *right-down path* is a se... | [
"我們對 $N$ 進行數學歸納法。假設命題對 $N-1$ 成立。考慮最左上角的那一格所在的路徑 $P$。若 $P$ 是右上行路徑，表示 $P$ 的所有格子都在最上方一橫行或最右方一直欄中。這意味著當我們移除最上方一橫行或最右方一直欄時，剩下的 $(N-1) \\times (N-1)$ 方格紙與其上的對應分拆仍符合題意，依據歸納假設至少被分為 $N-1$ 條路徑，從而原始的 $N \\times N$ 方格紙至少被分拆為 $N$ 塊。\n故僅需考慮 $P$ 為右下行路徑之情況。此處的關鍵觀察為：若 $P$ 包含最右下角的一格，則被 $P$ 分隔的兩區可以合併為一個 $(N-1) \\times (N-1)$ 的方格紙（如下圖... | Taiwan | IMO 3J, Mock Exam 1 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Chinese; English | proof only | null | |
05kr | Problem:
Déterminer tous les polynômes $P$ à coefficients entiers pour lesquels l'ensemble $P(\mathbb{N})$ contient une suite géométrique infinie de raison $a$ avec $a \notin\{-1,0,1\}$ et de premier terme non nul. | [
"Solution:\n\nTout d'abord, on note que si $P$ est un polynôme ayant les propriétés de l'énoncé, alors $-P$ les possède aussi (il suffit de changer le premier terme de la suite géométrique en son opposée). On peut donc supposer que le coefficient dominant de $P$ est strictement positif.\n\nSoit $\\left(u_{k}\\right... | France | Iran | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Factorization t... | null | proof and answer | All such polynomials are exactly P(x) = w (v x + u)^n with integers n > 0, v ≠ 0, and w ≠ 0. | |
03z4 | Let $A_1, A_2, \dots, A_n$ be $n$ non-empty subsets of a finite set $A$ of real numbers satisfying the following conditions:
(1) The sum of elements of $A$ is equal to $0$;
(2) Pick arbitrarily a number from each $A_i$, and their sum is strictly positive.
Prove that there exist sets $A_{i_1}, A_{i_2}, \dots, A_{i_k}$, ... | [
"Let $A = \\{a_1, \\dots, a_m\\}$ with $a_1 > \\dots > a_m$. By (1) we have $a_1 + \\dots + a_m = 0$. Consider the smallest element of each $A_i$, the sum of these numbers is greater than $0$. Assume that there are exactly $k_i$ sets among $A_1, \\dots, A_n$ whose minimal element is $a_i$, $i = 1, 2, \\dots, m$. Th... | China | China Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0jaj | Prove that there exists a real constant $c$ such that for any pair $(x, y)$ of real numbers, there exist relatively prime integers $m$ and $n$ satisfying the relation
$$\sqrt{(x-m)^2 + (y-n)^2} < c \log(x^2 + y^2 + 2).$$
(This problem was suggested by Daniel Kane.) | [
"** **(By Adam Hesterberg). Without loss of generality we may consider points $(x, y)$ with $x > y > 0$. For any $c$, let $d = \\frac{c}{2} \\log(x^2 + y^2 + 2)$. Choose $c$ large enough that\n$$\nc > \\frac{\\sqrt{2}}{\\log(2)} \\quad \\text{and} \\quad d \\ge \\max\\{9 \\cdot 20 \\cdot 21, 21 + 21 \\log(x)\\}.\n$... | United States | Team Selection Test Selection Test | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
04cp | Find the real number $A$, given that the coefficient with $x^{12}$ in the polynomial
$$
(1 + x^4)^{12} + A (x(1 - x^2)^2)^{12}
$$
equals 100. | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | final answer only | -120 | |
0jlo | Problem:
Mark and William are playing a game with a stored value. On his turn, a player may either multiply the stored value by 2 and add 1 or he may multiply the stored value by 4 and add 3. The first player to make the stored value exceed $2^{100}$ wins. The stored value starts at 1 and Mark goes first. Assuming bot... | [
"Solution:\n\nAnswer: 33\n\nWe will work in the binary system in this solution.\n\nLet multiplying the stored value by 2 and adding 1 be Move $A$ and multiplying the stored value by 4 and adding 3 be Move $B$. Let the stored value be $S$. Then, Move $A$ affixes one 1 to $S$, while Move $B$ affixes two 1s. The goal ... | United States | HMMT November 2014 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 33 | |
0ccx | The triangle $ABC$ has $\angle BAC = 90^\circ$ and $\angle ABC = 60^\circ$. The points $D$ and $E$ are taken on the sides $AC$, respectively $AB$, so that $CD = 2 \cdot DA$ and $DE$ is the bisector of the angle $\angle ADB$. Denote $M$ the intersection of the lines $CE$ and $BD$, and $P$ the intersection of the lines $... | [
"a) Let $AB = a$. Then $BC = 2a$, $AC = \\sqrt{BC^2 - AB^2} = a\\sqrt{3}$, $AD = \\frac{a}{3}\\sqrt{3}$, $BD = \\sqrt{AB^2 + AD^2} = \\frac{2a}{3}\\sqrt{3} = 2AD$. This yields $\\angle ABD = 30^\\circ$, hence $\\angle ADB = 60^\\circ$.\n\n\n\nThis gives $\\angle ADE = 30^\\circ = \\angle AC... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
07yd | Problem:
In ciascuna delle caselle di una tabella quadrata $4 \times 4$ è scritta la cifra $1$ o la cifra $2$. Si sa che la somma delle $9$ cifre contenute in ciascuno dei $4$ quadrati $3 \times 3$ contenuti nella tabella è multipla di $4$, mentre la somma di tutte le $16$ cifre non è multipla di $4$.
Determinare il ... | [
"Solution:\n\nIl massimo ed il minimo sono, rispettivamente, $30$ e $19$ e sono realizzati, per esempio, dalle seguenti tabelle\n\n| 2 | 2 | 2 | 2 |\n|---|---|---|---|\n| 2 | 2 | 1 | 2 |\n| 2 | 1 | 2 | 2 |\n| 2 | 2 | 2 | 2 |\n\n| 1 | 1 | 1 | 1 |\n|---|---|---|---|\n| 1 | 2 | 2 | 1 |\n| 1 | 2 | 1 | 1 |\n| 1 | 1 | 1 ... | Italy | null | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | maximum 30, minimum 19 | |
0blk | The points $T, A, B, C$ are non-coplanar and the straight lines $a, b, c$ are the parallels from the centroid $G$ of the triangle $ABC$ to the straight lines $TA, TB$, respectively $TC$. Let $\{A'\} = a \cap (TBC)$, $\{B'\} = b \cap (TAC)$ and $\{C'\} = c \cap (TAB)$.
a) Prove that $(A'B'C') \parallel (ABC)$.
b) Comp... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof and answer | a/3 | |
0hcf | Number $1000$ was split into $9$ (not necessarily different) positive integer additive terms. After that, we list all different numbers that can be obtained from adding some of these terms (from one to eight). What is the minimum number of numbers listed? | [
"Let us split $1000$ into $8$ numbers $100$ and one number $200$. In this case, the sum of some terms can have $9$ different values: $100 \\cdot 1$, $100 \\cdot 2$, $\\ldots$, $100 \\cdot 8$, and $200 + 100 \\cdot 7 = 900$.\n\nLet us prove that it is impossible to obtain less than $9$ different numbers. Since $9$ d... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 9 | |
05ck | Find all functions $f : \mathbb{R} \to \mathbb{R}$ that satisfy
$$
f(y^2f(x) - f(xy)) = f(y^2) + 2(x^2 - f(x))(f(y) - 1) + 1
$$
for all real numbers $x$ and $y$. | [
"Substituting $y = 1$ into the given equation, we obtain\n$$\nf(0) = f(1) + 2(x^2 - f(x))(f(1) - 1) + 1\n$$\nwhich is equivalent to\n$$\n2f(x)(f(1) - 1) = f(1) - f(0) + 2x^2(f(1) - 1). \\quad (3)\n$$\n\nIf $f(1) = 1$ then (3) implies $0 = 1 - f(0) + 1$, or equivalently, $f(0) = 2$.\nSubstituting now $y = 0$ into th... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | f(x) = x^2 + 1 | |
0ej3 | Problem:
Ana in Meta sta se hkrati odpeljali iz vasi Zabukovje v vas Zahrastje, Ana s kolesom in Meta z avtom. Ana je vozila s konstantno hitrostjo $30~\mathrm{km}/\mathrm{h}$, Meta pa s konstantno hitrostjo $70~\mathrm{km}/\mathrm{h}$. Ko je Meta prišla v Zahrastje, je bila tam $1~\mathrm{h}$, nato pa se je z enako hi... | [
"Solution:\nOznačimo z $x$ razdaljo, ki jo je prevozila Ana do srečanja z Meto. Potem je Meta do srečanja z Ano prevozila $x + 2 \\cdot 105~\\mathrm{km}$. Če čas, ki je pretekel do srečanja, zapišemo z Aninega in Metinega stališča v urah, dobimo\n$$\n\\frac{x}{30} = \\frac{x + 2 \\cdot 105}{70} + 1\n$$\nod koder sl... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | C | |
0f8f | Problem:
Prove that for any tetrahedron the radius of the inscribed sphere $r < \dfrac{ab}{2(a + b)}$, where $a$ and $b$ are the lengths of any pair of opposite edges. | [] | Soviet Union | 22nd ASU | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
089u | Problem:
Nell'isola dei Cavalieri (che dicono sempre la verità) e dei Furfanti (che mentono sempre) viene effettuato un sondaggio fra i 2013 abitanti, in cui ci sono tre domande: "Tifi per la squadra A?", "Tifi per la squadra B?" e "Tifi per la squadra C?". Sappiamo che ogni isolano risponde a tutte e tre le domande e... | [
"Solution:\n\nLa risposta è (C). Sia $C$ il numero dei cavalieri e $F$ il numero dei furfanti. Chiaramente, ogni cavaliere intervistato risponderà esattamente una volta \"Sì\" (alla domanda relativa alla squadra per la quale tifa) e due volte \"No\", mentre un furfante risponderà esattamente un \"No\" e due \"Sì\".... | Italy | Progetto Olimpiadi della Matematica - Gara di Febbraio | [
"Discrete Mathematics > Logic",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | C | |
05m1 | Problem:
Déterminer le plus grand nombre d'entiers que l'on peut extraire de l'ensemble $\{1,2, \ldots, 2014\}$ de sorte que la différence de deux quelconques de ces entiers soit différente de 17. | [
"Solution:\nCet exercice découle en fait d'une utilisation astucieuse du principe des tiroirs. Soit $E$ une partie de $\\{1,2, \\ldots, 2014\\}$ ne contenant que des entiers dont la différence n'est jamais égale à 17.\n\nTout d'abord, pour tout entier $a$, on note $S_{a}$ l'ensemble $\\{a, a+17\\}$, et $T_{a}$ l'en... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1011 | |
08ck | Problem:
Sono date tre circonferenze $\Gamma, \Gamma_{1}, \Gamma_{2}$ di raggi rispettivamente 6, 3, 2. $\Gamma_{1}$ e $\Gamma_{2}$ sono tangenti esternamente in $A$, mentre $\Gamma$ tange entrambe le altre circonferenze internamente, rispettivamente in $A_{1}$ ed $A_{2}$. Determinare il raggio della circonferenza cir... | [
"Solution:\n\nLa risposta è $\\mathbf{( E )}$. Siano $O, O_{1}, O_{2}$ i centri di $\\Gamma, \\Gamma_{1}, \\Gamma_{2}$ rispettivamente. Dal momento che $\\Gamma_{1}, \\Gamma_{2}$ sono tangenti esternamente, la distanza $O_{1} O_{2}$ è uguale alla somma dei raggi di $\\Gamma_{1}, \\Gamma_{2}$, ovvero $O_{1} O_{2}=5$... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | E | |
0aj1 | Let $m, n, p$ be fixed positive real numbers which satisfy $mnp = 8$. Depending on these constants, find the minimum of
$$
x^2 + y^2 + z^2 + mxy + nxz + pyz
$$
where $x, y, z$ are arbitrary positive real numbers satisfying $xyz = 8$. When is the equality attained?
a) $m = n = p = 2$
b) arbitrary (but fixed) positive r... | [
"a)\nUse AM-GM and $xyz=8$ to get\n$$\nx^2 + y^2 + z^2 + xy + xy + xz + xz + yz + yz \\geq 9\\sqrt{x^6 y^6 z^6} = 36.\n$$\n\nWe have equality for $x = y = z = 2$.\n\nb)\nUsing $xyz = 8$, we can transform the given expression:\n$$\nx^2 + y^2 + z^2 + mxy + nxz + pyz = x^2 + \\frac{8p}{x} + y^2 + \\frac{8n}{y} + z^2 +... | North Macedonia | European Mathematical Cup | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | a) Minimum value: 36, attained at x = y = z = 2. b) Minimum value: 6 * cube root of 2 * (cube root of m squared + cube root of n squared + cube root of p squared). Equality occurs at x = cube root of 4p, y = cube root of 4n, z = cube root of 4m. | |
0i1u | Express
$$
\sum_{k=0}^{n} (-1)^k (n-k)!(n+k)!
$$
in closed form. | [
"**First Solution.** (By Tiankai Liu) Let\n$$\nf(k) = (n+1-k)!(n+k)!\n$$\nfor integers $0 \\le k \\le n + 1$. Note that\n$$\n\\begin{aligned}\nf(k) + f(k + 1) &= (n+1-k)!(n+k)! + (n-k)!(n+k+1)! \\\\\n&= (n + 1 - k + n + k + 1)(n-k)!(n+k)! \\\\\n&= 2(n + 1)(n-k)!(n+k)!.\n\\end{aligned}\n$$\nTherefore,\n$$\n\\begin{a... | United States | USA IMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Generating functions"
] | English | final answer only | \sum_{k=0}^{n} (-1)^k (n-k)!(n+k)! = \frac{(n!)^2}{2} + \frac{(-1)^n (2n+1)!}{2(n+1)} | |
04xv | Show that for any real $x > 0$ and integer $n > 0$ we have
$$
x^n + \frac{1}{x^n} - 2 \ge n^2 \left(x + \frac{1}{x} - 2\right).
$$ | [
"Without loss of generality assume that $y = \\sqrt{x} > 1$. The identity\n$$\na^2 + \\frac{1}{a^2} - 2 = \\left(a - \\frac{1}{a}\\right)^2\n$$\nreduces the problem to showing that\n$$\ny^n - \\frac{1}{y^n} \\ge n\\left(y - \\frac{1}{y}\\right)\n$$\nor $y^{2n} - n(y^{n+1} - y^{n-1}) - 1 \\ge 0$. Upon division by $y... | Czech-Polish-Slovak Mathematical Match | Cesko-Slovacko-Poljsko 2013 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0dke | Let $ABC$ be a triangle with $\angle BAC = 90^\circ$ with the altitude $AH$ ($H \in BC$). A circle $(\omega)$ passes through $B, C$ and cuts the segments $AB, AC$ at $M, N$ respectively. Circle $(\omega)$ also cuts the line $AH$ at $D, E$ ($D$ lies between $A, H$). Suppose that $DE = AH\sqrt{5}$, prove that the circumc... | [
"Base on the power from $H$ to the circle $(\\omega)$, $HD \\cdot HE = HB \\cdot HC = AH^2$. Moreover, $HD + HE = DE = AH\\sqrt{5}$. Thus, the lengths $HD, HE$ will be the solutions of the quadratic equation\n$$\nx^2 - AH\\sqrt{5} \\cdot x + AH^2 = 0.\n$$\nSince $D$ is inside triangle $ABC$, $HD < AH$, entails $HE ... | Saudi Arabia | Saudi Booklet | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
08ja | Problem:
Let $G$ be the centroid of the the triangle $A B C$. Reflect point $A$ across $C$ at $A'$. Prove that $G, B, C, A'$ are on the same circle if and only if $G A$ is perpendicular to $G C$.
Problem:
Fie $G$ centrul de greutate al triunghiului $ABC$ şi $A'$ simetricul lui $A$ faţă de $C$. Demonstrați că punctele ... | [
"Solution:\n$$\nG A \\perp G C \\Leftrightarrow \\frac{4}{9} m_{a}^{2}+\\frac{4}{9} m_{c}^{2}=b^{2} \\Leftrightarrow 5 b^{2}=a^{2}+c^{2}\n$$\nMoreover,\n$$\nG B^{2}=\\frac{4}{9} m_{b}^{2}=\\frac{2 a^{2}+2 c^{2}-b^{2}}{9}=\\frac{9 b^{2}}{9}=b^{2}\n$$\nhence $G B=A C=C A'$ (1). Let $C'$ be the intersection point of t... | JBMO | 7th JBMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous >... | null | proof only | null | |
07u5 | Each square of an *n* × *n* grid is coloured either blue or red, where *n* is a positive integer. There are *k* blue cells in the grid. Pat adds the sum of the squares of the numbers of blue cells in each row to the sum of the squares of the numbers of blue cells in each column to form $S_B$. He then performs the same ... | [
"If the $i$-th row of the grid has $b_i$ blue cells, then the contribution to $S_B - S_R$\nfrom this row is $b_i^2 - (n - b_i)^2 = 2nb_i - n^2$. Adding these contributions over\nall rows yields $2nk - n^3$ and similarly the columns also contribute $2nk - n^3$;\nthus we have\n$$\nS_B - S_R = 2(2nk - n^3) = 2n(2k - n... | Ireland | IRL_ABooklet | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 15 and 313 | |
0d3u | Let $p$ be a prime number. Prove that there exist infinitely many positive integers $n$ such that $p$ divides
$$
1^{n}+2^{n}+\cdots+(p+1)^{n}
$$ | [
"Let $k$ be a positive integer. Using Fermat's little theorem we have\n$$\n1^{k(p-1)}+2^{k(p-1)}+\\cdots+(p+1)^{k(p-1)} \\equiv \\underbrace{1^{k}+1^{k}+\\cdots+1^{k}}_{p-1 \\text{ times}}+0^{k}+1^{k} \\equiv 0 \\pmod{p} .\n$$\nTherefore, for $n=k(p-1)$, and $k=1,2,3, \\ldots$, the prime number $p$ divides\n$$\n1^{... | Saudi Arabia | SAMC | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English, Arabic | proof only | null | |
0hva | Problem:
Let $a$, $b$, $c$ be pairwise distinct integers. Prove that
$$
\frac{a^{3}+b^{3}+c^{3}}{3} \geq a b c+\sqrt{3(a b+b c+c a+1)} .
$$ | [
"Solution:\nLet $3 k^{2}-1=a b+b c+c a$, so we need $a^{3}+b^{3}+c^{3} \\geq 3(a b c+3 k)$. Now, we have\n$$\n(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \\geq 2^{2}+1^{2}+1^{2}=6 .\n$$\nIn particular, we get\n$$\n(a+b+c)^{2}=\\frac{1}{2}\\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\\right]+3\\left(3 k^{2}-1\\right) \\geq 9 k^{2}\n$$\nTh... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0brh | Consider triangle $ABC$ with $m(\angle B) = 30^\circ$, $m(\angle C) = 15^\circ$ and $M$ the midpoint of the side $[BC]$. Let $N \in (BC)$ be such that $[NC] = [AB]$. Show that $[AN]$ is the bisector of the angle $MAC$. | [
"Let $P$ be the point where the perpendicular bisector of the segment $[BC]$ meets $AB$. Then $m(\\angle PCB) = 30^\\circ$, $m(\\angle PCA) = 15^\\circ$ and $m(\\angle MPC) = 60^\\circ$.\n\n\nSince $PC = PB$ and $NC = AB$, it follows that $\\frac{AP}{NC} = \\frac{BP}{BC}$, that is $\\frac{P... | Romania | 67th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
068g | 1. Let $ABC$ be an acute-angled triangle with $AB < AC < BC$, inscribed in the circle $c(O,R)$. The circle $c_1$ with center $A$ and radius $AC$ intersects the circle $c(O,R)$ at point $D$ and the extension of the side $CB$ at $E$. The line $AE$ intersects the circle $c(O,R)$ at point $F$ and $G$ is the symmetric point... | [
"Since the quadrilateral $AFBC$ is inscribed in the circle $(c)$, we have: $\\angle F_1 = \\angle ACB = \\angle C$. Since triangle $AEC$ is isosceles we have $\\angle E_1 = \\angle ACB = \\angle C$. Therefore $\\angle F_1 = \\angle E_1$, and hence the triangle $BEF$ is isosceles and hence\n$$\nBE = BF \\qquad (1).\... | Greece | 34th Hellenic Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting tw... | English | proof only | null | |
03r8 | In a planar rectangular coordinate system $xOy$, the area enclosed by the graph of function $f(x) = a\sin ax + \cos ax$ ($a > 0$) defined on an interval with the least positive period and by the graph of function $g(x) = \sqrt{a^2 + 1}$ is ______. | [
"We rewrite function $f(x)$ as $f(x) = \\sqrt{a^2+1}\\sin(ax + \\varphi)$, where $\\varphi = \\arctan\\frac{1}{a}$. Its least positive period is $\\frac{2\\pi}{a}$, and its amplitude is $\\sqrt{a^2+1}$. By symmetry of the figure enclosed by the graphs of the functions $f(x)$ and $g(x)$, we can change the figure int... | China | China Mathematical Competition (Hainan) | [
"Calculus > Integral Calculus > Applications",
"Precalculus > Trigonometric functions"
] | English | final answer only | 2π√(a^2+1)/a | |
0hmn | Problem:
As shown in the diagram above, the vertices of a regular decagon are colored alternately black and white. We would like to draw colored line segments between the points in such a way that
a. Every black point is connected to every white point by a line segment.
b. No two line segme... | [
"Solution:\n\nNote that the five main diagonals of the decagon must all be different colors since they all intersect at the decagon's center. Thus, at least $5$ colors are needed. To see that $5$ colors are also sufficient, we can simply assign each black point a color and use that color to connect it with all the ... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 5 | |
0ch5 | We will say that the positive integers $m$ and $n$ have property $\mathcal{P}$ if for every divisor $d_1$ of $m$ and every divisor $d_2$ of $n$, the number $d_1 + d_2$ is a prime.
a) Prove that if $m$ and $n$ have property $\mathcal{P}$ and are different, then $m+n$ is odd.
b) Find all the pairs $(m, n)$ of positive ... | [
"a) Without loss of generality, we may suppose $1 \\le m < n$.\nIf $m$ and $n$ are odd, then $m \\ge 1$ and $n \\ge 3$. Taking $d_1 = 1$ and $d_2 = n$ yields $d_1 + d_2 = n + 1$, which is an even number at least $4$, so it is composite, contradiction.\nIf $m$ and $n$ are even, taking $d_1 = 2$ and $d_2 = 2$ yields ... | Romania | 74th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | If the two numbers are distinct, their sum is odd. The pairs with the property are (1,1), (1,2), and (1,4). | |
0gx4 | Diagonals $AC$ and $BD$ of the quadrangle $ABCD$ intersect at point $O$. We know that diagonal $BD$ is perpendicular to the side $AD$, $\angle BAD = \angle BCD = 60^\circ$, $\angle ADC = 135^\circ$. Find the ratio $DO:OB$.
Fig. 1
**Answer:** 1:2. | [
"Under the problem statement we can easily find the following angles (fig.1): $\\angle ABD = 30^\\circ$, $\\angle BDC = 45^\\circ$, $\\angle DBC = 75^\\circ$. Let's draw rays $ADE$ and $ABF$. Then $\\angle EDC = 45^\\circ$, $\\angle FBC = 75^\\circ$. Therefore $BC$ is a bisector of $\\angle DBF$ and $DC$ is a bisec... | Ukraine | Ukrajina 2008 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof and answer | 1:2 | |
0dwv | Problem:
Tangenti iz točke $P$ na krožnico $k$ se krožnice dotikata v točkah $A$ in $B$. Naj bo $X$ poljubna točka na krajšem loku $\widehat{A B}$. Označimo s $C$ pravokotno projekcijo točke $P$ na premico $A X$ in z $D$ pravokotno projekcijo točke $P$ na premico $B X$. Dokaži, da premica $C D$ poteka skozi neko točko... | [
"Solution:\n\nNaj bo $O$ središče krožnice $k$, $A'$ razpolovišče daljice $P A$, $B'$ razpolovišče daljice $P B$ in $Y$ razpolovišče $A' B'$. Označimo še $\\angle A P B = \\varphi$ in $\\angle X A P = \\alpha$. Potem je $\\angle O A X = \\frac{\\pi}{2} - \\alpha$, $\\angle B O A = \\pi - \\varphi$, $\\angle A X B =... | Slovenia | 49. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0c7a | a) Prove that there exist differentiable functions $f : (0, \infty) \to (0, \infty)$, such that $f(f'(x)) = x$, for any $x > 0$.
b) Prove that there do not exist differentiable functions $f : \mathbb{R} \to \mathbb{R}$, such that $f(f'(x)) = x$, for any $x \in \mathbb{R}$. | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof only | null | |
0bgc | Problem:
Să se calculeze limita $\lim _{n \rightarrow \infty} \int_{0}^{1} \mathrm{e}^{x^{n}} \mathrm{~d} x$. | [] | Romania | Olimpiada Naţională de Matematică, Etapa judeţeană şi a municipiului Bucureşti | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Limits"
] | null | proof and answer | 1 | |
06ab | Let $n > 4$, be a positive integer, which is divided by $4$. We denote by $A_n$ the sum of all odd positive divisors of $n$. We denote by $B_n$ the sum of all even positive divisors of $n$, with the exclusion of $n$. Find the smaller possible value of $f(n) = B_n - 2A_n$. For which values of the positive integer $n$ is... | [
"Let $d_1, \\dots, d_k$ be the odd positive divisors of $n$. Then the numbers $2d_1, \\dots, 2d_k$ are even divisors of $n$. Also each of them is not divisible by $4$, and so none of them can be equal to $n$. Moreover, none of them is equal to $4$. Therefore we have:\n$$\nA_n = d_1 + \\dots + d_k \\text{ and } B_n ... | Greece | 39th Hellenic Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | Minimum value is 4; achieved exactly for n = 4p where p is prime (including n = 8). | |
0dej | Let $ABC$ be a triangle with $AB = AC$ and $M$ is the midpoint of the altitude $AD$. Consider $(\omega)$ as the circle of center $M$ and tangent to $AB$, $AC$. From some point $T$ on the line $BC$ (outside triangle $ABC$), construct two tangents of $(\omega)$ and they cut $AB$ at $P$, $Q$, cut $AC$ at $R$, $S$. Prove t... | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0hr3 | Problem:
Does there exist a convex polygon that can be partitioned into non-convex quadrilaterals? | [
"Solution:\n\nThe answer is no. Assume that, on the contrary, it is possible to partition a polygon $P$ into non-convex quadrilaterals. Let $n$ be the number of quadrilaterals. Denote by $S$ the total sum of all internal angles of all the quadrilaterals. Since the sum of internal angles of each quadrilateral is $36... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | No | |
0ikj | Problem:
Let $a$, $b$, $c$ be the roots of $x^{3}-9 x^{2}+11 x-1=0$, and let $s=\sqrt{a}+\sqrt{b}+\sqrt{c}$. Find $s^{4}-18 s^{2}-8 s$. | [
"Solution:\n\nFirst of all, as the left side of the first given equation takes values $-1, 2, -7$, and $32$ when $x=0, 1, 2$, and $3$, respectively, we know that $a$, $b$, and $c$ are distinct positive reals. Let $t=\\sqrt{a b}+\\sqrt{b c}+\\sqrt{c a}$, and note that\n$$\n\\begin{aligned}\ns^{2} & = a+b+c+2 t = 9+2... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | null | proof and answer | -37 | |
01fj | Let $S$ be a finite set with $n \ge 2$ elements. Two players, $A$ and $B$, alternately choose nonempty proper subsets of $S$, where
(1) it is not allowed to choose a set that contains a set previously chosen by any player,
(2) it is not allowed to choose a set that is contained in any previously chosen set,
(3) it i... | [
"It is straightforward to verify that the following is a winning strategy for $A$. Player $A$ first chooses a singleton subset $\\{x\\} \\subset S$. After that, he responds to $B$ choosing a set $T$ by choosing $(S \\setminus \\{x\\}) \\setminus T$."
] | Baltic Way | Baltic Way 2019 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | Player A has a winning strategy. | |
0d0q | Point $P$ lies inside quadrilateral $ABCD$ such that $\widehat{APD} = \widehat{BPC} = 90^\circ$ and $AP \cdot DP = BP \cdot CP$. Let $O$ denote the circumcenter of triangle $CDP$. Prove that line $OP$ bisects segment $AB$. | [
"Let $M$ be the midpoint of $AB$, and let $E$ be the point on line $BP$ such that $AE \\parallel MP$. Then $P$ is the midpoint of $EB$. Since $\\frac{AP}{PB} = \\frac{CP}{PD}$, we have $\\frac{AP}{PE} = \\frac{CP}{PD}$. Also, note that\n\n$\\widehat{EPC} = \\widehat{APD} = 90^\\circ$, so $\\widehat{EPA} = \\widehat... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions an... | English | proof only | null | |
08eh | Problem:
Sia $N$ il numero di sestuple ordinate di interi $(a, b, c, d, e, f)$ tali che $a^{3}+b^{3}+c^{3}+d^{3}+e^{3}+f^{3}=168$ e $-202120212021^{9}<a b c d e f<202120212021^{9}$, dove $abcdef$ è il prodotto dei sei interi. Quale dei seguenti è il resto di $N$ nella divisione per 6?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4 | [
"Solution:\n\nLa risposta è (C). Osserviamo che se esiste una soluzione con $a<b<c<d<e<f$, allora questa contribuisce con $6!$ sestuple ordinate di soluzioni, quindi possiamo ignorarla dato che stiamo contando il numero di soluzioni modulo 6.\n\nAnalogamente per ogni partizione $m_{1}+\\cdots+m_{k}=6$ con $m_{i} \\... | Italy | Italian Mathematical Olympiad - February Round | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, in... | null | MCQ | C | |
0hat | Solve the equation $\frac{\sqrt{x+2}}{\cos 2x+3} = \frac{\sqrt{x+1}}{\cos 2x+1}$. | [
"The equation can be written the following way:\n$$\n\\frac{\\sqrt{x+2}}{\\sqrt{x+1}} = \\frac{\\cos 2x + 3}{\\cos 2x + 1}.\n$$\nConsider two functions: $f(t) = \\frac{t+2}{t+1}$, $t \\ge 0$ and $g(y) = \\frac{y+3}{y+1}$, $y \\in (-1; 1]$.\n\nIf $t \\ge 0$, then $f(t) = \\frac{t+2}{t+1} < 2 \\Leftrightarrow t+2 < 2... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Other"
] | English | proof and answer | x = 0 | |
0k3j | Problem:
The endpoints of a chord $S T$ with constant length are moving along a semicircle with diameter $A B$. Let $M$ be the midpoint of $S T$ and $P$ the foot of the perpendicular from $S$ to $A B$. Prove that the angle $S P M$ is independent of the location of $S T$. | [
"Solution:\n\nDraw the other half of the circle, and extend $S P$ until it hits the circle again at $S'$. Note that $S'$ is the reflection of $S$ across $A B$. Then by SAS, $\\triangle P S M \\sim \\triangle S' S T$, so $\\angle S P M = \\angle S S' T$. But $\\angle S S' T$ is constant, since it is inscribed in $\\... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
07t6 | Let $N$, $k$ be positive integers with $k \le N$. An $(N, k)$-Mountain Tetris mini-game is played on an $N \times N$ grid. An *ascending path contour* is any path on the grid made of horizontal and vertical segments, which starts at $(0, 0)$ and reaches $(N, N)$ without ever going down or back.
Every $(N, k)$-Mountain... | [
"Given an initial ascending path contour, let $d_1, d_2, \\dots, d_k$ denote the lengths of the horizontal segments, then $d_i > 0$ and $\\sum_i d_i = N$. Because the contour starts in horizontal direction and has exactly $k$ horizontal and $k$ vertical segments, it ends with a vertical segment and so no horizontal... | Ireland | IRL_ABooklet_2020 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | (N-k)/(k+1) | |
0i8f | Problem:
$r$ and $s$ are integers such that
$$
3 r \geq 2 s - 3 \text{ and } 4 s \geq r + 12.
$$
What is the smallest possible value of $r / s$? | [
"Solution:\n\nWe simply plot the two inequalities in the $s r$-plane and find the lattice point satisfying both inequalities such that the slope from it to the origin is as low as possible. We find that this point is $(2,4)$ (or $(3,6)$), as circled in the figure, so the answer is $2 / 4 = 1 / 2$.\n\n![](attached_i... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 1/2 | |
0hmj | Problem:
Two bikers, Bill and Sal, simultaneously set off from one end of a straight road. Neither biker moves at a constant rate, but each continues biking until he reaches one end of the road, at which he instantaneously turns around. When they meet at the opposite end from where they started, Bill has traveled the ... | [
"Solution:\n\nDefine a pass $(\\mathrm{P})$ to be an time when Bill and Sal pass one another moving in opposite directions and a turn (T) to be a time when one of the bikers turns around. If both bikers turn around simultaneously, we may alter their speeds slightly, causing one turn to happen before the other, with... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 8 | |
0dfn | We consider all partitions of a positive integer $n$ into a sum of (non-negative integer) exponents of $2$ (i.e. $1, 2, 4, 8, \dots$). A number in the sum is allowed to repeat an arbitrary number of times (e.g. $7 = 2 + 2 + 1 + 1 + 1$) and two partitions differing only in the order of summands are considered to be equa... | [
"Let $D(n) = E(n) - O(n)$. We trivially have $O(1) = 1$ and $E(1) = 0$, thus $D(1) = -1$, and $E(2) = O(2) = 1$ (respectively $2 = 1+1$ and $2 = 2$), hence $D(2) = 0$. We will show by total induction that $D(n) = 0$ for all $n > 2$. Assume it holds for all numbers from $2$ to $n-1$. If $n$ is odd, a partition must ... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | E(n) - O(n) = -1 for n = 1, and E(n) - O(n) = 0 for all n > 1 | |
06j1 | Let $ABC$ be a triangle with $\angle ABC > \angle BCA$ and $\angle BCA \ge 30^\circ$. The angle bisectors of $\angle ABC$ and $\angle BCA$ meet the opposite sides of the triangle at the points $D$ and $E$, respectively. The line $BD$ intersects the line $CE$ at $P$. Assume that $PD = PE$ and that the incircle of the tr... | [
"The largest possible value of $BC$ is $3 + \\sqrt{3}$.\nSince $PD = PE$ and $AP$ bisects $\\angle EAD$, the quadrilateral $AEPD$ is cyclic or it is a kite. The latter case is impossible since $\\angle B > \\angle C$. Thus, we have\n$$\n180^{\\circ} = \\angle EAD + \\angle DPE = A + \\left(90^{\\circ} + \\frac{A}{2... | Hong Kong | 1997-2023 IMO HK TST | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimiza... | null | proof and answer | 3 + sqrt(3) | |
0k9u | Problem:
In Middle-Earth, nine cities form a $3$ by $3$ grid. The top left city is the capital of Gondor and the bottom right city is the capital of Mordor. How many ways can the remaining cities be divided among the two nations such that all cities in a country can be reached from its capital via the grid-lines witho... | [
"Solution:\n\nFor convenience, we will center the grid on the origin of the coordinate plane and align the outer corners of the grid with the points $(\\pm 1, \\pm 1)$, so that $(-1,1)$ is the capital of Gondor and $(1,-1)$ is the capital of Mordor.\n\nWe will use casework on which nation the city at $(0,0)$ is par... | United States | HMMT November 2019 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 30 | |
09w4 | Determine all pairs $(a, b)$ of positive integers for which
$$
a + b = \varphi(a) + \varphi(b) + \gcd(a, b).
$$
Here, $\varphi(n)$ is the number of integers $k \in \{1, 2, \dots, n\}$ satisfying $\gcd(n, k) = 1$. | [
"First suppose that $a = 1$. Then $\\varphi(1) = 1$. For all positive integers $b$ we have $\\gcd(a, b) = 1$. Therefore in this case the equation is $1 + b = 1 + \\varphi(b) + 1$, or equivalently, $\\varphi(b) = b - 1$. This is equivalent to the statement that there exists a unique integer from $\\{1, 2, \\dots, b\... | Netherlands | IMO Team Selection Test 2, June 2020 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequ... | English | proof and answer | (1, p) and (p, 1) where p is prime; and (2^k, 2^k) for integers k ≥ 1 | |
05yt | Problem:
Montrer que pour tous réels $a$, $b$, $c$ strictement positifs:
$$
\frac{b c}{a^{2}+2 b c}+\frac{c a}{b^{2}+2 c a}+\frac{a b}{c^{2}+2 a b} \leqslant 1 \leqslant \frac{a^{2}}{a^{2}+2 b c}+\frac{b^{2}}{b^{2}+2 c a}+\frac{c^{2}}{c^{2}+2 a b}
$$ | [
"Solution:\nCommençons par résoudre une des inégalités. Dans ce problème, l'inégalité la plus simple à étudier est celle de droite. On applique l'inégalité arithmético-géométrique sur les dénominateurs pour avoir $2 b c \\leqslant b^{2}+c^{2}$ par exemple, ce qui donne\n$$\n\\frac{a^{2}}{a^{2}+2 b c}+\\frac{b^{2}}{... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOi 2 : Algèbre | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0irh | Problem:
Let $ABC$ be a triangle with $\angle BAC = 90^\circ$. A circle is tangent to the sides $AB$ and $AC$ at $X$ and $Y$ respectively, such that the points on the circle diametrically opposite $X$ and $Y$ both lie on the side $BC$. Given that $AB = 6$, find the area of the portion of the circle that lies outside t... | [
"Solution:\n\nLet $O$ be the center of the circle, and $r$ its radius, and let $X'$ and $Y'$ be the points diametrically opposite $X$ and $Y$, respectively. We have $OX' = OY' = r$, and $\\angle X' O Y' = 90^\\circ$. Since triangles $X' O Y'$ and $BAC$ are similar, we see that $AB = AC$. Let $X''$ be the projection... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | π - 2 | |
035u | Problem:
In a group of $B$ boys and $G$ girls it is known that $G \geq 2 B-1$. Some boys know some girls. Prove that it possible to arrange a dance in pairs in such a way that all boys will dance and every boy who does not know the girl in his pair knows only girls who do not dance. | [
"Solution:\n\nIf for every $s=1,2, \\ldots, B$ any $s$ boys know together at least $s$ girls then the Hall (marriages') theorem implies that every boy can dance with a known girl and the condition is satisfied.\n\nLet us assume now the converse and choose the largest $s \\leq B$, such that there are $s$ boys who kn... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof only | null | |
0774 | Problem:
Let $ABC$ be a triangle with $\angle BAC > 90^\circ$. Let $D$ be a point on the segment $BC$ and $E$ be a point on the line $AD$ such that $AB$ is tangent to the circumcircle of triangle $ACD$ at $A$ and $BE$ is perpendicular to $AD$. Given that $CA = CD$ and $AE = CE$, determine $\angle BCA$ in degrees. | [
"Solution:\n\nLet $\\angle C = 2\\alpha$. Then $\\angle CAD = \\angle CDA = 90^\\circ - \\alpha$. Moreover, $\\angle BAD = 2\\alpha$ as $AB$ is tangent to the circumcircle of $\\triangle CAD$. Since $AE = AD$, it gives $\\angle AEC = 2\\alpha$. Thus $\\triangle AEC$ is similar to $\\triangle ACD$. Hence\n$$\n\\frac... | India | Indian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 45 | |
04ad | The incircle of the acute triangle $ABC$ touches the segments $BC$, $CA$, $AB$ at points $D$, $E$, $F$ respectively. Let $S$ be the incenter and $P$ be the intersection of the line $DS$ and the segment $EF$. If $M$ is the midpoint of the segment $BC$ prove that the points $A$, $P$ and $M$ are collinear. | [
"Since quadrilaterals $BFSD$ and $CDSE$ are cyclic, we have $\\angle FSD = 180^\\circ - \\beta$ and $\\angle ESD = 180^\\circ - \\gamma$ so $\\angle FSP = \\beta$ and $\\angle ESP = \\gamma$.\n\nLet $x = \\angle FAP$ and $y = \\angle BAM$. Then $\\angle EAP = \\alpha - x$ and $\\angle CAM =... | Croatia | CroatianCompetitions2011 | [
"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 > Triangles > Triangle tr... | null | proof only | null | |
00j3 | We are given a tetrahedron with 5 edges of length $2$ and one of length $1$. A point $P$ either in the interior of the tetrahedron or on its surface (but not outside the tetrahedron) has distances from the surfaces of the tetrahedron we name $a, b, c$ and $d$. For which points $P$ is the value of $a+b+c+d$ minimal and ... | [
"The tetrahedron has two equilateral faces whose sides are of length $2$, and two isosceles faces with two sides of length $2$ and one of length $1$. Let $F$ be the area of each equilateral face and $G$ the area of each isosceles face. It is obvious that $F > G$ holds. Further, let $a$ and $b$ be the distances of $... | Austria | AustriaMO2011 | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | English | proof and answer | Minimum: all points on the short edge common to the two isosceles faces. Maximum: all points on the edge common to the two equilateral faces. | |
0fh6 | Problem:
Cada punto de un plano está pintado de un color elegido entre tres distintos. ¿Existen necesariamente dos puntos de ese plano que disten $1~\mathrm{cm}$ y que estén pintados del mismo color? | [
"Solution:\n\nLa respuesta es sí. Consideremos puntos $A$, $B$, $C$, $D$, $E$, $F$, $G$ de manera que $ABFD$ es un rombo, de lados $AB$, $AD$, $BD$, $BF$, y $FD$ de longitud $1$; y $AEGC$ es también un rombo, de lados $AE$, $AC$, $EC$, $EG$ y $GC$ de longitud $1$. Además unimos $F$ con $G$ por una arista de longitu... | Spain | OME 26 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | Yes | |
0gds | 令 $\mathbb{R}$ 為全體實數所成之集合。試找出所有的函數 $f: \mathbb{R} \to \mathbb{R}$ 使得對任意的實數 $x, y$, 都有
$$
f(x + f(y)) + f(xy) = y f(x) + f(y) + f(f(x)).
$$
Let $\mathbb{R}$ be the set of all real numbers. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for any $x, y \in \mathbb{R}$, there holds
$$
f(x + f(y)) + f(xy) = y f... | [
"將原關係記為 (*)，並定義 $P(a, b)$ 為將 $x = a, y = b$ 代入函數 $f$ 的條件所得到的性質。我們依下列步驟分析：\n\n(甲). 操作 $P(x, 1)$ 得到 $f(x + f(1)) = f(1) + f(f(x))$。由此得到 $f(f(1 - f(1))) = 0$。整理如下：\n$$\nf(1 - f(1)) = a, \\quad f(a) = 0. \\tag{1}\n$$\n另外，操作 $P(a, a)$ 並由 (1) 得到\n$$\nf(a^2) = f(0). \\tag{2}\n$$\n再操作 $P(0, a^2)$ 以及 $P(0, x)$，我們依序可整理得到\n$$... | Taiwan | 2020 Taiwan IMO 1J | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = 0 for all real x; f(x) = x for all real x | |
0foz | Let $p_1, p_2, \dots, p_{n+1}$ denote the first $n+1$ primes. Suppose that $\{A, B\}$ is a partition of the set $X = \{p_1, p_2, \dots, p_n\}$, where $A = \{q_1, q_2, \dots, q_s\}$ and $B = \{r_1, r_2, \dots, r_t\}$. Prove that if $m = q_1q_2\dots q_s + r_1r_2\dots r_t < p_{n+1}^2$, then $m$ is a prime. | [
"Assume to the contrary, that $m$ is not a prime number. Then $m = ab$ for some integers $a$ and $b$ with $1 < a < m$ and $1 < b < m$. Let $p$ be the smallest prime that divides $a$ and let $q$ be the smallest prime that divides $b$. WLOG we may assume that $p \\le q$. We now consider two cases according to whether... | Spain | BARCELONA TECH MATHCONTEST | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Spanish | proof only | null | |
0a5t | Problem:
For any positive integer $n$ let $n! = 1 \times 2 \times 3 \times \dots \times n$. Do there exist infinitely many triples $(p, q, r)$, of positive integers with $p > q > r > 1$ such that the product
$$p! \cdot q! \cdot r!$$
is a perfect square? | [
"Solution:\nYes. Let $t$ be an arbitrary positive integer and consider the following perfect square:\n$$(t!)^{2} = (t!) \\cdot (t! - 1)! \\cdot t!.$$ \nSo if we consider $(p,q,r) = (t!, t! - 1, t)$ then $p!q!r! = (t!)^{2}$ which is a perfect square. Since the choice of $t$ is arbitrary, there must be infinitely ma... | New Zealand | NZMO Round Two | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0jbk | Let $P$ be a point in the plane of $ABC$, and $\gamma$ a line passing through $P$. Let $A'$, $B'$, $C'$ be the points where the reflections of lines $PA$, $PB$, $PC$ with respect to $\gamma$ intersect lines $BC$, $AC$, $AB$, respectively. Prove that $A'$, $B'$, $C'$ are collinear.
 | [
"There are several possible configurations depending on the location of $P$ and the orientation of $\\gamma$. We will consider the configuration above but will use directed lengths and angles so our arguments apply to all diagram configurations. By the law of sines on triangles $A_1PB$ and $A_1CP$, we have\n$$\nBP ... | United States | USAMO | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00bu | Let $ABC$ be an acute-angled triangle with $AC > AB$. Let $\Gamma$ be the circumference circumscribed about the triangle $ABC$ and $D$ the midpoint of the smaller arc $BC$ of $\Gamma$. Let $E$ and $F$ be points in the segments $AB$ and $AC$ respectively such that $AE = AF$. Let $P \ne A$ be the second intersection poin... | [
"Let $M$ be the midpoint of the segment $BC$.\n\n\n\nLet $\\alpha = \\angle AEF$. Since $AE = AF$, we have that $\\angle AFE = \\angle AEF = \\alpha$. Considering the cyclic quadrilaterals $APEF$ and $APDH$, we have that\n$$\n\\alpha = \\angle AEF = \\angle APF = \\angle APH = \\angle ADH.\... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0gjl | A $k$-set is a set with exactly $k$ elements. For a 6-set $A$ and any collection $\mathcal{F}$ of 4-sets, we say that $A$ is $\mathcal{F}$-good if there are exactly three elements $B_1, B_2, B_3$ in $\mathcal{F}$ that are subsets of $A$, and they furthermore satisfy
$$
(A \setminus B_1) \cup (A \setminus B_2) \cup (A \... | [
"**答案為 $n = 6, 7, 8$。** 注意到對於 $6 \\le m \\le n$, 若我們對 $\\{1, 2, \\dots, n\\}$ 可構造滿足題意的 $\\mathcal{F}$, 則必然可以對 $\\{1, 2, \\dots, m\\}$ 構造滿足題意的 $\\mathcal{F}$。故我們只需證明兩點：\n\n1. $n = 9$ 時不存在滿足題意的 $\\mathcal{F}$\n**證明**：反證法，假設這樣的 $\\mathcal{F}$ 存在。則對於任何 $A \\subset \\{1, 2, \\dots, n\\}$, 存在三個 $B \\in \\mathcal{F}$ 滿足題意... | Taiwan | IMO 1J, Independent Study 2 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | Chinese; English | proof and answer | n = 6, 7, 8 | |
0ejb | Problem:
Diagonali $AC$ in $BD$ trapeza $ABCD$ se sekata v točki $E$ in razdelita trapez na 4 trikotnike s ploščinami $25~\mathrm{cm}^2$, $36~\mathrm{cm}^2$, $X~\mathrm{cm}^2$ in $X~\mathrm{cm}^2$ (glej sliko). Kolikšna je vrednost $X$?
(A) 25
(B) 30
(C) 32
(D) 36
(E) 61
 | [
"Solution:\n\nKer imata trikotnika $AED$ in $ABE$ enaki višini iz točke $A$, velja $X : 36 = |DE| : |EB|$. Podobno imata trikotnika $CDE$ in $CEB$ enaki višini iz točke $C$, zato velja $25 : X = |DE| : |EB|$. Sledi $X : 36 = 25 : X$, od koder izračunamo $X = \\sqrt{25 \\cdot 36} = 30$."
] | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles"
] | null | MCQ | B | |
0fqj | Let $a$, $b$ and $c$ be real numbers such that $p(x) = x^4 + a x^3 + b x^2 + a x + c$ has exactly three different real roots; these roots are $\tan y$, $\tan 2y$ and $\tan 3y$ for some real number $y$. Find all possible values of $y$, $0 \le y < \pi$. | [
"that is,\n$$\n\\tan(2ky) + \\tan(my) + \\tan(ny) = 0\n$$\nprovided that $r$, $s$, $t$ are $\\tan(ky)$, $\\tan(my)$ and $\\tan(ny)$ respectively.\nWe consider now the following cases:\n* If $r = \\tan y$, $s = \\tan 2y$ and $t = \\tan 3y$ then $\\tan 2y + \\tan 5y = 0$, and\n$$\ny \\in \\left\\{ \\frac{\\pi}{7}, \\... | Spain | SPANISH MATHEMATICAL OLYMPIAD (FINAL ROUND) | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | y ∈ {π/7, 2π/7, 3π/7, 4π/7, 5π/7, 6π/7} ∪ {π/8, 3π/8, 5π/8, 7π/8} ∪ {π/9, 2π/9, π/3, 4π/9, 5π/9, 2π/3, 7π/9, 8π/9} | |
0aco | Prove that if $\left|\frac{a+b}{2}\right| + \left|\frac{a-b}{2}\right| < c$, for $a, b, c \in \mathbb{R}$, then $|a| < c$ and $|b| < c$. | [
"By the properties of absolute value, we have\n$$\n|a| = 2\\left|\\frac{a}{2}\\right| = \\left|\\frac{a}{2} + \\frac{a}{2}\\right| = \\left(\\frac{a}{2} + \\frac{b}{2}\\right) + \\left(\\frac{a}{2} - \\frac{b}{2}\\right) \\le \\left|\\frac{a+b}{2}\\right| + \\left|\\frac{a-b}{2}\\right| < c \\text{ i.e. } |a| < c.\... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0jzj | Problem:
A polynomial $P$ of degree $2015$ satisfies the equation $P(n)=\frac{1}{n^{2}}$ for $n=1,2, \ldots, 2016$. Find $\lfloor 2017 P(2017)\rfloor$. | [
"Solution:\n\nLet $Q(x)=x^{2} P(x)-1$. Then $Q(n)=n^{2} P(n)-1=0$ for $n=1,2, \\ldots, 2016$, and $Q$ has degree $2017$. Thus we may write\n$$\nQ(x)=x^{2} P(x)-1=(x-1)(x-2) \\ldots(x-2016) L(x)\n$$\nwhere $L(x)$ is some linear polynomial. Then $Q(0)=-1=(-1)(-2) \\ldots(-2016) L(0)$, so $L(0)=-\\frac{1}{2016!}$.\n\n... | United States | February 2017 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | -9 |
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