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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0izl | Problem:
How many nondecreasing sequences $a_{1}, a_{2}, \ldots, a_{10}$ are composed entirely of at most three distinct numbers from the set $\{1,2, \ldots, 9\}$ (so $1,1,1,2,2,2,3,3,3,3$ and $2,2,2,2,5,5,5,5,5,5$ are both allowed)? | [
"Solution:\n\nAnswer: 3357\n\nFrom any sequence $a_{1}, a_{2}, \\ldots, a_{10}$, construct a sequence $b_{1}, b_{2}, \\ldots, b_{9}$, where $b_{i}$ counts the number of times $i$ occurs in the sequence. There is a correspondence from all possible sequences $b_{1}, b_{2}, \\ldots, b_{9}$ with at most 3 nonzero terms... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 3357 | |
09i4 | There are pebbles at the center of each cell on a $12 \times 12$ board and two players, Painter and Flicker, play the following game.
Painter chooses a number of colors and paints each pebble with one of the colors. Flicker chooses two pebbles of the same color that has no blocking pebble in between them, and replaces ... | [] | Mongolia | Round 2 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 4 | |
0c4j | The sequence $(x_n)_{n \ge 1}$ fulfils $x_{n+3} = x_n + x_{n+1} + x_{n+2}$, for every $n \in \mathbb{N}^*$. Find $x_1, x_2, x_3 \in \mathbb{Q} \cap (0, 1)$, so that $(x_n)_{n \ge 1}$ has a limit. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | No such triple exists. | |
01oc | Given a quadrilateral $ABCD$ with $AB + CD = 6$, $BC + DA = 8$.
Find the area of $ABCD$ if it has the greatest area among all quadrilaterals with the mentioned sums of the opposite sides. | [
"Answer: $12$.\nWe use the following obvious inequalities. If the quadrilateral $ABCD$ has the sides $AB = a$, $BC = b$, $CD = c$, $DA = d$, and the area $S$, then\n$$S \\le \\frac{ab + cd}{2} \\text{ because}$$\n$$\nS = S(ABC) + S(ACD) = \\\\\n= \\frac{1}{2}ab \\sin \\beta + \\frac{1}{2}cd \\sin \\delta \\le \\fra... | Belarus | Belorusija 2012 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof and answer | 12 | |
05xc | Problem:
Exercice 16. Soit $p$ un nombre premier. Martin la grenouille est situé en position $0$ sur la droite réelle. À chaque seconde, Martin effectue un mouvement : il peut rester à sa position, faire un saut de $1$ sur la droite, ou faire un saut de $1$ sur la gauche. Martin souhaite être revenu au bout de $p-1$ m... | [
"Solution:\n\nNotons $a_{n}(k)$ le nombre de suites de $n$ mouvements tel que si Martin était initialement à la position $0$, il se retrouve à la position $k$ (par exemple, pour $n=2$, $a_{2}(0)=2$ et $a_{2}(2)=a_{2}(-2)=1$). Cela revient donc à compter le nombre de $n$-uplets à valeurs dans $\\{-1,0,1\\}$ de somme... | France | Préparation Olympique Française de Mathématiques - ENVOI 5 : Pot-POURRI | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Pri... | null | proof and answer | s_p ≡ 0 if p = 3; s_p ≡ 1 if p ≡ 1 mod 3; s_p ≡ p − 1 if p ≡ 2 mod 3 | |
0cb1 | Find all integers $n \ge 0$ for which there exist integers $a$ and $b$ such that $a + 2^b = n^{2022}$ and $a^2 + 4^b = n^{2023}$. | [
"The only number is $n = 1$.\nTo this end, notice that $2(a^2 + 4^b) \\ge (a + 2^b)^2$ to conclude that $2n^{2023} \\ge n^{4044}$, hence $n = 0$ or $n = 1$. The case $n = 0$ leads to no solution, while $n = 1$ holds for $a = 0$ and $b = 0$."
] | Romania | THE Sixteenth STARS OF MATHEMATICS Competition | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 1 | |
095h | Problem:
Să se arate că oricare ar fi numerele reale pozitive $a$, $b$ şi $c$, este adevărată inegalitatea:
$$
\sqrt{\frac{a}{b+c}} + \sqrt{\frac{b}{c+a}} + \sqrt{\frac{c}{a+b}} > 2.
$$ | [
"Solution:\nConsiderăm numerele reale pozitive $\\frac{a}{b+c}$ şi $1$. Din inegalitatea dintre media geometrică şi media armonică, avem\n$$\n\\sqrt{\\frac{a}{b+c} \\cdot 1} \\geq \\frac{2 \\frac{a}{b+c} \\cdot 1}{\\frac{a}{b+c} + 1} = \\frac{2a}{a+b+c}.\n$$\nIdem $\\sqrt{\\frac{b}{c+a}} \\geq \\frac{2b}{a+b+c}$ şi... | Moldova | A 62-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0j8r | Problem:
The integers from $1$ to $n$ are written in increasing order from left to right on a blackboard. David and Goliath play the following game: starting with David, the two players alternate erasing any two consecutive numbers and replacing them with their sum or product. Play continues until only one number on t... | [
"Solution:\n\nAnswer: $4022$\n\nIf $n$ is odd and greater than $1$, then Goliath makes the last move. No matter what two numbers are on the board, Goliath can combine them to make an even number. Hence Goliath has a winning strategy in this case.\n\nNow suppose $n$ is even. We can replace all numbers on the board b... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 4022 | |
0476 | A positive integer $N$ is called a *good number* if the decimal representation of $N$ can be divided into at least 5 segments of digits, each segment containing at least one non-zero digit, and these segments (ignoring any leading zeros) can be viewed as positive integers which can be divided into two groups, with each... | [
"Let $p = 1 + a + a^2 + \\cdots + a^m$ be a prime number, where $a > 1$ and $m > 2$ are integers. Prove that $\\frac{10^{p-1}-1}{p}$ is a good number.\n\n*Proof*. Since $p = 1 + a + \\cdots + a^m = \\frac{a^{m+1}-1}{a-1}$ is a prime number, $q = m + 1$ must be a prime number. Given $q = m + 1 > 3$, we have $q \\ge ... | China | The 65th IMO China National Team Selection Test | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof only | null | |
00n0 | We are given an arbitrary acute-angled triangle $ABC$ and its altitudes $AD$ and $BE$ where $D$ and $E$ denote their feet on sides $BC$ and $AC$, respectively. Let furthermore $F$ and $G$ be two points on segments $AD$ and $BE$, respectively, such that
$$
\frac{AF}{FD} = \frac{BG}{GE}.
$$
The line through $C$ and $F$ i... | [
"The two right-angled triangles $ADC$ and $BEC$ are inversely similar to each other, see Figure 6. Here, the sides $AD$ and $BE$ correspond to each other.\nBut the condition\n$$\n\\frac{AF}{FD} = \\frac{BG}{GE}\n$$\nmeans: The two points $F$ and $G$ divide the two sides $AD$ and $BE$, respectively, in equal ratios.... | Austria | AUT_ABooklet_2020 | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0gzf | Find the least natural number $n > 100$ for which the next statement is true: maximal sum of digits amongst the numbers $n - 100, n - 99, \ldots, n, \ldots, n + 99, n + 100$ has $n$. Justify your answer. | [
"Consider an arbitrary number. If the two last digits differ from $99$ then replacing them by $99$ we obtain a number with bigger sum of digits. Numbers $199, 299, \\ldots, 899$ don't satisfy this condition, because for each $n \\in 199, 299, \\ldots, 899$ a number $n + 100$ has the sum of digits bigger by $1$. And... | Ukraine | 50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010) | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | 999 | |
01oq | Solve the equation in prime numbers $p$ and $q$: $q^3 = p^2 - p + 1$. | [
"Answer: $p = 19$, $q = 7$.\n\nWe have\n$$\nq^3 = p^2 - p + 1 \\Leftrightarrow (q-1)(q^2 + q + 1) = p(p-1). \\quad (1)\n$$\nIf $(q-1) \\nmid p$, then $q \\ge p+1$, so $q^3 > p^2 - p + 1$.\nHence, $(q^2 + q + 1) \\nmid p$, i.e.\n$$\nq^2 + q + 1 = k p \\quad (2)\n$$\nfor some $k \\in \\mathbb{N}$. It follows that $k(... | Belarus | Belarusian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | p = 19, q = 7 | |
052h | Find all remainders which one can get when dividing by $6$ an integer $n$ which satisfies $n^3 = m^2 + m + 1$ for some integer $m$. | [
"Numbers $n$ and $n^3$ give the same remainder when dividing by $6$. Also, $m^2 + m + 1$ is odd and gives the remainder $0$ or $1$ when dividing by $3$. The only possibility to get $0$ as the remainder is when $m = 3k+1$, but then\n$$\nn^3 = (9k^2 + 6k + 1) + (3k + 1) + 1 = 9k^2 + 9k + 3 = 3(3k^2 + 3k + 1)\n$$\nwhi... | Estonia | Open Contests | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 1 | |
0eto | Consider two positive integers $a$ and $b$ such that $a + 2b = 2020$.
a. Determine the largest possible value of the greatest common divisor of $a$ and $b$.
b. Determine the smallest possible value of the least common multiple of $a$ and $b$. | [
"We observe that $2020 = 2^2 \\cdot 5 \\cdot 101$. Whenever $a + 2b = 2020$, $d = \\gcd(a, b)$ must divide $2020$. So, to find the largest possible value of $d$, we have to look at the large divisors of $2020$. $d = 2020$ is not possible, since in this case $a + 2b \\ge 2020 + 2 \\cdot 2020 > 2020$. The next larges... | South Africa | The South African Mathematical Olympiad Third Round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a) 505; b) 808 | |
07zf | Problem:
Se $x$ è un numero reale positivo si denoti con $[x]$ la parte intera di $x$, cioè il massimo intero $n \leq x$. Si calcoli la somma $\sum_{n=1}^{1000000} [\sqrt{n}] = [\sqrt{1}] + [\sqrt{2}] + \ldots + [\sqrt{999999}] + [\sqrt{1000000}]$.
[Lo studente può utilizzare, se crede, la seguente formula: $\sum_{i=... | [
"Solution:\n\nSi ha $[\\sqrt{n}] = k$ se e solo se $k^2 \\leq n < (k+1)^2 = k^2 + 2k + 1$, ossia per $n = k^2 + j$ con $0 \\leq j \\leq 2k$, ovvero per $2k + 1$ valori di $j$.\n\nPertanto,\n$$\n\\begin{gathered}\n\\sum_{n=1}^{K^2-1} [\\sqrt{n}] = \\sum_{k=1}^{K-1} k(2k+1) = \\\\\n= 2 \\sum_{k=1}^{K-1} k^2 + \\sum_{... | Italy | GARA NAZIONALE di MATEMATICA | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 666167500 | |
0ist | Problem:
Point $D$ lies inside the triangle $ABC$. If $A_1$, $B_1$, and $C_1$ are the second intersection points of the lines $AD$, $BD$, and $CD$ with the circles circumscribed about $\triangle BDC$, $\triangle CDA$, and $\triangle ADB$, prove that
$$
\frac{AD}{AA_1} + \frac{BD}{BB_1} + \frac{CD}{CC_1} = 1
$$ | [
"Solution:\n\nLet $k$ be the circle with center $D$ and radius $1$. Consider the inversion with respect to the circle $k$ and denote by $A^*$, $B^*$, $C^*$, $A_1^*$, $B_1^*$, and $C_1^*$ the images of $A$, $B$, $C$, $A_1$, $B_1$, and $C_1$, respectively.\n\n\n\nThe point $C_1^*$ belongs to ... | United States | Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Inversion"
] | null | proof only | null | |
04va | In the convex pentagon $ABCDE$ $|\angle CBA| = |\angle BAE| = |\angle AED|$ holds. On the sides $AB$ and $AE$, there are points $P$ and $Q$, respectively, such that $|AP| = |BC| = |QE|$ and $|AQ| = |BP| = |DE|$. Prove that $CD \parallel PQ$. (Patrik Bak) | [
"Since $|BC| = |AP| = |EQ|$, $|BP| = |AQ| = |ED|$ and $|\\angle CBP| = |\\angle PAQ| = |\\angle QED|$, the triangles $PBC$, $QAP$ and $DEQ$ are congruent by the condition SAS.\n\nHence $|CP| = |PQ| = |QD|$ and also\n$$\n|\\angle CPQ| = 180^\\circ - |\\angle BPC| - |\\angle APQ| = 180^\\circ - |\\angle PQA| - |\\ang... | Czech Republic | 72nd Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03kl | Problem:
We denote an arbitrary permutation of the integers $1, \ldots, n$ by $a_{1}, \ldots, a_{n}$. Let $f(n)$ be the number of these permutations such that
(i) $a_{1}=1$;
(ii) $|a_{i}-a_{i+1}| \leq 2, \quad i=1, \ldots, n-1$.
Determine whether $f(1996)$ is divisible by $3$. | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | Yes | |
0b5u | Prove that given any two permutations $\sigma, \tau \in S_n$, there exists some function $f : \{1, \dots, n\} \to \{-1, 1\}$ such that we simultaneously have, for any indices $1 \le i \le j \le n$,
$$
\left| \sum_{k=i}^{j} f(\sigma(k)) \right| \le 2 \quad \text{and} \quad \left| \sum_{k=i}^{j} f(\tau(k)) \right| \le 2.... | [
"We will first prove it for the even case, $n = 2m$. Consider the graph whose set of vertices is $\\{1, \\dots, 2m\\}$, and with red edges between $\\sigma(2k-1)$ and $\\sigma(2k)$, and blue edges between $\\tau(2k-1)$ and $\\tau(2k)$, where $1 \\le k \\le m$ (notice that multiple edges may occur).\n\nClearly, each... | Romania | Local Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0792 | Problem:
All the squares of a $2024 \times 2024$ board are coloured white. In one move, Mohit can select one row or column whose every square is white, choose exactly 1000 squares in this row or column, and colour all of them red. Find the maximum number of squares that Mohit can colour red in a finite number of moves... | [
"Solution:\n\nLet $n=2024$ and $k=1000$. We claim that the maximum number of squares that can be coloured in this way is $k(2n-k)$, which evaluates to $3048000$.\n\nIndeed, call a row/column bad if it has at least one red square. After the first move, there are exactly $k+1$ bad rows and columns: if a row was picke... | India | INMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 3048000 | |
0hj8 | Problem:
Let $p$, $q$, and $r$ be distinct primes. Prove that $p+q+r+p q r$ is composite. | [
"Solution:\nNote that at most one of $p$, $q$, and $r$ can equal $2$. If $p=2$, then $q$ and $r$ are both odd, but $p q r$ is even. Therefore their sum is even. Similarly, if $q$ or $r$ is $2$, then $p+q+r+p q r$ is even. Finally, if $p$, $q$, and $r$ are all odd, then so is $p q r$, and $p+q+r+p q r$ is even. Thus... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
05il | Problem:
Déterminer tous les entiers naturels $m, n$ et tous les nombres premiers $p$ tels que
$$
m\left(4 m^{2}+m+12\right)=3\left(p^{n}-1\right)
$$ | [
"Solution:\nOn récrit l'équation sous la forme\n$$\n3 p^{n}=4 m^{3}+m^{2}+12 m+3=(4 m+1)\\left(m^{2}+3\\right)\n$$\nOn en déduit qu'il existe $u, v, a, b$ entiers naturels tels que $4 m+1=3^{u} p^{a}$ et $m^{2}+3=3^{v} p^{b}$ avec $u+v=1$ et $a+b=n$.\nSi $m=0$, l'équation devient $3 p^{n}=3$, donc $n=0$ et tout nom... | France | Test EGMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | m=0, n=0, p any prime; and m=12, n=4, p=7 | |
0bfa | Prove that, if $a, b, c$ are the complex coordinates of the vertices of a triangle and $|a + 2b - 3c| = |a - 4b + 3c|$, then the triangle is right. | [
"Let $a$, $b$, $c$ be the complex coordinates of the vertices of the triangle. The given condition is:\n$$\n|a + 2b - 3c| = |a - 4b + 3c|\n$$\nLet $z_1 = a + 2b - 3c$ and $z_2 = a - 4b + 3c$. Then $|z_1| = |z_2|$.\n\nConsider the difference:\n$$\n|z_1|^2 = |z_2|^2\n$$\nExpanding both sides:\n$$\n(a + 2b - 3c)(\\ove... | Romania | Shortlisted Problems for the 64th NMO | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0237 | Problem:
As férias de Tomás - Durante suas férias, Tomás teve 11 dias com chuva. Durante esses 11 dias, se chovia pela manhã havia sol sem chuva à tarde, e se chovia à tarde, havia sol sem chuva pela manhã. No total, Tomás teve 9 manhãs e 12 tardes sem chuva. Quantos dias duraram as férias de Tomás? | [
"Solution:\n\n16 dias"
] | Brazil | null | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 16 | |
0bnk | Let $p$ be a prime greater than $3$. For a positive integer $k$, let $R(k)$ denote the remainder of $k$ when divided by $p$. Determine all positive integers $a < p$ such that $m + R(ma) > a$ for every $m = 1, 2, \dots, p-1$. | [
"The required integers are $p-1$ along with all the numbers of the form $\\lfloor p/q \\rfloor$, $q = 2, \\dots, p-1$. In other words, these are $p-1$, along with the numbers $1, 2, \\dots, \\lfloor \\sqrt{p} \\rfloor$, and also the (distinct) numbers $\\lfloor p/q \\rfloor$, $q = 2, \\dots, \\lfloor \\sqrt{p} - 1/... | Romania | THE 2015 Seventh ROMANIAN MASTER OF MATHEMATICS | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | All a with a = p − 1 and all a of the form a = ⎣p/q⎦ for integers q = 2, 3, …, p − 1 (duplicates ignored). Equivalently: a = p − 1 together with the integers 1, 2, …, ⎣√p⎦ and the distinct values ⎣p/q⎦ for q = 2, …, ⎣√p − 1/2⎦. | |
0f72 | Problem:
Two equal squares, one with blue sides and one with red sides, intersect to give an octagon with sides alternately red and blue. Show that the sum of the octagon's red side lengths equals the sum of its blue side lengths. | [] | Soviet Union | 20th ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
00eb | We say that a positive integer $N$ is *rioplatense* if it satisfies the following two conditions:
* It is possible to find 34 consecutive integers such that their product is divisible by $N$ but none of them is divisible by $N$.
* It is *not* possible to find 30 consecutive integers such that their product is divisible... | [
"Notice $N > 34$ so that the first condition can hold.\n\nFirst we will show that $N$ must be a prime power. Assume the contrary, so $N$ has two or more distinct prime factors. In this case, we can write $N = ab$ with $a, b$ relatively prime integers greater than 1. By the Chinese Remainder Theorem, there exists an... | Argentina | Rioplatense Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | All integers of the form 31^k with k ≥ 2 | |
05rm | Problem:
Soit $m$ et $n$ deux entiers naturels. Démontrer que $n!\neq m^{2}+2019$. | [
"Solution:\n\nProcédons par l'absurde, et supposons que l'on dispose de deux entiers naturels $m$ et $n$ tels que $n! = m^{2} + 2019$. Tout d'abord, on sait que $n! \\geqslant 2019 > 6! = 720$, donc que $n \\geqslant 7$. Par conséquent, $n!$ est divisible par $7! = 2^{4} \\times 3^{2} \\times 5 \\times 7$, et l'on ... | France | Préparation Olympique Française de Mathématiques - Test du 15 Mai 2019 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequali... | null | proof only | null | |
04ma | Determine all triples $(x, y, z)$ of real numbers that satisfy
$$
\begin{align*}
x + y - z &= -1 \\
x^2 - y^2 + z^2 &= 1 \\
-x^3 + y^3 + z^3 &= -1.
\end{align*}
$$ | [
"From the first equation we get $x + y = z - 1$. By plugging this into the second equation we get:\n$$\n\\begin{aligned}\nx^2 - y^2 &= 1 - z^2, \\\\\n(x + y)(x - y) &= (1 - z)(1 + z), \\\\\n(z - 1)(x - y) &= -(z - 1)(1 + z), \\\\\n(z - 1)(x - y + z + 1) &= 0.\n\\end{aligned}\n$$\nThere are now two cases: $z = 1$ or... | Croatia | Croatia_2018 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | {(1, -1, 1), (-1, -1, -1)} | |
0gje | 給定任一質數 $p$，當一個集合恰包含三個元素 $a, b, c$ 且 $a + b \equiv c \pmod p$，則我們稱這個集合為 $p$-good。
找出所有質數 $p$，使得 $\{1, 2, \dots, p-1\}$ 可以被全部分割成許多 $p$-good 集合。
Given a prime number $p$, a set is said to be *p-good* if the set contains exactly three elements $a, b, c$ and $a + b \equiv c \pmod p$.
Find all prime number $p$ such that $\{... | [
"找出所有質數 $p$，使得 $\\{1, 2, \\dots, p-1\\}$ 可以被全部分割成許多 $p$-good 集合。\n\n**解.** Clearly, we must have $p \\equiv 1 \\pmod 6$. To show that all such $p$ satisfies the condition, choose $g$ to be a primitive root modulo $p$. Let $x = g^{(p-1)/6}$. Can verify that $x^3 \\equiv -1 \\pmod p$ thus $x^2 - x + 1 \\equiv 0 \\pmo... | Taiwan | IMO 1J, Independent Study 1 | [
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | Chinese; English | proof and answer | All primes congruent to 1 modulo 6 | |
0gol | Let $A = \{1, 2, \dots, 2012\}$, $B = \{1, 2, \dots, 19\}$ and $S$ be the set of all subsets of $A$. Determine the number of functions $f : S \to B$ satisfying the condition $f(A_1 \cap A_2) = \min\{f(A_1), f(A_2)\}$ for all $A_1, A_2 \in S$. | [
"The answer is $1^{2012} + 2^{2012} + \\dots + 19^{2012}$.\n\nWe first observe that the minimum element of $U \\cup V$ is the minimum of the minimum elements of $U$ and $V$ for all finite sets $U$ and $V$ of integers.\n\nLet the value of $f(A)$ be $n$. Then by the given property of $f$ we have $f(X) \\le n$ for all... | Turkey | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Functional equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 1^{2012} + 2^{2012} + \dots + 19^{2012} | |
09cn | $\forall x, y \in \mathbb{R}^+ = (0, +\infty) : f\left(\sqrt{\frac{x^2 + xy + y^2}{2012}}\right) = \frac{f(x) - f(y)}{2}$ байх бүх $f : \mathbb{R}^+ \to \mathbb{R}$ функцийг ол. | [
"$\\forall a, b \\in \\mathbb{R}^+: a * b = \\sqrt{\\frac{a^2 + ab + b^2}{2012}}$ болог. Тэгвэл\n$$\nf(a * b) = \\frac{f(a) + f(b)}{2} \\text{ байна. } \\forall x_1, x_2, x_3, x_4 \\in \\mathbb{R}^+ :\n$$\n$$\nf((x_1 * x_2) * (x_3 * x_4)) = \\frac{f(x_1 * x_2) + f(x_3 * x_4)}{2} = \\frac{f(x_1) + f(x_2) + f(x_3) + ... | Mongolia | ОУМО-53 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | Mongolian | proof and answer | All constant functions f(x) = c for x > 0, where c is a real constant. | |
0hlz | Problem:
Let $k$ be a rational number greater than $1$ (correction by Fengning Ding). Prove that there exist positive integers $a, b, c$ satisfying the equations
$$
\begin{aligned}
a^{2}+b^{2} & =c^{2} \\
\frac{a+c}{b} & =k .
\end{aligned}
$$ | [
"Solution:\n\nLet $k = x / y$, where $x$ and $y$ are positive integers. We find that $x > y$. It suffices to note that\n$$\na = x^{2} - y^{2}, \\quad b = 2 x y, \\quad c = x^{2} + y^{2}\n$$\nare positive integers satisfying both of the given equations."
] | United States | Berkeley Math Circle | [
"Number Theory > Diophantine Equations > Pythagorean triples"
] | null | proof only | null | |
0iw3 | Problem:
Compute $e^{A}$ where $A$ is defined as
$$
\int_{3 / 4}^{4 / 3} \frac{2 x^{2}+x+1}{x^{3}+x^{2}+x+1} d x
$$ | [
"Solution:\nWe can use partial fractions to decompose the integrand to $\\frac{1}{x+1}+\\frac{x}{x^{2}+1}$, and then integrate the addends separately by substituting $u=x+1$ for the former and $u=x^{2}+1$ for latter, to obtain $\\ln (x+1)+\\left.\\frac{1}{2} \\ln \\left(x^{2}+1\\right)\\right|_{3 / 4} ^{4 / 3}=\\le... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | proof and answer | 16/9 | |
0je6 | Problem:
An isosceles trapezoid $ABCD$ with bases $AB$ and $CD$ has $AB = 13$, $CD = 17$, and height $3$. Let $E$ be the intersection of $AC$ and $BD$. Circles $\Omega$ and $\omega$ are circumscribed about triangles $ABE$ and $CDE$. Compute the sum of the radii of $\Omega$ and $\omega$. | [
"Solution:\n\nLet $\\Omega$ have center $O$ and radius $R$ and let $\\omega$ have center $P$ and radius $M$. Let $Q$ be the intersection of $AB$ and $OE$. Note that $OE$ is the perpendicular bisector of $AB$ because the trapezoid is isosceles. Also, we see $OE$ is the circumradius of $\\Omega$.\n\nOn the other hand... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 39 | |
03os | The function $f(x) = \frac{x}{1-2^x} - \frac{x}{2}$ is ( ).
(A) an even but not odd function
(B) an odd but not even function
(C) a both even and odd function
(D) a neither even nor odd function | [
"It is easy to see that the domain of $f(x)$ is $(-\\infty, 0) \\cup (0, +\\infty)$. When $x \\in (-\\infty, 0) \\cup (0, +\\infty)$, we have\n$$\n\\begin{aligned}\nf(-x) &= \\frac{-x}{1-2^{-x}} - \\frac{x}{2} \\\\\n&= \\frac{-x \\cdot 2^x}{2^x - 1} + \\frac{x}{2} \\\\\n&= \\frac{x}{1-2^x} - x + \\frac{x}{2} \\\\\n... | China | China Mathematical Competition | [
"Precalculus > Functions"
] | English | MCQ | A | |
099p | $k$ circles with radius $0.5$ were fit into a $n \times n$ square without any overlaps. Suppose that maximum value of $k$ is $S(n)$. Find the minimum value of $n$ such that $S(n) - n^2 \geq 100$.
(proposed by Davii-Od) | [
"Answer $n = 29$. Observe that thus position (big circle) better than this position (small circle). Hence by easy calculation about circles and empty areas of square, we get satisfying above condition $n = 29$."
] | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 29 | |
0gjo | 對於任意正整數 $n$ 考慮其二進位表示。定義 $f(n)$ 為將其二進位表示中所有 0 移除後得到的數，而 $g(n)$ 代表二進位表示中 1 的數量。例如，$f(19) = 7$ 而 $g(19) = 3$。找出所有的正整數 $n$ 使得
$$
n = f(n)^{g(n)}.
$$
For any positive integer $n$, consider its binary representation. Denote by $f(n)$ the number we get after removing all the 0's in its binary representation, and $g(n)$ the n... | [
"解. 令 $\\Gamma_1$ 和 $\\Gamma_2$ 的圓心分別為 $P$ 和 $Q$。設 $\\Gamma_2$ 與 $AB$ 的另外一個交點為 $G$、$\\Gamma_1$ 與 $AC$ 的另外一個交點為 $H$, $B$ 對 $\\Gamma_2$ 的另外一個切點為 $X_1$, $C$ 對 $\\Gamma_1$ 的切點為 $Y$, $Y_1$。\nLemma1. $G, H, D$ 三點共線且平行直線 $BC$。\nProof: 我們有 $\\angle AGD = \\angle AFD = \\angle AFC = \\angle ABC$, 所以 $GD \\parallel BC$。同理 $H... | Taiwan | IMO 2J, Independent Study 2 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Chinese; English | proof and answer | [1, 9] | |
0kvg | Problem:
Let $PABC$ be a tetrahedron such that $\angle APB = \angle APC = \angle BPC = 90^{\circ}$, $\angle ABC = 30^{\circ}$, and $AP^{2}$ equals the area of triangle $ABC$. Compute $\tan \angle ACB$. | [
"Solution:\n\nObserve that\n$$\n\\begin{aligned}\n\\frac{1}{2} \\cdot AB \\cdot AC \\cdot \\sin \\angle BAC & = [ABC] = AP^{2} \\\\\n& = \\frac{1}{2}\\left(AB^{2} + AC^{2} - BC^{2}\\right) \\\\\n& = AB \\cdot AC \\cdot \\cos \\angle BAC\n\\end{aligned}\n$$\nso $\\tan \\angle BAC = 2$. Also, we have $\\tan \\angle A... | United States | HMMT February 2023 | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 8 + 5*sqrt(3) | |
0fdg | Problem:
Determinar todos los posibles valores enteros no negativos que puede tomar la expresión $\frac{m^{2}+m n+n^{2}}{m n-1}$, siendo $m$ y $n$ enteros no negativos tales que $m n \neq 1$. | [
"Solution:\nSea $\\frac{m^{2}+m n+n^{2}}{m n-1}=k,\\ k \\in \\mathbb{N}$ (naturales con el $0$).\n\nEn el caso $m=n$, el número $k=3+\\frac{3}{m^{2}-1}$ es un entero positivo si $m=0$ ó $m=2$; de donde $k=0$ ó $k=4$ respectivamente.\n\nEl caso $n=0$ lleva a que $k=-m^{2}$ y por tanto $m=k=0$.\n\nConsideremos ahora ... | Spain | null | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | 0, 4, 7 | |
0col | Given a positive integer $n \ge 3$. Find the minimal integer $k$ satisfying the following property.
For every $n$ points $A_i = (x_i, y_i)$ on the plane, no three of them being collinear, and for every real numbers $c_i$ ($1 \le i \le n$) there exists a polynomial $P(x, y)$ such that $P(x_i, y_i) = c_i$ for all $i = 1,... | [
"Ответ. $k = [n/2]$.\n\n**Лемма.** Для любых точек $A_i = (x_i, y_i)$ ($1 \\le i \\le n$) на плоскости, никакие три из которых не лежат на одной прямой, найдётся такой многочлен $P(x, y)$ степени не больше $[n/2]$, что $P(x_n, y_n) = 1$ и $P(x_i, y_i) = 0$ при $i = 1, \\dots, n-1$.\n\n**Доказательство.** Заметим, ч... | Russia | Final round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English; Russian | proof and answer | floor(n/2) | |
05y8 | Problem:
Un certain nombre de diagonales divisent un polygone convexe en triangles. Ces segments ne peuvent s'intersecter que sur un sommet du polygone. Sur chaque sommet du polygone on écrit le nombre de triangles qui touchent ce sommet. Est-il possible de reconstruire les diagonales en connaissant seulement les nomb... | [
"Solution:\n\nIl est toujours possible de reconstruire les diagonales en utilisant ce procédé :\n- On considère un sommet \"visible\" $v$ qui ne touche qu'un seul triangle.\n- On trace la diagonale qui relie les deux sommets $x$ et $y$ adjacents à $v$.\n- On \"cache\" $v$ et on retire 1 au nombre de triangles adjac... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0gh7 | 設 $ABC$ 為銳角三角形，其中 $AC > AB$，且點 $O$ 為其外心。設點 $D$ 為 $BC$ 線段上一點。過點 $D$ 引一條與 $BC$ 垂直的直線，設該線分別與直線 $AO, AC, AB$ 交於點 $W, X, Y$。三角形 $AXY$ 的外接圓與三角形 $ABC$ 的外接圓再交於點 $Z \neq A$。
已知 $D \neq W$ 且 $OW = OD$。證明：$DZ$ 與三角形 $AXY$ 的外接圓相切。
Let $ABC$ be an acute-angled triangle with $AC > AB$, let $O$ be its circumcenter, and let $D$ be a p... | [
"設 $AO$ 與 $BC$ 交於點 $E$。由於 $EDW$ 是直角三角形且 $O$ 在 $WE$ 上，條件 $OW = OD$ 說明了 $O$ 是三角形 $EDW$ 的外心。於是 $OD = OE$，得到 $D, E$ 兩點對稱於 $BC$ 邊的中垂線。\n觀察有：\n$$\n180^\\circ - \\angle DXZ = \\angle ZXY = \\angle ZAY = \\angle ZCD,\n$$\n所以 $CDXZ$ 四點共圓。\n\n\n接著證明 $AZ \\parallel BC$。為此，引入圓 $ABC$ 上的輔助點 $Z'$ 滿足 $AZ' ... | Taiwan | 2023 數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-... | Chinese (Traditional) | proof only | null | |
0dne | Problem:
Два играча играју следећу игру. Играчи наизменично записују по један природан број већи од један, при чему није дозвољено записати линеарну комбинацију претходно записаних бројева са ненегативним целим коефицијентима. Игру губи играч који не може да запише нови број. Да ли неко од играча има победничку страте... | [
"Solution:\n\nЗа $a, b \\in \\mathbb{N}$ дефинишимо $\\mathcal{L}\\{a, b\\} = \\left\\{ a x + b y \\mid x, y \\in \\mathbb{N}_0 \\right\\}$. Прво ћемо доказати једно помоћно тврђење.\n\nЛема. Нека су $a > 1$ и $b > 1$ узајамно прости природни бројеви.\n\n(a) $N = a b - a - b$ је највећи природан број ван скупа $\\m... | Serbia | 8. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | First player | |
00tm | Let $ABC$ be a scalene triangle and let $I$ be its incenter. The projections of $I$ on $BC$, $CA$ and $AB$ are $D$, $E$ and $F$ respectively. Let $K$ be the reflection of $D$ over the line $AI$, and let $L$ be the second point of intersection of the circumcircles of the triangles $BFK$ and $CEK$. If $\frac{1}{3}BC = AC... | [
"Writing $AE = AF = x$, $BF = BD = y$ and $CE = CD = z$, the condition $\\frac{1}{3}BC = AC - AB$ translates to $y+z = 3(z-y)$ giving $z = 2y$, i.e. $CD = 2BD$.\nLetting $B'$ be the reflection of $B$ on $AI$ we have that $B'$ belongs on $AC$ with $B'E = BF = BD = \\frac{1}{2}CD = \\frac{1}{2}CE$ therefore $B'$ is t... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations... | null | proof only | null | |
0byc | Let $M$ be the midpoint of the side $[AB]$ of the square $ABCD$, let $P$ be the projection of $B$ on $CM$ and let $N$ be the midpoint of the segment $[CP]$. The bisector of the angle $DAN$ meets line $DP$ in $Q$. Show that the quadrilateral $BMQN$ is a parallelogram.
Adrian Bud
 | [
"From $\\triangle BMC \\sim \\triangle PBC$ follows $\\frac{BM}{PB} = \\frac{BC}{PC}$, hence $CP = 2BP$, so $[BP] = [PN] = [NC]$.\n\n\n\nThen $\\Delta NCD \\equiv \\Delta PBC$ (SAS) leads to $\\widehat{DNC} = \\widehat{BPC}$, that is $DN \\perp CP$, so $[DN]$ is a median and an altitude in ... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0gcs | 設 $ABC$ 為等腰三角形, 其中 $AB = AC$, 並令 $M$ 為 $BC$ 邊的中點。令 $P$ 為平面上一點異於 $A$, 滿足 $PB < PC$, 且 $PA$ 與 $BC$ 平行。設點 $X$ 在直線 $PB$ 上, 點 $Y$ 在直線 $PC$ 上, 使得 $B$ 落在線段 $PX$ 上, $C$ 落在線段 $PY$ 上, 並且 $\angle PXM = \angle PYM$。證明: $A, P, X, Y$ 四點共圓。 | [
"因為 $AB = AC$，知 $AM$ 為 $BC$ 邊的中垂線，故\n$$\n\\angle PAM = \\angle AMC = 90^\\circ\n$$\n\n\n\n現過點 $Y$ 作與 $PC$ 垂直的直線, 設其與直線 $AM$ 交於點 $Z$。 (注意: 點 $M$ 界於 $A, Z$ 兩點之間。) 可知\n$$\n\\angle PAZ = \\angle PYZ = 90^\\circ.\n$$\n故 $P, A, Y, Z$ 四點共圓。\n\n$$\n\\angle CMZ = \\angle CYZ = 90^\\circ,\n$$\n得 $C, ... | Taiwan | 二〇一九數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
00mh | Let $S = \{1, 2, \dots, 2017\}$.
Find the maximal $n$ with the property that there exist $n$ distinct subsets of $S$ such that for no two subsets their union equals $S$. | [
"Answer: $n = 2^{2016}$.\n\nProof:\nThere are $2^{2016}$ subsets of $S$ which do not contain $2017$. The union of any two such subsets does not contain $2017$ and is thus a proper subset of $S$. Thus $n \\ge 2^{2016}$.\n\nTo show the other direction, we group the subsets of $S$ into $2^{2016}$ pairs in such a way t... | Austria | 48th Austrian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2^2016 | |
0d2w | Find all positive integers $n < 589$ for which $589$ divides $n^2 + n + 1$. | [
"Because $589 = 19 \\times 31$, we will find all positive integers $n < 589$ such that both $19$ and $31$ divide $n^2 + n + 1$.\nLet $n$ be such an integer. We have\n$$\n0 \\equiv n^2 + n + 1 \\equiv n^2 + 20n + 1 \\equiv (n + 10)^2 - 2^2 \\equiv (n + 8)(n + 12) \\pmod{19}.\n$$\nHence, $19$ divides $n^2 + n + 1$ if... | Saudi Arabia | Selection tests for the Balkan Mathematical Olympiad 2013 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 87, 273, 315, 501 | |
0l74 | Problem:
Compute the number of ways to arrange 3 copies of each of the 26 lowercase letters of the English alphabet such that for any two distinct letters $x_{1}$ and $x_{2}$, the number of $x_{2}$'s between the first and second occurrences of $x_{1}$ equals the number of $x_{2}$'s between the second and third occurre... | [
"Solution:\n\nFirst, we prove such a string can be divided into blocks where each block consists of the same substring written three times. We prove the following lemma.\n\nLemma 1. For any letter $x_{1}$, the strings between the first and second occurrences of $x_{1}$ and between the second and third occurrences o... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 2^{25} * 26! | |
0fy5 | Problem:
Finde alle ganzen Zahlen $n$, sodass $2^{n}+3^{n}+6^{n}$ das Quadrat einer rationalen Zahl ist. | [
"Solution:\n\nWir setzen $A(n)=2^{n}+3^{n}+6^{n}$. Es gilt $A(-1)=1^{2}, A(0)=3, A(1)=6$ und $A(2)=7^{2}$. Wir betrachten nun zwei Fälle:\n\n- Sei $n \\geq 2$ und $A(-n)$ ein rationales Quadrat. Dann ist auch\n$$\n6^{2 n} A(-n)=6^{n}\\left(1+2^{n}+3^{n}\\right)=6^{n} B(n)\n$$\nein Quadrat und zwar sogar einer natür... | Switzerland | IMO Selektion | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | n = -1 and n = 2 | |
009x | Find all $a \in \mathbb{N}$ such that $n(a+n)$ is not a perfect square for any $n \in \mathbb{N}$. | [
"The answer is $a = 1, 2, 4$. For all $n \\in \\mathbb{N}$ we have\n$$\nn^2 < n(n+1) < n(n+2) < (n+1)^2, \\quad n^2 < n(n+4) < (n+2)^2, \\quad n(n+4) \\neq (n+1)^2.\n$$\nHence $n(a+n)$ is never a perfect square for $a \\in \\{1, 2, 4\\}$.\n\nOn the contrary, for each $a \\ne 1, 2, 4$ there is an $n \\in \\mathbb{N}... | Argentina | Argentine National Olympiad 2015 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | a = 1, 2, 4 | |
0cy7 | In triangle $ABC$ with centroid $G$ let $M \in (AB)$ and $N \in (AC)$ be points on two of its sides. Prove that points $M$, $G$, $N$ are collinear if and only if
$$
\frac{MB}{MA} + \frac{NC}{NA} = 1
$$ | [] | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordi... | English | proof only | null | |
04kp | Determine all triples $(x, y, z)$ of positive real numbers that satisfy
$$
\begin{aligned}
3\lfloor x \rfloor - \{y\} + \{z\} &= 20.3 \\
3\lfloor y \rfloor + 5\lfloor z \rfloor - \{x\} &= 15.1 \\
\{y\} + \{z\} &= 0.9.
\end{aligned}
$$
*For a real number $t$, $\lfloor t \rfloor$ denotes the largest integer not greater t... | [] | Croatia | Mathematical competitions in Croatia | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (7.9, 2.8, 2.1) | |
06ji | Find all natural numbers $n$ such that $n$, $n^2 + 10$, $n^2 - 2$, $n^3 + 6$ and $n^5 + 36$ are all prime numbers. | [
"The only solution is $n = 7$.\nObserve that\n$$\n\\begin{aligned}\nn &\\equiv n \\quad (\\text{mod } 7), \\\\\nn^2 + 10 &\\equiv (n + 2)(n - 2) \\quad (\\text{mod } 7), \\\\\nn^2 - 2 &\\equiv (n + 3)(n - 3) \\quad (\\text{mod } 7), \\\\\nn^3 + 6 &\\equiv (n - 1)(n^2 + n + 1) \\quad (\\text{mod } 7), \\\\\nn^5 + 36... | Hong Kong | Year 2016 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | 7 | |
0jo3 | Problem:
Let $m, n$ be positive integers with $m \geq n$. Let $S$ be the set of pairs $(a, b)$ of relatively prime positive integers such that $a, b \leq m$ and $a+b>m$.
For each pair $(a, b) \in S$, consider the nonnegative integer solution $(u, v)$ to the equation $a u-b v=n$ chosen with $v \geq 0$ minimal, and let $... | [
"Solution:\nThe fact that $I(a, b) \\subseteq (0,1)$ follows from the small-ness of $n \\leq m < a+b$: the smallest solution $(u, v)$ has $0 \\leq v \\leq a-1$, so $u = \\frac{n + b v}{a} < \\frac{(a+b) + b(a-1)}{a} = b+1$ forces $1 \\leq u \\leq b$.\n\nFor the main part of the problem, it suffices (actually, is eq... | United States | HMMT February 2015 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",... | null | proof only | null | |
0c2m | Consider two integers $n$ and $q$, $n \ge 2$, $q \ge 2$, and $q \ne 1 \pmod 4$. Let $K$ be a finite field having exactly $q$ elements. Show that for any $a$ from $K$, there is $x$ and $y$ in $K$, such that $a = x^2 + y^2$. (It is known that any finite field is commutative.) | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Field Theory",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof only | null | |
031m | Problem:
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f\left(x^{2}+y+f(y)\right)=2 y+(f(x))^{2}
$$
for any $x, y \in \mathbb{R}$. | [
"Solution:\nIt follows by\n$$\nf\\left(x^{2}+y+f(y)\\right)=2 y+(f(x))^{2}\n$$\nthat the function $f$ is surjective. Note also that $(f(x))^{2}=(f(-x))^{2}$. In particular, we may choose $a$ such that $f(a)=f(-a)=0$. Setting $x=0, y= \\pm a$ in (1) gives $0=f( \\pm a)=(f(0))^{2} \\pm 2 a$, i.e., $a=0$. Plugging $y=... | Bulgaria | Team selection test for 44. IMO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = x | |
01qh | Prove that for all even positive integers $n$ the following inequality holds
$$
a) \{n\sqrt{6}\} > \frac{1}{n}; \quad b) \{n\sqrt{6}\} > \frac{1}{n - 1/(5n)}.
$$ | [
"Let $m = [n\\sqrt{6}]$. Then $n\\sqrt{6} > m$, or $6n^2 - m^2 > 0$. Note that $6n^2 - m^2 \\ne 1$ or $4 \\pmod{3}$, $6n^2 - m^2 \\ne 2 \\pmod{4}$ (recall that $n$ is even), $6n^2 - m^2 \\neq 3 \\pmod{9}$, so $6n^2 - m^2 \\geq 5$. Now we have\n$$\n\\{n\\sqrt{6}\\} = n\\sqrt{6} - m = \\frac{6n^2 - m^2}{n\\sqrt{6} + ... | Belarus | Selection and Training Session | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Other"
] | English | proof only | null | |
096p | Problem:
Numerele reale $x, y, z$ satisfac condițiile $x-2y+z=2$ și $x+y-2z=5$. Să se afle aceste numere astfel, încât valoarea expresiei $E=xy+yz+xz$ să fie cea mai mică; să se afle această valoare. | [
"Solution:\n\nDin relațiile date, $\\left\\{\\begin{array}{l}x-2y+z=2, \\\\ x+y-2z=5 ;\\end{array}\\right.$ se pot exprima $y$ și $z$ prin $x$.\n\nÎnmulțind prima egalitate cu $2$ și adunând-o cu a doua, se obține $y=x-3$.\n\nApoi, înmulțind egalitatea a doua cu $2$ și adunând-o cu prima, se obține $z=x-4$.\n\nSubs... | Moldova | A 63-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | x = 7/3, y = -2/3, z = -5/3; minimum E = -13/3 | |
0kqh | Problem:
Let $x_{1}, x_{2}, \ldots, x_{2022}$ be nonzero real numbers. Suppose that $x_{k}+\frac{1}{x_{k+1}}<0$ for each $1 \leq k \leq 2022$, where $x_{2023}=x_{1}$. Compute the maximum possible number of integers $1 \leq n \leq 2022$ such that $x_{n}>0$.
Proposed by: Akash Das | [
"Solution:\nLet the answer be $M$. If $M>1011$, there would exist two consecutive positive terms $x_{k}, x_{k+1}$ which contradicts the assumption that $x_{k}+\\frac{1}{x_{k+1}}<0$. Thus, $M \\leq 1011$. If $M=1011$, then the $2022 x_{i}$ s must alternate between positive and negative. WLOG, assume $x_{2 k-1}>0$ an... | United States | HMMT February 2022 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1010 | |
057y | Two regular polygons have a common circumcircle. The sum of the areas of the incircles of these polygons equals the area of their common circumcircle. Find all possibilities of how many vertices can the two polygons have. | [
"The ratio of the inradius and the circumradius of a regular $n$-gon is $\\cos\\frac{180^\\circ}{n}$. Hence the ratio of the areas of the incircle and the circumcircle of a regular $n$-gon is $\\cos^2\\frac{180^\\circ}{n}$.\n\nLet a regular $n$-gon and a regular $m$-gon with a common circumcircle be given. W.l.o.g.... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof and answer | The polygons are either a triangle and a hexagon, or two squares. | |
02ht | Problem:
Se $m$ e $n$ são inteiros maiores do que zero com $m < n$, definimos $m \nabla n$ como a soma dos inteiros entre $m$ e $n$, incluindo $m$ e $n$. Por exemplo, $5 \nabla 8 = 5 + 6 + 7 + 8 = 26$.
Então o valor de $\frac{22 \nabla 26}{4 \nabla 6}$ é:
A) 4
B) 6
C) 8
D) 10
E) 12 | [
"Solution:\n\nDe acordo com a definição de $\\nabla$, temos\n$$\n\\frac{22 \\nabla 26}{4 \\nabla 6} = \\frac{22 + 23 + 24 + 25 + 26}{4 + 5 + 6} = \\frac{120}{15} = 8.\n$$"
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | C | |
082x | Problem:
Quanto misura il raggio della circonferenza piccola nella figura a fianco, se il lato del quadrato è lungo $1$?
(A) $3-2 \sqrt{2}$
(B) $\frac{1}{4}$
(C) $\frac{\sqrt{2}}{4}$
(D) $\frac{\sqrt{2}-1}{2}$
(E) $4-3 \sqrt{2}$.
 | [
"Solution:\n\nLa risposta è (A). Con riferimento alla figura a fianco, detto $r$ il raggio richiesto, si ha $1 + r + r \\sqrt{2} = \\sqrt{2}$, da cui $r = \\frac{\\sqrt{2} - 1}{\\sqrt{2} + 1}$ e infine, razionalizzando, $r = 3 - 2 \\sqrt{2}$.\n\n"
] | Italy | Progetto Olimpiadi di Matematica 2003 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | A | |
0jy6 | Problem:
Find the number of integers $n$ with $1 \leq n \leq 2017$ so that $(n-2)(n-0)(n-1)(n-7)$ is an integer multiple of $1001$. | [
"Solution:\n\nNote that $1001 = 7 \\cdot 11 \\cdot 13$, so the stated product must be a multiple of $7$, as well as a multiple of $11$, as well as a multiple of $13$.\n\nThere are $4$ possible residues of $n$ modulo $11$ for which the product is a multiple of $11$; similarly, there are $4$ possible residues of $n$ ... | United States | HMMT November 2017 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 99 | |
0bil | Let $x_1, x_2, \dots, x_5$ be real numbers. Find the least positive integer $n$ with the following property: if there exist $n$ distinct sums of the form $x_p + x_q + x_r$ (with $1 \le p < q < r \le 5$) which are equal to $0$, then $x_1 = x_2 = \dots = x_5 = 0$.
Bulgaria, 2003 | [
"If the numbers are $1, 1, 1, 1, -2$, (or $1, 1, 1, -2, -2$) there are $6$ sums that are equal to $0$ without the numbers themselves being $0$. Therefore, knowing $6$ sums (or less) to be $0$ is not enough to conclude that the numbers are all $0$.\n\nNow we prove that $7$ is enough. Suppose we are given that $7$ su... | Romania | 65th NMO Selection Tests for JBMO | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 7 | |
0ku1 | Problem:
A real number $x$ is chosen uniformly at random from the interval $(0,10)$. Compute the probability that $\sqrt{x}$, $\sqrt{x+7}$, and $\sqrt{10-x}$ are the side lengths of a non-degenerate triangle. | [
"Solution:\nFor any positive reals $a, b, c$, numbers $a, b, c$ are the side lengths of a triangle if and only if\n$$\n(a+b+c)(-a+b+c)(a-b+c)(a+b-c)>0 \\Longleftrightarrow \\sum_{\\mathrm{cyc}}\\left(2 a^{2} b^{2}-a^{4}\\right)>0\n$$\n(to see why, just note that if $a \\geq b+c$, then only the factor $-a+b+c$ is ne... | United States | HMMT November 2023 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 22/25 | |
03cl | There are $M$ countries and $N$ towns on a planet. Some of the towns are connected by roads. It is known that:
(1) there are at least three towns in any country;
(2) any town in a country is connected by roads with at least half of the towns in this country;
(3) any town is connected with road to exactly one town in an... | [
"Consider a graph $G = (V, E)$ where the vertices are the towns and the edges are the roads. Denote the towns in the countries by $V_1, V_2, \\dots, V_M$. It follows from the condition of the problem that\n$$\nV = V_1 \\cup V_2 \\cup \\dots \\cup V_M, \\quad V_i \\cap V_j = \\emptyset, \\quad |V_i| \\ge 3 \\quad (1... | Bulgaria | Bulgarian National Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory"
] | English | proof only | null | |
0jus | Problem:
Let the sequence $\{a_{i}\}_{i=0}^{\infty}$ be defined by $a_{0}=\frac{1}{2}$ and $a_{n}=1+(a_{n-1}-1)^{2}$. Find the product
$$
\prod_{i=0}^{\infty} a_{i} = a_{0} a_{1} a_{2} \ldots
$$ | [
"Solution:\nLet $\\{b_{i}\\}_{i=0}^{\\infty}$ be defined by $b_{n}=a_{n}-1$ and note that $b_{n}=b_{n-1}^{2}$. The infinite product is then\n$$\n(1+b_{0})(1+b_{0}^{2})(1+b_{0}^{4}) \\ldots (1+b_{0}^{2^{k}}) \\ldots\n$$\nBy the polynomial identity\n$$\n(1+x)(1+x^{2})(1+x^{4}) \\ldots (1+x^{2^{k}}) \\cdots = 1+x+x^{2... | United States | HMMT November 2016 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 2/3 | |
0bwp | a) If $ABCA'B'C'$ is a right prism and the points $M \in (BC)$, $N \in (CA)$, $P \in (AB)$ are such that the straight lines $A'M$, $B'N$ and $C'P$ are pairwise perpendicular and concurrent, then the prism is regular.
b) If $ABCA'B'C'$ is a regular prism and $\frac{AA'}{AB} = \frac{\sqrt{6}}{4}$, then one can find the ... | [
"a) Let $\\{O\\} = A'M \\cap B'N \\cap C'P$. The plane $(B'N, C'P)$ intersects the parallel planes $(ABC)$ and $(A'B'C')$ along parallel straight lines, hence $NP \\parallel B'C'$, so $PN \\parallel BC$. In the same way $MP \\parallel AC$, $NM \\parallel AB$. Then $PNCM$ and $PNMB$ are parallelograms, hence $M$ is ... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0k07 | Problem:
You have $128$ teams in a single elimination tournament. The Engineers and the Crimson are two of these teams. Each of the $128$ teams in the tournament is equally strong, so during each match, each team has an equal probability of winning.
Now, the $128$ teams are randomly put into the bracket.
What is the... | [
"Solution:\n\nThere are $\\binom{128}{2} = 127 \\cdot 64$ pairs of teams. In each tournament, $127$ of these pairs play.\n\nBy symmetry, the answer is $\\frac{127}{127 \\cdot 64} = \\frac{1}{64}$."
] | United States | February 2017 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 1/64 | |
0c4s | Let $n \ge 2$ be an integer and $f : [a, b] \to [a, b]$ be a continuous function, differentiable on $(a, b)$, with continuous positive derivative. It is known that the equation $f'(x) = 1$ has a solution. Prove that there exists $c \in (a, b)$ such that
$$
f^{[n]}(b) - f^{[n]}(a) = (b - a) \cdot (f'(c))^{n+1},
$$
where... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | English | proof only | null | |
0hdg | Define $H_a$, $H_b$, $H_c$ as projections of $A$, $B$, $C$ on $BC$, $AC$, $AB$ respectively. $H$ is the orthocenter of $\triangle ABC$ and $K$ is a reflection of $A$ over $BC$. The line parallel to $H_bH_c$ through $H$ meets $AB$ and $AC$ in points $X$ and $Y$. Prove that inscribed circles of $\triangle ABC$ and $\tria... | [
"Define new points: $M$ and $N$ - points symmetric to $H$ over $AB$ and $AC$ respectively, $S$ and $L$ - midpoints of $XH$ and $HY$ respectively. Let's prove a few facts (Fig 34).\n\n\nFig. 34\n\n1. $H_c$, $S$, $H_a$ are collinear.\nIt can be seen from $\\angle SH_cH = \\angle SHH_c = \\ang... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tang... | null | proof only | null | |
02l1 | Problem:
Numa folha de papel cabem 100 caracteres na largura e 100 na altura. Nessa folha são escritos sucessivamente os números $1, 2, 3, \ldots$ com um espaço entre cada um. Quando no final de uma linha não há espaço para escrever um número este é escrito na linha seguinte. Qual é o ultimo número escrito na folha? | [
"Solution:\n\nNa $1^\\text{a}$ linha escrevemos os números de $1$ a $9$, cada um seguido de um espaço, ocupando $18$ espaços, e sobram $82$ espaços. Cada número de $2$ algarismos mais um espaço ocupa $3$ lugares na linha. Como $82 = 27 \\times 3 + 1$, completamos a $1^\\text{a}$ linha com $27$ números de dois algar... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 2220 | |
0guv | Let $n$ be a positive integer and $k$ be the number of positive divisors of $n$. If any two positive divisors of $n$ have different remainders when divided by $k + 1$, then show that either $k + 1$ is a prime number or equal to $4$. | [
"We consider two cases.\n\nCase 1: $x = 0$ is one of the remainders.\n\nThen $k+1$ divides exactly one of the divisors. Thus $k+1$ divides $n$ and indeed $k+1 = n$. So $k+1$ has $k$ positive divisors. Then $k|k+1$ or $k-1|k+1$. Therefore, $k=1,2$ or $3$, that is $k+1=2,3$ or $4$.\n\nCase 2: $0$ is not one of the re... | Turkey | Team Selection Test for JBMO 2024 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0jfa | Problem:
Triangle $A B C$ is inscribed in a circle centered at $O$, and $M$ is the midpoint of $B C$. Suppose that $A, M$, and $O$ are collinear. Prove that $\triangle A B C$ is either right or isosceles (or both). | [
"Solution:\n\nIf $M$ and $O$ coincide, then $B C$ is a diameter. Then $\\angle A$ is right (this follows from the well-known theorem that an angle inscribed in a semicircle is a right angle).\n\nIf $M$ and $O$ do not coincide, then line $M O$ must be the perpendicular bisector of $B C$ since $M$ and $O$ are both eq... | United States | Berkeley Math Circle | [
"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 only | null | |
0cz2 | Let $ABC$ be a non-isosceles triangle with circumcenter $O$, incenter $I$, and orthocenter $H$. Prove that angle $\widehat{OIH}$ is obtuse. | [
"Recall some preliminary facts. The nine-point circle of a triangle $ABC$ passes through the midpoints of the sides, the midpoints of the segments joining its vertices to the orthocenter $H$ and the pedal point (i.e., the feet of its altitudes to the sides). Its center is the midpoint $O_{9}$ of the segment joining... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"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"
] | English | proof only | null | |
04c7 | Calculate $144^{\log_5 1000} : 10^{6\log_5 12}$. | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | final answer only | 1 | |
0d96 | Does there exist an integer $n \geq 3$ and an arithmetic sequence $a_{0}, a_{1}, \ldots, a_{n}$ such that the polynomial $a_{n} x^{n}+\ldots+a_{1} x+a_{0}$ has $n$ roots which also form an arithmetic sequence? | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | English | proof and answer | Yes, with n = 3 | |
063e | Problem:
Es sei $ABCD$ ein konvexes Sehnenviereck mit $|AD| = |BD|$. Seine Diagonalen $\overline{AC}$ und $\overline{BD}$ mögen sich in $E$ schneiden. Der Mittelpunkt des Inkreises des Dreiecks $BCE$ heiße $I$. Der Umkreis des Dreiecks $BIE$ schneide das Innere der Strecke $\overline{AE}$ im Punkt $N$.
Man beweise, da... | [
"Solution:\n\nSetzt man $\\varphi = \\Varangle DAB$, so muss auch $\\Varangle ABD = \\varphi$ sein, denn das Dreieck $ABD$ wurde als gleichschenklig vorausgesetzt. Hieraus ergibt sich $\\Varangle BDA = 180^{\\circ} - 2\\varphi$ und mit Hilfe des Peripheriewinkelsatzes folgt $\\Varangle BCE = \\Varangle BCA = 180^{\... | Germany | Germany TST | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
01fd | The two women Inger and Ellen play the following game. On a large piece of paper are $2019$ six-pointed snowflakes $*$, placed in general position (and not touching). Each of the six arms of the snowflake has an extreme end, called its *apex*.
A move consists of drawing a curve connecting two apices, possibly belongin... | [
"**Inger has a winning strategy.**\n\nLet Inger select an arbitrary snowflake $S$ and connect a pair of opposite apices. By the Jordan Curve Theorem, this curve (along with $S$) will divide the plane into two domains, the curve (along with $S$) being their common boundary. It is obvious that Inger may draw her curv... | Baltic Way | Baltic Way 2019 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | English | proof and answer | Inger has a winning strategy. | |
0dc6 | Let $p$ be a prime number. Show that $7p + 3^{p} - 4$ is not a perfect square. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residu... | English | proof only | null | |
0azb | Problem:
We say that the constant $a$ is a fixed point of a function $f$ if $f(a)=a$. Find all values of $c$ such that $f(x)=x^{2}-2$ and $g(x)=2x^{2}-c$ share a common fixed point. | [
"Solution:\n\nNote that if $f(x)=x$, then $x^{2}-x-2=0$, or $(x-2)(x+1)=0$. Thus, the fixed points of $f$ are $2$ and $-1$, and so we wish for either of these to be a fixed point of $g$.\n\nIf $2$ is a fixed point of $g$, we must have $g(2)=2=8-c$, and so $c=6$.\n\nIf $-1$ is a fixed point of $g$, then $2-c=-1$, an... | Philippines | 20th Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | c = 3 or 6 | |
01ix | Let $n > 1$ be an integer and let an $2 \times n$ grid be given, where the cells are labelled as $(x, y)$ with $1 \le x \le 2$ and $1 \le y \le n$. A grasshopper lives on the grid and occasionally jumps from one cell to another. The *length* of a jump from a cell $(x_1, y_1)$ to $(x_2, y_2)$ is defined as $|x_1 - x_2| ... | [
"Answer: The solution is $n^2 + 2n - 3$.\n\nCombining the x-bound and the y-bound, we get in both cases the upper bound $n^2 + 2n - 2$. This can still be improved by observing that it is not possible to satisfy both bounds with equality. For $n = 2k + 1$, in order to get $n^2 - 1$ units in y-direction, the two end ... | Baltic Way | Baltic Way 2023 Shortlist | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | n^2 + 2n - 3 | |
02su | Problem:
Escreve-se, em ordem crescente, cada um dos múltiplos de $3$ cuja soma com $1$ é um quadrado perfeito:
$$
3, 15, 24, 48, \ldots
$$
a) Qual é o próximo número que aparecerá, nesta sequência, depois do $48$?
b) Qual é o oitavo número desta sequência?
c) Qual é o número que aparecerá, nesta sequência, na $2013^{... | [
"Solution:\n\na) Note que $48$ é um termo da sequência que é um múltiplo de $3$ que satisfaz que $48+1=49=7^{2}$. Assim, para encontrar o seu sucessor, tentaremos obter um múltiplo de três que, acrescido de uma unidade, seja igual a $8^{2}=64$. Subtraindo uma unidade de $64$, obtemos $63$, que é realmente um múltip... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | a) 63; b) 168; c) 9120399 | |
0je4 | Problem:
The rightmost nonzero digit in the decimal expansion of $101!$ is the same as the rightmost nonzero digit of $n!$, where $n$ is an integer greater than $101$. Find the smallest possible value of $n$. | [
"Solution:\n$101!$ has more factors of $2$ than $5$, so its rightmost nonzero digit is one of $2, 4, 6, 8$. Notice that if the rightmost nonzero digit of $101!$ is $2k$ ($1 \\leq k \\leq 4$), then $102!$ has rightmost nonzero digit $102(2k) \\equiv 4k \\pmod{10}$, and $103!$ has rightmost nonzero digit $103(4k) \\e... | United States | HMMT November 2013 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | null | proof and answer | 103 | |
0fj4 | Problem:
Halla el número natural $n$ que es el producto de los primos $p, q$ y $r$, sabiendo que
$$
r-q=2p \quad \text{y} \quad rq+p^{2}=676
$$ | [
"Solution:\nTomamos $x = r - p = q + p$. Entonces,\n$$\nx^{2} = (r - p)(q + p) = rq + (r - q)p - p^{2} = rq + 2p^{2} - p^{2} = rq + p^{2} = 676\n$$\nluego $x = 26$. Y $p$ es un primo tal que $26 - p$ y $26 + p$ son primos. Probamos con los posibles primos $p$ menores que $26$ y se ve que eso sólo se cumple para $p ... | Spain | Viernes 19 de enero de 2001 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 2001 | |
0eb1 | Find all positive integers $n$ for which there exists a rectangle, such that the lengths of its sides are positive integers while its perimeter equals $n$ and is equal to the area of the rectangle. | [
"Let $a$ and $b$ be the lengths of the sides of a rectangle that satisfies the conditions. Then $2(a+b) = n = ab$. This equality can be rewritten as $a(b-2) = 2b$. Now we can express $a$ in terms of $b$:\n$$\na = \\frac{2b}{b-2} = \\frac{2(b-2)}{b-2} + \\frac{4}{b-2} = 2 + \\frac{4}{b-2}.\n$$\n\nSince $a$ and $b$ a... | Slovenia | National Math Olympiad in Slovenia | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 16, 18 | |
05hl | Problem:
Trouver tous les polynômes à coefficients réels $P$ tels que le polynôme
$$
(X+1) P(X-1)-(X-1) P(X)
$$
soit constant. | [
"Solution:\nOn présente deux solutions : une qui utilise une factorisation polynomiale, et une autre plus calculatoire.\n\nCommençons par la solution par factorisation. Supposons que $(X+1) P(X-1)-(X-1) P(X)$ est constant égal à $2k$. En prenant $X=-1$, on obtient $2P(-1)=2k$. En prenant $X=1$, on obtient $2P(0)=2k... | France | Préparation Olympique Française de Mathématiques | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All solutions are P(X) = a X(X+1) + c with real a, c; for these, (X+1)P(X-1) − (X-1)P(X) equals the constant 2c. | |
0i8s | Problem:
How many positive rational numbers less than $\pi$ have denominator at most $7$ when written in lowest terms? (Integers have denominator $1$.) | [
"Solution:\n\nWe can simply list them. The table shows that there are $3+3+6+6+12+6+18=54$.\n\n| Denominator | Values |\n| ---: | :--- |\n| 1 | $\\frac{1}{1}, \\frac{2}{1}, \\frac{3}{1}$ |\n| 2 | $\\frac{1}{2}, \\frac{3}{2}, \\frac{5}{2}$ |\n| 3 | $\\frac{1}{3}, \\frac{2}{3}, \\frac{4}{3}, \\frac{5}{3}, \\frac{7}{3... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | final answer only | 54 | |
0l01 | Cyclic quadrilateral $ABCD$ has lengths $BC = CD = 3$ and $DA = 5$ with $\angle CDA = 120^\circ$. What is the length of the shorter diagonal of $ABCD$?
(A) $\frac{31}{7}$ (B) $\frac{33}{7}$ (C) 5 (D) $\frac{39}{7}$ (E) $\frac{41}{7}$ | [
"Place the figure in the coordinate plane with $D = (0, 0)$ and $A = (5, 0)$. Because $\\angle CDA = 120^\\circ$ and $CD = 3$, it follows that $C = \\left(-\\frac{3}{2}, \\frac{3}{2}\\sqrt{3}\\right)$. The perpendicular bisectors of $\\overline{CD}$ and $\\overline{AD}$ intersect at the center of the circumscribing... | United States | AMC 12 A | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | MCQ | D | |
0cy0 | Let $a$, $b$, $c$ be positive real numbers such that $a b c = 8$. Prove that
$$
\frac{a-2}{a+1} + \frac{b-2}{b+1} + \frac{c-2}{c+1} \leq 0.
$$ | [] | Saudi Arabia | SAMC | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
01tj | An infinite sequence $(a_n)$, $n \in \mathbb{N}$, of positive numbers is called *lacunar* if there exists a number $q > 1$ such that $a_{n+1}/a_n \ge q$ for all $n \in \mathbb{N}$. Also, the sequence is called *solitary* if there exists a number $r > 1$ such that the interval $(x, rx)$ contains at most one term of this... | [
"a) Let the sequence $(a_n)$, $n \\in \\mathbb{N}$, be lacunar. Then there exists a number $q > 1$ such that\n$$\na_{n+1} \\ge q a_n \\quad \\forall n \\in \\mathbb{N}. \\quad (1)\n$$\nIn particular, any lacunar sequence is increasing. From (1) it follows that any interval $(x, qx)$ contains at most one term of thi... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series"
] | English | proof and answer | a) Yes; b) No | |
01gk | In a school with $n$ students, a team of students must be chosen to clean the school. A team of students is called *good* if the school principal believes that they can manage to clean the school. It is known that at least one team is good, and that if a team is good, then it is also good after adding any other student... | [
"Let $S$ be the set students and let $G$ be the set of good teams. For any student $s$ consider the trivial injection from the set of good teams without $s$ to the set of good teams containing $s$ by taking a team $T$ and mapping it to $T \\cup \\{s\\}$. Since this is an injection we conclude that there are at leas... | Baltic Way | Baltic Way 2020 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof only | null | |
0e3o | Problem:
V dveh sadovnjakih so prvo leto nabrali skupaj 315 ton sadja. Naslednje leto se je skupni pridelek povečal za $40\ \%$. V prvem sadovnjaku se je pridelek povečal za $25\ \%$, v drugem pa za $50\ \%$. Koliko ton sadja so v vsakem sadovnjaku nabrali prvo leto? | [
"Solution:\n\nZapišemo zvezo $315 + 40\\ \\%$ od $315$ in izračunamo, da je to $441$.\n\nUpoštevamo pridelek v 1. sadovnjaku $x + \\frac{25}{100} x = x + \\frac{x}{4} = \\frac{5x}{4}$,\nter pridelek v drugem sadovnjaku $(315 - x) + \\frac{50}{100}(315 - x) = (315 - x) + 0,5(315 - x) = 1,5(315 - x)$.\n\nSeštejemo pr... | Slovenia | 10. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | First orchard: 126 tons; Second orchard: 189 tons | |
0j1q | Problem:
Consider an infinite grid of unit squares. An $n$-omino is a subset of $n$ squares that is connected. Below are depicted examples of 8-ominoes. Two $n$-ominoes are considered equivalent if one can be obtained from the other by translations and rotations. What is the number of distinct 15-ominoes? Your score w... | [
"Solution:\n\nAnswer: 3426576\n\nWe claim that there are approximately $\\frac{3^{n-1}}{4}$ $n$-ominoes. First, we define an order on the squares in an $n$-omino, as follows: we order the squares from left to right, and within a column, we order the squares from top to bottom.\n\nWe construct an $n$-omino by starti... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 3426576 | |
0hvv | Problem:
Let $ABC$ be an acute triangle with circumcenter $O$, incenter $I$, orthocenter $H$. If $OI = HI$, what are the possible values of the angles of triangle $ABC$? | [
"Solution:\n\nAnswer: this occurs if and only if some angle is $60$ degrees.\n\nOne direction is immediate; if $\\angle A = 60^{\\circ}$ then $BHOIC$ are cyclic since $\\angle BHC = \\angle BIC = \\angle BOC = 120^{\\circ}$.\n\nFor the other direction, note that we have an \"SSA congruence\" of triangles $AIH$ and ... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"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",
"Geometry > Plane Geometry > Miscellaneous > C... | null | proof and answer | Some angle is 60 degrees. | |
0jou | Problem:
Find the shortest distance between the lines $\frac{x+2}{2}=\frac{y-1}{3}=\frac{z}{1}$ and $\frac{x-3}{-1}=\frac{y}{1}=\frac{z+1}{2}$ | [
"Solution:\nFirst we find the direction of a line perpendicular to both of these lines. By taking the cross product $(2,3,1) \\times (-1,1,2) = (5,-5,5)$ we find that the plane $x-y+z+3=0$ contains the first line and is parallel to the second. Now we take a point on the second line, say the point $(3,0,-1)$ and fin... | United States | HMMT November 2015 | [
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | 5√3/3 | |
0bab | Show that there is an infinite number of positive integers $t$ such that none of the equations $x^2 + y^6 = t$, $x^2 + y^6 = t + 1$, $x^2 - y^6 = t$, $x^2 - y^6 = t + 1$ has solutions $(x, y) \in \mathbb{Z} \times \mathbb{Z}$. | [
"If $x$ is a positive integer, then either $x^{12} \\equiv 0 \\pmod{13}$ or $x^{12} \\equiv 1 \\pmod{13}$, hence $x^6$ is congruent with $-1$, $0$ or $1$ modulo $13$. Therefore, if $t$ is congruent with $6 \\pmod{13}$, then $\\pm x^6 + t$ is congruent with $5$, $6$ or $7 \\pmod{13}$, while $\\pm x^6 + t + 1$ is con... | Romania | 62nd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequ... | null | proof only | null | |
096m | Problem:
Determinați valoarea minimă a modulului numărului complex $z$, care verifică relația
$$
|z+12|+|z-5 i|=13
$$ | [
"Solution:\n\nDacă $P(a_{1}, b_{1})$ este imaginea numărului complex $z_{1}=a_{1}+b_{1} i$, iar $Q(a_{2}, b_{2})$ - imaginea numărului complex $z_{2}=a_{2}+b_{2} i$, atunci $|z_{1}-z_{2}|=PQ$.\nFie $M$ imaginea lui $z_{1}=-12$, $N$ imaginea lui $z_{2}=5 i$ și $A$ imaginea lui $z$. Atunci $|z+12|=AM$, $|z-5 i|=AN$, ... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 60/13 |
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