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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
06st | For a sequence $x_{1}, x_{2}, \ldots, x_{n}$ of real numbers, we define its price as
$$
\max_{1 \leqslant i \leqslant n}\left|x_{1}+\cdots+x_{i}\right| .
$$
Given $n$ real numbers, Dave and George want to arrange them into a sequence with a low price. Diligent Dave checks all possible ways and finds the minimum possibl... | [
"If the initial numbers are $1,-1,2$, and $-2$, then Dave may arrange them as $1,-2,2,-1$, while George may get the sequence $1,-1,2,-2$, resulting in $D=1$ and $G=2$. So we obtain $c \\geqslant 2$.\n\nTherefore, it remains to prove that $G \\leqslant 2 D$. Let $x_{1}, x_{2}, \\ldots, x_{n}$ be the numbers Dave and... | IMO | 55th International Mathematical Olympiad Shortlist | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 2 | |
0l2u | Problem:
Alf, the alien from the 1980s TV show, has a big appetite for the mineral apatite. However, he's currently on a diet, so for each integer $k \geq 1$, he can eat exactly $k$ pieces of apatite on day $k$. Additionally, if he eats apatite on day $k$, he cannot eat on any of days $k+1, k+2, \ldots, 2k-1$. Compute... | [
"Solution:\n\nIf Alf doesn't eat on day $100$, he could have changed his diet so that he eats on all the same days except the last day is changed to $100$. This attains strictly more apatite, and therefore an optimal diet must have Alf eating on day $100$.\n\nKnowing this, Alf must not have eaten anything on days $... | United States | HMMT November 2024 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 197 | |
05cp | Find all triples $(x, y, z)$ of positive integers, such that
$$3 \cdot x! + 4 \cdot y! = 5 \cdot z!.$$ | [
"**Solution 1:** If $x \\le y$, then $4 \\cdot y! < 3 \\cdot x! + 4 \\cdot y! \\le 7 \\cdot y!$, so $\\frac{4}{5} \\cdot y! < z! < \\frac{7}{5} \\cdot y!$. Two distinct factorials differ by a factor of at least 2, so $z < y$ would yield $z! \\le \\frac{1}{2} \\cdot y! < \\frac{4}{5} \\cdot y! < z!$, contradiction. ... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (2, 1, 2) and (2, 3, 3) | |
02l7 | Problem:
O Riquinho distribuiu $R\$ 1000,00$ reais entre os seus amigos: Antônio, Bernardo e Carlos da seguinte maneira: deu, sucessivamente, 1 real ao Antônio, 2 reais ao Bernardo, 3 reais ao Carlos, 4 reais ao Antônio, 5 reais ao Bernardo, etc. Quanto que o Bernardo recebeu? | [
"Solution:\n\nO dinheiro foi repartido em parcelas na forma\n$$\n1+2+3+\\cdots+n \\leq 1000\n$$\nComo $1+2+3+\\cdots+n$ é a soma $S_{n}$ dos $n$ primeiros números naturais a partir de $a_{1}=1$ temos:\n$$\nS_{n}=\\frac{\\left(a_{1}+a_{n}\\right) n}{2}=\\frac{(1+n) n}{2} \\leq 1000 \\Longrightarrow n^{2}+n-2000 \\le... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 345 | |
0lcr | Let $a_1, a_2, \dots, a_{14}$ be some positive real numbers. Prove that
$$
\frac{a_1}{a_2+a_3} + \frac{a_2}{a_3+a_4} + \dots + \frac{a_{14}}{a_1+a_2} \ge \frac{a_1}{a_{14}+a_1} + \frac{a_2}{a_1+a_2} + \dots + \frac{a_{14}}{a_{13}+a_{14}}.
$$
When does the equality occur? | [] | Vietnam | MOCK TEST FOR VMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof and answer | Equality occurs if and only if all the numbers are equal. | |
04k1 | Let $n \ge 2$ be a positive integer and $p$ a prime number. If the number $p-1$ is divisible by $n$, and the number $n^3-1$ is divisible by $p$, prove that $4p-3$ is a square of an integer. | [
"Since $n$ divides $p-1$, there exists a positive integer $a$ such that $p-1 = an$. We also have $p-1 \\ge n$.\n\nFrom the condition that $n^3-1 = (n-1)(n^2+n+1)$ is divisible by the prime number $p$, it follows that $n^2+n+1$ is divisible by $p$. Indeed, $1 \\le n-1 < n+1 \\le p$, so $n-1$ cannot be divisible by $... | Croatia | Croatian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
055k | Let $n$ be a positive integer. In how many ways can an $n \times n$ table be filled with integers from $0$ to $5$ such that
a) the sum of each row is divisible by $2$ and the sum of each column is divisible by $3$;
b) the sum of each row is divisible by $2$, the sum of each column is divisible by $3$ and the sum of e... | [
"a) Let's fill the top left $(n-1) \\times (n-1)$ subtable arbitrarily; this can be done in $6^{(n-1)^2}$ ways. Now there are $3$ ways to fill each of the top $n-1$ cells of the rightmost column and $2$ ways to fill each of the left $n-1$ cells of the bottom row to satisfy the requirements. The value for the last e... | Estonia | IMO Team Selection Contest II | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | English | proof and answer | a) 6^{n^2 - n}; b) if n = 1: 1; if n = 2: 6; if n ≥ 3: 6^{n^2 - n - 2} | |
0ekc | Problem:
Vsota dveh naravnih števil je enaka trikratniku njune razlike, njun zmnožek pa je enak štirikratniku njune vsote. Koliko je vsota teh dveh naravnih števil?
(A) 9
(B) 10
(C) 12
(D) 15
(E) 18 | [
"Solution:\n\nOznačimo ti dve naravni števili z $m$ in $n$. Tedaj je $m+n=3(m-n)$ in $m n=4(m+n)$. Iz prve enakosti dobimo $4 n=2 m$ oziroma $m=2 n$. Ko slednje vstavimo v drugo enakost, dobimo $2 n^{2}=12 n$, od koder sledi $n=6$. Torej je $m=12$ in $m+n=18$."
] | Slovenia | 66. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | MCQ | E | |
00bz | A sequence of natural numbers is *admissible* if its terms are less or equal to $100$ and its sum is greater than $1810$. Find the least $d$ such that each admissible sequence has a subsequence sum in the interval $[1810-d, 1810+d]$. | [
"Consider the sequence $\\alpha$ with $17$ terms equal to $98$ and $2$ terms equal to $96$. Its sum is $17 \\cdot 98 + 2 \\cdot 96 = 1858 > 1810$, so $\\alpha$ is admissible. Note that $\\alpha$ has exactly two subsequence sums in the interval $[1810-48, 1810+48] = [1762, 1858]$. They are its extremes: $1858$ the s... | Argentina | Argentina_2018 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 48 | |
0jax | Problem:
Let $S$ be a set of size $3$. How many collections $T$ of subsets of $S$ have the property that for any two subsets $U \in T$ and $V \in T$, both $U \cap V$ and $U \cup V$ are in $T$? | [
"Solution:\nAnswer: $74$\n\nLet us consider the collections $T$ grouped based on the size of the set $X = \\bigcup_{U \\in T} U$, which we can see also must be in $T$ as long as $T$ contains at least one set. This leads us to count the number of collections on a set of size at most $3$ satisfying the desired proper... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 74 | |
0dje | Let $ABC$ be a triangle with $\angle B = 2\angle C$ and angle bisector $BD$. The symmedian of vertex $B$ in triangles $DAB$, $BCD$ cuts the corresponding circumcircle at $M$, $N$. Denote $P$ as the reflection of $B$ over $C$. Prove that the circle $(MNP)$ is tangent to $BC$ and both of circles $(DAB), (BCD)$. | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0hu8 | Problem:
Pentagon $A B C D E$ is cyclic, i.e., inscribed in a circle. Diagonals $A C$ and $B D$ meet at $P$, and diagonals $A D$ and $C E$ meet at $Q$. Triangles $A B P$, $A E Q$, $C D P$, $C D Q$, and $A P Q$ have equal areas. Prove that the pentagon is regular. | [
"Solution:\n\nAdding the area of triangle $B C P$ to those of $\\triangle A B P$, $\\triangle C D P$, we see that $\\triangle A B C$, $\\triangle D B C$ have equal areas, so $A D$, $B C$ are parallel. Then $A B C D$ is a cyclic trapezoid, so it is isosceles and $A B = C D$. Similarly, $A C D E$ is a trapezoid and $... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
08wc | Let $a$ and $b$ be positive integers. Suppose an $a \times b$ square grid is given and $N$ of the $ab$ square boxes of the grid are marked by $\checkmark$. It was possible to mark all of the $ab$ boxes by repeating the following procedure:
Procedure: If you find a row or a column of the boxes for which all but one of ... | [
"In the sequel, we assume that all of $ab$ boxes of the original grid can be marked by repeating the given procedure a certain number of times after we reach the situation where $N$ of the boxes are marked, and we show that $N \\geq (a - 1)(b - 1)$.\n\nSuppose the last marked box to attain the goal of marking all o... | Japan | Japan Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Algorithms"
] | English | proof and answer | (a - 1)(b - 1) | |
0akl | Find all primes $p$ and $q$ such that $3p^{q-1}$ divides $11^p+17^p$. | [
"For $p=2$ it is directly checked that there are no solutions. Assume that $p>2$. Observe that $N=11^p+17^p \\equiv 4 \\pmod 8$, so $8 \\nmid 3p^{q-1}+1 > 4$. Consider an odd prime divisor $r$ of $3p^{q-1}+1$. Obviously, $r \\notin \\{3,11,17\\}$. There exist $b$ such that $17b \\equiv 1 \\pmod r$. Then $r|b^pN \\e... | North Macedonia | Balkan Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (3,3) | |
0g8n | 設 $\triangle ABC$ 的內切圓圓心為 $I$, 且該內切圓分別與 $CA, AB$ 邊切於點 $E, F$。令點 $E, F$ 對 $I$ 的對稱點分別為 $G, H$。設點 $Q$ 為 $GH$ 與 $BC$ 的交點, 並設點 $M$ 為 $BC$ 的中點。證明 $IQ$ 與 $IM$ 垂直。
Let $I$ be the incenter of the triangle $ABC$, and let the incircle touch the sides $CA, AB$ at the points $E, F$, respectively. Let the reflection points of $E, F... | [
"如圖，內切圓與 $BC$ 邊切於點 $D$, $GH$ 分別與 $BI, CI$ 交於點 $C'$, $B'$，$D'$ 為 $GH$ 上一點使得 $ID' \\perp GH$。\n\n1. \n\na. 因 $\\angle C'ID' = \\frac{1}{2}(\\angle A + \\angle B)$, $\\angle IC'B' = \\frac{1}{2}\\angle C$。同理, $\\angle IB'C' = \\frac{1}{2}\\angle B$。由此知 $\\triangle IBC \\sim \\triangle IB'C'$。\n\nb. 因 $G, H$ 分別為 $E, F$... | Taiwan | 二〇一四數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilateral... | null | proof only | null | |
0081 | Let $r_2, r_3, \dots, r_{1000}$ be the remainders of an odd positive integer upon division by $2, 3, \dots, 1000$. It is known that they are pairwise distinct and one of them is $0$. Find all values of $k$ for which it is possible that $r_k = 0$. | [
"Let $N$ be the odd integer; then the first remainder $r_2$ equals $1 = 2 - 1$. Next, $r_j = j - 1$ cannot hold for all $j$ or else no $r_j$ is $0$. Let $k > 2$ be the first number such that $r_k \\ne k - 1$. Then $r_j = j - 1$ for $j = 2, \\dots, k - 1$, so $r_k \\ne 1, 2, \\dots, k - 2$ because the $r_j$ are pair... | Argentina | Mathematical Olympiad Rioplatense | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | all primes p with 500 < p < 1000 | |
0h4w | Andriy and Olesia play such game. Firstly, Andriy chooses a chessman and place it on the chessboard. Then they moves in turn by the rules of the chosen chessman. However, it is not allowed to put the chessman on the field that Andriy began from or was already been used. Looser is the one who can not move. Who wins if b... | [
"Olesia always wins.\n\nFor every chessman the chessboard is divided into couples of squares that are connected by the move of chosen chessman. Then the win strategy of Olesia is as follows: Andriy moves the chessman to the square of some couple (it also concerns to the first Andriy's choice of placing the chessman... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round (Second Tour) | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Olesia wins in both cases (knight and bishop). | |
0jri | Problem:
Rosencrantz plays $n \leq 2015$ games of question, and ends up with a win rate (i.e. $\frac{\# \text{ of games won }}{\# \text{ of games played }}$) of $k$. Guildenstern has also played several games, and has a win rate less than $k$. He realizes that if, after playing some more games, his win rate becomes hi... | [
"Solution:\n\nAnswer: $\\frac{1}{2015}$\n\nWrite $k=\\frac{m}{n}$, for relatively prime integers $m, n$. For the property not to hold, there must exist integers $a$ and $b$ for which\n$$\n\\frac{a}{b}<\\frac{m}{n}<\\frac{a+1}{b+1}\n$$\n(i.e. at some point, Guildenstern must \"jump over\" $k$ with a single win)\n$$\... | United States | HMMT November 2015 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequenc... | null | proof and answer | 1/2015 | |
074r | Let $ABCD$ be a cyclic quadrilateral and let $E$ be the point of intersection of its diagonals $AC$ and $BD$. Suppose $AD$ and $BC$ meet in $F$. Let the midpoints of $AB$ and $CD$ be $G$ and $H$ respectively. If $\Gamma$ is the circumcircle of triangle $EGH$, prove that $FE$ is tangent to $\Gamma$. | [
"Let $AB$ and $DC$ meet in $M$; let $FE$ meet $HG$ in $T$, $AB$ in $P$ and $DC$ in $Q$ respectively. We use the following fact: in a triangle $ABC$ with circumcircle $\\gamma$, given a point $D$ outside $\\gamma$, $AD$ is tangent to $\\gamma$ if and only if\n\n\n$$\n\\frac{DB}{DC} = \\frac{... | India | Indija TS 2010 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Ad... | null | proof only | null | |
028c | Problem:
Mistério das bolas - Henrique têm duas urnas. A primeira urna contém somente bolas pretas e a segunda somente bolas brancas. Henrique retirou um número de bolas da primeira urna e as colocou na segunda. Em seguida, retirou o mesmo número de bolas da segunda urna e as colocou na primeira. Depois disso o número... | [
"Solution:\n\nSeja $m$ o número de bolas pretas na primeira urna e $n$ o de bolas brancas na segunda urna. Inicialmente, Henrique retirou $k$ bolas pretas da primeira urna e as colocou na segunda urna. Nesse ponto a situação é a seguinte:\n- na 1a urna temos:\n$$\n\\underbrace{m-k}_{\\text{pretas}}\n$$\n- na $2^{\\... | Brazil | Nível 2 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | equal | |
0kf3 | Problem:
While waiting for their food at a restaurant in Harvard Square, Ana and Banana draw 3 squares $\square_{1}, \square_{2}, \square_{3}$ on one of their napkins. Starting with Ana, they take turns filling in the squares with integers from the set $\{1,2,3,4,5\}$ such that no integer is used more than once. Ana's... | [
"Solution:\n\nRelabel $a_{1}, a_{2}, a_{3}$ as $a, b, c$. This is minimized at $x=\\frac{-b}{2a}$, so $M = c - \\frac{b^{2}}{4a}$.\n\nIf in the end $a=5$ or $b \\in \\{1,2\\}$, then $\\frac{b^{2}}{4a} \\leq 1$ and $M \\geq 0$. The only way for Ana to block this is to set $b=5$, which will be optimal if we show that... | United States | HMMO 2020 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 451 | |
0198 | Decide, whether there exists a set $M$ consisting of five integers such that for any integer $k$ not divisible by $5$ there exist $a, b \in M$ such that $a - b + k$ is divisible by $25$. | [
"**Answer.** There does not exist such a set.\n\n**Proof.** Assume that $M = \\{a, b, c, d, e\\}$ were such a set. As there are $20$ differences of distinct members from $M$ and $20$ residue classes modulo $25$ whose members are not divisible by $5$, the two lines\n$$\n1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | No; such a set does not exist. | |
06rz | Let $AB C$ be a triangle with $AB \neq AC$ and circumcenter $O$. The bisector of $\angle BAC$ intersects $BC$ at $D$. Let $E$ be the reflection of $D$ with respect to the midpoint of $BC$. The lines through $D$ and $E$ perpendicular to $BC$ intersect the lines $AO$ and $AD$ at $X$ and $Y$ respectively. Prove that the q... | [
"The bisector of $\\angle BAC$ and the perpendicular bisector of $BC$ meet at $P$, the midpoint of the minor arc $\\widehat{BC}$ (they are different lines as $AB \\neq AC$). In particular $OP$ is perpendicular to $BC$ and intersects it at $M$, the midpoint of $BC$.\n\nDenote by $Y'$ the reflection of $Y$ with respe... | IMO | 53rd International Mathematical Olympiad Shortlisted Problems with Solutions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneo... | null | proof only | null | |
0k44 | Problem:
Find the sum of all positive integers $n$ such that $1+2+\cdots+n$ divides
$$
15\left[(n+1)^2+(n+2)^2+\cdots+(2 n)^2\right] .
$$ | [
"Solution:\nAnswer: 64\nWe can compute that $1+2+\\cdots+n=\\frac{n(n+1)}{2}$ and $(n+1)^2+(n+2)^2+\\cdots+(2 n)^2=\\frac{2 n(2 n+1)(4 n+1)}{6}-\\frac{n(n+1)(2 n+1)}{6}=\\frac{n(2 n+1)(7 n+1)}{6}$, so we need $\\frac{15(2 n+1)(7 n+1)}{3(n+1)}=\\frac{5(2 n+1)(7 n+1)}{n+1}$ to be an integer. The remainder when $(2 n+... | United States | HMMT November 2018 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 64 | |
0er6 | 14. I accidentally decreased a number by $60\%$ instead of increasing it by $60\%$. This incorrect value now needs to be increased by $k\%$ to get to the correct value. What is the value of $k$?
15. Exactly two years ago the Benson family had 4 members, and their average age was $19$. The Bensons then adopted another ... | [
"14. $300$\nI ended up with $40\\%$ of the number instead of $160\\%$. So I need to multiply this new result by $4$, or add it three times to itself, which is an increase of $300\\%$.\n\n15. $11$\n2 years ago the sum of all the family's ages was $4 \\times 19 = 76$. That should have increased by $2 \\times 4 = 8$, ... | South Africa | South African Mathematics Olympiad Second Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | final answer only | k = 300; adopted child's present age = 11 | |
0ix1 | Problem:
A kite is a quadrilateral whose diagonals are perpendicular. Let kite $ABCD$ be such that $\angle B = \angle D = 90^{\circ}$. Let $M$ and $N$ be the points of tangency of the incircle of $ABCD$ to $AB$ and $BC$ respectively. Let $\omega$ be the circle centered at $C$ and tangent to $AB$ and $AD$. Construct ano... | [
"Solution:\nLet's focus on the right triangle $ABC$ and the semicircle inscribed in it since the situation is symmetric about $AC$. First we find the radius $a$ of circle $O$. Let $AB = x$ and $BC = y$. Drawing the radii $OM$ and $ON$, we see that $AM = x - a$ and $\\triangle AMO \\sim \\triangle ABC$. In other wor... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | (1 + sqrt(5))/2 | |
02kc | Problem:
Qual é o menor número inteiro positivo $N$ tal que $\frac{N}{3}$, $\frac{N}{4}$, $\frac{N}{5}$, $\frac{N}{6}$ e $\frac{N}{7}$ são números inteiros?
A) 420
B) 350
C) 210
D) 300
E) 280 | [
"Solution:\n\nPara que $\\frac{N}{3}$, $\\frac{N}{4}$, $\\frac{N}{5}$, $\\frac{N}{6}$ e $\\frac{N}{7}$ sejam números inteiros, $N$ deve ser múltiplo comum de $3, 4, 5, 6$ e $7$. Como queremos o menor $N$ possível, ele deve ser o menor múltiplo comum de $3, 4, 5, 6$ e $7$. Sendo o MMC entre $3, 4, 5, 6$ e $7$ igual ... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | MCQ | A | |
07ag | Consider the second degree polynomial $x^2 + a x + b$ with real coefficients. We know that the necessary and sufficient condition for this polynomial to have roots in real numbers is that its discriminant, $a^2 - 4b$, be greater than or equal to zero. Note that the discriminant is also a polynomial with variables $a$ a... | [
"If we put $a = c = 0$, polynomial $x^4 + b x^2 + d$ can be written as product of four linear terms if and only if quadratic polynomial $y^2 + b y + d$ has two nonnegative roots. Therefore $P(0, b, 0, d) \\ge 0$ if and only if $b \\le 0$, $d \\ge 0$ and $b^2 - 4d \\ge 0$. For a fixed $b \\le 0$ let $Q_b(d) = P(0, b... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof only | null | |
02t4 | Problem:
Uma técnica muito usada para calcular somatórios é a Soma Telescópica. Ela consiste em "decompor" as parcelas de uma soma em partes que se cancelem. Por exemplo,
$$
\begin{aligned}
& \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}= \\
& \left(\frac{1}{1}-\frac{1}{2}\right)+\lef... | [
"Solution:\n\na) Veja que o primeiro número ímpar é $2 \\cdot 1-1$ e, sabendo que os números ímpares crescem de 2 em 2, podemos concluir que o número ímpar que estará na posição $m$ em nossa contagem é\n$$\n\\begin{aligned}\n2 \\cdot 1-1+\\underbrace{2+2+\\ldots+2}_{m-1 \\text{ vezes }} & =2 \\cdot 1-1+2(m-1) \\\\\... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 764049 | |
0614 | Problem:
Gegeben sei ein hinreichend großer Vorrat von gleichseitigen Dreiecken und Quadraten, alle mit der gleichen Seitenlänge. Aus diesen Bausteinen lassen sich konvexe* Polygone bilden, indem man sie in der Ebene lückenlos und überschneidungsfrei aneinander legt. (Die Figur zeigt drei Möglichkeiten für ein Sechsec... | [] | Germany | Auswahlwettbewerb zur IMO 2000 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 12 | |
0307 | Problem:
No desenho ao lado, o triângulo $A B C$ é equilátero e $B D = C E = A F = \frac{A B}{3}$. A razão $\frac{E G}{G D}$ pode ser escrita na forma $\frac{m}{n}$, $\operatorname{mdc}(m, n) = 1$. Quanto vale $m+n$ ?
 | [
"Solution:\n\nVeja que $\\frac{E G}{G D} = \\frac{[E B G]}{[B G D]}$. Agora, $\\frac{[E B G]}{[B C F]} = \\frac{\\frac{1}{2} E B \\cdot B G \\cdot \\operatorname{sen}(\\angle E B G)}{\\frac{1}{2} B C \\cdot B F \\cdot \\operatorname{sen}(\\angle C B F)}$. Como $\\angle E B G = \\angle C B F$, temos\n$$\n\\frac{[E B... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 5 | |
0cwb | There are 16 coins — eight heavy ones of weight $11$ g each, and eight light ones of weight $10$ g each, but it is unknown which coin is of which type. One of the coins is commemorative. Determine whether the commemorative coin is light or heavy, by performing three weighings on a two-pan scales. (K. Knop) | [
"Let us denote the commemorative coin by $Y$. Set aside two non-commemorative coins $A$ and $B$, and distribute the remaining $14$ coins into two groups of $7$ each so that $Y$ is placed on the left pan. We will call a coin \"left\" or \"right\" if it was placed on the left or right pan, respectively, in this weigh... | Russia | Final round | [
"Discrete Mathematics > Logic"
] | English; Russian | proof only | null | |
0d34 | Let $F_{0}=0$, $F_{1}=1$ and $F_{n+1}=F_{n}+F_{n-1}$, for all positive integer $n$, be the Fibonacci sequence. Prove that for any positive integer $m$ there exist infinitely many positive integers $n$ such that
$$
F_{n}+2 \equiv F_{n+1}+1 \equiv F_{n+2} \quad \bmod m
$$ | [
"Let $m$ be a positive integer and consider the infinite set of pairs $(F_{k}, F_{k+1})$, for $k \\in \\mathbb{N}$. By the pigeonhole principle, there exists a pair $(a, b)$ of integers $0 \\leq a, b \\leq m-1$ and an infinite sequence of integers $0<k_{1}<k_{2}<\\cdots$ such that\n$$\n(F_{k_{i}}, F_{k_{i}+1}) \\eq... | Saudi Arabia | Selection tests for the Gulf Mathematical Olympiad 2013 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
058c | The incircle of triangle $ABC$ touches the sides $AB$ and $AC$ at points $K$ and $L$, respectively. The line $BL$ intersects the incircle of triangle $ABC$ at point $M$ ($M \neq L$). A circle passing through point $M$ touches the lines $AB$ and $BC$ at points $P$ and $Q$, respectively, and intersects the incircle of tr... | [
"Note that the circles of the problem can be obtained from each other by homothetic transformation with center $B$ since both are tangent to sides $BA$ and $BC$. Let $X$ be the other intersection point of the line $BL$ with the circumcircle of the triangle $MPQ$.\n\n\n\nThe aforementioned h... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
01sz | Find all real numbers $a$ such that there exists a function $f: \mathbb{R} \to \mathbb{R}$ satisfying the following conditions:
1) $f(f(x)) = x f(x) - a x$ for all real $x$;
2) $f$ is not constant;
3) $f$ takes the value $a$. | [
"Answer: $a = 0$ or $a = -1$.\n\nBy condition, there is $c$ such that $f(c) = a$. Substituting $c$ for $x$ in the equality\n$$\nf(f(x)) = x f(x) - a x, \\quad (1)\n$$\nwe obtain $f(a) = f(f(c)) = c f(c) - a c = c a - a c = 0$.\n\nSubstituting $a$ for $x$ in (1), we obtain\n$$\nf(0) = f(f(a)) = a f(a) - a^2 = -a^2 \... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | a = 0 or a = -1 | |
0e26 | Find all prime numbers $p$, $q$ and $r$ such that $15p + 7pq + qr = pqr$. | [
"The identities $qr = pqr - 15p - 7pq = p(qr - 15 - 7q)$ imply that $qr$ is divisible by $p$. Since $p$, $q$ and $r$ are prime numbers, there are only two possibilities: $p = q$ or $p = r$.\n\nIf $p = q$ we get $15 + 7q + r = qr$, which can be rewritten as\n$$\n22 = qr - 7q - r + 7 = (q - 1)(r - 7).\n$$\nExactly on... | Slovenia | National Math Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (p, q, r) = (2, 2, 29), (13, 3, 13), (11, 5, 11) | |
05bv | a. Every side and diagonal of a regular $2025$-gon is coloured either red or blue. Can it happen that the same number of red and blue line segments meet at each vertex?
b. The same question if only the diagonals are coloured. | [
"a. Each side and diagonal connects vertices that are $1$ to $1012$ side lengths apart as you move along the polygon. Exactly $2$ sides or diagonals of a given length meet at each vertex. We color the sides and diagonals that have an odd number of sides between their endpoints red and all other diagonals blue. Sinc... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | a: Yes. b: No. | |
008y | Let $N$ be a positive integer such that the only perfect square number which divides it is $1$. Find the amount of positive multiples of $N$ which have exactly $N$ positive divisors. | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | the factorial of the number of distinct prime divisors of N | |
0j1y | Problem:
Let $S_{0}=0$ and let $S_{k}$ equal $a_{1}+2 a_{2}+\ldots+k a_{k}$ for $k \geq 1$. Define $a_{i}$ to be 1 if $S_{i-1}<i$ and -1 if $S_{i-1} \geq i$. What is the largest $k \leq 2010$ such that $S_{k}=0$ ? | [
"Solution:\nAnswer: 1092\n\nSuppose that $S_{N}=0$ for some $N \\geq 0$. Then $a_{N+1}=1$ because $N+1 \\geq S_{N}$. The following table lists the values of $a_{k}$ and $S_{k}$ for a few $k \\geq N$ :\n\n| $k$ | $a_{k}$ | $S_{k}$ |\n| :--- | ---: | :--- |\n| $N$ | | 0 |\n| $N+1$ | 1 |... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 1092 | |
01n1 | Consider a polynomial $P(x) = (x+d_1)(x+d_2) \cdots (x+d_9)$, where $d_1, d_2, \ldots, d_9$ are nine distinct integers.
Prove that there exists an integer $N$ such that for all integers $x \ge N$ the number $P(x)$ is divisible by a prime number greater than 20. | [] | Belarus | Belorusija 2012 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0ewo | Problem:
a. The two diagonals of a quadrilateral each divide it into two parts of equal area. Prove it is a parallelogram.
b. The three main diagonals of a hexagon each divide it into two parts of equal area. Prove they have a common point. [If $ABCDEF$ is a hexagon, then the main diagonals are $AD$, $BE$ and $CF$.] | [
"Solution:\n\na. Let the quadrilateral be $ABCD$ and let the diagonals $AC$, $BD$ meet at $E$. Then area $ABC = AC \\cdot EB \\cdot \\sin CEB / 2$, and area $ADC = AC \\cdot ED \\cdot \\sin CEB / 2$, so $E$ is the midpoint of $BD$. Similarly, it is the midpoint of $AC$. Hence the triangles $AEB$ and $CED$ are congr... | Soviet Union | 3rd ASU | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
067j | Determine the values of the parameter $\alpha \in \mathbb{R}$ for which the equation $x^2 + (\alpha - 2)x - (\alpha - 1)(2\alpha - 3) = 0$ has two roots such that the one of them is equal to the square of the other. | [
"1. We have the discriminant $D = (\\alpha - 2)^2 + 4(\\alpha - 1)(2\\alpha - 3) = (3\\alpha - 4)^2$, and so the equation has the roots $x_1 = \\alpha - 1$, $x_2 = -2\\alpha + 3$.\n\nTherefore we seek the values of $\\alpha$ for which:\n$$\n\\begin{align*} \nx_1 = x_2^2 \\text{ or } x_2 = x_1^2 &\\Leftrightarrow \\... | Greece | Hellenic Mathematical Olympiad ARCHIMEDES | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | 2, 5/4, sqrt(2), -sqrt(2) | |
07gb | A connected simple graph $G$ with maximum degree $d > 3$ is given. Suppose that
$$
x_d \geq x_{d-1} + 2x_{d-2} + \dots + (d-1)x_1,
$$
where $x_i$ is the number of vertices of degree $i$. Prove that there is a vertex of degree $d$ in $G$ such that after removing it the graph remains connected. | [
"We shall prove the statement by induction on the number of vertices. The base case is clear. Let $v$ be a vertex of degree $d$, and denote by $C_1, C_2, \\dots, C_k$ the connected components of $G - \\{v\\}$. Assume that $v$ is chosen such that $|C_1|$ is maximum among all connected components obtained from removi... | Iran | 38th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0gln | Determine all polynomials $P \in \mathbb{R}[x, y]$ such that
$$
P(a, b^2 - ac) + P(b, c^2 - ab) + P(c, a^2 - bc) = 0
$$
for all $a, b, c \in \mathbb{R}$. | [] | Thailand | T3MO 2017 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | P(x, y) = k x y for some real constant k | |
04u8 | Let $k$ be a semicircle with diameter $PQ$. Consider a chord $BC$ of fixed length $d$ whose endpoints are distinct from $P$, $Q$. A ray of light emanating from $B$ reaches point $C$ after reflecting from $PQ$ at such a point $A$ that $\angle PAB = \angle QAC$. Prove that $\angle BAC$ doesn't depend on the position of t... | [
"Reflect $k$ and $C$ about $PQ$ to get $l$ and $C'$, respectively (Fig. 1). Then $C'$ lies on $l$ and since $\\angle QAC' = \\angle QAC = \\angle PAB$ it also lies on $BA$. Triangle $C'CA$ is isosceles, hence\n$$\n\\angle BAC = \\angle AC'C + \\angle ACC' = 2 \\cdot \\angle BC'C\n$$\nThe chord $BC$ of circle $k \\c... | Czech Republic | 67th Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0fdc | Problem:
En un tablero de damas ($8 \times 8$), colocamos las 24 fichas del juego de modo que llenen las 3 filas de arriba. Podemos cambiar la posición de las fichas según el siguiente criterio: una ficha puede saltar por encima de otra a un hueco libre, ya sea horizontal (a izquierda o derecha), vertical (hacia arriba... | [
"Solution:\nNo podemos lograrlo:\nClasificamos (o coloreamos) las casillas del tablero en cuatro tipos, según la paridad de la fila y la columna que ocupan. Cada ficha se mueve siempre por el mismo tipo de casilla. Pero el número de casillas de cada tipo que están ocupadas en las posiciones inicial y final es disti... | Spain | null | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | No | |
04sb | Nice prime *is a prime equal to the difference of two cubes of positive integers.*
Find last digits of all nice primes. | [
"Firstly, let us note that $5^3 - 4^3 = 61$, $2^3 - 1^3 = 7$ and $3^3 - 2^3 = 19$ are nice primes, so 1, 7 and 9 belong to desired digits. We show that they are all desired digits.\nLet $p = m^3 - n^3$ be a nice prime, where $m > n$ are positive integers. Second factor in rewriting\n$$\np = m^3 - n^3 = (m-n)(m^2 + ... | Czech Republic | 65th Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof and answer | 1, 7, 9 | |
095w | Problem:
Fie $a>0$ şi $x, y \in \mathbb{R}$ astfel, încât $|x|<\frac{1}{a}$ şi $|y|<\frac{1}{a}$. Să se arate, că $\left|\frac{x+y}{1+a^{2} x y}\right|<\frac{1}{a}$. | [
"Solution:\n\nSe stabileşte consecutiv:\n\n1) $\\left\\{\\begin{array}{l}|x|<\\frac{1}{a}, \\\\ |y|<\\frac{1}{a} ;\\end{array}\\right. \\Leftrightarrow \\left\\{\\begin{array}{r}-\\frac{1}{a}<x<\\frac{1}{a}, \\\\ -\\frac{1}{a}<y<\\frac{1}{a} ;\\end{array}\\right. \\Leftrightarrow \\left\\{\\begin{array}{l}a x+1>0 \... | Moldova | A 62-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0g6h | 正方形 $ABCD$ 內部有一點 $P$, 已知 $\overline{PA} = x$, $\overline{PB} = z$, $\overline{PC} = y$, 試證:
$$
(x - y)^2 < 2z^2 < (x + y)^2
$$ | [
"令正方形 $ABCD$ 邊長為 $a$, $\\overline{PA}$ 在 $\\overline{BC}$ 投影長分別為 $t$, $\\overline{PB}$ 在 $\\overline{CD}$ 投影長為 $v$. 如圖所示\n\n則可將 $x, y, z$ 表示如下：\n$$\nx^2 = t^2 + (a - v)^2 \\qquad (1)\n$$\n$$\ny^2 = v^2 + (a - t)^2 \\qquad (2)\n$$\n$$\nz^2 = t^2 + v^2 \\qquad (3)\n$$\n\n由(1)和(3)解得\n$$\nv = \... | Taiwan | 二〇一二數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
0dt3 | What is the maximum number of integers that can be chosen from $1, 2, \ldots, 99$ so that the chosen integers can be arranged in a circle with the property that the product of every pair of neighbouring integers is a 3-digit number? | [
"Since $31 \\times 32 = 992$ and $31 \\times 33 = 1023$, any two numbers larger than $31$ cannot be neighbours. So there must be a number $< 32$ between a pair of such numbers. Also $1$ cannot be chosen. Also the two neighbours of $31$ are $32$ or less. So the maximum number of chosen integers is $\\le 30 \\times 2... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 59 | |
0511 | Find all four-digit numbers, which after deleting any one digit turn into a three-digit number that is a divisor of the original number. | [
"Let $\\overline{abcd}$ be such a number. Since $\\overline{abcd}$ is divisible by $\\overline{abc}$, we have $d = 0$. Since $\\overline{abcd} = \\overline{abc0}$ is divisible by $\\overline{abd} = \\overline{ab0}$, we have $c = 0$. Since $\\overline{abcd} = \\overline{ab00}$ is divisible by $\\overline{acd} = \\ov... | Estonia | Estonian Math Competitions | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 1100, 1200, 1500, 2200, 2400, 3300, 3600, 4400, 4800, 5500, 6600, 7700, 8800, 9900 | |
0c7u | In the interior of non-zero angle $\widehat{AOD}$ consider points $B$ and $C$ such that $OA = OB$, $OD = OC$, the segments $AC$ and $BD$ meet in $P$, and the semi-line ($PO$ is the angle bisector of $\widehat{APD}$). Prove that the angles $\widehat{AOB}$ and $\widehat{COD}$ are equal. | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0kk7 | Problem:
Kevin writes down the positive integers $1, 2, \ldots, 15$ on a blackboard. Then, he repeatedly picks two random integers $a, b$ on the blackboard, erases them, and writes down $\operatorname{gcd}(a, b)$ and $\operatorname{lcm}(a, b)$. He does this until he is no longer able to change the set of numbers writt... | [
"Solution:\n\nSince $v_{p}(\\operatorname{gcd}(a, b))=\\min \\left(v_{p}(a), v_{p}(b)\\right)$ and $v_{p}(\\operatorname{lcm}(a, b))=\\max \\left(v_{p}(a), v_{p}(b)\\right)$, we may show the following:\n\nClaim. For any prime $p$ and non-negative integer $k$, the number of numbers $n$ on the board such that $v_{p}(... | United States | HMMT November 2021 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 360854 | |
0ja1 | Problem:
You are given two line segments of length $2^{n}$ for each integer $0 \leq n \leq 10$. How many distinct nondegenerate triangles can you form with three of the segments? Two triangles are considered distinct if they are not congruent. | [
"Solution:\n\nAnswer: 55\n\nFirst, observe that if we have three sticks of distinct lengths $2^{a} < 2^{b} < 2^{c}$, then $2^{a} + 2^{b} < 2^{b+1} \\leq 2^{c}$, so we cannot form a triangle. Thus, we must have (exactly) two of our sticks the same length, so that our triangle has side lengths $2^{a}, 2^{a}, 2^{b}$. ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 55 | |
09e8 | If $\angle CBA = 90^\circ$, $\angle BAP = \angle PAR = \angle RAC$, $\angle BCQ = \angle QCR = \angle RCA$, $\angle QRC = 142^\circ$ then find the angle $\angle BAC$.
 | [
"Denote $\\angle BAP = \\angle PAR = \\angle RAC = \\alpha$. Then $\\angle BAC = 3\\alpha$. Denote $\\angle BCQ = \\angle QCR = \\angle RCA = \\gamma$. Then $\\angle ACB = 3\\gamma$. Since $\\angle ABC = 90^\\circ$, $\\angle ACB + \\angle BAC = 180^\\circ - 90^\\circ = 90^\\circ \\Rightarrow 3\\alpha + 3\\gamma = 9... | Mongolia | Mongolian 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 and answer | 66° | |
0bkm | Determine all positive integers $n$ such that all positive integers less than $n$ and coprime to $n$ are powers of primes. | [
"Notice that $6$ is the first index $k$ such that $p_1p_2 \\cdots p_{k-2} > p_{k-1}p_k$. Now, if $p_1p_2 \\cdots p_{k-2} > p_{k-1}p_k$ for some index $k \\ge 6$, then (by Bertrand-Tchebysheff) $p_1p_2 \\cdots p_{k-1} > p_{k-1}^2p_k > 2p_{k-1} \\cdot 2p_k > p_kp_{k+1}$, so $p_1p_2 \\cdots p_{k-2} > p_{k-1}p_k$ for a... | Romania | 65th NMO Selection Tests for BMO and IMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 18, 20, 24, 30, 42, 60 | |
0787 | In a school, every pair of students are either friends or strangers. A sequence of (not necessarily distinct) students $A_1, A_2, \dots, A_{2023}$ is called *mischievous* if
* Total number of friends of $A_1$ is odd.
* $A_i$ and $A_{i+1}$ are friends for $i = 1, 2, \dots, 2022$.
* Total number of friends of $A_{2023}$... | [
"We first put the problem in graph theoretic terms:\nConsider a finite simple graph $G$. A walk of length $2022$ is called *mischievous* if the degree of its starting vertex is odd, and the degree of its ending vertex is even. Prove that the total number of *mischievous* walks is even.\nLet $2022 = m$. We prove the... | India | IMO TST Day 3 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
035j | Problem:
Let $\mathbb{R}^*$ be the set of non-zero real numbers. Find all functions $f: \mathbb{R}^* \rightarrow \mathbb{R}^*$ such that
$$
f\left(x^{2}+y\right)=f^{2}(x)+\frac{f(x y)}{f(x)}
$$
for all $x, y \in \mathbb{R}^*,\ y \neq -x^{2}$. | [
"Solution:\nSet $\\alpha=f(1)$. Then setting $y=1$ and $x=1$ in\n$$\nf\\left(x^{2}+y\\right)=f^{2}(x)+\\frac{f(x y)}{f(x)}\n$$\ngives\n$$\nf\\left(x^{2}+1\\right)=f^{2}(x)+1\n$$\nand\n$$\nf(y+1)=\\alpha^{2}+\\frac{f(y)}{\\alpha}\n$$\nrespectively. Using (3), we consecutively get\n$$\n\\begin{gathered}\nf(2)=\\alpha... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x for every nonzero real number x | |
0kbp | Problem:
Sean enters a classroom in the Memorial Hall and sees a $1$ followed by $2020$ $0$'s on the blackboard. As he is early for class, he decides to go through the digits from right to left and independently erase the $n$th digit from the left with probability $\frac{n-1}{n}$. (In particular, the $1$ is never eras... | [
"Solution:\n\nSuppose Sean instead follows this equivalent procedure: he starts with $M = 10\\ldots 0$ on the board, as before. Instead of erasing digits, he starts writing a new number on the board. He goes through the digits of $M$ one by one from left to right, and independently copies the $n$th digit from the l... | United States | HMMO 2020 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 681751 | |
0579 | Integers $a$, $b$, $c$ and $n$ are given such that $1 \le a < b < c \le n$. Juku and Miku play the following game on a strip of size $1 \times n$: In the beginning, squares number $a$, $b$, $c$ contain one piece each, whereby the squares are numbered from the right to the left by consecutive integers starting from $1$.... | [
"Firstly, note that, in any position where the number of empty squares between the leftmost and the middle piece differs from the number of empty squares in the right from the rightmost piece, one can make a move that makes these two quantities equal. Indeed, if the number of empty squares between the leftmost and ... | Estonia | Open Contests | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Juku wins if c − b ≠ a; Miku wins if c − b = a. | |
0gpi | Find all polynomials $P(x)$ with integer coefficients satisfying $P(n!) = |P(n)|!$ for all positive integers $n$. | [
"The answer is $P(x) = 1$, $P(x) = 2$ and $P(x) = x$.\nFirst recall that for two polynomials $P(x)$ and $Q(x)$ if we have $P(x) = Q(x)$ for infinitely many $x$, then $P(x) = Q(x)$ for every $x$.\nPlugging in $n = 1, 2$ gives $P(1) = |P(1)|!$ and $P(2) = |P(2)|!$ which implies that $P(1), P(2) \\in \\{1, 2\\}$.\n\nC... | Turkey | 20th Turkish Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | P(x) = 1; P(x) = 2; P(x) = x | |
00qc | Consider a lattice of side length $1$ equilateral triangles forming a regular hexagon of side length $n$. Show that the number of ways of simultaneously selecting six vertices of the lattice to form the vertices of a regular hexagon is a perfect square. | [
"By a lattice hexagon we will mean a regular hexagon whose sides run along edges of the lattice. Given any regular hexagon $H$, we construct a lattice hexagon whose edges pass through the vertices of $H$, as shown in the figure, which we will call the enveloping lattice hexagon of $H$. Given a lattice hexagon $G$ o... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlist | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | English | proof only | null | |
07za | Problem:
Una sfera di raggio $r = 15~\mathrm{cm}$ è appoggiata su due binari distanti fra loro $24~\mathrm{cm}$ come in figura. Se la sfera fa una rotazione completa, di quanto avanza sui binari?
(A) $24~\mathrm{cm}$
(B) $30~\mathrm{cm}$
(C) $15\pi~\mathrm{cm}$
(D) $18\pi~\mathrm{cm}$
(E) $30\pi~\mathrm{cm}$
![](atta... | [] | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | MCQ | D | |
04j3 | On the side $AC$ of the triangle $ABC$ the points $D$ and $E$ are given such that $D$ is between $C$ and $E$. Let $F$ be the intersection of the circumcircle of the triangle $ABD$ and the line through the point $E$ parallel to $BC$ such that $E$ and $F$ are on different sides of the line $AB$. Let $G$ be the intersecti... | [
"Let $F'$ be the intersection of the circumcircle of the triangle $ABD$ and the line $BG$ (different from $B$).\n\nThe quadrilateral $DAF'B$ is cyclic, so we have $\\angle BF'D = \\angle BAD = \\angle BAC$. Since $GE \\parallel AB$, we have $\\angle BAC = \\angle GEC$. Hence $\\angle GF'D =... | Croatia | Croatia Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0dwy | Problem:
Lukcu je bilo med uro matematike dolgčas, zato je najprej narisal krog in nato naokrog po obodu še $n$ praznih polj, kjer je $n \geq 3$, ter vanje zapisal po 1 pozitivno število. Kasneje je ta števila pobrisal, v vsako polje pa zapisal kvadratni koren zmnožka dveh števil, ki sta prej ležali na temu polju sose... | [
"Solution:\n\n1. način\nOznačimo z $a_{1}, a_{2}, \\ldots, a_{n}$ števila, ki so v poljih ležala na začetku, z $b_{1}, b_{2}, \\ldots, b_{n}$ pa tista, ki ležijo na koncu. Po neenakosti med aritmetično in geometrijsko sredino velja\n$$\na_{1}+a_{2}+\\cdots+a_{n}=\\frac{a_{1}+a_{3}}{2}+\\frac{a_{2}+a_{4}}{2}+\\cdots... | Slovenia | 49. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0eok | Lewis Hamilton completes a 72-lap race travelling at an average speed of 288 km/h. Each lap is 6 km in length. The time taken, in hours, for him to complete the race is
(A) 1 (B) 2 (C) 2.5 (D) 3 (E) 1.5 | [
"Time = Distance/Speed, so the time taken is equal to $$(72 \\times 6)/288 = 1.5$$ hours."
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | English | MCQ | E | |
09sf | Problem:
Je mag elk van de getallen 1 tot en met 2014 een kleur geven, waarbij precies de helft rood moet worden en de andere helft blauw. Vervolgens bekijk je het aantal $k$ van positieve gehele getallen die te schrijven zijn als de som van een rood en een blauw getal. Bepaal de maximale waarde van $k$ die je kunt be... | [
"Solution:\n\nNoem $n=2014$. We gaan bewijzen dat de maximale $k$ gelijk is aan $2n-5$. Het kleinste getal dat je zou kunnen schrijven als de som van een rood en een blauw getal is $1+2=3$ en het grootste getal is $(n-1)+n=2n-1$. Er zijn dus hoogstens $2n-3$ getallen te schrijven als de som van een rood en een blau... | Netherlands | IMO-selectietoets II | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 4023 | |
08e9 | Problem:
In ogni casella di una tabella $8 \times 8$ abita un cavaliere o un furfante. Come da tradizione, i cavalieri dicono sempre la verità, mentre i furfanti mentono sempre. Tutti gli abitanti della tabella affermano che "il numero dei furfanti nella mia colonna è maggiore (strettamente) del numero dei furfanti ne... | [] | Italy | XXXVI Olimpiade Italiana di Matematica | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 255 | |
0jl9 | Problem:
In quadrilateral $A B C D$, $\angle D A C = 98^{\circ}$, $\angle D B C = 82^{\circ}$, $\angle B C D = 70^{\circ}$, and $B C = A D$. Find $\angle A C D$.
 | [
"Solution:\n\nAnswer: $28$\n\nLet $B'$ be the reflection of $B$ across $C D$. Note that $A D = B C$, and $\\angle D A C + \\angle C B' D = 180^{\\circ}$, so $A C B' D$ is a cyclic trapezoid. Thus, $A C B' D$ is an isosceles trapezoid, so $\\angle A C B' = 98^{\\circ}$. Note that $\\angle D C B' = \\angle B C D = 70... | United States | HMMT 2014 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 28 | |
0esp | If $a$ and $b$ are positive integers such that $\frac{a}{b} = \frac{2}{3}$ and $a + b = 80$, then the product $ab$ is equal to
(A) 1 500 (B) 1 599 (C) 1 667 (D) 1 596 (E) 1 536 | [
"Since $a = \\frac{2}{3}b$ and $a + b = 80$, we have\n$$\n80 = \\frac{2}{3}b + b = \\frac{5}{3}b\n$$\nso\n$$\nb = \\frac{3}{5} \\times 80 = 48.\n$$\nThen $a = 80 - 48 = 32$ and $ab = 32 \\times 48 = 1536$."
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | E | |
0coa | Points $B$ and $D$ lie on a circle $\omega$. The tangent lines to $\omega$ at $B$
and $D$ intersect at $P$. A line passing through $P$ intersects $\omega$ at $A$ and $C$. Let $\ell$ be an arbitrary line parallel to $BD$ and intersecting the polygonal lines $ABC$ and $ADC$. Prove that $\ell$ divides the lengths of thes... | [
"Треугольники $PBA$ и $PCB$ подобны, так как $\\angle BPC$ — общий,\nа $\\angle PBA = \\angle PCB = \\frac{1}{2} \\overline{AB}$. Значит, $\\frac{BA}{BC} = \\frac{PB}{PC}$. Аналогично, из подобия треугольников $PDA$ и $PCD$ следует, что $\\frac{DA}{DC} = \\frac{PD}{PC}$.\nТак как $PB = PD$, то $\\frac{BA}{BC} = \\f... | Russia | Regional round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof only | null | |
0ixf | Problem:
A polyhedron has faces that are all either triangles or squares. No two square faces share an edge, and no two triangular faces share an edge. What is the ratio of the number of triangular faces to the number of square faces? | [
"Solution:\nLet $s$ be the number of square faces and $t$ be the number of triangular faces. Every edge is adjacent to exactly one square face and one triangular face. Therefore, the number of edges is equal to $4s$, and it is also equal to $3t$. Thus $4s = 3t$ and $\\frac{t}{s} = \\frac{4}{3}$."
] | United States | Harvard-MIT November Tournament | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | 4/3 | |
02ut | Problem:
João estava estudando para as Olimpíadas de Matemática e se deparou com a seguinte equação
$$
\frac{1}{x} + \frac{1}{y} = \frac{1}{2} + \frac{1}{z}
$$
onde $x$, $y$ e $z$ são inteiros positivos. Após tentar encontrar todas as soluções sem sucesso, ele pediu ajuda para o professor Piraldo, que decidiu dar algu... | [
"Solution:\n\na. Substituindo $x=1$ na equação, temos\n$$\n\\begin{aligned}\n& \\frac{1}{1} + \\frac{1}{y} = \\frac{1}{2} + \\frac{1}{z} \\\\\n& \\frac{1}{2} + \\frac{1}{y} = \\frac{1}{z}\n\\end{aligned}\n$$\nSe $z \\geq 2$, então $\\frac{1}{2} + \\frac{1}{y} = \\frac{1}{z} \\leq \\frac{1}{2}$. Daí, $\\frac{1}{y} \... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a) With x = 1: (y, z) = (2, 1).
b) With x = 2 and y ≥ 2: (y, z) = (n, n) for any integer n ≥ 2.
c) With x = 3 and y ≥ 3: (y, z) ∈ {(3, 6), (4, 12), (5, 30)}.
d) With x, y ≥ 4: no solutions.
By symmetry in x and y, swapping x and y in any listed triple yields another solution. Consequently, all solutions are obtained fr... | |
09nl | The sum of nine different natural numbers is $111$. Show that the sum of four of them is at least $61$. | [] | Mongolia | MMO2025 Round 2 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof only | null | |
09cm | $ n \mid 53^{\frac{n-1}{2}} + 1 $ байх сондгой $ n $ тоо төгсгөлгүй олон олдохыг батал. | [
"I арга:\n\n$n(k) = \\frac{53^{2k} + 1}{2}$, $k \\ge 1$ тоонууд бүгд хариу болж чадна: $A$ сондгой үед\n\n$n(k) \\mid 53^{2k} \\cdot A + 1$ байх тул $\\frac{n(k)-1}{2} = 2^k \\cdot A$, $A$ сондгой гэж харуулъя.\n\n$$\n\\frac{n(k)-1}{2} \\equiv 0 \\pmod{2^k} \\to 53^{2k} \\equiv 1 \\pmod{2^{k+2}} \\Rightarrow 53^{2k... | Mongolia | ОУМО-53 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity"
] | Mongolian | proof only | null | |
02fg | Let $p(n)$ be the largest prime which divides $n$. Show that there are infinitely many positive integers $n$ such that $p(n) < p(n + 1) < p(n + 2)$. | [
"Let $q$ be an odd prime and take $n + 1 = q^{2^k}$. Then $p(q^{2^k}) = q$. Since $\\gcd(q^{2^k} + 1, q^{2^l} + 1) = 2$ for $k \\neq l$ (indeed, if $d$ is such $\\gcd$ and $k < l$,\n\n$$q^{2k} \\equiv -1 \\pmod d \\implies q^{2l} \\equiv 1 \\pmod d \\iff d \\mid 2$$\n\n$p(q^{2k} + 1)$ can be arbitrarily large. So l... | Brazil | XVII OBM | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
08pr | Problem:
Let $ABC$ be a triangle with side-lengths $a$, $b$, $c$, inscribed in a circle with radius $R$ and let $I$ be its incenter. Let $P_{1}$, $P_{2}$ and $P_{3}$ be the areas of the triangles $ABI$, $BCI$ and $CAI$, respectively. Prove that
$$
\frac{R^{4}}{P_{1}^{2}}+\frac{R^{4}}{P_{2}^{2}}+\frac{R^{4}}{P_{3}^{2}} ... | [
"Solution:\nLet $r$ be the radius of the inscribed circle of the triangle $ABC$. We have that\n$$\nP_{1}=\\frac{r c}{2}, \\quad P_{2}=\\frac{r a}{2}, \\quad P_{3}=\\frac{r b}{2}\n$$\nIt follows that\n$$\n\\frac{1}{P_{1}^{2}}+\\frac{1}{P_{2}^{2}}+\\frac{1}{P_{3}^{2}}=\\frac{4}{r^{2}}\\left(\\frac{1}{c^{2}}+\\frac{1}... | JBMO | Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0gw8 | A point $S$ lies on the side $PM$ of a trapezoid $MPQ$ with bases $PM$ and $RQ$ ($PQ < PM$, and $S$ is distinct from the vertices). The bisectors of the angles $MSQ$ and $MPQ$ meet at point $O$. It is known that the segment $OI$, where $I$ is the incenter of the triangle $PQR$, is parallel to the bases of the trapezoid... | [
"Нехай $\\angle SPO = \\alpha$. Очевидно, що $\\angle SPO = \\angle QPO = \\angle PQI = \\angle RQI = \\angle POI$.\nЗокрема, оскільки $\\angle PQI = \\angle POI$, то чотирикутник $POQI$ циклічний, причому,\nз урахуванням паралельності прямих $PO$ і $QI$, — рівнобічна трапеція. Таким\nчином, $OI = PQ$. Якщо $\\angl... | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
08co | Problem:
a. Trovare tutti gli interi positivi $n$ di due cifre che godano della seguente proprietà: entrambi gli interi che si ottengono cancellando una delle due cifre della rappresentazione decimale di $n$ sono divisori (interi positivi) di $n$.
b. Sia $n>10$ un intero che si scrive con $k$ cifre decimali, tutte di... | [
"Solution:\n\na. Scriviamo $n=10 a+b$ con $a$ e $b$ cifre decimali, ossia $1 \\leq a \\leq 9$ e $1 \\leq b \\leq 9$ : per ipotesi $a=0$ non è possibile (dato che $n>10$ ), e $b=0$ non è possibile perché in tal caso cancellando la prima cifra di $n$ si troverebbe 0, che non divide $n$. Le condizioni sono allora che ... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a) 11, 22, 33, 44, 55, 66, 77, 88, 99, 12, 24, 36, 48, 15. b) The number must have exactly two digits. | |
0ez8 | Problem:
Given any positive integer $n$, show that we can find infinitely many integers $m$ such that $m$ has no zeros (when written as a decimal number) and the sum of the digits of $m$ and $mn$ is the same. | [] | Soviet Union | 4th ASU | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0cn6 | $k$ rooks are placed on the $10 \times 10$ board. All the squares beaten by at least one rook are marked (a rook in particular beats its own square). It occurs that after removing any rook from the board, at least one marked square becomes not beaten. Find the greatest possible value of $k$. (S. Berlov) | [
"Ответ. $16$.\n\nРассмотрим расстановку $k$ ладей, удовлетворяющую условию. Возможны два случая.\n\n1. Пусть в каждом столбце стоит хотя бы по одной ладье. Тогда вся доска находится под боем, и можно убрать ладью из любого столбца, в котором их хотя бы две. Значит, в этом случае в каждом столбце стоит ровно по одно... | Russia | Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English; Russian | proof and answer | 16 | |
0ic1 | For what real values of $k > 0$ is it possible to dissect a $1 \times k$ rectangle into two similar, but noncongruent, polygons? | [
"**First Solution:** We will show that a dissection satisfying the requirements of the problem is possible if and only if $k \\neq 1$.\n\nWe first show by contradiction that such a dissection is not possible when $k=1$. Assume that we have such a dissection. The common boundary of the two dissecting polygons must b... | United States | USA IMO | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | k ≠ 1 | |
02ip | Problem:
Se $\frac{\sqrt{x}}{\sqrt{y}}=5$ então $\frac{x+y}{2y}$ é igual a:
(A) $5/2$
(B) $3\sqrt{2}$
(C) $13y$
(D) $\frac{25y}{2}$
(E) $13$ | [
"Solution:\n\nElevando ao quadrado ambos os membros de $\\frac{\\sqrt{x}}{\\sqrt{y}}=5$, obtemos $\\frac{x}{y}=25$. Agora,\n$$\n\\frac{x+y}{2y} = \\frac{1}{2} \\times \\frac{x+y}{y} = \\frac{1}{2} \\times \\left(\\frac{x}{y} + \\frac{y}{y}\\right) = \\frac{1}{2} \\times \\left(\\frac{x}{y} + 1\\right) = \\frac{1}{2... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | E | |
09cl | $(1+x)^n$ олон гишүүнтийн тэгш коэффициенттэй гишүүдийг дарахад үлдэх олон гишүүнтийг $Q_n(x)$ гэе. $Q_{2012}(1)$-ийг ол. | [
"$p \\in \\mathbb{P}$ бол $(a+b)p^n = ap^n + bp^n$ (мод $p$) чанар болон $2012 = 2^{10} + 2^9 + 2^8 + 2^7 + 2^6 + 2^5 + 2^4 + 2^3 + 2^2$ байхгы ашиглан\n$$\n(1+x)^{2012} = (1+x)^{2010}(1+x)^{29}(1+x)^{28}(1+x)^{27}(1+x)^{26}(1+x)^{24}(1+x)^{23}(1+x)^{22}\n$$\n$$\n\\equiv (1+x^{2^{10}})(1+x^{2^9})(1+x^{2^8})(1+x^{2^... | Mongolia | ОУМО-53 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | Mongolian | proof and answer | ∑_{j=0}^{31} ∑_{i=0}^{7} \binom{2012}{4i + 64j} | |
0ebc | For how many positive integers $n$, $n \le 2015$ is the fraction $\frac{3n-1}{2n^2+1}$ reducible? | [
"Suppose that the fraction $\\frac{3n-1}{2n^2+1}$ is reducible. Then there exists a positive integer $a$ different from $1$ which divides $3n-1$ and $2n^2+1$. It follows that $a$ divides also $3(2n^2+1)-2n(3n-1) = 2n+3$ and hence also $3(2n+3)-2(3n-1) = 11$. Since $11$ is a prime number it follows $a=11$, hence $11... | Slovenia | National Math Olympiad 2015 – Final Round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 183 | |
06t6 | Let $a_{1} < a_{2} < \cdots < a_{n}$ be pairwise coprime positive integers with $a_{1}$ being prime and $a_{1} \geqslant n+2$. On the segment $I = [0, a_{1} a_{2} \cdots a_{n}]$ of the real line, mark all integers that are divisible by at least one of the numbers $a_{1}, \ldots, a_{n}$. These points split $I$ into a nu... | [
"Let $A = a_{1} \\cdots a_{n}$. Throughout the solution, all intervals will be nonempty and have integer end-points. For any interval $X$, the length of $X$ will be denoted by $|X|$.\n\nDefine the following two families of intervals:\n$$\n\\begin{aligned}\n\\mathcal{S} & = \\{[x, y]: x < y \\text{ are consecutive m... | IMO | 55th International Mathematical Olympiad Shortlist | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization... | null | proof only | null | |
0cj6 | Let $n \in \mathbb{N}$, $n \ge 3$. For any $x \in U(\mathbb{Z}_n)$ we denote $o_1(x)$ and $o_2(x)$ the orders of the element $x$ in the groups $(\mathbb{Z}_n, +)$, respectively $(U(\mathbb{Z}_n), \cdot)$, and $a(x) = o_1(x) + o_2(x)$. We consider the set $T_n = \{a(x) \mid x \in U(\mathbb{Z}_n)\}$.
a) Determine $T_8$.... | [] | Romania | 75th NMO | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Algebra > Abstract Algebra > Group Theory"
] | English | proof and answer | {9, 10} | |
0gos | Show that
$$
a + b + c + \sqrt{3} \geq 8abc \left( \frac{1}{a^2 + 1} + \frac{1}{b^2 + 1} + \frac{1}{c^2 + 1} \right)
$$
for all positive real numbers $a, b, c$ satisfying $ab + bc + ca \leq 1$. | [
"We first observe that $a^2 + 1 \\ge a^2 + ab + bc + ca \\ge 4a\\sqrt{bc}$ where the second inequality results from $A.M. \\ge G.M.$. Therefore we have $2\\sqrt{bc} \\ge \\frac{8abc}{a^2+1}$. Summing this up with similar inequalities for $b$ and $c$ gives that it suffices to show that\n$$\na + b + c + \\sqrt{3} \\g... | Turkey | Team Selection Test | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0gp0 | Let $a_{n+1} = a_n^3 - 2a_n^2 + 2$ for all $n \ge 1$ and $a_1 = 5$. Prove that if $p \equiv 3 \pmod 4$ is a prime divisor of $a_{2011} + 1$, then $p = 3$. | [
"Observe that $a_{n+1} - 2 = a_n^2(a_n - 2)$ for all $n \\ge 1$. By induction on $n$ we obtain $a_{n+1} - 2 = 3a_n^2a_{n-1}^2 \\cdots a_1^2$ for all $n \\ge 1$. Therefore $a_{2011} + 1 = 3(a_{2010}^2a_{2009}^2 \\cdots a_1^2 + 1) = (a_{2010}a_{2009} \\cdots a_1)^2 + 1$.\n\nLet $p \\equiv 3 \\pmod 4$ be a prime divis... | Turkey | 19th Turkish Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof only | null | |
03hi | Problem:
Given a circle with diameter $AB$ and a point $X$ on the circle different from $A$ and $B$, let $t_{a}$, $t_{b}$ and $t_{x}$ be the tangents to the circle at $A$, $B$ and $X$ respectively. Let $Z$ be the point where line $AX$ meets $t_{b}$ and $Y$ the point where line $BX$ meets $t_{a}$. Show that the three li... | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | null | proof only | null | |
0i30 | Problem:
What is the area of a square inscribed in a semicircle of radius $1$, with one of its sides flush with the diameter of the semicircle? | [
"Solution:\n\nCall the center of the semicircle $O$, a point of contact of the square and the circular part of the semicircle $A$, the closer vertex of the square on the diameter $B$, and the side length of the square $x$. We know $OA = 1$, $AB = x$, $OB = \\frac{x}{2}$, and $\\angle ABO$ is right. By the Pythagore... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 4/5 | |
0aaw | Find the number by which the sum of the numbers $54863$ and $30608$ has to be decreased in order to obtain their difference? | [
"We solve the equation $(54863 + 30608) - x = 54863 - 30608$.\n\nIts solution is $x = 85471 - 24255 = 61216$.\n\nThe sum has to be decreased by $61216$."
] | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 61216 | |
0geb | 設 $n$ 為正整數。對於滿足 $\sum_{i=1}^{2n} a_i = \sum_{j=1}^{2n} b_j = n$ 的 $4n$ 個非負實數 $a_1, \dots, a_{2n}$ 及 $b_1, \dots, b_{2n}$, 定義兩集合
$$
A := \left\{ \sum_{j=1}^{2n} \frac{a_i b_j}{a_i b_j + 1} : i \in \{1, \dots, 2n\} \text{ 滿足 } \sum_{j=1}^{2n} \frac{a_i b_j}{a_i b_j + 1} \neq 0 \right\},
$$
$$
B := \left\{ \sum_{i=1}^{2n}... | [
"The maximum is $\\frac{n}{2}$. This is achieved when exactly half of $a_i$ and exactly half of $b_j$ are $1$, and the others are $0$.\n\nTo show that this is the maximum possible, WLOG assume that $a_1, \\dots, a_s$ and $b_1, \\dots, b_t$ are nonzero, and the rest are zero. Then we have $a_1 + \\cdots + a_s = b_1 ... | Taiwan | 2021 數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | n/2 | |
0jkc | Problem:
Let $f(n)$ and $g(n)$ be polynomials of degree $2014$ such that $f(n)+(-1)^{n} g(n)=2^{n}$ for $n=1,2, \ldots, 4030$. Find the coefficient of $x^{2014}$ in $g(x)$. | [
"Solution:\nAnswer: $\\frac{3^{2014}}{2^{2014} \\cdot 2014!}$\n\nDefine the polynomial functions $h_{1}$ and $h_{2}$ by $h_{1}(x)=f(2 x)+g(2 x)$ and $h_{2}(x)=f(2 x-1)-g(2 x-1)$. Then, the problem conditions tell us that $h_{1}(x)=2^{2 x}$ and $h_{2}(x)=2^{2 x-1}$ for $x=1,2, \\ldots, 2015$.\n\nBy the Lagrange inte... | United States | HMMT 2014 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 3^{2014} / (2^{2014} · 2014!) | |
0kou | There is a polynomial $P(x)$ with integer coefficients such that
$$
P(x) = \frac{(x^{2310} - 1)^6}{(x^{105} - 1)(x^{70} - 1)(x^{42} - 1)(x^{30} - 1)}
$$
holds for every $0 < x < 1$. Find the coefficient of $x^{2022}$ in $P(x)$. | [] | United States | AIME II | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | 220 | |
0d8d | Let $m, n, k$ and $l$ be positive integers with $n \neq 1$ such that $n^{k} + m n^{l} + 1$ divides $n^{k+l} - 1$. Prove that either $m = 1$ and $l = 2k$; or $l \mid k$ and $m = \frac{n^{k-l} - 1}{n^{l} - 1}$. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
042w | Let $z$ be a complex number. If $\frac{z-2}{z-i}$ is a real number ($i$ is the imaginary unit), then the minimum of $|z+3|$ is ______. | [
"Suppose $z = a + bi$ ($a, b \\in \\mathbb{R}$). By the given condition we can find\n$$\n\\begin{aligned} \\operatorname{Im} \\left( \\frac{z-2}{z-i} \\right) &= \\operatorname{Im} \\left( \\frac{(a-2)+bi}{a+(b-1)i} \\right) \\\\ &= \\frac{-(a-2)(b-1)+ab}{a^2+(b-1)^2} \\\\ &= \\frac{a+2b-2}{a^2+(b-1)^2} = 0, \\end{... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | sqrt(5) | |
06kw | Given that $\{a_n\}$ and $\{b_n\}$ are two sequences of integers defined by
$$
a_1 = 1,\ a_2 = 10,\ a_{n+1} = 2a_n + 3a_{n-1} \quad \text{for } n = 2, 3, 4, \dots,
$$
$$
b_1 = 1,\ b_2 = 8,\ b_{n+1} = 3b_n + 4b_{n-1} \quad \text{for } n = 2, 3, 4, \dots
$$
Prove that, besides the number ‘1’, no two numbers in the sequen... | [
"The two sequences are $a_n = 1, 10, 23, 76, \\dots$ and $b_n = 1, 8, 28, 116, \\dots$. Considering modulo $9$, we have\n$$\na_n \\equiv 1, 1, 5, 4, 5, 4, \\dots \\pmod{9},\n$$\n$$\nb_n \\equiv 1, 8, 1, 8, \\dots \\pmod{9}.\n$$\nSince each term of the two sequences only depends on the two previous terms, we can sho... | Hong Kong | CHKMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
04zc | In a regular $n$-gon, either $0$ or $1$ is written at each vertex. Using non-intersecting diagonals, Juku divides this polygon into triangles. Then he writes into each triangle the sum of the numbers at its vertices. Prove that Juku can choose the diagonals in such a way that the maximal and minimal number written into... | [
"If all numbers written at the vertices of the polygon are equal, then the claim holds trivially. Hence assume that there are both zeros and ones among the numbers at the vertices. We prove by induction that, for every convex polygon, the partition into triangles can be chosen in such a way that Juku writes either ... | Estonia | Estonija 2010 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0gdc | 令 $n \ge 1$ 為一整數。在 $n \times n$ 的表格中, 每個格子填入一個整數。假設下列兩條件成立:
(i) 方格上的整數, 除以 $n$ 的餘數都是 $1$。
(ii) 每一列的總和, 以及每一行的總和, 除以 $n^2$ 的餘數都是 $n$。
設 $R_i$ 為第 $i$ 列所有數字的乘積, 而 $C_j$ 為第 $j$ 行所有數字的乘積。
試證 $n^4$ 整除 $\sum_{i=1}^n R_i - \sum_{j=1}^n C_j$。 | [
"Let $A_{i,j}$ be the entry on row $i$ and column $j$. Let $P$ be the product of all $n^2$ entries. Denote $a_{i,j} = A_{i,j} - 1$ and $r_i = R_i - 1$.\nBy condition (i), the number $n$ divides $a_{i,j}$. So every product of two or more $a_{i,j}$ is divisible by $n^2$, hence\n$$\nR_i = \\prod_{j=1}^n (1 + a_{i,j}) ... | Taiwan | 二〇一九數學奧林匹亞競賽第三階段選訓營 | [
"Number Theory > Modular Arithmetic"
] | null | proof only | null |
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