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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0c4u | Find all the perfect squares $n = \overline{abcb}$, so that $a + c = b$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 70th NMO | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 3969 | |
0aq7 | Problem:
Sharon has a chandelier containing $n$ identical candles. She lights up the candles for $n$ consecutive Sundays in the following manner: the first Sunday, she lights up one candle for one hour; the second Sunday, she lights up two candles, conveniently chosen, for one hour; and continues in the same fashion, ... | [
"Solution:\n\nall positive odd integers\n\nSince the candles are identical and they are of equal lengths right after the $n$th Sunday, they are used equal number of times for all the $n$ Sundays. The total number of times they are used for the $n$ Sundays is\n$$\n1+2+3+\\cdots+n=\\frac{n(n+1)}{2}\n$$\nIt follows th... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | all positive odd integers | |
0iis | Problem:
What is the probability that two cards randomly selected (without replacement) from a standard 52-card deck are neither of the same value nor the same suit? | [
"Solution:\n\nAfter choosing a first card, the second needs to be in one of the other three suits and of a different value. Hence, the answer is $\\frac{3 \\cdot 12}{52-1} = \\frac{12}{17}$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 12/17 | |
0f64 | Problem:
$ABC$ and $A'B'C'$ are equilateral triangles and $ABC$ and $A'B'C'$ have the same sense (both clockwise or both counter-clockwise). Take an arbitrary point $O$ and points $P$, $Q$, $R$ so that $OP$ is equal and parallel to $AA'$, $OQ$ is equal and parallel to $BB'$, and $OR$ is equal and parallel to $CC'$. Sh... | [] | Soviet Union | 18th ASU | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0kr3 | Problem:
A group of 101 Dalmathians participate in an election, where they each vote independently on either candidate $A$ or $B$ with equal probability. If $X$ Dalmathians voted for the winning candidate, the expected value of $X^{2}$ can be expressed as $\frac{a}{b}$ for positive integers $a, b$ with $\operatorname{g... | [
"Solution:\nClaim: with 101 replaced with $2k+1$, the expectation of $X^{2}$ is\n$$\n\\frac{\\binom{2k}{k}}{2^{2k+1}}(2k+1)^{2}+\\frac{(2k+1)(2k+2)}{4}.\n$$\nThe answer is this value taken modulo 103, which can be calculated by noting that the integers modulo 103 form a finite field. Note that the multiplicative in... | United States | HMMT November 2022 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | final answer only | 51 | |
0hbh | There is a group of $2n$ people, among whom there are couples of friends. It is known that every person in this group has exactly $k \ge 1$ friends (if "A" is familiar with "B", then conversely, "B" is familiar with "A"). Find such a number $k$ for which this group can always be divided into two subgroups of $n$ people... | [
"For $k=1$ everything it is obvious. All friends are divided into pairs. In order to be able to split everyone properly, each pair must fall into one subgroup. But each subgroup must have the same number of people. Therefore, if there is an even number of such pairs, that is $n=2m$, it can be done by adding exactly... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | k ≥ 3 | |
0g4h | Problem:
The wizards Albus and Brian are playing a game on a square of side length $2n+1$ metres surrounded by lava. In the centre of the square there sits a toad. In a turn, a wizard chooses a direction parallel to a side of the square and enchants the toad. This will cause the toad to jump $d$ metres in the chosen d... | [
"Solution:\n\nBrian wins, irrespective of $n$. Suppose Brian plays with the following strategy: for every move Albus makes, Brian makes a move in the opposite direction. If Brian were to lose, this would mean the toad was at the edge of the board two jumps prior, before Albus's final move, as otherwise Brian's fina... | Switzerland | Swiss Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Brian wins for all n | |
0953 | Problem:
Să se arate că $8 \cdot \cos^{2} x \cdot (1-\operatorname{ctg} x) < 1$, $\forall x \in \left[\frac{\pi}{4}, \frac{\pi}{2}\right]$. | [
"Solution:\n\nFie $f(x) = \\cos x$ și $g(x) = \\sin x - \\cos x$, $\\forall x \\in \\left[\\frac{\\pi}{4}, \\frac{\\pi}{2}\\right]$. Avem $f(x) \\geq 0$ și $g(x) \\geq 0$, $\\forall x \\in \\left[\\frac{\\pi}{4}, \\frac{\\pi}{2}\\right]$. Aplicând inegalitatea mediilor, obținem $2 \\sqrt{f(x) g(x)} \\leq f(x) + g(x... | Moldova | A 61-a OLIMPIADA DE MATEMATICA A REPUBLICII MOLDOVA | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0g7g | 在銳角三角形 $ABC$ 中, $D, E, F$ 三點分別是由 $A, B, C$ 所引的高的垂足。
設 $I_1, I_2$ 分別為三角形 $AEF$ 及三角形 $BDF$ 的內心; $O_1, O_2$ 分別為三角形 $ACI_1$ 及三角形 $BCI_2$ 的外心。試證：直線 $I_1I_2$ 與直線 $O_1O_2$ 平行。 | [
"\n\n設 $\\angle CAB = \\alpha, \\angle ABC = \\beta, \\angle BCA = \\gamma$。我們先證 $A, B, I_1, I_2$ 四點共圓。由於 $AI_1$ 與 $BI_2$ 分別平分 $\\angle CAB$ 及 $\\angle ABC$, 它們的延長線會交在 $\\triangle ABC$ 的內心 $I$。$E, F$ 兩點位於以 $BC$ 為直徑的圓上, 故有 $\\angle AEF = \\angle ABC$ 以及 $\\angle AFE = \\angle ACB$。於是 $\\tria... | Taiwan | 二〇一三數學奧林匹亞競賽第二階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Coaxal... | null | proof only | null | |
05uo | Problem:
Montrer que pour tout entier $n$, le nombre $n^{3}-7 n$ est divisible par $6$. | [
"Solution:\n\nUne première approche est de regarder l'expression modulo $6$ :\n$$\nn^{3}-7 n \\equiv n^{3}-n \\equiv (n-1) n (n+1) \\quad \\bmod 6\n$$\ndonc il suffit de montrer que $6$ divise $(n-1) n (n+1)$. Or parmi trois nombres consécutifs, au moins un est divisible par $2$ et au moins un est divisible par $3$... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
046x | Given a positive integer $n$, find the smallest positive integer $k$ with the following property: After coloring any $k$ unit squares black and the remaining squares white in a $2n \times 2n$ grid, it is always possible to make all squares black using a finite number of the following two operations:
(Operation 1) Sele... | [
"*Proof.* The desired value of $k$ is $n^2 + n + 1$.\n\nFirst, we provide an example of coloring $n^2 + n$ cells black, such that it is impossible to perform operation 1. This implies that $k \\ge n^2 + n + 1$.\n\nColor the anti-diagonal cells black, and then color black all the cells $(i, j)$ in the top-left part ... | China | China-TST-2023B | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | n^2 + n + 1 | |
0hx5 | Problem:
Let $N$ be a positive integer. Define a sequence $a_{n}$, $n \geq 0$, by
$$
a_{0}=0, \quad a_{1}=1, \quad a_{n+1}+a_{n-1}=a_{n}\left(2-\frac{1}{N}\right) \quad(n \geq 1)
$$
Prove that $a_{n}<\sqrt{N+1}$ for all $n \geq 0$. | [
"Solution:\nLemma. For all $n \\geq 0$, $a_{n}$ and $a_{n+1}$ are related by a quadratic equation, namely\n$$\na_{n+1}^{2}+a_{n}^{2}-\\left(2-\\frac{1}{N}\\right) a_{n} a_{n+1}=1\n$$\nProof. By induction. The case $n=0$ is obvious. Suppose that for some $n \\geq 0$, (1) holds. Then\n$$\n\\begin{aligned}\na_{n+2}^{2... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
09vb | We consider sequences $a_1, a_2, \dots, a_n$ consisting of $n$ integers. For given $k \le n$, we can partition the numbers of the sequence into $k$ groups as follows: $a_1$ goes in the first group, $a_2$ in the second group, and so on until $a_k$ which goes in the $k$-th group. Then $a_{k+1}$ goes in the first group ag... | [
"a.\nAn example of a correct sequence is $5, 7, 6, 3, 1, 2$. This sequence consists of six distinct numbers and is 2-composite since $5 + 6 + 1 = 7 + 3 + 2$. It is also 3-composite since $5 + 3 = 7 + 1 = 6 + 2$.\n\nb.\nA possible solution is $8, 17, 26, 27, 19, 10, 1$. This sequence consists of seven distinct integ... | Netherlands | Second Round, March 2019 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a) 5, 7, 6, 3, 1, 2
b) 8, 17, 26, 27, 19, 10, 1
c) Largest k is 50; example sequence: 1, 2, ..., 48, 49, 100, 99, 98, ..., 52, 51 | |
03i3 | Problem:
A rectangular city is exactly $m$ blocks long and $n$ blocks wide (see diagram). A woman lives in the southwest corner of the city and works in the northeast corner. She walks to work each day but, on any given trip, she makes sure that her path does not include any intersection twice. Show that the number $f(... | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0hpj | Problem:
Let $k$ be a positive integer. Prove that there is a positive integer $N$ with the following properties:
a. $N$ has $k$ digits, none of which is $0$.
b. No matter how the digits of $N$ are rearranged, the resulting number is not divisible by $13$. | [
"Solution:\n\nLet $N_{1}$ be the number consisting of $k$ ones. If $N_{1}$ is not divisible by $13$, it is the number $N$ we seek, since no matter how its digits are rearranged, it is still the same number.\n\nIf $N_{1}$ is divisible by $13$, consider $N_{1} + 1$, that is to say the number consisting of $k-1$ ones ... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0jmo | Problem:
For any positive integers $a$ and $b$, define $a \oplus b$ to be the result when adding $a$ to $b$ in binary (base 2), neglecting any carry-overs. For example, $20 \oplus 14=10100_{2} \oplus 1110_{2}=11010_{2}=26$. (The operation $\oplus$ is called the exclusive or.) Compute the sum
$$
\sum_{k=0}^{2^{2014}-1}... | [
"Solution:\n\nAnswer: $\\quad 2^{2013}\\left(2^{2014}-1\\right)$ OR $2^{4027}-2^{2013}$\n\nLet $k=a_{2013} a_{2012} \\ldots a_{0}$ in base 2. Then $\\left\\lfloor\\frac{k}{2}\\right\\rfloor=\\overline{0 a_{2013} \\ldots a_{1}}$ in base 2. So the leftmost digit of $k \\oplus\\left\\lfloor\\frac{k}{2}\\right\\rfloor$... | United States | HMMT November 2014 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2^{2013}(2^{2014}-1) | |
003l | Sea $ABC$ un triángulo rectángulo e isósceles, con $AB = AC$. Consideramos los puntos $M$ y $N$ en $AB$ tales que $AM = BN$. Se traza desde $A$ la perpendicular a $CM$ que corta a $BC$ en $P$. Si $\angle APC = 62^\circ$, calcular la medida del ángulo $BNP$. | [] | Argentina | Argentina 2006 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Español | proof and answer | 73° | |
02kb | Problem:
Manoel testou sua pontaria lançando cinco flechas no alvo reticulado de quadrados de comprimento $1~\mathrm{cm}$, ilustrado na figura. Uma flecha que acerta dentro do círculo menor conta 300 pontos; na região sombreada conta 100 pontos, entre a região sombreada e o círculo maior conta 50 pontos e fora do círc... | [
"Solution:\n\na. Marcamos os pontos, conforme mostra a figura:\n\n\n\nb. No círculo menor temos apenas o ponto $A$. Portanto, Manoel acertou apenas uma vez neste círculo, o que lhe dá 300 pontos.\n\nc. Para calcular o total de pontos, observe que no ponto $B$ ele ganha 100 pontos, no $C$ ga... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | b: 1; c: 500 | |
0j3v | Problem:
What is the sum of all numbers between $0$ and $511$ inclusive that have an even number of $1$'s when written in binary? | [
"Solution:\nAnswer: $65408$\n\nCall a digit in the binary representation of a number a bit. We claim that for any given $i$ between $0$ and $8$, there are $128$ numbers with an even number of $1$'s that have a $1$ in the bit representing $2^{i}$. To prove this, we simply make that bit a $1$, then consider all possi... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 65408 | |
0i6a | Problem:
Determine the number of subsets $S$ of $\{1,2,3, \ldots, 10\}$ with the following property: there exist integers $a < b < c$ with $a \in S$, $b \notin S$, $c \in S$. | [
"Solution:\n\nThere are $2^{10} = 1024$ subsets of $\\{1,2, \\ldots, 10\\}$ altogether. Any subset without the specified property must be either the empty set or a block of consecutive integers. To specify a block of consecutive integers, we either have just one element (10 choices) or a pair of distinct endpoints ... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics"
] | null | proof and answer | 968 | |
0hfe | $AH_a$, $BH_b$ and $CH_c$ are the altitudes of the triangle $ABC$. Prove that if $\frac{H_b C}{AC} = \frac{H_c A}{AB}$, then the line, symmetric to the line $BC$ with respect to the line $H_b H_c$, is tangent to the circumscribed circle of $\triangle H_b H_c A$. | [] | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Concurrenc... | English | proof only | null | |
0dpa | In convex quadrilateral $ABCD$
$$
\angle ADB + \angle ACB = \angle CAB + \angle DBA = 30^\circ \text{ and } AD = BC.
$$
Prove that segments $DB$, $CA$ and $DC$ form the sides of the right triangle. | [] | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0jc0 | Problem:
Let rectangle $A B C D$ have lengths $A B = 20$ and $B C = 12$. Extend ray $B C$ to $Z$ such that $C Z = 18$. Let $E$ be the point in the interior of $A B C D$ such that the perpendicular distance from $E$ to $\overline{A B}$ is $6$ and the perpendicular distance from $E$ to $\overline{A D}$ is $6$. Let line ... | [
"Solution:\n\nAnswer: $72$\n\nDraw the line parallel to $\\overline{A D}$ through $E$, intersecting $\\overline{A B}$ at $F$ and $\\overline{C D}$ at $G$. It is clear that $X F E$ and $Y G E$ are congruent, so the area of $A X Y D$ is equal to that of $A F G D$. But $A F G D$ is simply a $12$ by $6$ rectangle, so t... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 72 | |
0k6j | Problem:
Let $\mathbb{R}$ be the set of real numbers. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that for all real numbers $x$ and $y$, we have
$$
f\left(x^{2}\right)+f\left(y^{2}\right)=f(x+y)^{2}-2 x y
$$
Let $S=\sum_{n=-2019}^{2019} f(n)$. Determine the number of possible values of $S$. | [
"Solution:\nLetting $y=-x$ gives\n$$\nf\\left(x^{2}\\right)+f\\left(x^{2}\\right)=f(0)^{2}+2 x^{2}\n$$\nfor all $x$. When $x=0$ the equation above gives $f(0)=0$ or $f(0)=2$.\nIf $f(0)=2$, then $f(x)=x+2$ for all nonegative $x$, so the LHS becomes $x^{2}+y^{2}+4$, and RHS becomes $x^{2}+y^{2}+4 x+4 y+4$ for all $x+... | United States | HMMT February 2019 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 2039191 | |
0b8q | Let $x$, $y$ be positive real numbers and $n$ be a positive integer. Prove that if $x^{2n+1} + y^{2n+1} \ge 2$ then also $x^{n+1} + y^{n+1} \ge x^n + y^n$. | [
"LEMMA. $D_{n,k+1} D_{n,k}^{-1} \\le D_{n,1}^2$.\n*Proof.* By simple calculations we get\n$$\nD_{n,k+1} D_{n,k}^{-1} = \\frac{x^{2n+2} + y^{2n+2} + xy(x^{n+k} y^{n-k} + x^{n-k} y^{n+k})}{x^{2n} + y^{2n} + x^{n+k} y^{n-k} + x^{n-k} y^{n+k}}\n$$\nand\n$$\nD_{n,1}^2 = \\frac{x^{2n+2} + y^{2n+2} + 2(xy)^{n+1}}{x^{2n} +... | Romania | Local Mathematical Competitions | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof only | null | |
0evb | A continent has a finite number of castles and each castle belongs to exactly one of the two countries $A$ and $B$. Each castle has one general. We say two castles are *neighboring* if there is a path between those two castles. We also say that a castle $P$ and a set of castles $Q$ is *neighboring* if $P$ is neighborin... | [
"The implication $(1) \\Rightarrow (2)$ is obvious. We prove the converse. Let $G = (V, E)$ be a graph where $V$ is the set of castles and each edge corresponds to a neighboring relation. We set $A = \\{v_1, v_2, \\dots, v_n\\}$. Now let $P_i \\subseteq V - A$ be the set of castles whose generals decide to attack c... | South Korea | Korean Mathematical Olympiad Final Round | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof only | null | |
09wc | Determine all pairs of integers $(x, y)$ such that $2xy$ is a perfect square and $x^2 + y^2$ is a prime number. | [
"We have $2xy = a^2$ for some nonnegative integer $a$, and $x^2 + y^2 = p$ for some prime number $p$.\nSince a prime number is never a perfect square, we see that $x, y \\neq 0$. Since $2xy$ is a perfect square, it follows that $x$ and $y$ must both be positive, or both be negative. If $(x, y)$ is a solution, then ... | Netherlands | Final Round | [
"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 | (1, 2), (2, 1), (-1, -2), (-2, -1) | |
0iaf | Problem:
For any subset $S \subseteq \{1,2, \ldots, 15\}$, a number $n$ is called an "anchor" for $S$ if $n$ and $n+|S|$ are both members of $S$, where $|S|$ denotes the number of members of $S$. Find the average number of anchors over all possible subsets $S \subseteq \{1,2, \ldots, 15\}$. | [
"Solution:\nWe first find the sum of the numbers of anchors of all subsets $S$; this is equivalent to finding, for each $n$, the number of sets for which $n$ is an anchor, and then summing over all $n$. Suppose that $n$ is an anchor for $S$, and $S$ has $k$ elements. Then $n, n+k \\in S \\Rightarrow k \\geq 2$, and... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 13/8 | |
04jk | Let $n$ be a positive integer. Determine all positive real numbers $x$ for which
$$
\frac{2^2}{x+1} + \frac{3^2}{x+2} + \dots + \frac{(n+1)^2}{x+n} + n x^2 = n x + \frac{n(n+3)}{2}
$$ | [
"Since\n$$\nnx = \\underbrace{x + \\cdots + x}_{n} \\quad \\text{and} \\quad \\frac{n(n+3)}{2} = (1 + \\cdots + n) + n,\n$$\nthe given equation is equivalent to\n$$\n\\sum_{k=1}^{n} \\frac{(k+1)^2}{x+k} + n x^2 - \\sum_{k=1}^{n} x - \\sum_{k=1}^{n} k - n = 0,\n$$\ni.e.\n$$\n\\sum_{k=1}^{n} \\left[ \\frac{(k+1)^2}{x... | Croatia | Croatia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | x = 1 | |
0f37 | Problem:
Let $S$ be the set $\{0, 1\}$. Given any subset of $S$ we may add its arithmetic mean to $S$ (provided it is not already included -- $S$ never includes duplicates). Show that by repeating this process we can include the number $1/5$ in $S$. Show that we can eventually include any rational number between $0$ a... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
05hr | Problem:
Soient $a_{1}, \ldots, a_{2019}$ des entiers positifs. Montrer qu'il y a équivalence entre:
(i) il existe un réel $x$ tel que pour tout $i \in \{1, \ldots, 2019\}$, on a : $a_{i}=\lfloor i x\rfloor$
(ii) pour tous $i, j \in \{1, \ldots, 2019\}$ vérifiant $i+j \leqslant 2019$, on a : $a_{i}+a_{j} \leqslant a... | [
"Solution:\n\nProuvons tout d'abord le sens direct : s'il existe $x$ réel tel que pour tout $i$ entre $1$ et $2019$, $a_{i}=\\lfloor i x\\rfloor$, et si on se donne $i, j$ entre $1$ et $2019$ tels que $i+j \\leqslant 2019$, alors $(i+j)x = i x + j x \\geqslant \\lfloor i x\\rfloor + \\lfloor j x\\rfloor = a_{i} + a... | France | ENVOI 2 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0916 | Problem:
Let $n \geqslant 2$ be an integer. There are $n$ positive integers written on a blackboard. In each step we choose two of the numbers on the blackboard and replace each of them by their sum. Determine all values of $n$ for which it is always possible to get $n$ identical integers in a finite number of steps. | [
"Solution:\n\nStarting from the $n$-tuple $(2,2,1,1, \\ldots, 1)$ with any $n \\geqslant 3$, we get always an $n$-tuple in which the number of maximal values is even. Hence no odd $n \\geqslant 3$ is as required.\n\nLet us show by induction that any even $n \\geqslant 2$ is satisfactory, which is obvious if $n=2$. ... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | all even n (n ≥ 2) | |
05qh | Problem:
On fixe un entier naturel $n \geqslant 2$. Déterminer tous les nombres réels $x \geqslant -1$ tels que pour tous nombres réels $a_{1}, \ldots, a_{n} \geqslant 1$ on ait
$$
\frac{a_{1}+x}{2} \times \cdots \times \frac{a_{n}+x}{2} \leqslant \frac{a_{1} a_{2} \cdots a_{n}+x}{2} .
$$ | [
"Solution:\n\nEn prenant $a_{i}=1$ pour tout $i$, on obtient la condition $\\left(\\frac{1+x}{2}\\right)^{n} \\leqslant \\frac{1+x}{2}$. Notons $y=\\frac{1+x}{2}$. Par hypothèse, $y \\geqslant 0$. Comme $y^{n} \\leqslant y$, on a $y^{n-1} \\leqslant 1$ donc $y \\leqslant 1$, ce qui implique $x \\leqslant 1$.\n\nRéc... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | [-1, 1] | |
0iwz | Problem:
How many functions $f:\{1,2,3,4,5\} \rightarrow \{1,2,3,4,5\}$ satisfy $f(f(x))=f(x)$ for all $x \in \{1,2,3,4,5\}$? | [
"Solution:\n\nA fixed point of a function $f$ is an element $a$ such that $f(a)=a$. The condition is equivalent to the property that $f$ maps every number to a fixed point. Counting by the number of fixed points of $f$, the total number of such functions is\n$$\n\\begin{aligned}\n\\sum_{k=1}^{5} \\binom{5}{k} k^{5-... | United States | $12^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 196 | |
0cnw | Given an $8 \times 8$ chessboard. Choose one of its diagonals and call the 8 cells of this diagonal *a fence*. The Rook starts from an arbitrary square outside the fence and makes some moves so that it does not get to one square twice, and it does not stand on the squares of the fence. Find the maximal possible number ... | [
"Разделим доску на четыре квадраты $4 \\times 4$. Заметим, что, если ладья прыгает через забор, то либо начальная, либо конечная клетка прыжка отмечена серым на рис. 22. Так как серых клеток 24 и через каждую может проходить максимум два прыжка, то всего может оказаться не более 48 прыжков.\n\nПри этом, если их ров... | Russia | Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English; Russian | proof and answer | 47 | |
0ckq | a) Let $a, b \in \mathbb{R}$ be two real numbers, with $a < b$, and $f : [a, b] \to \mathbb{R}$ a strict monotone function such that $\int_a^b f(x) dx = 0$. Show that $f(a) \cdot f(b) < 0$.
b) Determine the convergent sequences $(a_n)_{n \ge 1}$ of real numbers, for which there is a strict monotone function $f : \math... | [
"a) If $f([a, b]) \\subseteq [0, \\infty)$ or $f([a, b]) \\subseteq (-\\infty, 0]$, then $m = |f(\\frac{a+b}{2})| > 0$, and on one of the intervals $(a, \\frac{a+b}{2})$ or $(\\frac{a+b}{2}, b)$ the inequality $|f(x)| > m$ holds for any $x$ in that interval. Then\n$$\n\\begin{aligned}\n0 &= \\left| \\int_a^b f(x) \... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series"
] | English | proof and answer | null | |
0adx | Докажи дека, за секој непарен број $x$ изразот $x^3 + 3x^2 - x - 3$ е делив со 48. | [
"Изразот $x^3 + 3x^2 - x - 3$ можеме да го запишеме како\n$$\nx^3 + 3x^2 - x - 3 = x^2(x + 3) - (x + 3) = (x + 3)(x - 1)(x + 1).\n$$\nБидејќи $x$ е непарен, имаме $x = 2k - 1$, $k \\in N$. Тогаш\n$$\nx^3 + 3x^2 - x - 3 = (2k - 1 + 3)(2k - 1 - 1)(2k - 1 + 1) = 8(k - 1)k(k + 1).\n$$\nЈасно изразот е делив со 8, а $(k... | North Macedonia | Републички натпревар по математика за основно образование | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | Macedonian, English | proof only | null | |
09l8 | Prove that the sum of the squares of seven consecutive terms in an arithmetic progression of positive integers cannot be a perfect square. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
09ih | Let $ABC$ be a triangle and let $E$ denote the midpoint of $BC$. The circle passing through the points $A$ and $B$ and is tangent to the perpendicular bisector of $AC$ at point $D$ meets the side $AC$ again at point $K$ different from $A$. Let $F$ denote the midpoint of the segment $KC$. Prove that $DE$ is perpendicula... | [] | Mongolia | Mongolian Mathematical Olympiad Round 2 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0jxg | Problem:
Find all ordered triples $(a, b, c)$ of positive integers with $a^{2}+b^{2}=4c+3$. | [
"Solution:\nThere are no such ordered triples. Since the right side is odd, one of $a$ and $b$ is odd and the other even. But the square of any even number is a multiple of $4$, and the square of any odd number has a remainder of $1$ when divided by $4$. But the right side leaves a remainder of $3$ when divided by ... | United States | Berkeley Math Circle | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | There are no such ordered triples. | |
07uz | Prove that
$$
\prod_{k=1}^{n-1} \sin \frac{k\pi}{n} = \frac{n}{2^{n-1}}, \quad \text{for } n = 2, 3, \dots
$$ | [
"Note that for any complex number $\\zeta$ with $|\\zeta| = 1$ we have\n$$\n|1 - \\zeta|^2 = (1 - \\zeta)(1 - \\bar{\\zeta}) = 2 - (\\zeta + \\bar{\\zeta}) = 2(1 - \\operatorname{Re}(\\zeta)).\n$$\nIf $\\zeta = e^{2i\\theta} = \\cos(2\\theta) + i\\sin(2\\theta)$, we can use $1 - \\cos(2\\theta) = 2\\sin^2\\theta$ t... | Ireland | IRL_ABooklet | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof only | null | |
04jr | For all positive real numbers $x$, $y$ and $z$ prove the following inequality
$$
\frac{x^2}{xy+z} + \frac{y^2}{yz+x} + \frac{z^2}{zx+y} \ge \frac{(x+y+z)^3}{3[x^2(y+1) + y^2(z+1) + z^2(x+1)]}.
$$ | [
"By the CSB inequality we have\n$$\n\\left( \\frac{x^2}{xy+z} + \\frac{y^2}{yz+x} + \\frac{z^2}{zx+y} \\right) [x(xy+z) + y(yz+x) + z(zx+y)] \\geq (x\\sqrt{x} + y\\sqrt{y} + z\\sqrt{z})^2. \\quad (1)\n$$\n\n$$\n\\left( \\frac{x\\sqrt{x} + y\\sqrt{y} + z\\sqrt{z}}{3} \\right)^{\\frac{2}{3}} \\ge \\frac{x+y+z}{3},\n$... | Croatia | Croatian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0b50 | Problem:
Fixăm un număr întreg $n \geq 3$. Pentru fiecare submulţime nevidă a mulţimii $\{1,2, \ldots, n\}$, considerăm media aritmetică a elementelor sale. Fie $S$ mulţimea valorilor distincte ale acestor medii aritmetice. Determinaţi cea mai mica valoare absolută $|a-b|$, când $a$ şi $b$ parcurg mulţimea $S$ şi $a \... | [] | Romania | TESTUL 1 | [
"Discrete Mathematics > Other",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 1/((n-1)(n-2)) | |
04bv | Determine the number of quadratic functions with coefficients from the set
$$
S = \{1, 2, 4, 8, 16, 32\},
$$
that have no real roots. | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 135 | |
0fkp | Problem:
Sean $p$ y $q$ dos números primos positivos diferentes. Prueba que existen enteros positivos $a$ y $b$, tales que la media aritmética de todos los divisores positivos del número $n = p^{a} q^{b}$ es un número entero. | [
"Solution:\n\nLa suma de todos los divisores de $n$ viene dada por la fórmula\n$$\n\\left(1 + p + p^{2} + \\ldots + p^{a}\\right)\\left(1 + q + q^{2} + \\ldots + q^{b}\\right)\n$$\ncomo se puede comprobar desarrollando los paréntesis. El número $n$ tiene $(a+1)(b+1)$ divisores positivos y la media aritmética de tod... | Spain | XLIV Olimpiada Matemática Española | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
05q6 | Problem:
Soit $n \geq 3$ un entier et considérons $n$ droites en position générale (c'est-à-dire que trois droites ne sont jamais concourantes et deux droites jamais parallèles). Combien de triangles sont formés par ces droites?
N.B. Par exemple, dans la figure ci-dessous, il y a 4 triangles.
![](attached_image_1.pn... | [
"Solution:\n\nSi on numérote de 1 à $n$ les $n$ droites, on remarque que chaque triangle est codé par la donnée d'un sous-ensemble de $\\{1,2, \\ldots, n\\}$ à trois éléments. Il y a donc $\\binom{n}{3} = \\frac{n(n-1)(n-2)}{6}$ triangles."
] | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | C(n,3) = n(n-1)(n-2)/6 | |
03da | Let $P$ be a polynomial with real coefficients and such that for any positive integer $n$ the number $P(n)$ is an integer. There exist distinct prime integers $p_1, p_2, \dots, p_k$ such that for any positive integer $n$ the number $P(n)$ is divisible by at least one of $p_1, \dots, p_k$. Prove that there exists $i$ su... | [
"We prove first that $P$ has rational coefficients. Let $d$ be the degree of $P$. Consider the numbers $a_i = P(i)$ for $i = 1, 2, \\dots, d+1$. Let\n$$\nQ(x) = \\sum_{l=1}^{d+1} a_l \\prod_{i \\neq l, 1 \\le i \\le d+1} \\frac{x-i}{l-i}.\n$$\nThen the degree of $Q$ is at most $d$ and $Q(i) = a_i$ for $1 \\le i \\l... | Bulgaria | Bulgaria 2022 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
01fl | A *palindrome* is a word which is build using 2 letters and is equal to its reverse, for example: ABBA and ABABABABA are palindromes. Prove that any 2019-letter word (which uses 2 letters) is *build* by at most 808 palindromes. | [
"Note that any 5-letter word may be build with at most two palindromes (case-by-case checking). Any 2019-letter word can be divided into 404 5-letter words, hence it can be divided into at most 808 palindromes."
] | Baltic Way | Baltic Way 2019 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
073v | Prove that three distinct non-zero integers $a$, $b$, $c$ satisfy the equation
$$
\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = 3
$$
if and only if $a$, $b$, $c$ are given by
$$
a = kuv^2, \quad b = -ku^2(u+v), \quad c = kv(u+v)^2,
$$
(up to cyclic permutations) for some integers $u$, $v$, $k$. | [
"Let $a$, $b$, $c$ be a solution of the given equation. Putting\n$$\n\\frac{a}{b} = X, \\quad \\frac{b}{c} = Y,\n$$\nwe have $c/a = (1/XY)$. Note that $X$ and $Y$ are rational numbers. The equation takes the form\n$$\nX^2Y + XY^2 + 1 = 3XY. \\quad (1)\n$$\nThis is a cubic curve. We observe that $(1, 1)$ is a point ... | India | Indija TS 2008 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | All solutions are, up to cyclic permutation, a = k u v^2, b = −k u^2(u+v), c = k v (u+v)^2 for some integers u, v, k. | |
0j4c | Problem:
For all real numbers $x$, let
$$
f(x) = \frac{1}{\sqrt[2011]{1 - x^{2011}}}
$$
Evaluate $(f(f(\ldots(f(2011)) \ldots)))^{2011}$, where $f$ is applied 2010 times. | [
"Solution:\nDirect calculation shows that $f(f(x)) = \\frac{\\sqrt[2011]{1 - x^{2011}}}{-x}$ and $f(f(f(x))) = x$. Hence $(f(f(\\ldots(f(x)) \\ldots))) = x$, where $f$ is applied 2010 times. So $(f(f(\\ldots(f(2011)) \\ldots)))^{2011} = 2011^{2011}$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 2011^{2011} | |
0cl1 | Fix an integer $n \ge 3$. Determine the smallest positive integer $k$ satisfying the following condition:
For any tree $T$ with vertices $v_1, v_2, \dots, v_n$ and any pairwise distinct complex numbers $z_1, z_2, \dots, z_n$, there is a polynomial $P(X, Y)$ with complex coefficients of total degree at most $k$ such tha... | [
"*First solution.* First we provide a proof that $k \\ge n - 1$. Let $T$ be the path where $v_i$ and $v_{i+1}$ are adjacent for all $1 \\le i \\le n - 1$. Let $\\omega$ be a primitive root of unity of order $n$ and let $a_i = \\omega^i$ for all $1 \\le i \\le n$.\nIf $f(X) = P(X, \\omega X)$, then for all $1 \\le i... | Romania | Seventeenth ROMANIAN MASTER OF MATHEMATICS | [
"Discrete Mathematics > Graph Theory",
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathemat... | English | proof and answer | n - 1 | |
05yq | Problem:
Soit $n \geqslant 1$ un entier. Morgane dispose initialement de $n$ piles dont chacune contient une pièce. Elle s'autorise ensuite des opérations de la forme suivante : elle choisit deux piles, prélève autant de pièces de la première pile que de la deuxième, et forme une nouvelle pile avec les pièces qu'elle ... | [
"Solution:\n\nSi $n$ est une puissance de $2$, Morgane peut se débrouiller pour n'obtenir qu'une seule pile, en procédant comme suit : elle fusionne les $n$ piles de taille $1$ en $n / 2$ piles de taille $2$, puis fusionne ces $n / 2$ piles en $n / 4$ piles de taille $4$, et ainsi de suite.\n\nRéciproquement, lorsq... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | The minimal number is 1 if the starting number is a power of two, and 2 otherwise. | |
05qk | Problem:
Soit $n \geqslant 2$ un entier naturel. On se donne $2 n$ boules. Sur chacune de ces boules est écrit un nombre. On suppose que, à chaque fois que l'on regroupe les boules en $n$ paires, deux de ces paires ont la même somme.
a. Montrer que quatre de ces boules comportent le même nombre.
b. Montrer que le no... | [
"Solution:\n\na. On note $a_{1} \\geqslant a_{2} \\geqslant \\cdots \\geqslant a_{2 n}$ les valeurs prises par les $2 n$ boules dans l'ordre décroissant. On les apparie ainsi : $(a_{1}, a_{2}), (a_{3}, a_{4}), \\ldots$ Comme il existe $i<j$ impairs tels que $a_{i}+a_{i+1}=a_{j}+a_{j+1}$, et comme $a_{i} \\geqslant ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
03op | Find all positive integers $k$ such that, for any positive numbers $a$, $b$ and $c$ satisfying the inequality $k(ab + bc + ca) > 5(a^2 + b^2 + c^2)$, there must exist a triangle with $a$, $b$ and $c$ as the length of its three sides respectively. (posed by Qian Zhangwang) | [
"$$\na^2 + b^2 + c^2 \\geq ab + bc + ca,\n$$\nso $k > 5$. Hence $k \\ge 6$.\n\nrespectively, by the assumption in the problem, we have\n$$\nk(1 \\times 1 + 1 \\times 2 + 1 \\times 2) \\le 5(1^2 + 1^2 + 2^2),\n$$\nthat is, $k \\le 6$.\nWe will prove that $k = 6$ satisfies the requirement below. There is no harm in a... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | k = 6 | |
0f8y | Problem:
Two walkers are at the same altitude in a range of mountains. The path joining them is piecewise linear with all its vertices above the two walkers. Can they each walk along the path until they have changed places, so that at all times their altitudes are equal? | [] | Soviet Union | 23rd ASU | [
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Yes | |
0772 | Let $ABC$ be an acute-angled triangle with incentre $I$. Draw a line perpendicular to $BI$ at $I$ and let it intersect $BC$ and $BA$ at $D$ and $E$ respectively. Let $P$ and $Q$ be respectively the incentres of the triangles $BIA$ and $BIC$. Suppose the four points $D, E, P, Q$ are concyclic. Prove that $BA = BC$. | [
"Join $DE$ and $PQ$. Let $S$ be the centre of the circle $\\Gamma$ passing through $E, P, Q, D$. Let $P'$ denote the reflection of $P$ in $BI$. Since $\\angle PBI = \\angle QBI$, it follows that $P'$ lies on $BQ$. Since $E$ and $D$ are symmetric about the line $EI$, and $P$ and $P'$ are also symmetric about $BI$, i... | India | India_2017 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0j04 | Problem:
The rank of a rational number $q$ is the unique $k$ for which $q = \frac{1}{a_{1}} + \cdots + \frac{1}{a_{k}}$, where each $a_{i}$ is the smallest positive integer such that $q \geq \frac{1}{a_{1}} + \cdots + \frac{1}{a_{i}}$. Let $q$ be the largest rational number less than $\frac{1}{4}$ with rank $3$, and s... | [
"Solution:\n\nAnswer: $(5,21,421)$\n\nSuppose that $A$ and $B$ were rational numbers of rank $3$ less than $\\frac{1}{4}$, and let $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}$ be positive integers so that $A = \\frac{1}{a_{1}} + \\frac{1}{a_{2}} + \\frac{1}{a_{3}}$ and $B = \\frac{1}{b_{1}} + \\frac{1}{b_{2}} + \\fra... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | (5,21,421) | |
01xj | For a positive integer $n$ write down all its positive integer divisors in increasing order: $1 = d_1 < d_2 < \dots < d_k = n$.
Find all positive integers $n$ divisible by $2019$ such that $n = d_{19} \cdot d_{20}$. | [
"Answer: $n = 3 \\cdot 673^{18}$; $n = 3^{18} \\cdot 673$.\n\nNote that the equality $n = d_{19} \\cdot d_{20}$ implies that $n$ has exactly $38$ divisors. Indeed,\n$$\nn = d_{19} \\cdot d_{20} = d_{18} \\cdot d_{21} = d_{17} \\cdot d_{22} = \\dots = d_1 \\cdot d_{38} = d_{38}.\n$$\nHence either $n = p^\\alpha$, wh... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | n = 3*673^18 or n = 3^18*673 | |
0ikr | Problem:
What is the smallest positive integer $n$ such that $n^{2}$ and $(n+1)^{2}$ both contain the digit 7 but $(n+2)^{2}$ does not? | [
"Solution:\nThe last digit of a square is never 7. No two-digit squares begin with 7. There are no 3-digit squares beginning with the digits $17, 27, 37$, or $47$. In fact, the smallest square containing the digit 7 is $576 = 24^{2}$. Checking the next few numbers, we see that $25^{2} = 625$, $26^{2} = 676$, $27^{2... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | final answer only | 27 | |
0i57 | Problem:
Real numbers $a, b, c$ satisfy the equations $a+b+c=26,\ 1/a+1/b+1/c=28$. Find the value of
$$
\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a}{c}+\frac{c}{b}+\frac{b}{a} .
$$ | [
"Solution:\nMultiplying the two given equations gives\n\n$$\n\\frac{a}{a}+\\frac{a}{b}+\\frac{a}{c}+\\frac{b}{a}+\\frac{b}{b}+\\frac{b}{c}+\\frac{c}{a}+\\frac{c}{b}+\\frac{c}{c}=26 \\cdot 28=728,\n$$\n\nand subtracting $3$ from both sides gives the answer, $725$."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | 725 | |
07kr | Suppose that $x$, $y$ and $z$ are positive real numbers such that $xyz \ge 1$.
a. Prove that
$$
27 \le (1+x+y)^2 + (1+y+z)^2 + (1+z+x)^2,
$$
with equality if and only if $x = y = z = 1$.
b. Prove that
$$
(1+x+y)^2 + (1+y+z)^2 + (1+z+x)^2 \le 3(x+y+z)^2,
$$
with equality if and only if $x = y = z = 1$. | [
"**Solution to Part (a).** One can establish this in several ways. For instance, using the convexity of $t \\to t^2$ we have\n$$\n\\left(\\frac{3+2(x+y+z)}{3}\\right)^2 = \\left(\\frac{(1+x+y)+(1+y+z)+(1+z+x)}{3}\\right)^2 \n\\le \\frac{(1+x+y)^2+(1+y+z)^2+(1+z+x)^2}{3},\n$$\nwith equality iff\n$$\n1+x+y=1+y+z=1+z+... | Ireland | Irish Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
00rr | A positive integer $n$ is *downhill* if its decimal representation $\overline{a_k a_{k-1} \dots a_0}$ satisfies $a_k \ge a_{k-1} \ge \dots \ge a_0$. A real-coefficient polynomial $P$ is *integer-valued* if $P(n)$ is an integer for all integer $n$, and *downhill-integer-valued* if $P(n)$ is an integer for all downhill p... | [
"A downhill number can always be written as $a - b_1 - b_2 - \\dots - b_9$, where $a$ is of the form $\\overline{99\\dots99}$ and each $b_i$ either equals $0$ or is of the form $\\overline{11\\dots11}$.\nLet $n$ be a positive integer. The numbers of the form $\\overline{99\\dots99}$ yield at most $n$ different rema... | Balkan Mathematical Olympiad | BMO 2016 Short List Final | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | No | |
0e4c | Let $O$ be the circumcentre of the acute triangle $ABC$. Let $H$ denote the orthocentre and let $D$ be the foot of the altitude from $C$. The line perpendicular to $OD$ at $D$ intersects the segment $BC$ at $E$. The circumcircle of the triangle $BCH$ intersects the line $AB$ at $B$ and $F$. Prove that the points $E, F$... | [] | Slovenia | Selection Examinations for the IMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Simson ... | null | proof only | null | |
030z | Problem:
Fie $a$, $b$ și $c$ trei numere reale pozitive cu suma $3$. Arătați că:
$$
\frac{a b}{a b+a+b}+\frac{b c}{b c+b+c}+\frac{c a}{c a+c+a}+\frac{1}{9}\left(\frac{(a-b)^{2}}{a b+a+b}+\frac{(b-c)^{2}}{b c+b+c}+\frac{(c-a)^{2}}{c a+c+a}\right) \leqslant 1
$$ | [
"Solution:\nSoluţia 1. Avem\n$$\n\\frac{a b}{a b+a+b}+\\frac{(a-b)^{2}}{9(a b+a+b)}=\\frac{a^{2}+7 a b+b^{2}}{9(a b+a+b)}\n$$\nVom arăta că\n$$\n\\frac{a^{2}+7 a b+b^{2}}{a b+a+b} \\leqslant a+b+1\n$$\nAceastă inegalitate se scrie echivalent\n$$\na^{2}+7 a b+b^{2} \\leqslant a^{2}+3 a b+b^{2}+a^{2} b+a b^{2}+a+b \\... | Brazil | Al doilea baraj de selecție pentru OBMJ | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
07py | Show that the reciprocals of the altitudes of a triangle of area $\Delta$, and semi-perimeter $s$, are the side lengths of another triangle whose area is $1/4\Delta$, and whose perimeter is $s/\Delta$. | [
"Since the area of a triangle is \"half the base by the height\", then, in the usual notation, $2\\Delta = ah_a = bh_b = ch_c$. This can be rewritten as follows\n$$\n\\frac{1}{h_a} = \\frac{1}{2\\Delta}a, \\quad \\frac{1}{h_b} = \\frac{1}{2\\Delta}b, \\quad \\frac{1}{h_c} = \\frac{1}{2\\Delta}c,\n$$\nhence the tria... | Ireland | Ireland | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | The reciprocals of the altitudes form the sides of a triangle similar to the original with similarity factor 1/(2Δ). Its perimeter is s/Δ and its area is 1/(4Δ). | |
06ib | Let $f(x) = ax + b$ where $a$ and $b$ are integers. If $f(f(0)) = 0$ and $f(f(f(4))) = 9$, find the value of $f(f(f(1))) + f(f(f(2))) + \dots + f(f(f(2014)))$. | [
"From $0 = f(f(0)) = f(b) = ab + b = (a+1)b$, we get $a = -1$ or $b = 0$.\nIf $b = 0$, i.e. $f(x) = ax$, then we have $9 = f(f(f(4))) = f(f(4a)) = f(4a^2) = 4a^3$, which has no solution as $a$ is an integer.\n\nSo we must have $a = -1$. Then $f(x) = -x + b$, and hence $f(f(x)) = -(-x + b) + b = x$. It follows that ... | Hong Kong | Hong Kong Preliminary Selection Contest | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 2029105 | |
0k5y | Problem:
Do there exist five points in the plane, not all collinear, such that the distance between any pair is one of $\{1,2,3, \ldots, 9\}$? | [
"Solution:\n\nYes. Take the five points $(0,0)$, $(\\pm 3,0)$ and $(0, \\pm 4)$. The distances which appear are $3$, $4$, $5$, $6$, and $8$."
] | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | Yes; for example, the points (0,0), (3,0), (-3,0), (0,4), (0,-4) work. | |
0glb | Let $\mathbb{N}_0$ be the set of nonnegative integers. Find all functions $f : \mathbb{N}_0 \to \mathbb{N}_0$ satisfying the equation
$$
f^{f(m)}(n) = n + 2f(m)
$$
for all $m, n \in \mathbb{N}_0$ such that $m \le n$. | [
"Observe that since $f^0(n) = n$ by definition, $f(n) \\equiv 0$ is a solution. Now suppose that for some $c \\in \\mathbb{N}_0$, $f(c) \\ge 1$.\nAs $f(f^{f(c)-1}(n - 2f(c))) = n$ for all $n \\ge 2f(c)$, $f$ is onto for all $n \\ge 2f(c)$. Therefore $f^m(n) = n + 2m$ for all $n \\ge m \\ge 2f(c)$.\nSince we have $f... | Thailand | The first T3MO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | Either f is identically zero, or there exists an integer t ≥ 0 such that f(n) = 0 for n < t and f(n) = n + 2 for n ≥ t. | |
0i4n | Problem:
$p$ and $q$ are primes such that the numbers $p+q$ and $p+7q$ are both squares. Find the value of $p$. | [
"Solution:\n\nWriting $x^2 = p + q$, $y^2 = p + 7q$, we have $6q = y^2 - x^2 = (y - x)(y + x)$. Since $6q$ is even, one of the factors $y - x$, $y + x$ is even, and then the other is as well; thus $6q$ is divisible by $4 \\Rightarrow q$ is even $\\Rightarrow q = 2$ and $6q = 12$. We may assume $x, y$ are both taken... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 2 | |
0gex | 設 $E$, $F$ 分別為三角形 $ABC$ 的邊 $CA$, $AB$ 上兩點。令 $X$ 為三角形 $AEF$ 的外接圓和三角形 $ABC$ 的外接圓 $\Gamma$ 異於 $A$ 的交點，$K$ 為三角形 $ABE$ 的外接圓和三角形 $ACF$ 的外接圓異於 $A$ 的交點。設 $AK$ 和 $\Gamma$ 異於 $A$ 的交點為 $M$，$M$ 對於 $BC$ 的對稱點為 $N$。作 $XN$ 與 $\Gamma$ 異於 $X$ 的交點 $S$。
證明：$SM$ 平行於 $BC$。
Let $ABC$ be a triangle with circumcircle $\Gamma$, and points $E$ ... | [
"在 $\\odot(ABC)$ 上取一點 $S'$ 使得 $S'M$ 平行 $BC$，則 $NBS'C$ 為平行四邊形，因此 $S = S'$ 若且唯若 $XS'$ 平分 $\\overline{BC}$。注意到\n$$\n\\angle KBF = \\angle KFC, \\quad \\angle BFK = \\angle ECK \\Rightarrow \\triangle KBF \\sim \\triangle KEC,\n$$\n$$\n\\angle XBF = \\angle XCE, \\quad \\angle BFX = \\angle CEX \\Rightarrow \\triangle ... | Taiwan | 2021 數學奧林匹亞競賽第二階段選訓營, 獨立研究(二) | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscell... | null | proof only | null | |
0gge | 某國有 $n$ 座城市, 任兩座城市間都有唯一的一條道路, 且都被規定只能單向行駛。一條從城市 $X$ 到城市 $Y$ 的路徑為一系列的單向道路, 使得一個人可以從 $X$ 經由這些道路移動到 $Y$, 且中間不會重複拜訪相同的城市。一組路徑被稱為獨立的, 若且唯若這其中的任兩條路徑都沒有使用相同的道路。
對於一個城市 $X$, 令 $n_X$ 為由 $X$ 向外的單向道路總數。對於兩個相異城市 $X$ 與 $Y$, 令 $N_{XY}$ 為在所有的獨立路徑組中, 從 $X$ 到 $Y$ 路徑數的最大可能值。證明: $N_{XY} = N_{YX}$ 若且唯若 $n_X = n_Y$。 | [
"讓我們以 $X \\to Y$ 表示 $X$ 有單向道路通往 $Y$, 以 $X \\rightsquigarrow Y$ 表示 $X$ 到 $Y$ 的路徑。令\n$$\n\\mathcal{F}_X := \\{C : X \\to C\\} \\quad \\mathcal{F}_X := \\{C : X \\leftarrow C\\}\n$$\n為所有單向道路來自 (通往) $X$ 的城市所成集合; 注意到 $n_X = |\\mathcal{F}_X|$。\n\n我們稱一條路徑是短的, 若且唯若其道路數 $\\le 2$; 反之則稱為長路徑。我們先證明以下 Lemma:\n\n**Lemma:** 令 $\\m... | Taiwan | 2022 數學奧林匹亞競賽第一階段選訓營, 國際競賽實作(二) | [
"Discrete Mathematics > Graph Theory > Menger's theorem / max-flow, min-cut"
] | Chinese; English | proof only | null | |
0908 | Let $ABC$ be an isosceles triangle with $AB = AC = 5$. Let $D$ be a point on side $AB$ satisfying $AD = 3$, and let $E$ be a point on side $BC$ (excluding the endpoints $B$ and $C$). Let $\omega$ be the circle passing through $E$ and tangent to line $AB$ at $B$. Suppose that $\omega$ is tangent to the circumcircle of t... | [
"$\\frac{14\\sqrt{65}}{13}$\n\nBy the alternate segment theorem and the assumption $AB = AC$, we have $\\angle EFB = \\angle EBA = \\angle ACB$. Therefore, the four points $A$, $B$, $F$, $C$ are concyclic.\n\nLet $X$ be any point on the tangent to $\\omega$ at $E$, lying on the same side of line $AF$ as $B$. Then, ... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 14√65/13 | |
0bqh | Problem:
În triunghiul $ABC$ se consideră punctele $M \in (AB)$, $N \in (BC)$, $P \in (CA)$ astfel încât $AM = BN = CP$. Dacă $G_1, G_2, G_3$ sunt centrele de greutate ale triunghiurilor $AMP$, $BMN$, respectiv $CNP$, să se arate că triunghiurile $ABC$ şi $G_1G_2G_3$ au acelaşi centru de greutate dacă şi numai dacă tr... | [] | Romania | Olimpiada Națională de Matematică - Etapa Locală | [
"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 > Transformations > Rotation"
] | null | proof only | null | |
06u0 | Define $P(n) = n^{2} + n + 1$. For any positive integers $a$ and $b$, the set
$$
\{P(a), P(a+1), P(a+2), \ldots, P(a+b)\}
$$
is said to be fragrant if none of its elements is relatively prime to the product of the other elements. Determine the smallest size of a fragrant set. | [
"We have the following observations.\n(i) $(P(n), P(n+1)) = 1$ for any $n$.\nWe have $(P(n), P(n+1)) = \\left(n^{2} + n + 1, n^{2} + 3n + 3\\right) = \\left(n^{2} + n + 1, 2n + 2\\right)$. Noting that $n^{2} + n + 1$ is odd and $\\left(n^{2} + n + 1, n + 1\\right) = (1, n + 1) = 1$, the claim follows.\n\n(ii) $(P(n... | IMO | IMO 2016 Shortlisted Problems | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof and answer | 6 | |
0ezs | Problem:
(1) $A_1A_2A_3$ is a triangle. Points $B_1$, $B_2$, $B_3$ are chosen on $A_1A_2$, $A_2A_3$, $A_3A_1$ respectively and points $D_1$, $D_2$, $D_3$ on $A_3A_1$, $A_1A_2$, $A_2A_3$ respectively, so that if parallelograms $A_iB_iC_iD_i$ are formed, then the lines $A_iC_i$ concur. Show that $A_1B_1 : A_2B_2 : A_3B_... | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
04c9 | The lengths of all sides of a quadrilateral are integers, and each of them is a divisor of the sum of the other three lengths. Prove that at least two of the sides of that quadrilateral have equal lengths. | [
"On the contrary, let's assume that all sides are of different lengths; i.e. $a < b < c < d$ ($a$, $b$, $c$, $d$ are lengths of the sides, ordered by their length). Each of the lengths is a divisor of $S = a + b + c + d$, by assumption.\nAlso, it must be $a + b + c > d$ or equivalently $S > 2d$ and finally $\\frac{... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Divisibility / Factorization",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Geometry > Plane Geometry > Quadrilaterals"
] | English | proof only | null | |
0kfo | Problem:
Let $p > 5$ be a prime number. Show that there exists a prime number $q < p$ and a positive integer $n$ such that $p$ divides $n^{2} - q$. | [
"Solution:\n\nNote that the condition $p \\mid n^{2} - q$ just means that $q$ is a quadratic residue modulo $p$, or that the Legendre symbol $\\left(\\frac{q}{p}\\right)$ is $1$. We use these standard facts about the Legendre symbol:\n- If $p \\equiv \\pm 1 \\pmod{8}$, then $\\left(\\frac{2}{p}\\right) = 1$.\n- For... | United States | HMMT February 2020 | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity",
"Number Theory > Algebraic Number Theory > Quadratic forms",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
03hm | Problem:
For a positive number such as $3.27$, $3$ is referred to as the integral part of the number and $.27$ as the decimal part. Find a positive number such that its decimal part, its integral part, and the number itself form a geometric progression. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (1 + sqrt(5))/2 | |
01bi | There is a big crowd of boys and girls. Is it always possible to give them hats of $100$ colors (everybody gets one hat) such that if some boy is familiar with at least $2014$ girls then all these girls have hats of at least $2$ colors and the same for girls holds: if some girl is familiar with at least $2014$ boys the... | [
"**Answer:** No.\n\nLet $D = 100$, $p = 2014$ for clarity. We take a set $S_1$ consisting of $(p-1)D+1$ elements as the first part of graph $G$ (\"boys\"). As the second part $S_2$ of $G$ (\"girls\"), we take the set of all $p$-element samplings from $S_1$ and join every such sampling with all its elements in $S_1$... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | No | |
04e5 | A $6 \times 6$ table is given.
a) If any 9 fields of the table are labeled, prove that it is possible to select three rows and three columns that contain all of the labeled fields.
b) Label 10 fields of the table so that for any three rows and three columns that we select there is at least one labeled field that is con... | [
"a) We consider three different cases (note that these are the only possibilities):\n1. There are three rows that contain at least two labeled fields. In this case we select these three rows and we are left with at most 3 labeled fields, so we can select three columns that contain the remaining labeled fields.\n2. ... | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | null | |
0h09 | Solve the equation $\sin \frac{\pi\sqrt{x}}{4} + \cos \frac{\pi\sqrt{2-x}}{4} = \sqrt{2}$. | [
"We have $x \\in [0, 2]$, for such $x$ both functions $\\sin \\frac{\\pi\\sqrt{x}}{4}$ and $\\cos \\frac{\\pi\\sqrt{2-x}}{4}$ are increasing, thus, their sum is also increasing function. Hence, our equation has at most one real root. From the other hand, one can easily check that $x = 1$ is, indeed, the solution."
... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 24, 2010) | [
"Precalculus > Trigonometric functions",
"Precalculus > Functions"
] | English | proof and answer | x = 1 | |
0ii8 | Problem:
Four circles with radii $1, 2, 3$, and $r$ are externally tangent to one another. Compute $r$. (No proof is necessary.) | [
"Solution:\n\nLet $A, B, C, P$ be the centers of the circles with radii $1, 2, 3$, and $r$, respectively. Then, $ABC$ is a $3$-$4$-$5$ right triangle. Using the law of cosines in $\\triangle PAB$ yields\n$$\n\\cos \\angle PAB = \\frac{3^2 + (1 + r)^2 - (2 + r)^2}{2 \\cdot 3 \\cdot (1 + r)} = \\frac{3 - r}{3(1 + r)}... | United States | Harvard-MIT Mathematics Tournament, Team Round B | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | final answer only | 6/23 | |
0a7k | Problem:
Let $F$ be an increasing real function defined for all $x$, $0 \leq x \leq 1$, satisfying the conditions
(i) $F\left(\frac{x}{3}\right)=\frac{F(x)}{2}$,
(ii) $F(1-x)=1-F(x)$.
Determine $F\left(\frac{173}{1993}\right)$ and $F\left(\frac{1}{13}\right)$. | [
"Solution:\nCondition (i) implies $F(0)=\\frac{1}{2} F(0)$, so $F(0)=0$. Because of condition (ii), $F(1)=1-F(0)=1$. Also $F\\left(\\frac{1}{3}\\right)=\\frac{1}{2}$ and $F\\left(\\frac{2}{3}\\right)=1-F\\left(\\frac{1}{3}\\right)=\\frac{1}{2}$. Since $F$ is an increasing function, this is possible only if $F(x)=\\... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 7 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | F(173/1993) = 3/16; F(1/13) = 1/7 | |
069v | Determine all functions $f : (0, +\infty) \to \mathbb{R}$ satisfying
$$
(y^2 + 1)f(x) - y f(xy) = y f\left(\frac{x}{y}\right), \text{ for all } x, y > 0.
$$
(IMO 2018 shortlist, modified) | [] | Greece | SELECTION EXAMINATION | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | All solutions are f(x) = a x + b/x for real constants a, b. | |
07kn | Determine, with proof, all integers $x$ for which $x(x+1)(x+7)(x+8)$ is a perfect square. | [
"We would like to find all pairs of integers $(x, y)$ which satisfy\n$$\nx(x+1)(x+7)(x+8) = y^2.\n$$\nWith $z = x + 4$ this equation translates into the equivalent equations\n$$\n\\begin{align*}\n& (z-4)(z-3)(z+3)(z+4) = y^2 \\\\\n& (z^2 - 9)(z^2 - 16) = y^2 \\\\\n& z^4 - 25z^2 + 12^2 = y^2 \\quad \\text{now multip... | Ireland | Irish Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | [-9, -8, -7, -4, -1, 0, 1] | |
0bvu | Let $n$ be an integer greater than or equal to $3$, and let $\mathcal{P}_n$ be the collection of all planar (simple) $n$-gons no two distinct sides of which are parallel or lie along some line. For each member $P$ of $\mathcal{P}_n$, let $f_n(P)$ be the least cardinal a cover of $P$ by triangles formed by lines of supp... | [
"The required maximum is $n - 2$. This follows from the fact that the image of $f_n$ consists of the first $n - 2$ positive integers.\n\nInduct on $n$ to show that $f_n(P) \\le n - 2$ for all $P$ in $\\mathcal{P}_n$. The base case $n = 3$ is clear. If $P$ is convex, then $f_n(P) = 1$, since $P$ has three sides whos... | Romania | Fifteenth IMAR Mathematical Competition | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | n - 2 | |
0jc7 | Problem:
Consider the function
$$
f(x) = \frac{(x-2)(x+1)(2x-1)}{x(x-1)}
$$
Suppose that $u$ and $v$ are real numbers such that
$$
f(u) = f(v).
$$
Suppose that $u$ is rational. Prove that $v$ is rational. | [
"Solution:\n\nIt is not hard to see that (for all real $x$ except $0$ and $1$)\n$$\nf(1-x) = -f(x) \\quad \\text{and} \\quad f\\left(\\frac{1}{x}\\right) = -f(x)\n$$\nTherefore\n$$\nf(x) = f\\left(\\frac{1}{1-x}\\right) = f\\left(1 - \\frac{1}{x}\\right)\n$$\nIf $u \\in \\mathbb{Q}$ is given, then $v = u$, $v = 1/(... | United States | Berkeley Math Circle Monthly Contest | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
02hi | Problem:
1) Os valores positivos de $x$ para os quais $(x-1)(x-2)(x-3)<0$ formam o conjunto:
$(1,3)$ $(2,3)$
$(0,3)$
$(0,1) \cup (2,3)$
$(1,2)$ | [
"Solution:\n\n1. Para que um produto de três fatores seja negativo, devemos ter dois fatores positivos e um fator negativo, ou os três negativos.\n$(+)(+)(-) = -$ \\ $(+)(-)(+) = -$ \\ $(-)(+)(+) = -$ \\ $(-)(-)(-) = -$\n\nAs possibilidades são:\n\n1) $\\underbrace{(x-1)}_{+}(\\underbrace{x-2}_{-})(\\underbrace{x-3... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | (0,1) \cup (2,3) | |
0c6u | Let $A_1A_2\ldots A_n$ be a regular polygon. Find the number of subsets $\{A_i, A_j, A_k, A_l\}$, whose elements are the vertices of a trapezoid.
Cătălin Gherghe | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | If n is odd: n(n − 1)(n − 3)/8. If n is even: n(n − 2)(n − 3)/8. | |
0ju0 | Problem:
On the blackboard, Amy writes $2017$ in base-$a$ to get $133201_{a}$. Betsy notices she can erase a digit from Amy's number and change the base to base-$b$ such that the value of the number remains the same. Catherine then notices she can erase a digit from Betsy's number and change the base to base-$c$ such ... | [
"Solution:\n\n$2017 = 133201_{4} = 13201_{6} = 1201_{12}$"
] | United States | HMMT November | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 22 | |
08ie | Problem:
Let $x^{2} + b x + c = 0$ be the equation, where $b$ and $c$ are two consecutive triangular numbers and $c > b \geq 10$. Prove that this equation has two irrational solutions. (The number $m$ is triangular, if $m = n(n-1)/2$ for certain positive integer $n \geq 1$.) | [
"Solution:\n\nLet $b$ and $c$ be two consecutive triangular numbers with $c > b \\geq 10$.\n\nA triangular number is of the form $m = n(n-1)/2$ for some integer $n \\geq 1$.\n\nLet $b = n(n-1)/2$ and $c = (n+1)n/2$ for some $n \\geq 5$ (since $b \\geq 10$).\n\nThe quadratic equation is:\n$$\nx^2 + b x + c = 0\n$$\n... | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Other"
] | null | proof only | null | |
00rp | Find all the integer solutions $(x, y, z)$ of the equation
$$
(x + y + z)^5 = 80xyz(x^2 + y^2 + z^2).
$$ | [
"We directly check the identity\n$$\n(x + y + z)^5 - (-x + y + z)^5 - (x - y + z)^5 - (x + y - z)^5 = 80xyz(x^2 + y^2 + z^2).\n$$\nTherefore, if integers $x$, $y$ and $z$ satisfy the equation from the statement, we then have\n$$\n(-x + y + z)^5 + (x - y + z)^5 + (x + y - z)^5 = 0.\n$$\nBy Fermat's theorem at least ... | Balkan Mathematical Olympiad | BMO 2016 Short List Final | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | {(0, t, -t), (t, 0, -t), (t, -t, 0) for all integers t} | |
0e8f | We want to cover a table of size $4 \times 4$ with dominoes of the following shape
(can also be reflected or rotated),
where the dominoes may overlap or extend over the edges of the table.
At least how many dominoes do we need? | [
"We need at least 5 dominoes. If it is possible to cover the table with four dominoes, then they cannot overlap and cannot extend over the edges of the table since four dominoes cover exactly 16 squares. In this case there are only two possible ways of covering the top left corner as shown in the figures. But in bo... | Slovenia | National Math Olympiad 2013 - Final Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 5 | |
0lbj | Có 42 thí sinh tham gia Kì thi chọn đội tuyển Olympic Toán Quốc tế. Biết rằng mỗi thí sinh quen đúng 20 thí sinh khác. Chứng minh rằng có thể chia các thí sinh hoặc thành 2 nhóm hoặc thành 21 nhóm, sao cho trong mỗi cách chia số người của các nhóm bằng nhau và hai người bất kì trong cùng nhóm thì quen nhau. | [] | Vietnam | Kì thi chọn học sinh vào Đội tuyển Quốc gia Dự thi IMO | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | Vietnamese | proof only | null | |
0fjw | Problem:
Probar que para todo entero positivo $n$, la expresión decimal de
$$
\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}
$$
es periódica mixta. | [
"Solution:\nTenemos\n$$\n\\frac{1}{n}+\\frac{1}{n+1}+\\frac{1}{n+2}=\\frac{3 n^{2}+6 n+2}{n(n+1)(n+2)}\n$$\nSabemos que para que una fracción origine un decimal periódico mixto, una vez reducida debe tener en el denominador algún factor primo del conjunto 2,5 y alguno que no sea ni 2 ni 5.\n\nVeamos primero que la ... | Spain | Spanish National Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
029e | Problem:
Artur e Dinah vão disputar o jogo do par ou ímpar maluco. Dinah escolhe "par" e Artur escolhe "ímpar". Em seguida, cada um escreve um número inteiro positivo em uma folha de papel sem que o outro a veja. Emílio recolhe as duas folhas, multiplica os números e declara Dinah vencedora se o resultado for par e Ar... | [
"Solution:\n\na) Temos que o produto de número par por um número ímpar é sempre par! Portanto, se Dinah pedir par e escrever no papel um número par, ela certamente ganhará.\n\nb) Dinah escolhe um número natural, digamos $3q_1 + r_1$, onde $q_1$ e $r_1$ são naturais e $r_1$ é o resto na divisão desse número por três... | Brazil | null | [
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
09hz | Let $ABCD$ be a trapezoid with obtuse angle at vertex $D$. The diagonals $AC$ and $BD$ meet at $O$ and the line through $O$ parallel to $AB$ meets the circumcircle of $BCO$ again at $P$. The circumcircles of $ADO$ and $BCO$ intersect again at $Q$. Prove that the line $PQ$ bisects the segment $BC$. | [
"\nSince $OP \\parallel AB$ and $P$, $C$, $O$, $Q$, $B$ are cyclic, we have $\\angle POC = \\angle PQC = \\angle PBC = \\alpha$ and $\\angle POB = \\angle PCB = \\angle PQB$. It follows that $\\triangle OAB \\sim \\triangle BPC$. Hence\n$$\n\\frac{BP}{PC} = \\frac{OA}{OB}.\n$$\nBecause $ABC... | Mongolia | Round 3 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity"
] | null | proof only | null | |
0gtv | Let $ABC$ be an acute angled triangle and $K, L$ be points on $AC, BC$ respectively such that $\angle AKB = \angle ALB$. Let $P$ be the intersection of $AL, BK$ and $Q$ be the midpoint of segment $KL$. Let $T, S$ be the intersection $AL, BK$ with the circumcircle of $ABC$, respectively. Prove that $TK, SL, PQ$ are conc... | [
"**3. Answer:** All constant functions.\nLet $a < b$ be two arbitrary numbers. Consider a sufficiently large $x$ so that $x > -a$ and $x > b - f(-a)$, thus $-x < a$ and $x + f(x) > b - f(-a) + f(-a) = b$ (here we used $f(x) \\ge f(-a)$ since $x > -a$). Now $-x < a < b < x + f(x)$ while $f(-x) = f(x+f(x))$, hence $f... | Turkey | Team Selection Test for JBMO 2023 | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
01ry | Given a cyclic quadrilateral $ABCD$ with $AB = AD$. Points $M$ and $N$ are marked on the sides $CD$ and $BC$, respectively, so that $DM + BN = MN$.
Prove that the circumcenter of the triangle $AMN$ belongs to the segment $AC$. | [
"On the prolongation of the segment $CD$ over $D$ we mark the point $K$ such that $DK = BN$. Then the triangles $KDA$ and $NBA$ are equal since $\\angle KDA = \\angle NBA$, $KD = BN$, $DA = AB$. Hence $KA = NA$, $KM = DK + DM = BN + DM = NM$, so that the triangles $KMA$ and $NMA$ are equal. It easily follows that $... | Belarus | SELECTION and TRAINING SESSION | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | English | proof only | null | |
01cp | A family wears three colours of clothing: red, blue and green, with a separate laundry bin for each colour. Each week, the family generates a total of $K$ kilogrammes of laundry (the proportion of each colour is subject to variation). The laundry is first sorted by colour and disposed of in the bins. Next, the heaviest... | [
"Answer: $\\frac{5}{2}K$.\nEach week, the accumulation of laundry increases the total amount by $K$, after which the washing decreases it by at least one third, because, by the pigeon-hole principle, the bin with the most laundry must contain at least a third of the total. Hence the amount of laundry post-wash afte... | Baltic Way | Baltic Way 2015 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 5/2 K |
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