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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
06g2 | How many odd coefficients are there in the expansion of $(x^2 - x + 1)^{2009}$? | [
"In the following, we consider polynomials in $\\mathbb{F}_2[x]$. This means all coefficients are taken modulo $2$. Let $f(n)$ be the number of odd coefficients in $(x^2 - x + 1)^n$.\n\n**Claim 1.** We have $(x^2 - x + 1)^{2k} = x^{2k+1} - x^{2k} + 1$ for any nonnegative integer $k$.\n*Proof.* It suffices to note t... | Hong Kong | IMO HK TST | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 645 | |
0hgw | $A$ is one of the points of intersection of circles $\omega_1$ and $\omega_2$, $T_1T_2$ is the external common tangent to these circles, which tangents $\omega_1$ at $T_1$ and tangents $\omega_2$ at $T_2$. $B_1$ is any point on the circle $\omega_1$, $B_2$ is any point on the circle $\omega_2$, such that $A$, $B_1$ and... | [
"Let $X$ be the second intersection point of $\\omega$ and $(AT_1T_2)$. Observe that $\\angle (AX, XC_1) = \\angle (AB_1, B_1C_1) = \\angle (AB_1, T_1B_1) = \\angle (AT_1, T_1T_2) = \\angle (AX, XT_2)$ and then $X \\in C_1T_2$. Similarly, $X \\in C_2T_1$ and hence $C_1T_2$ and $C_2T_1$ intersect on $\\omega$."
] | Ukraine | Problems from Ukrainian Authors | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
09kc | Let $Q_n(x) = 1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}$ for $n \ge 0$ integer and $x$ real.
Prove that we have $\frac{Q_n(b) - Q_n(a)}{b-a} \ge Q_{n-1}\left(\frac{a+b}{2}\right)$ for any $n \ge 1$ and $b > a > 0$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
0hw6 | Problem:
Let $ABCDEF$ be a hexagon circumscribing a circle $\omega$. The sides $AB$, $BC$, $CD$, $DE$, $EF$, $FA$ touch $\omega$ at $U$, $V$, $W$, $X$, $Y$, and $Z$ respectively; moreover, $U$, $W$, and $Y$ are the midpoints of sides $AB$, $CD$, and $EF$, respectively. Prove that $UX$, $VY$, and $WZ$ are concurrent. | [
"Solution:\n\nSince $U$ is the midpoint of $AB$, we have $ZA = AU = UB = BV$ and so (letting $O$ be the center of $\\omega$) $\\triangle OZA \\cong \\triangle OUA \\cong \\triangle OUB \\cong \\triangle OVB$. Thus arcs $ZU$ and $UV$ are equal, and so $UX$ is the bisector of $VXZ$. Similarly, $VY$ and $WZ$ are the b... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
06ss | Let $z_{0} < z_{1} < z_{2} < \cdots$ be an infinite sequence of positive integers. Prove that there exists a unique integer $n \geqslant 1$ such that
$$
z_{n} < \frac{z_{0} + z_{1} + \cdots + z_{n}}{n} \leqslant z_{n+1} .
$$ | [
"For $n = 1, 2, \\ldots$ define\n$$\nd_{n} = \\left(z_{0} + z_{1} + \\cdots + z_{n}\\right) - n z_{n}\n$$\nThe sign of $d_{n}$ indicates whether the first inequality in (1) holds; i.e., it is satisfied if and only if $d_{n} > 0$.\nNotice that\n$$\nn z_{n+1} - \\left(z_{0} + z_{1} + \\cdots + z_{n}\\right) = (n+1) z... | IMO | 55th International Mathematical Olympiad Shortlist | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
05ux | Problem:
Trouver tous les entiers $n \geqslant 1$ tels que
$$
6^{n}-1 \mid 7^{n}-1
$$ | [
"Solution:\nSoit $n$ un éventuel entier vérifiant $6^{n}-1 \\mid 7^{n}-1$.\nOn a $5=6-1 \\mid 6^{n}-1^{n}$, donc $5 \\mid 7^{n}-1$. On calcule donc les puissances de $7$ modulo $5$ :\n\n| | $n$ | 0 | 1 | 2 | 3 | 4 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $7^{n}(\\bmod 5)$ | 1 | 2 | 4 | 3 | 1... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | no such positive integer n | |
090o | Let $a_1, a_2, \dots$ and $b_1, b_2, \dots$ be two sequences consisting of positive integers such that, for any positive integer $n$,
$$
(a_{n+1}, b_{n+1}) = \left( \frac{a_n}{2}, b_n + \frac{a_n}{2} \right) \text{ or } (a_{n+1}, b_{n+1}) = \left( a_n + \frac{b_n}{2}, \frac{b_n}{2} \right)
$$
holds. How many initial pa... | [
"\\boxed{1064}\n$$\n\\text{For a positive integer } k, \\text{ denote the maximum nonnegative integer } i \\text{ such that } 2^i \\text{ divides } k \\text{ by } v_2(k). \\text{ Note that } v_2(kl) = v_2(k) + v_2(l) \\text{ holds for any positive integers } k \\text{ and } l, \\text{ and that } v_2\\left(\\frac{k}... | Japan | The 35th Japanese Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | 1064 | |
0bxj | If $a, b, c \in [-1, 1]$ satisfy $a + b + c + abc = 0$, prove that
$$
a^2 + b^2 + c^2 \ge 3(a + b + c).
$$
When does the equality hold? | [
"*First solution.* If $a + b + c \\le 0$, the inequality is obviously satisfied, with equality occurring if and only if $a = b = c = 0$. If $a + b + c > 0$, then $abc < 0$. It is not possible for all the variables to be negative (their sum would be negative), therefore one of them is negative and the other two are ... | Romania | THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | Equality holds exactly for (a, b, c) = (0, 0, 0) and for permutations of (−1, 1, 1). | |
036l | Problem:
Find the maximal cardinality of a set of phone numbers satisfying the following three conditions:
a) all of them are five-digit numbers (the first digit can be $0$);
b) each phone number contains at most two different digits;
c) the deletion of an arbitrary digit in two arbitrary phone numbers (possibly in... | [] | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 212 | |
0gia | Let $f(x)$ be a polynomial with real coefficients such that the inequality
$$
x + 1 \le f(x) \le 3x^2 - 5x + 4
$$
holds for all real numbers $x$. Find all possible values for $f(11)$.
設 $f(x)$ 為實係數多項式，滿足：
$$
x + 1 \le f(x) \le 3x^2 - 5x + 4.
$$
對所有實數 $x$ 均成立。試求 $f(11)$ 的所有可能值。 | [
"$f(11) \\in [12, 312]$\n\n易知 $f(x)$ 至多為二次多項式。注意到，$y = x + 1$ 為二次函數，$y = 3x^2 - 5x + 4$ 在點 $(1, 2)$ 處的切線。由於 $3x^2 - 5x + 4 = 3(x - 1)^2 + x + 1$，所以存在實數 $a \\in [0, 3]$ 使得\n$$\nf(x) = a(x - 1)^2 + x + 1.\n$$\n故 $f(11)$ 的所有可能值為 $[11 + 1, 3(11 - 1)^2 + 11 + 1] = [12, 312]$。"
] | Taiwan | APMO Taiwan Preliminary Round 2 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | Chinese; English | proof and answer | [12, 312] | |
0dxo | Problem:
V trimestnem številu so stotice večje od desetic in desetice večje od enic. Če števke tega trimestnega števila zapišemo v obratnem vrstem redu in dobljeno število prištejemo prvotnemu, dobimo število, ki vsebuje samo lihe števke. Določi vsa trimestna števila, za katera to velja. | [
"Solution:\n\nOznačimo trimestno število z $\\overline{a b c}$. Veljati mora $a > b > c$, poleg tega pa je število $\\overline{a b c} + \\overline{c b a} = 10^{2}(a + c) + 10(2b) + (a + c)$ sestavljeno iz samih lihih števk. Če je $a + c < 10$, je števka na mestu desetic soda, kar ni možno. Zato mora biti $a + c \\g... | Slovenia | 51. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 843, 932, 942 | |
0jav | Problem:
You have a twig of length $1$. You repeatedly do the following: select two points on the twig independently and uniformly at random, make cuts on these two points, and keep only the largest piece. After $2012$ repetitions, what is the expected length of the remaining piece? | [
"Solution:\n\nThe answer is $\\left(\\dfrac{11}{18}\\right)^{2012}$.\n\nFirst, let $p(x)$ be the probability density of $x$ being the longest length.\nLet $a_n$ be the expected length after $n$ cuts. Then\n$$\na_n = \\int_0^1 p(x) \\cdot (x a_{n-1}) \\, dx = a_{n-1} \\int_0^1 x p(x) \\, dx = a_1 a_{n-1}.\n$$\nIt fo... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other",
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | final answer only | (11/18)^{2012} | |
07hh | Given $1400$ real numbers, prove that among them there are at least three numbers $x$, $y$, $z$ such that
$$
\left| \frac{(x-y)(y-z)(z-x)}{1+x^4+y^4+z^4} \right| < \frac{9}{1000} .
$$ | [
"Putting $C = \\frac{9}{1000}$, $n = 1400$. Assume to the contrary that for all $z \\ge y \\ge x$ we have\n$$\n(z - y)(y - x)(z - x) \\ge C(1 + x^4 + y^4 + z^4).\n$$\nNotice that $4(y-x)(z-y) \\le (y-x+z-y)^2 = (z-x)^2$. Whence,\n$$(z-x)^3 \\ge 4(z-y)(y-x)(z-x) \\ge 4C(1+x^4+y^4+z^4) \\ge 4C(1+x^4+z^4) \\ge 4C.$$ \... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0bwc | Show that the equation $x^2 + y^2 - z^2 + xy = 0$ has infinitely many positive integer solutions. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0aic | Find all $n \in \mathbb{N}$ divisible by $11$, such that all numbers that can be obtained from $n$ by an arbitrary rearrangement of its digits are again divisible by $11$.
Најди ги сите $n \in \mathbb{N}$ деливи со $11$, такви што сите броеви кои се добиваат со произволна прераспределба на цифрите на бројот $n$ повтор... | [
"From the condition $11|n$, the number $n$ must have at least two digits. Let $n = \\overline{a_k a_{k-1} \\dots a_0}$ where $a_i$, $0 \\le i \\le k$ are digits and $a_k \\ne 0$. From the former discussion we have $k \\ge 1$.\n\nWe will show that all digits in the number $n$ are equal. Namely, from the condition of... | North Macedonia | Macedonian Junior Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | Exactly the numbers whose decimal digits are all equal and whose length is even; equivalently n equals a times a repunit of even length with a from one to nine. | |
0g8f | 給定正整數 $k$, 試求所有整係數多項式 $f(x)$, 使得對於所有正整數 $n$ 都有 $f(n)$ 整除 $(n!)^k$, 此處 $n! = 1 \cdot 2 \cdots n$.
Given a positive integer $k$, find all polynomials $f(x)$ of integral coefficients such that $f(n)$ divides $(n!)^k$ for all positive integers $n$, here $n! = 1 \cdot 2 \cdots n$. | [
"引理：$p$ 為一質數，若 $p|f(n)$，則 $p|n$.\n\n證明：假設 $p$ 不整除 $n$，則可設 $n = kp + q$，其中 $k$ 為正整數且 $0 < q < p$.\n\n由於 $f(x)$ 是整係數多項式，因此我們有\n$$\nkp|f(n) - f(n - kp) \\Rightarrow p|f(n) - f(n - kp)\n$$\n又 $p|f(n)$ 且 $n = kp + q$，故有\n$$\np|f(n) - f(n - kp) \\Rightarrow p|f(q)\n$$\n由題設知 $f(q)|(q!)^k$，因此有 $p|(q!)^k$.\n\n但因為 $0 < p < q... | Taiwan | 二〇一四數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | f(x) = x^r for some integer r with 0 ≤ r ≤ k | |
0gxc | A line $l_1$ intersects the parabola $y = ax^2 + bx + c$ ($a \neq 0$) at the two points $A$ and $B$. A line $l_2$ is parallel to the line $l_1$ and tangent to this parabola at the point $C$. Prove that the arithmetic mean of abscissas of points $A$ and $B$ equals the abscissa of point $C$. | [
"Let the equation of the line $l_1$ be $y = kx + d$. Then abscissas of points $A$ and $B$ are defined using the equality $ax^2 + bx + c = kx + d$. Therefore these abscissas $x_1, x_2$ satisfy that quadratic equation, then according to Vieta's formula we have equality $x_1 + x_2 = \\frac{-b + k}{a}$.\n\nThe abscissa... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof only | null | |
08z7 | In isosceles triangle $ABC$ with $AB = AC$, point $O$ is in its interior (not including circumference) and the circle $\omega$ centered $O$ and passing through $C$ intersects the sides (excluding end points) $BC$ and $AC$ at $D$ and $E$, respectively. Let $\Gamma$ be the circumcircle of triangle $AEO$ and intersect wit... | [
"Let $\\angle YXZ$ denote the directed angle between lines $XY$ and $XZ$, measured modulo $180^\\circ$.\n\nSince $OE = OF$, we have $\\angle FAO = \\angle OAE = \\angle OAC$. Hence\n$$\n\\begin{aligned}\n\\angle AOF &= 180^\\circ - \\angle FAO - \\angle OFA \\\\\n&= 180^\\circ - \\angle OAC - \\angle OEC \\\\\n&= 1... | Japan | Japan 2022 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03pw | Suppose $x, y \in (-2, 2)$ and $xy = -1$. Then the minimum value of $u = \frac{4}{4-x^2} + \frac{9}{9-y^2}$ is ( ).
(A) $\frac{8}{5}$
(B) $\frac{24}{11}$
(C) $\frac{12}{7}$
(D) $\frac{12}{5}$ | [
"**Solution I** We have\n$$\n\\begin{aligned}\nu &= \\frac{4}{4-x^2} + \\frac{9x^2}{9x^2-1} = 1 + \\frac{35x^2}{-9x^4 + 37x^2 - 4} \\\\\n&= 1 + \\frac{35}{37 - \\left( \\left( 3x - \\frac{2}{x} \\right)^2 + 12 \\right)}\n\\end{aligned}\n$$\nSince $x \\in (-2, -\\frac{1}{2}) \\cup (\\frac{1}{2}, 2)$, so $u$ reaches ... | China | China Mathematical Competition (Shaanxi) | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | MCQ | D | |
05qt | Problem:
Soit $n$ un entier strictement positif. Montrer qu'il existe $n$ entiers 2 à 2 distincts $r_{1}, \ldots, r_{n}$ tels que chaque $r_{i}$ divise $r_{1}+\cdots+r_{n}$. | [
"Solution:\n\nLa solution s'inspire des fractions égyptiennes. Une fraction égyptienne est un $n$-uplet d'entiers distincts $a_{1}, \\ldots, a_{n}$ vérifiant $1 / a_{1} + \\cdots + 1 / a_{n} = 1$, par exemple $1 / 2 + 1 / 3 + 1 / 6 = 1$.\n\nAlors en posant $r_{i} := \\prod_{j \\neq i} a_{j} = \\frac{a_{1} \\cdots a... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0ib7 | Problem:
Find all numbers $n$ with the following property: there is exactly one set of 8 different positive integers whose sum is $n$. | [
"Solution:\n36, 37\n\nThe sum of 8 different positive integers is at least $1+2+3+\\cdots+8=36$, so we must have $n \\geq 36$. Now $n=36$ satisfies the desired property, since in this case we must have equality - the eight numbers must be $1, \\ldots, 8$.\n\nAnd if $n=37$ the eight numbers must be $1,2, \\ldots, 7,... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 36, 37 | |
0eaa | Let $\mathcal{K}$ be the circumcircle of the acute triangle $ABC$ with $|AB| < |AC|$. Let $p$ be the reflection of the line $BC$ over the line $AB$. The line $p$ intersects the circle $\mathcal{K}$ at $B$ and $E$. The tangent to $\mathcal{K}$ at $A$ intersects the line $p$ at $D$. Let $F$ be the reflection of the point... | [
"By the tangent-chord angle theorem we have $\\angle CAF = \\angle CBA$. Since $p$ is the reflection of the line $BC$ over the line $AB$, we have $\\angle CBA = \\angle ABD$. The points $A, B, E, C$ are concyclic, so $\\angle ABD = \\angle ACE$. Hence, $\\angle CAF = \\angle ACE$, and the line $CE$ is parallel to t... | Slovenia | National Math Olympiad in Slovenia | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0gbj | 令 $n$ 為一正整數。求最小的正整數 $k$, 滿足: 存在一個將 $2n \times 2n$ 白色棋盤上的 $k$ 格塗黑的方法, 使得我們只有唯一一種用 $1 \times 2$ 和 $2 \times 1$ 格骨牌覆蓋棋盤的方法, 使得:
(1) 所有骨牌都貼齊棋盤格線且不超出棋盤;
(2) 任兩個骨牌都不重疊;
(3) 任一骨牌最多都只蓋到 1 格黑色格子。 | [
"1. 我們首先構造 $k=2n$ 時的畫法。將棋盤的格子以 $(i, j)$ 標記, 並將所有 $\\{(i, j) : j = i \\text{ or } j = i + 1, i \\le n\\}$ 塗黑, 如下圖:\n\n以下證明只有一 encode方式排列骨牌。考慮棋盤上由兩條對角線切割出 a的上、下、左、右四個區域。\n- 首先注意到, 對於 pr $i \\le n$, 覆蓋 $(i, i)$ 的骨牌都必須覆蓋 $(i, i+1)$, 也就是它們是直放的。這會迫使在左區域的骨牌都必須直放 (可用歸納法證明。)\n- 同理, 對於所有 $i \\le n$, ... | Taiwan | 二〇一七數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2n | |
0bxb | The sides of an equilateral triangle are divided into $n$ equal parts by $n-1$ points on each side. Through these points one draws parallel lines to the sides of the triangle. Thus, the initial triangle is divided into $n^2$ equal equilateral triangles. In every vertex of such a triangle there is a beetle. The beetles ... | [
"It is easy to see (by induction or otherwise) that, for $n \\ge 3$ odd, the set of the vertices of the small triangle can be partitioned into groups of 3 and 4 vertices that form either an equilateral triangle or a rhombus formed by gluing together two such triangles. On each of these polygonal lines, the beetles ... | Romania | THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | All positive integers except 2, 4, and 6 | |
04wr | There are $n$ children around the round-table. Erika is the oldest among them and she has $n$ candies. No other child has any candy. Erika decided to distribute the candies and determined following rules. In every round all the children with at least two candies show. Erika chooses one of them and he/she sends one by o... | [
"First we show for $n$ even the distribution never ends with every child having one candy. In every round only two candies change position and they move in two opposite directions. This leads us to studying the entire sum of distances of candies from one child, say Erika. We label the seats in clockwise direction b... | Czech-Polish-Slovak Mathematical Match | Cesko-Slovacko-Poljsko 2006 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | All odd n ≥ 3 | |
0lal | The sequence of real numbers $(a_n)$ is defined by
$$
a_1 = 5 \quad \text{and} \quad a_n = \sqrt[n]{a_{n-1}^{n-1} + 2^{n-1} + 2 \cdot 3^{n-1}} \quad \text{for all } n \ge 2.
$$
Prove that the sequence $(a_n)$ is decreasing. | [
"From the determining formula for $(a_n)$ it is easily to find that\n$$\na_n = (2^n + 3^n)^{\\frac{1}{n}} \\quad \\forall n \\ge 1.\n$$\nFor all $n \\ge 1$, we have:\n$$\n\\begin{align*}\n2^n + 3^n > 3^n &\\Rightarrow (2^n + 3^n)^{n+1} > 3^n (2^n + 3^n)^n > (2^{n+1} + 3^{n+1})^n \\\\\n&\\Rightarrow (2^n + 3^n)^{\\f... | Vietnam | Vietnamese Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
01r5 | Given triangle $ABC$ with $AB = c$, $BC = a$, $CA = b$. The pairs of points $C_1$ and $C_2$, $A_1$ and $A_2$, $B_1$ and $B_2$ are marked on the sides $AB$, $BC$, $CA$, respectively so that the following equalities are valid:
$$
\begin{aligned}
\frac{CA_1}{a} &= \frac{CB_2}{b} = \frac{a+b}{a+... | [
"Let $K, L, N$ be points of intersection of the bisectors $AK, BL, CN$ and the circumcircle of a triangle $ABC$, respectively (see the Fig.). Let $I$ be the incenter of $\\triangle ABC$ and let the segment $NL$ meet $AC$ at $B'_2$. Let $M$ be the intersection point of $CN$ and $AB$. By the trefoil theorem $AL = IL$... | Belarus | Final Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
03ac | Find the least positive integer which divides $2^n + 15$ for some positive integer $n$ and has the form $3x^2 - 4xy + 3y^2$ for some integers $x$ and $y$. | [
"Let $d = 3x^2 - 4xy + 3y^2$ for some integers $x$ and $y$ and suppose that $d$ divides $2^n + 15$ for some positive integer $n$. Obviously $d$ is odd and this implies that $x$ and $y$ have different parity. Then we have $d \\equiv 3 \\pmod{4}$. Moreover, it follows from $3d = (3x - 2y)^2 + 5y^2$ that $3d \\equiv (... | Bulgaria | Fall Mathematical Competition | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 23 | |
046l | For positive integers $m, n$, define
$$
S(m, n) = \{(a, b) \in \mathbb{Z}^2 \mid 1 \le a \le m, 1 \le b \le n, \gcd(a, b) = 1\}.
$$
Prove: for any positive integers $d, r$, there exist integers $m, n$ not less than $d$, such that $|S(m, n)| \equiv r \pmod{d}$. Here, $|A|$ represents the number of elements in the finite... | [
"Let $n = d + r$ and $m = d \\cdot (d+r) + 1$. Then, for $1 \\le b \\le d+r$, the number of integers in the range $1, 2, \\dots, m$ that are coprime to $b$ is given by\n$$\n\\frac{\\varphi(b)}{b} \\cdot d \\cdot (d+r)! + 1.\n$$\nNote that $b$ divides $(d+r)!$, so this number is congruent to $1$ modulo $b$. Therefor... | China | 2023 Chinese IMO National Team Selection Test | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
05e3 | Problem:
We are given a positive integer $s \geqslant 2$. For each positive integer $k$, we define its twist $k^{\prime}$ as follows: write $k$ as $a s+b$, where $a, b$ are non-negative integers and $b<s$, then $k^{\prime}=b s+a$. For the positive integer $n$, consider the infinite sequence $d_{1}, d_{2}, \ldots$ wher... | [
"Solution:\n\nFirst, we consider the difference $k-k^{\\prime \\prime}$. If $k=a s+b$ as in the problem statement, then $k^{\\prime}=b s+a$. We write $a=l s+m$ with $m, l$ non-negative numbers and $m \\leq s-1$. This gives $k^{\\prime \\prime}=m s+(b+l)$ and hence $k-k^{\\prime \\prime}=(a-m) s-l=l\\left(s^{2}-1\\r... | European Girls' Mathematical Olympiad (EGMO) | EGMO 2023 | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0bqw | Let $n \in \mathbb{N}, n \ge 4$, and $A_n = \{1, 2, 3, \dots, n\}$. Find the number of solutions in the set $A_n \times A_n \times A_n \times A_n$ of the system
$$
\begin{cases}
x + z = 2y \\
y + t = 2z
\end{cases}
$$ | [] | Romania | 67th NMO Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | Let n = 3q + r with r in {0,1,2}. The number of solutions is:
- if r = 0: 3q^2
- if r = 1: 3q^2 + 2q + 1
- if r = 2: 3q^2 + 4q + 2
Equivalently, if m = floor((n−1)/3), the count is n + 2mn − 3m^2 − 3m. | |
09ys | Five distinct positive integers are in a sequence ordered from small to large. The middle number is $20$. The difference between the smallest two numbers equals the difference between the largest two numbers. The fourth number is four times as large as the first number, and the fifth number is twice as large as the sec... | [] | Netherlands | Junior Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | MCQ | C | |
05kd | Problem:
Trouver tous les entiers naturels $k>0$ tels que l'équation en $x$ et $y$ :
$$
x(x+k)=y(y+1)
$$
ait une solution en entiers strictement positifs. | [
"Solution:\nL'équation de l'énoncé s'écrit encore $\\left(x+\\frac{k}{2}\\right)^{2}=\\left(y+\\frac{1}{2}\\right)^{2}+\\frac{k^{2}-1}{4}$, soit, en factorisant:\n$$\n\\left(x-y+\\frac{k-1}{2}\\right) \\cdot\\left(x+y+\\frac{k+1}{2}\\right)=\\frac{k^{2}-1}{4} .\n$$\nDistinguons deux cas selon la parité de $k$.\n\nS... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | k = 1 and all integers k ≥ 4 | |
0de8 | Prove that the set of all divisors of a positive integer which is not a perfect square can be divided into pairs so that in each pair, one number is divided by another. | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0a6l | Problem:
Let $ABC$ be a right-angled triangle with $\angle BAC = 90^{\circ}$, $\angle ABC = 70^{\circ}$, and $AB = 1$. Let $M$ be the midpoint of $BC$. Let $D$ be the point on the extension of $AM$ beyond $M$ such that $\angle CDA = 110^{\circ}$. Find the length of $CD$. | [
"Solution:\n\nConstruct point $E$ so that $ABEC$ is a rectangle. The diagonals of any rectangle bisect each other, that is, they meet at each other's midpoints. Hence $AE$ and $BC$ meet at $M$, i.e. $E$ lies on line $AM$.\n\n\n\nBy symmetry in rectangle $ABEC$, we have\n$$\n\\angle CEA = \\... | New Zealand | NZMO Round One | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 1 | |
0iu3 | Problem:
Paul starts with the number $19$. In one step, he can add $1$ to his number, divide his number by $2$, or divide his number by $3$. What is the minimum number of steps Paul needs to get to $1$? | [
"Solution:\n\nAnswer: $6$\n\nOne possible path is $19 \\rightarrow 20 \\rightarrow 10 \\rightarrow 5 \\rightarrow 6 \\rightarrow 2 \\rightarrow 1$."
] | United States | 2nd Annual Harvard-MIT November Tournament | [
"Discrete Mathematics > Algorithms",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 6 | |
07pt | Find, with proof, the greatest positive integer which cannot be expressed in the form $17x + 127y$, with $x$ and $y$ non-negative integers. | [
"The answer is $2015$. To prove this, first, observe (using Euclid's algorithm, for example), that\n$$\n15 \\cdot 17 - 2 \\cdot 127 = 1 \\qquad (1)\n$$\nNext, note that\n$$\n\\begin{align*}\n2016 &= (17 - 1)(127 - 1) = 16 \\cdot 127 - 17 + 1 \\\\\n&= 16 \\cdot 127 - 17 + 15 \\cdot 17 - 2 \\cdot 127 \\tag{2} \\\\\n&... | Ireland | Ireland | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 2015 | |
05pe | Problem:
a) Déterminer toutes les fonctions $f: \mathbb{R} \longrightarrow \mathbb{Z}$ telles que
$$
f(f(y)-x)^{2}+f(x)^{2}+f(y)^{2}=f(y)(1+2 f(f(y)))
$$
pour tous réels $x$ et $y$.
b) Déterminer toutes les fonctions $f: \mathbb{R} \longrightarrow \mathbb{R}$ telles que
$$
f(f(y)-x)^{2}+f(x)^{2}+f(y)^{2}=f(y)(1+2 f(f(y... | [
"Solution:\na) Soit $f$ une éventuelle solution du problème. Notons $P(x, y)$ l'égalité de l'énoncé pour les valeurs $x$ et $y$. On pose $c=f(0)$.\nPour tout $x$, de $P(0, x)$ on déduit que $f(f(x))^{2}+c^{2}+f(x)^{2}=f(x)+2 f(f(x)) f(x)$, d'où $f(x)=c^{2}+(f(f(x))-f(x))^{2}$. Il s'ensuit que $f(x) \\geqslant c^{2}... | France | Olympiades Françaises de Mathématiques - Envoi 2 (Algèbre) | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | For both parts (a) and (b), the only solutions are the constant functions f(x) = 0 for all real x and f(x) = 1 for all real x. | |
0jsu | Problem:
Point $P_{1}$ is located 600 miles West of point $P_{2}$. At 7:00 AM a car departs from $P_{1}$ and drives East at a speed of 50 miles per hour. At 8:00 AM another car departs from $P_{2}$ and drives West at a constant speed of $x$ miles per hour. If the cars meet each other exactly halfway between $P_{1}$ an... | [
"Solution:\n\nEach car meets having traveled 300 miles. Therefore the first car traveled for $300 / 50 = 6$ hours, and so the second car traveled for 5 hours. The second car must have traveled $300 / 5 = 60$ miles per hour."
] | United States | HMMT November 2016 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 60 | |
06qp | Let the sides $AD$ and $BC$ of the quadrilateral $ABCD$ (such that $AB$ is not parallel to $CD$) intersect at point $P$. Points $O_{1}$ and $O_{2}$ are the circumcenters and points $H_{1}$ and $H_{2}$ are the orthocenters of triangles $ABP$ and $DCP$, respectively. Denote the midpoints of segments $O_{1}H_{1}$ and $O_{... | [
"We keep triangle $ABP$ fixed and move the line $CD$ parallel to itself uniformly, i.e. linearly dependent on a single parameter $\\lambda$ (see Figure 1). Then the points $C$ and $D$ also move uniformly. Hence, the points $O_{2}, H_{2}$ and $E_{2}$ move uniformly, too. Therefore also the perpendicular from $E_{2}$... | IMO | IMO Problem Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordin... | English | proof only | null | |
0eqs | What is the difference between the largest and the second largest odd factors of $2016$? | [
"$2016$ is $2^5 \\times 3^2 \\times 7$. The largest odd factor is $3^2 \\times 7 = 63$; and the next largest odd factor is $3 \\times 7 = 21$. $63 - 21 = 42$"
] | South Africa | South African Mathematics Olympiad Second Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | final answer only | 42 | |
08y9 | Find all triplets $(a, b, c)$ of positive integers satisfying the following identity:
$$
a^2 + b + 3 = (b^2 - c^2)^2.
$$ | [
"Let $N = |b^2 - c^2|$. Then, from $N^2 = a^2 + b + 3 \\ge 1 + 1 + 3 = 5$ it follows that $N \\ge 3$. In particular, since $b \\ne c$, we get\n$$\nN = |b + c||b - c| \\ge (b + 1) \\cdot 1 = b + 1,\n$$\nso we have $b \\le N - 1$. Since from $a^2 < N^2$ we have $a \\le N - 1$, we get\n$$\n(1) \\quad 0 = N^2 - (a^2 + ... | Japan | 2019 Japanese Mathematical Olympiad, Final Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (2, 2, 1) | |
0h5f | It is known that $\frac{a}{b+c+d} + \frac{b}{c+d+a} + \frac{c}{d+a+b} + \frac{d}{a+b+c} = 1$. Find the value of the expression
$$
\frac{a^2}{b+c+d} + \frac{b^2}{c+d+a} + \frac{c^2}{d+a+b} + \frac{d^2}{a+b+c}.
$$ | [
"Multiplying the given equality by $(a+b+c+d)$, we obtain\n$$\n\\begin{aligned}\n& \\frac{a(a+b+c+d)}{b+c+d} + \\frac{b(a+b+c+d)}{c+d+a} + \\frac{c(a+b+c+d)}{d+a+b} + \\frac{d(a+b+c+d)}{a+b+c} = \\\\\n& = \\frac{a^2+a(b+c+d)}{b+c+d} + \\frac{b^2+b(c+d+a)}{c+d+a} + \\frac{c^2+c(d+a+b)}{d+a+b} + \\frac{d^2+d(a+b+c)}{... | Ukraine | Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 0 | |
0hqq | Problem:
Let $ABC$ be a triangle, and let the bisector of $\angle BAC$ meet $BC$ at $D$. Let $O, O_{1}, O_{2}$ be the circumcenters of triangles $ABC$, $ABD$, and $ADC$, respectively. Prove that $OO_{1} = OO_{2}$. | [
"Solution:\nWe use the standard fact that the circumcenter of a triangle is the intersection of the perpendicular bisectors of the three sides. Note that $OO_{1} \\perp AB$ since both $O$ and $O_{1}$ lie on the perpendicular bisector of $AB$. Similarly, $OO_{2} \\perp AC$ and $O_{1}O_{2} \\perp AD$. Hence the (coun... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fz2 | Problem:
Sei $n \geq 2$ eine natürliche Zahl. Finde in Abhängigkeit von $n$ die grösste natürliche Zahl $d$, sodass eine Permutation $a_{1}, a_{2}, \ldots, a_{n}$ der Zahlen $1,2, \ldots, n$ existiert mit
$$
\left|a_{i}-a_{i+1}\right| \geq d, \quad \text{ für } i=1,2, \ldots, n-1
$$ | [
"Solution:\n\nWir betrachten zuerst den Fall wo $n$ ungerade ist, also $n=2k+1$.\nDann gilt für jedes $m \\in \\{1,2, \\ldots, 2k+1\\}$ sicher $|k+1-m| \\leq k$, also $d \\leq k$. Mit der folgenden Permutation sehen wir, dass $d=k$ auch möglich ist:\n$$\n\\begin{aligned}\na_{2i+1} & = k+1+i, \\text{ für } i=0,1, \\... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | floor(n/2) | |
0kk8 | Problem:
Compute the number of complex numbers $z$ with $|z|=1$ that satisfy
$$
1+z^{5}+z^{10}+z^{15}+z^{18}+z^{21}+z^{24}+z^{27}=0
$$ | [
"Solution:\nLet the polynomial be $f(z)$. One can observe that\n$$\nf(z)=\\frac{1-z^{15}}{1-z^{5}}+z^{15} \\frac{1-z^{15}}{1-z^{3}}=\\frac{1-z^{20}}{1-z^{5}}+z^{18} \\frac{1-z^{12}}{1-z^{3}}\n$$\nso all primitive 15th roots of unity are roots, along with $-1$ and $\\pm i$.\n\nTo show that there are no more, we can ... | United States | HMMT Spring 2021 Guts Round | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 11 | |
0an7 | Problem:
Determine all values of $k \in \mathbb{R}$ for which the equation
$$
\frac{4\left(2015^{x}\right)-2015^{-x}}{2015^{x}-3\left(2015^{-x}\right)}=k
$$
admits a real solution. | [] | Philippines | 18th PMO Area Stage | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (-∞, 1/3) ∪ (4, ∞) | |
0koq | Let $a$, $b$, $x$, and $y$ be real numbers with $a > 4$ and $b > 1$ such that
$$
\frac{x^2}{a^2} + \frac{y^2}{a^2 - 16} = \frac{(x - 20)^2}{b^2 - 1} + \frac{(y - 11)^2}{b^2} = 1.
$$
Find the least possible value for $a + b$. | [] | United States | AIME II | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | 23 | |
00c7 | The faces of a cube of size $1 \times 1 \times 1$ are painted in black, grey and white, two faces in each color, so that opposite faces have the same color. We have a squared $2018 \times 2018$ board divided into $2018^2$ cells of $1 \times 1$.
Ana and Beatriz play the following game. First, Ana puts the cube on a cel... | [
"Beatriz has a winning strategy. In order to win, Beatriz has to split the board into $2018^2/2$ horizontal dominoes and, whenever Ana occupies one cell of a domino, Beatriz in its turn must occupy the other cell of the same domino. To see that this strategy works, we may ensure that if $A$ and $\\bar{A}$ are the t... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Beatriz | |
01ke | Points $A_1, B_1$ are marked on the sides $AC$ and $BC$ of the triangle $ABC$, respectively, so that $A_1B_1 \parallel AB$. Points $A_2, B_2$ are the feet of perpendiculars from $A_1, B_1$ onto $AB$, respectively.
Prove that $AC = AB_2 + CB_1$ if and only if $BC = BA_2 + CA_1$.
(I. Voronovich) | [
"Let $A_1A_2 = B_1B_2 = 2x$, $\\angle A = \\alpha$, $\\angle B = \\beta$, $AB = c$, $AC = b$, $BC = a$. Then we have (see Fig. 1)\n$$\nAC = AB_2 + CB_1 \\iff b = c - BB_2 + a - BB_1 = c + a - 2x \\operatorname{ctg} \\beta - \\frac{2x}{\\sin \\beta} \\iff\n$$\n$$\na + c - b = 2x \\left( \\operatorname{ctg} \\beta + ... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance cha... | English | proof only | null | |
0a5u | Problem:
There are $2023$ employees in the office, each of them knowing exactly $1686$ of the others. For any pair of employees they either both know each other or both don't know each other. Prove that we can find $7$ employees each of them knowing all $6$ others. | [
"Solution:\nIf every person knows $1686$ others then for each person, there are $2023 - 1686 - 1 = 336$ people that they don't know. Now consider any group of $p$ people from the office. There will be at most $336p$ people who don't know someone in the group ($336$ for each person in the group). Therefore there are... | New Zealand | NZMO Round One | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
09pj | Problem:
Cirkels $\Gamma_{1}$ en $\Gamma_{2}$ snijden elkaar in $P$ en $Q$. Zij $A$ een punt op $\Gamma_{1}$ niet gelijk aan $P$ of $Q$. De lijnen $A P$ en $A Q$ snijden $\Gamma_{2}$ nogmaals in respectievelijk $B$ en $C$.
Bewijs dat de hoogtelijn uit $A$ in driehoek $A B C$ door een punt gaat dat onafhankelijk is van... | [
"Solution:\n\nDoor het tekenen van verscheidene nette plaatjes hebben we het vermoeden gekregen dat de genoemde hoogtelijn altijd door het middelpunt van $\\Gamma_{1}$ gaat. Dat dat ook daadwerkelijk zo is, gaan we nu bewijzen.\nHet voetpunt van de hoogtelijn uit $A$ op (het verlengde van) $B C$ noemen we $K$, en h... | Netherlands | UITWERKINGEN TOETS TRAININGSKAMP | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06z4 | Problem:
Let $k^{3} = 2$ and let $x$, $y$, $z$ be any rational numbers such that $x + y k + z k^{2}$ is non-zero. Show that there are rational numbers $u$, $v$, $w$ such that $(x + y k + z k^{2})(u + v k + w k^{2}) = 1$. | [
"Solution:\n\nWe need $x u + 2 z v + 2 y w = 1$, $y u + x v + 2 z w = 0$, $z u + y v + x w = 0$. This is just a straightforward set of linear equations. Solving, we get:\n\n$$\n\\begin{align*}\nu &= \\frac{x^{2} - 2 y z}{d}, \\\\\nv &= \\frac{2 z^{2} - x y}{d}, \\\\\nw &= \\frac{y^{2} - x z}{d},\n\\end{align*}\n$$\... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Algebra > Abstract Algebra > Field Theory",
"Algebra > Linear Algebra > Determinants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
06f2 | Is there a polynomial $f$ of degree $2007$ with integer coefficients, such that $f(n), f(f(n)), f(f(f(n))), \dots$ are pairwise relatively prime for every integer $n$? Justify your claim. | [
"Yes. For example, we can take $f(x) = x^{2007} - x^{2006} + 1$.\nIt suffices to show that $(m, f^k(m)) = 1$ for any positive integer $k$, since we can replace $m$ by any $f^j(n)$. Consider any prime $p$ dividing $m$. Then we have $f(m) \\equiv 0 \\pmod{p} \\equiv -0 + 1 = 1 \\pmod{p}$. Whenever $f^i(m) \\equiv 1 \... | Hong Kong | CHKMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | Yes; for example f(x) = x^{2007} − x^{2006} + 1. | |
01fy | Let $X$, $Y$ be points on $AB$, $AC$ of triangle $ABC$, respectively, such that $B$, $C$, $X$, $Y$ lie on one circle.
The median of triangle $ABC$ from $A$ intersects the perpendicular bisector of $XY$ at $P$. Find $\angle BAC$, if $PXY$ is equilateral. | [
"Since $AP$ is the symmedian of $AXY$ and $P$ lies on the perpendicular bisector of $XY$, then $PX$ and $PY$ are tangent to the circumcircle of triangle $AXY$. Therefore, we can easily find that $\\angle YAX = 60^\\circ$ or $\\angle YAX = 120^\\circ$.\n\nSince $B$, $C$, $X$, $Y$ lie on a circle, $XY$ is antiparalle... | Baltic Way | Baltic Way 2019 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 60 or 120 degrees | |
0g79 | 設 $x, y$ 為正整數, $x > y$ 且 $(x-y)^{xy} = x^y \cdot y^x$, 試求數對 $(x, y)$。 | [
"令 $x = dp$, $y = dq$, 其中 $d = (x, y)$ 為 $x, y$ 的最大公因數, $p, q \\in \\mathbb{N}$, $(p, q) = 1$, $p > q$, 則\n$$(d(p-q))^{d^2pq} = (dp)^{dq}(dq)^{dp} \\Leftrightarrow (d(p-q))^{dpq} = (dp)^q(dq)^p \\Leftrightarrow d^{dpq}(p-q)^{dpq} = d^{p+q}p^q q^p.$$\n\n欲證：$p+q < dpq$,\n假設 $p+q \\ge dpq$, 則 $(p-q)^{dpq} = d^{p+q-dpq... | Taiwan | 二0一三數學奧林匹亞競賽第一階段選訓營 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | (4, 2) | |
091e | Problem:
In each vertex of a regular $n$-gon there is a fortress. At the same moment each fortress shoots at one of the two nearest fortresses and hits it. The result of the shooting is the set of the hit fortresses; we do not distinguish whether a fortress was hit once or twice. Let $P(n)$ be the number of possible r... | [
"Solution:\n\nLet us denote each hit fortress by a black dot and each undamaged one with a white dot. Then $P(n)$ is the number of colourings of $n$ dots distributed on the circle with black and white colours in such a way, that no two white dots have exactly one dot in between them. The proof of this bijectivity i... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0flr | Let us consider a triangle with side lengths $a, b, c$ such that
$$
b(a+b)(b+c) = a^3 + b(a^2 + c^2) + c^3.
$$
Let us call $A, B$ and $C$ the values, measured in radians, of the angles of the triangle. Prove that the equality $\frac{1}{\sqrt{A}+\sqrt{B}} + \frac{1}{\sqrt{B}+\sqrt{C}} = \frac{2}{\sqrt{A}+\sqrt{C}}$ hold... | [] | Spain | Spanija 2012 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0iiu | Problem:
Compute the number of real solutions $(x, y, z, w)$ to the system of equations:
$$
\begin{array}{rlrl}
x & = z + w + z w x & z & = x + y + x y z \\
y & = w + x + w x y & w & = y + z + y z w
\end{array}
$$ | [
"Solution:\nThe first equation rewrites as $x = \\frac{w + z}{1 - w z}$, which is a fairly strong reason to consider trigonometric substitution. Let $x = \\tan(a)$, $y = \\tan(b)$, $z = \\tan(c)$, and $w = \\tan(d)$, where $-90^{\\circ} < a, b, c, d < 90^{\\circ}$. Under modulo $180^{\\circ}$, we find $a \\equiv c ... | United States | Harvard-MIT Mathematics Tournament | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | 5 | |
0342 | Problem:
Let $a, b, c > 0$ and $a + b + c = 1$. Prove that
$$
\frac{9}{10} \leq \frac{a}{1 + b c} + \frac{b}{1 + c a} + \frac{c}{1 + a b} < 1
$$ | [
"Solution:\nTo prove the right inequality, it is enough to use that the denominators are greater than $1$. Hence\n$$\n\\frac{a}{1 + b c} + \\frac{b}{1 + c a} + \\frac{c}{1 + a b} < a + b + c = 1\n$$\n\nTo show the left inequality, we may assume that $a \\leq b \\leq c$. Then\n$$\n\\frac{1}{1 + b c} \\leq \\frac{1}{... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0jdf | Problem:
Let $\mathcal{S}$ be a set of size $n$, and $k$ be a positive integer. For each $1 \leq i \leq k n$, there is a subset $S_{i} \subset \mathcal{S}$ such that $|S_{i}|=2$. Furthermore, for each $e \in \mathcal{S}$, there are exactly $2 k$ values of $i$ such that $e \in S_{i}$. Show that it is possible to choose ... | [
"Solution:\nConsider the undirected graph $G=(\\mathcal{S}, E)$ where the elements of $\\mathcal{S}$ are the vertices, and for each $1 \\leq i \\leq k n$, there is an edge between the two elements of $S_{i}$. (Note that there might be multiedges if two subsets are the same, but there are no self-loops.)\n\nConsider... | United States | HMIC | [
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
0d23 | Let $f(X) = a_{n} X^{n} + a_{n-1} X^{n-1} + \cdots + a_{1} X + p$ be a polynomial of integer coefficients where $p$ is a prime number. Assume that
$$
p > \sum_{i=1}^{n} \left| a_{i} \right| .
$$
Prove that $f(X)$ is irreducible. | [
"Assume that there exist two non-constant polynomials $g(X)$ and $h(X)$ with integer coefficients such that $f(X) = g(X) h(X)$. Because $p = g(0) h(0)$ is prime, we can assume that $|g(0)| = 1$.\n\nBecause the modulus of the product of the complex roots of $g(X)$ is equal to $1$, at least one of these roots, say $\... | Saudi Arabia | Selection tests for the Gulf Mathematical Olympiad 2013 | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof only | null | |
09za | The sides of a triangle have lengths $13$, $x$, and $2x$. Here $x$ is an integer.
How many possibilities are there for $x$?
A) 2 B) 6 C) 7 D) 8 E) 12 | [
"Let the sides be $13$, $x$, and $2x$.\n\nBy the triangle inequality, the sum of the lengths of any two sides must be greater than the third side.\n\nSo, we have:\n\n1. $13 + x > 2x$\n2. $13 + 2x > x$\n3. $x + 2x > 13$\n\nLet's solve each inequality:\n\n1. $13 + x > 2x \\implies 13 > x$\n2. $13 + 2x > x \\implies 1... | Netherlands | Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | MCQ | D) 8 | |
04o6 | One hundred people attended the party, some of whom were previously acquainted. All acquaintances were mutual and no new were made during the party.
A gong rang $100$ times during the party. After the first sounding of the gong, all the people not acquainted with anyone left the party. After the second sounding of the ... | [
"We show that $n$ can be $0, 1, 2, 3, \\dots, 98$. The following contains the description of a situation in which exactly $n$ people are present after the last chime (for $n = 0, 1, 2, \\dots, 98$):\nFor $n > 0$, we can divide all the people at the party into two groups, A and B. Let A contain $n$ people and assume... | Croatia | Croatian Junior Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 0, 1, 2, ..., 98 | |
0he3 | Give an example of a hexagon (not necessarily convex) that can be cut with one straight line into a triangle and a quadrilateral (not necessarily convex), but which cannot be cut into two triangles or two quadrilaterals. | [
"On Fig. 21, the dashed line shows how to cut the hexagon – one has to draw a segment $CF$ or $BD$.\n\nLet us now see where the line of separation of $ABCDEF$ can be drawn.\nIf it passes through a vertex of a hexagon and is different from lines $AC$ and $BE$, e.g. $AL$, then\non the side, a new point ($L$) appears,... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | null | |
05ca | Does there exist a geometric progression, among the members of which there are
a. $3$, $45$ and $2025$;
b. $3$, $\frac{45}{\sqrt{5}}$ and $2025$? | [
"a.\nLet the common ratio of the progression be $q$. W.l.o.g., assume that $q > 1$. One can also assume that the first term of the progression is $3$. Then $45 = 3 \\cdot q^k$ and $2025 = 3 \\cdot q^l$, where $k$ and $l$ are integers. This implies $q^k = 15$ and $q^l = 675$. Therefore $q^{kl} = 15^l$, as well as $q... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | a: no; b: yes | |
0bm5 | Show that there are positive odd integers $m_1 < m_2 < \dots$ and positive integers $n_1 < n_2 < \dots$ such that $m_k$ and $n_k$ are relatively prime, and $m_k^4 - 2n_k^4$ is a perfect square for each index $k$. | [
"Let $m$ and $n$ be relatively prime positive integers such that $m$ is odd and $m^4 - 2n^4$ is a perfect square, e.g., $m = 3$ and $n = 2$. Write $\\ell^2 = m^4 - 2n^4$, so $\\ell^4 = (m^4 - 2n^4)^2 = (m^4 + 2n^4)^2 - 8m^4n^4$, and $\\ell^4 - 8m^4n^4 - (m^4 + 2n^4)^2 = -16m^4n^4 = -(2mn)^4$. Multiply the latter by... | Romania | 2015 Ninth STARS OF MATHEMATICS Competition | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping"
] | English | proof only | null | |
0iuo | Problem:
Pick a random digit in the decimal expansion of $\frac{1}{99999}$. What is the probability that it is 0? | [
"Solution:\nThe decimal expansion of $\\frac{1}{99999}$ is $0.\\overline{00001}$.\n\nThe repeating block is $00001$, which has $5$ digits: four zeros and one $1$.\n\nThe probability that a randomly chosen digit is $0$ is $\\frac{4}{5}$."
] | United States | Harvard-MIT November Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 4/5 | |
0afn | Во триаголникот **АВС** аголот $\angle BAC = 70^\circ$, а аголот $\angle ABC = 50^\circ$. Точката $M$ се наоѓа во $\triangle ABC$ и притоа $\angle MAC = \angle MCA = 40^\circ$. Определи ги аглите $\angle AMB$ и $\angle BMC$.
 | [
"Од условите дадени на цртежот следува: $\\angle ACB = 60^\\circ$, $\\angle AMC = 100^\\circ$. Бидејќи $\\triangle AMC$ е рамнокрак и $\\angle ABC = \\frac{1}{2} \\angle AMC = 50^\\circ$ следува дека $M$ е центар на опишаната кружница околу $\\triangle ABC$. Оттука имаме: $\\angle AMB = 120^\\circ$ и $\\angle BMC =... | North Macedonia | Регионален натпревар по математика за основно образование | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Macedonian, English | proof and answer | ∠AMB = 120°, ∠BMC = 140° | |
08ps | Problem:
Given a rectangle $ABCD$ such that $AB = b > 2a = BC$, let $E$ be the midpoint of $AD$. On a line parallel to $AB$ through point $E$, a point $G$ is chosen such that the area of $GCE$ is
$$
(GCE) = \frac{1}{2}\left(\frac{a^{3}}{b} + ab\right)
$$
Point $H$ is the foot of the perpendicular from $E$ to $GD$ and a... | [
"Solution:\nLet $L$ be the foot of the perpendicular from $G$ to $EC$ and let $Q$ be the point of intersection of the lines $EG$ and $BC$. Then,\n$$\n(GCE) = \\frac{1}{2} EC \\cdot GL = \\frac{1}{2} \\sqrt{a^{2} + b^{2}} \\cdot GL\n$$\nSo, $GL = \\frac{a}{b} \\sqrt{a^{2} + b^{2}}$.\n\n\n\nO... | JBMO | Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0d9b | Let $ABC$ be an acute, non-isosceles triangle and $(O)$ be its circumcircle (with center $O$). Denote by $G$ the centroid of the triangle $ABC$, by $H$ the foot of the altitude from $A$ onto the side $BC$ and by $I$ the midpoint of $AH$. The line $IG$ intersects $BC$ at $K$.
1. Prove that $CK = BH$.
2. The ray $GH$ int... | [
"1)\nLet $M$ be the midpoint of $BC$ then $G \\in AM$ and $\\frac{AG}{AM} = \\frac{2}{3}$. Take the point $K'$ on $BC$ such that $M$ is the midpoint of $HK'$, then $AM$ is the median of triangle $AHK'$ and $G$ is its centroid.\nThen $K'G$ is the median of triangle $AHK'$ or $K'G$ passes through the midpoint of $AH$... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
09en | Let $a$, $b$, $c$, $d$, $e$ be not necessarily distinct divisors of $210$. Find all $5$-permutations $(a, b, c, d, e)$ which satisfy the condition $abcde > 44100$. | [
"Let $a$, $b$, $c$, $d$, $e$, $f$ be not necessarily distinct divisors of $210$. First we will find number of $6$-permutations $(a, b, c, d, e, f)$ that satisfy the condition $abcdef > 210^3$.\n$$\n\\text{Since } abcdef > 210^3 \\Leftrightarrow \\frac{210}{a} \\cdot \\frac{210}{b} \\cdot \\frac{210}{c} \\cdot \\fra... | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 1384768 | |
0g8u | 設 $P$ 為三角形 $ABC$ 內一點, 直線 $AP$, $BP$, $CP$ 分別與三角形 $ABC$ 的外接圓交於 $T$, $S$, $R$ 點 ($T \neq A$, $S \neq B$, $R \neq C$)。設 $U$ 為線段 $PT$ 內一點。過 $U$ 與 $AB$ 平行的直線分別與 $CR$ 交於 $W$ 點, 過 $U$ 與 $AC$ 平行的直線分別與 $BS$ 交於 $V$ 點。最後, 設過 $B$ 與 $CP$ 平行的直線, 與過 $C$ 與 $BP$ 平行的直線交於 $Q$ 點。已知 $RS$ 與 $VW$ 平行, 證明 $\angle CAP = \angle BAQ$。
Let $P$ be... | [
"(i) 設過 $U$ 與 $AC$ 平行的直線交 $CR$ 於 $X$, 過 $U$ 與 $AB$ 平行的直線交 $BS$ 於 $Y$。因 $UY \\parallel AB$, $\\triangle PUT \\sim \\triangle PAB$。由此得 $\\frac{PU}{AP} = \\frac{PY}{BP}$。同理 $\\frac{PU}{AP} = \\frac{PX}{CP}$。故 $\\frac{PY}{BP} = \\frac{PX}{CP}$, 因此 $XY \\parallel BC$。\n\n(ii) 因為 $\\angle VWP = \\angle XRS = \\angle PBC ... | Taiwan | 國際數學奧林匹亞競賽第二階段選訓營 模擬競賽(二) | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00eq | Let $ABC$ be an acute triangle. Denote by $D, E, F$ the midpoints of sides $BC, CA, AB$ respectively. The circle with diameter $AB$ intersects lines $AB$ and $AC$ again at $P$ and $Q$ respectively. The line through $P$ parallel to $BC$ meets line $DE$ at $R$, the line through $Q$ parallel to $BC$ meets line $DF$ at $S$... | [
"Since $D$ and $E$ are midpoints we know that $DE \\parallel AB$, and by definition $PR \\parallel BC$, hence $PRDB$ is a parallelogram. Analogously, $QSDC$ is a parallelogram. Thus $PR = BD = DC = SQ$.\n\n\n\nSince $AD$ is a diameter, we know that $AB \\perp PD$ and $AC \\perp QD$. Then, b... | Argentina | Cono Sur Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0d0m | Let $ABC$ be a triangle. Point $D$ lies on side $BC$. Let $O$, $O_1$, and $O_2$ be the circumcenters of triangle $ABC$, $ABD$, and $ACD$, respectively. Prove that circumcircles of triangles $BOO_1$ and $COO_2$ meet on line $BC$. | [
"Let $P$ be on $BC$ such that $OP$ is parallel to $AD$. If $O$ is on $BC$, then $P = O$ and the result is clear. We claim that the circumcircles of $BOO_1$ and $COO_2$ both pass through $P$. One of the angles $\\widehat{ADB}$ and $\\widehat{ADC}$ is not acute. Without loss of generality, assume that $\\widehat{ADB}... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0c3b | Let $D$ be a non-empty set of positive integers, let $d$ be the greatest common divisor of $D$, and let $d\mathbb{Z} = \{dn : n \in \mathbb{Z}\}$. Prove that there exists a bijection $f: \mathbb{Z} \to d\mathbb{Z}$ such that $|f(n) - f(n-1)|$ is a member of $D$ for all integers $n$.
*Amer. Math. Monthly* | [
"Reduce the problem to the case where $D$ is finite, by considering an element $d_0$ of $D$, representing each residue class in $D \\pmod{d_0}$ by some member of $D$, and collecting all these representatives to form a finite subset of $D$ whose greatest common divisor is again $d$.\n\nAssume henceforth $D$ finite a... | Romania | 69th NMO Selection Tests for BMO and IMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0i83 | A convex polygon $P$ in the plane is dissected into smaller convex polygons by drawing all of its diagonals. The lengths of all sides and all diagonals of the polygon $P$ are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers. | [
"Let $\\mathcal{P} = A_1A_2\\dots A_n$, where $n$ is an integer with $n \\ge 3$. The problem is trivial for $n=3$ because there are no diagonals and thus no dissections. We assume that $n \\ge 4$. Our proof is based on the following Lemma.\n\n**Lemma** Let $ABCD$ be a convex quadrilateral such that all its sides an... | United States | USA IMO 2003 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
057m | A circle $c$ with center $A$ passes through the vertices $B$ and $E$ of a regular pentagon $ABCDE$. The line $BC$ intersects the circle $c$ the second time at point $F$. Prove that lines $DE$ and $EF$ are perpendicular. | [
"The internal angles of a regular pentagon have size $108^\\circ$. Thus $\\angle EAB = 108^\\circ$ (Fig. 2), whence $\\angle EFC = \\angle EFB = \\frac{\\angle EAB}{2} = 54^\\circ$. As $\\angle CDE = 108^\\circ$ and $\\angle FCD = \\angle BCD = 108^\\circ$, from the quadrilateral $CDEF$ we obtain $\\angle DEF = 360... | Estonia | Open Contests | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
0dr1 | Let $ABC$ be an acute-angled triangle and let $D$, $E$ and $F$ be the midpoints of $BC$, $CA$ and $AB$ respectively. Construct a circle, centred at the orthocentre of triangle $ABC$, such that triangle $ABC$ lies in the interior of the circle. Extend $EF$ to intersect the circle at $P$, $FD$ to intersect the circle at ... | [
"Let the radius of the circle be $r$. Let $X$, $Y$ and $Z$ be the feet of the altitudes from $A$, $B$ and $C$ respectively. Let $PE$ intersect the altitude from $A$ at $U$. We have\n$$\nAP^2 = AU^2 + PU^2 = AU^2 + r^2 - UH^2 = r^2 + (AU+UH) \\cdot (AU-UH) = r^2 + AH \\cdot (AU-UH) = r^2 + AH \\cdot (UX - UH) = r^2 ... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0bm0 | Let $ABC$ be a triangle, let $O$ be its circumcentre, let $A'$ be the orthogonal projection of $A$ on the line $BC$, and let $X$ be a point on the open ray $AA'$ emanating from $A$. The internal bisectrix of the angle $BAC$ meets the circumcircle of $ABC$ again at $D$. Let $M$ be the midpoint of the segment $DX$. The l... | [
"Choose a point $Y$ such that $AONY$ is a parallelogram. Since the lines $AD$ and $ON$ are parallel, this point lies on the line $AD$ (see Fig. 1). We prove that the triangles $AOY$ and $AXD$ are similar. Since the line $AN$ bisects the segment $OY$ the conclusion follows.\nIt is well known that the internal bisect... | Romania | 66th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Mi... | null | proof only | null | |
0elo | Consider a triangle $ABC$ with $BC = 3$. Choose a point $D$ on $BC$ such that $BD = 2$. Find the value of
$$
AB^2 + 2AC^2 - 3AD^2.
$$ | [
"Drop the altitude from $A$ to $BC$, and let $F$ be its foot. Furthermore, suppose that $BF = x$ (if $F$ lies on the extension of $BC$ beyond $B$, assign a negative sign to $x$) and $AF = y$. Then, by the Pythagorean theorem,\n$$\nAB^2 = BF^2 + AF^2 = x^2 + y^2,\n$$\n$$\nAC^2 = CF^2 + AF^2 = (3-x)^2 + y^2,\n$$\n$$\... | South Africa | The South African Mathematical Olympiad Third Round | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 6 | |
02ko | Problem:
Uma linha de ônibus possui $12$ paradas numa rua em linha reta. A distância entre duas paradas consecutivas é sempre a mesma. Sabe-se que a distância entre a terceira e a sexta paradas é $3300$ metros. Qual é a distância entre a primeira e a última parada?
A) $8,4~\mathrm{km}$
B) $12,1~\mathrm{km}$
C) $9,9~\... | [
"Solution:\n\n\n\nComo a distância entre a $3^a$ e a $6^a$ paradas é $3300~\\mathrm{m}$, então a distância entre duas paradas consecutivas é $3300 \\div 3 = 1100~\\mathrm{m}$.\n\nPortanto, a distância entre a primeira e a última paradas é $1100~\\mathrm{m} \\times 11 = 12100~\\mathrm{m}$. C... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | MCQ | B | |
08up | Let $n$ be an integer greater than or equal to $2$. Suppose $n$ white points and $n$ black points are distributed on the circumference of a circle, and suppose we try to draw $2n$ line segments using these points so as to satisfy the following conditions:
(1) Each line segment has a white point and a black point as its... | [
"For $\\ell$ white points and $\\ell$ black points distributed on the circumference of a circle, call a method of drawing $2\\ell$ line segments a good method if the following three conditions are satisfied:\n(1) Each line segment has a white point and a black point as its end points.\n(2) By tracing these line seg... | Japan | Japan Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0d72 | Given a set of $2^{2016}$ cards with the numbers $1, 2, \ldots, 2^{2016}$ written on them. We divide the set of cards into pairs arbitrarily; from each pair, we keep the card with larger number and discard the other. We now again divide the $2^{2015}$ remaining cards into pairs arbitrarily; from each pair, we keep the ... | [
"Note that the remaining number is kept $1008$ times as the larger one of the pair. So it is bigger than at least $2^{1008}-1$ numbers.\n\nSimilarly, the remaining number is kept $1008$ times as the smaller one of the pair so it is smaller than at least $2^{1008}-1$ numbers.\n\nTherefore, the remaining number $x$ s... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | All integers x with 2^{1008} ≤ x ≤ 2^{2016} − 2^{1008} + 1 | |
052w | In a triangle $ABC$ the midpoints of $BC$, $CA$ and $AB$ are $D$, $E$ and $F$, respectively. Prove that the circumcircles of triangles $AEF$, $BFD$ and $CDE$ intersect all in one point. | [
"Let us first assume that triangle $ABC$ is not a right triangle – then the circumcenter $O$ of the triangle $ABC$ does not coincide with $D$, $E$, $F$ (see fig. 1). As the circumcenter is in the point of intersection of perpendicular bisectors of the sides,\n\n$\\angle AEO = 90^\\circ = \\angle AFO$, due to which ... | Estonia | Open Contests | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothet... | null | proof only | null | |
0i73 | Problem:
$$
\frac{3}{1^{2} \cdot 2^{2}}+\frac{5}{2^{2} \cdot 3^{2}}+\frac{7}{3^{2} \cdot 4^{2}}+\cdots+\frac{29}{14^{2} \cdot 15^{2}}.
$$ | [
"Solution:\n\nThe sum telescopes as\n\n$$\n\\left(\\frac{1}{1^{2}}-\\frac{1}{2^{2}}\\right)+\\left(\\frac{1}{2^{2}}-\\frac{1}{3^{2}}\\right)+\\cdots+\\left(\\frac{1}{14^{2}}-\\frac{1}{15^{2}}\\right)=\\frac{1}{1^{2}}-\\frac{1}{15^{2}}=\\frac{224}{225}.\n$$"
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 224/225 | |
0437 | Suppose that $A = \{1, 2, 3\}$, $B = \{2x + y \mid x, y \in A, x < y\}$, $C = \{2x + y \mid x, y \in A, x > y\}$. Then the sum of all the elements of $B \cap C$ is ______. | [
"By enumeration, we get $B = \\{4, 5, 7\\}$, $C = \\{5, 7, 8\\}$. Thus, $B \\cap C = \\{5, 7\\}$. Therefore, the sum of all the elements of $B \\cap C$ is $5 + 7 = 12$. $\\square$"
] | China | China Mathematical Competition | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 12 | |
0kx3 | Problem:
Each cell of a $3 \times 3$ grid is labeled with a digit in the set $\{1,2,3,4,5\}$. Then, the maximum entry in each row and each column is recorded. Compute the number of labelings for which every digit from $1$ to $5$ is recorded at least once. | [
"Solution:\n\nWe perform casework by placing the entries from largest to smallest.\n\n- The grid must have exactly one $5$ since an entry equal to $5$ will be the maximum in its row and in its column. We can place this in $9$ ways.\n- An entry equal to $4$ must be in the same row or column as the $5$; otherwise, it... | United States | HMMT February 2023 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | final answer only | 2664 | |
0dt7 | Find all positive integers $m, n$ satisfying $n! + 2^{n-1} = 2^m$. | [
"We first check that $n = 1, 2, 4$ are solutions among $n \\le 4$, with $m = 1, 2, 5$ respectively.\n\nWe now assume that $n \\ge 5$. By Legendre's formula, we know that $v_2(n!) = n - s_2(n)$ where $s_2(n)$ is the number of non-zero digits in the binary representation of $n$. Thus $v_2(n!) \\le n - 1$.\n\nIf $v_2(... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (m, n) = (1, 1), (2, 2), (5, 4) | |
06dy | For $n \ge 2$, let $a_1, a_2, \dots, a_n, a_{n+1}$ be positive and $a_2 - a_1 = a_3 - a_2 = \dots = a_{n+1} - a_n \ge 0$. Prove that
$$
\frac{1}{a_2^2} + \frac{1}{a_3^2} + \dots + \frac{1}{a_n^2} \le \frac{n-1}{2} \cdot \frac{a_1 a_n + a_2 a_{n+1}}{a_1 a_2 a_n a_{n+1}}
$$
Determine when equality holds. | [
"Let $d = a_j - a_{j-1} \\ge 0$. If $d > 0$, then\n$$\n\\frac{1}{a_k^2} < \\frac{1}{a_k^2 - d^2} = \\frac{1}{a_{k-1}a_{k+1}} = \\frac{a_{k+1} - a_{k-1}}{2d a_{k-1} a_{k+1}} = \\frac{1}{2d} \\left( \\frac{1}{a_{k-1}} - \\frac{1}{a_{k+1}} \\right)\n$$\nfor any $k > 1$. Therefore, we have\n$$\n\\begin{aligned}\n\\sum_... | Hong Kong | CHKMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | Equality holds if and only if all terms are equal. | |
0ldd | Find all real numbers $a$ such that there exists a function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying the following conditions
i) $f(1) = 2016$;
ii) $f(x + y + f(y)) = f(x) + ay$ for all real numbers $x, y$. | [
"For $a = 0$, we can check that the function $f(x) = 2016$ for all real numbers $x$ is satisfied.\n\nNow we consider the case $a \\neq 0$. By plugging $x = -f(y)$ in condition ii), we have\n$$\nf(y) = f(-f(y)) + ay\n$$\nfor all real numbers $y$. Hence, $f$ is injective.\n\nNext, by letting $y = 0$ in ii), we obtain... | Vietnam | VMO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | a = 0 or a = 4066272 | |
0k3f | Problem:
In the game of Connect Four, there are seven vertical columns which have spaces for six tokens. These form a $7 \times 6$ grid of spaces. Two players White and Black move alternately. A player takes a turn by picking a column which is not already full and dropping a token of their color into the lowest unoccu... | [
"Solution:\n\nAnswer: 0.0025632817"
] | United States | HMMT February 2018 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | final answer only | 0.0025632817 | |
0cg6 | Find all the positive integers $a$ and $b$, such that $\frac{7^a - 5^b}{8}$ is a prime number. | [
"For any natural number $k$, we have $5^{2k} = M_8 + 1$, $5^{2k+1} = M_8 + 5$, $7^{2k} = M_8 + 1$ and $7^{2k+1} = M_8 + 7$, therefore from $8 \\mid 7^a - 5^b$ we deduce that $a$ and $b$ are even. Denote $a = 2m$, $b = 2n$, with $m$ and $n$ positive integers. Then $7^{2m} - 5^{2n} = 8p$, i.e. $(7^m - 5^n)(7^m + 5^n)... | Romania | 74th NMO Selection Tests for JBMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (2, 2) | |
0jzr | Problem:
Let $p$ be a prime. A complete residue class modulo $p$ is a set containing at least one element equivalent to $k \pmod{p}$ for all $k$.
a. Show that there exists an $n$ such that the $n$th row of Pascal's triangle forms a complete residue class modulo $p$.
b. Show that there exists an $n \leq p^{2}$ such t... | [
"Solution:\n\nWe use the following theorem of Lucas:\nTheorem. Given a prime $p$ and nonnegative integers $a, b$ written in base $p$ as $a=\\{\\overline{a_{n}} a_{n-1} \\ldots a_{0}\\}_{p}$ and $b=\\overline{b_{n} b_{n-1} \\ldots b_{0}}$ respectively, where $0 \\leq a_{i}, b_{i} \\leq p-1$ for $0 \\leq i \\leq n$, ... | United States | February 2017 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
009o | Several coins are divided once into 200 groups, and then once again into 300 groups. Call a coin *special* if it is in a group of smaller size in the second division than in the first division. Find the minimum number of special coins. | [
"The least number of special coins is 101. Example with exactly 101 special coins: The first division has 200 groups with 101 coins each; the second division is obtained by dividing one of these groups into 101 groups of 1 coin.\nLet $x_1 \\le x_2 \\le \\dots \\le x_{200}$ be the sizes of the 200 groups in the firs... | Argentina | NATIONAL XXX OMA | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 101 | |
08c9 | Problem:
Il quadrilatero $ABCD$ ha le diagonali perpendicolari. Si sa inoltre che $AB=100$, $BC=120$, $CD=75$. Determinare la lunghezza di $AD$.
(A) 30
(B) $24 \sqrt{2}$
(C) $20 \sqrt{3}$
(D) 35
(E) $\frac{125}{2}$ | [
"Solution:\n\nLa risposta è $(\\mathbf{D})$. Siano, come in figura, $K$ il punto d'incontro delle diagonali, $x_{1}, x_{2}$ le lunghezze dei segmenti $AK, KC$ e $y_{1}, y_{2}$ quelle dei segmenti $BK, KD$ rispettivamente. Dal momento che tutti gli angoli in $K$ sono retti, per il teorema di Pitagora si ha\n$$\n\\le... | Italy | Gara di Febbraio | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | D | |
0gms | In a country with $n+1$ cities, there are two-way flights between some of these cities. A two-way flight between cities $A$ and $B$ means that within the same day there is a flight from $A$ to $B$ as well as one from $B$ to $A$, while there is no one-way flight from a city to another. There may be more than one two-way... | [] | Turkey | XIII. National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Maximum number of two-way flights: n(n+1)/2. The maximizing schedules are exactly those in which every pair of non-capital cities has exactly one two-way flight between them, and each non-capital city has exactly one two-way flight to the capital; no other flights occur. | |
04ni | Find all prime numbers $p$ for which there exists a positive integer $m$ such that the number $p^m + 4$ is a square of some positive integer. | [] | Croatia | Croatia_2018 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | p = 2 or p = 5 | |
0gsv | At least $d$ coefficients of a polynomial $P(x)$ of degree $d$ with real coefficients are equal to $1$. Find the maximal value of $d$ if $P(x)$ has $d$ real roots.
*Note: Roots of $P(x)$ need not be distinct.* | [
"The polynomial $x^4 + x^3 - 4x^2 + x + 1 = (x-1)^2(x^2 + 3x + 1)$ satisfies the conditions. Let us show that for $d \\ge 5$ there is no polynomial satisfying given conditions.\n\n**Solution 1.** Let $x_1, x_2, \\dots, x_d$ be the roots, $S_k$ be the sum of all $k$-tuple products of roots. Using Vieta theorem, we g... | Turkey | Turkish Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Descartes' Rule of Signs",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 4 | |
0h5r | Find all integers $a, b$ for which there exist integers $x, y$ such that the following equation holds:
$$
8x^4 + 8y^4 = a^4 + 6a^2b^2 + b^4.
$$ | [
"If $a, b$ have the same parity, define $x, y$ as:\n$$\nx = \\frac{a+b}{2}, \\quad y = \\frac{a-b}{2}.\n$$\nThen they are obviously integers, and by substituting them we verify that the equation does hold.\n\nIf $a, b$ don't have the same parity, the right-hand side of the equation is an odd integer. For example, i... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | All integer pairs a, b with the same parity. |
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