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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0c79 | Consider the set $A = \{1, 2, 3, \dots, 100\}$.
a) Give an example of a set $B$ having 11 elements from $A$, having the property: *from any two elements from $B$, the greatest common divisor of them is at least 9*.
b) Prove that, for any subset $C$ of $A$ having 11 elements, there are two distinct elements of $C$ suc... | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | {9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99} | |
01vh | The circle $S_1$ intersects the hyperbola $y = \frac{1}{x}$ at four points $A$, $B$, $C$ and $D$, and the other circle $S_2$ intersects the same hyperbola at four points $A$, $B$, $F$ and $G$. It is known that the radii of circles $S_1$ and $S_2$ are equal.
Prove that the points $C$, $D$, $F$ and $G$ are the vertices o... | [
"Let the abscissae of the points $A$, $B$, $C$, $D$, $F$ and $G$ be $a$, $b$, $c$, $d$, $f$ and $g$ respectively. If $O_1(\\alpha, \\beta)$ is the center of the circle $S_1$, the coordinates of the points $A$, $B$, $C$ and $D$ satisfy the system of equations\n$$\n\\begin{cases} (x - \\alpha)^2 + (y - \\beta)^2 = R^... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof only | null | |
0gno | Let $ABC$ be a triangle with $m(\angle B) = 90^\circ$. The incircle of $\triangle ABC$ is tangent to the side $BC$ at point $D$. Let the intersection of lines $XZ$ and $AD$ be $K$, where $X$ and $Z$ are the centers of incircles of $\triangle ABD$ and $\triangle ADC$, respectively. $XZ$ intersects the circumscribed circ... | [
"Let the incircle of $\\triangle ABD$ be tangent to $BC$ at $T$ and the incircle of $\\triangle ADC$ be tangent to $BC$ at $S$. The radii of incircles of $\\triangle ABD$ and $\\triangle ADC$ are $r_1$ and $r_2$, respectively. Then\n$$\n\\begin{aligned}\nDS &= \\frac{AD + DC - AC}{2} = \\frac{AD + BC - BD - AC}{2} ... | Turkey | Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | English | proof only | null | |
0jw6 | Problem:
A random number generator will always output $7$. Sam uses this random number generator once. What is the expected value of the output? | [
"Solution:\nThe only output is $7$, so the expected value is $7$."
] | United States | February 2017 | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 7 | |
06p9 | Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that
$$
f(x+f(y))=f(x+y)+f(y)
$$
for all $x, y \in \mathbb{R}^{+}$. (Symbol $\mathbb{R}^{+}$ denotes the set of all positive real numbers.) | [
"First we show that $f(y)>y$ for all $y \\in \\mathbb{R}^{+}$. Functional equation (1) yields $f(x+f(y))>f(x+y)$ and hence $f(y) \\neq y$ immediately. If $f(y)<y$ for some $y$, then setting $x=y-f(y)$ we get\n$$\nf(y)=f((y-f(y))+f(y))=f((y-f(y))+y)+f(y)>f(y),\n$$\ncontradiction. Therefore $f(y)>y$ for all $y \\in \... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | f(x) = 2x for all positive real x | |
0236 | Problem:
Considere a soma das três tabelas a seguir. A primeira representa $n$ linhas, sendo a primeira com $n$ números iguais a $n$, a segunda com $n-1$ números iguais a $n-1$ e assim por diante. Na segunda, temos uma distribuição de números parecida, mas em colunas em vez de linhas. Já na terceira, temos estes número... | [
"Solution:\n\na) Ao virar uma delas de ponta cabeça e juntar com outra, formamos um retângulo $n \\times (n+1)$. Como este retângulo possui o dobro de números de uma tabela, cada uma delas possui $\\frac{n(n+1)}{2}$ números.\n\nb) Cada número $k$ aparece $k$ vezes na primeira tabela, $k$ vezes na segunda e $k$ veze... | Brazil | NÍVEL 3 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1^2 + 2^2 + 3^2 + ... + n^2 = (1/6) n (n+1) (2n+1) | |
09x1 | Problem:
Laat $a, b \geq 2$ positieve gehele getallen met $\operatorname{ggd}(a, b)=1$ zijn. Zij $r$ de kleinste positieve waarde die aangenomen wordt bij een uitdrukking van de vorm $\frac{a}{b}-\frac{c}{d}$, met $c$ en $d$ positieve gehele getallen die voldoen aan $c \leq a$ en $d \leq b$. Bewijs dat $\frac{1}{r}$ g... | [
"Solution:\n\nWe laten eerst zien dat het mogelijk is om $c$ en $d$ zo te kiezen dat $\\frac{a}{b}-\\frac{c}{d}=\\frac{1}{bd}$.\n\nOmdat $\\operatorname{ggd}(a, b)=1$, bestaat er een multiplicatieve inverse $b^{-1}$ van $b$ modulo $a$. Kies nu $c$ met $1 \\leq c \\leq a$ zo dat $c \\equiv -b^{-1} \\bmod a$. Er geld... | Netherlands | IMO-selectietoets I | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0f4d | Problem:
Let $S$ be the set of points $(x, y)$ given by $y \leq -x^2$ and $y \geq x^2 - 2x + a$. Find the area of the rectangle with sides parallel to the axes and the smallest possible area which encloses $S$. | [] | Soviet Union | 15th ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | Let d = sqrt(1 - 2a). For a ≤ 1/2, the set S is nonempty and the minimal enclosing axis-parallel rectangle has area:
- If a ≤ 0: area = (1 - a) d = (1 - a) sqrt(1 - 2a).
- If 0 ≤ a ≤ 1/2: area = d^2 = 1 - 2a.
For a > 1/2, S is empty, so the minimal area is 0. | |
0f1a | Problem:
The finite sequence $\{a_n\}$ has each member $0$, $1$ or $2$. A move involves replacing any two unequal members of the sequence by a single member different from either. A series of moves results in a single number. Prove that no series of moves can terminate in a (single) different number. | [
"Solution:\n\nSuppose we start with $a$ $0$s, $b$ $1$s and $c$ $2$s. Each move changes the parity of all of $a$, $b$, $c$. Each move reduces the length of the sequence by $1$, so there must be $a + b + c - 1$ moves in all. Hence the total number of $0$s, the total number of $1$s and the total number of $2$s all hav... | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
08cj | Problem:
Quante sono le quaterne di interi positivi $(a, b, x, y)$ tali che $x+y=a \cdot b$ e $a+b=x \cdot y$ ?
(A) 1
(B) 5
(C) 9
(D) 17
(E) Infinite | [
"Solution:\n\nLa risposta è (C). Se $a, b, x, y$ sono tutti maggiori o uguali a $2$, allora $a b \\geq a+b = x y \\geq x+y = a b$, per cui ogni disuguaglianza deve essere un'uguaglianza e l'unica quaterna possibile è $(2,2,2,2)$.\n\nSe invece uno dei quattro numeri è uguale ad $1$, diciamo per esempio $b=1$, allora... | Italy | GARA di FEBBRAIO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | C | |
0iow | Problem:
Find the sum of all solutions for $x$ :
$$
\begin{aligned}
x y & = 1 \\
x + y & = 3
\end{aligned}
$$ | [
"Solution:\nSubstitute $3 - x$ in for $y$ into the first equation:\n$$\nx(3 - x) = 1 \\Leftrightarrow x^{2} - 3x + 1 = 0\n$$\nThis equation has two distinct roots, each of which corresponds to a possible solution $x$. The sum of the roots of the quadratic equation $a x^{2} + b x + c = 0$ is $\\frac{-b}{a}$, which i... | United States | 1st Annual Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 3 | |
061s | Problem:
In der Ebene liegen $n$ abgeschlossene Kreisscheiben $K_{1}, K_{2}, \ldots, K_{n}$ mit gleichem Radius $r$. Jeder Punkt der Ebene ist dabei in höchstens $2003$ dieser Kreisscheiben enthalten. Man beweise, dass jede Kreisscheibe $K_{i}$ höchstens $14020$ andere Kreisscheiben schneidet. | [
"Solution:\nWir führen einen Beweis durch Widerspruch. Dazu nehmen wir zusätzlich zur Voraussetzung an, dass eine Kreisscheibe (oBdA sei dies $K_{1}$) mindestens $14021$ andere Kreisscheiben schneidet. Die Mittelpunkte dieser Scheiben liegen dann offensichtlich alle in einer Kreisscheibe mit Radius $2r$ um den Mitt... | Germany | Auswahlwettbewerb zur IMO | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof only | null | |
0kb3 | Consider coins with positive real denominations not exceeding $1$. Find the smallest $C > 0$ such that the following holds: if we are given any $100$ such coins with total value $50$, then we can always split them into two stacks of $50$ coins each such that the absolute difference between the total values of the two s... | [
"The answer is $C = \\frac{50}{51}$. The lower bound is obtained if we have $51$ coins of value $\\frac{1}{51}$ and $49$ coins of value $1$. We now present two (similar) proofs that this $C = \\frac{50}{51}$ suffices.\n\n**First proof (original)** Let $a_1 \\le \\dots \\le a_{100}$ denote the values of the coins in... | United States | USA TSTST | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 50/51 | |
0eaq | In a zoo rabbits, parrots and snakes have 24 heads, 14 wings and 62 legs. How many snakes are there in this zoo?
(A) 5 (B) 6 (C) 7 (D) 8 (E) 9 | [
"Rabbits and snakes have no wings and parrots have two wings each, so there are $14 \\div 2 = 7$ parrots in the zoo. Thus, the parrots account for $7 \\cdot 2 = 14$ legs. Snakes have no legs, so the rabbits have $62 - 14 = 48$ legs altogether. Each rabbit has $4$ legs, so there are $48 \\div 4 = 12$ rabbits in the ... | Slovenia | National Math Olympiad in Slovenia | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | A | |
0gdo | 令 $\mathbb{R}$ 代表實數所成的集合。試找出所有函數 $f : \mathbb{R} \to \mathbb{R}$ 滿足對所有實數 $x, y \in \mathbb{R}$，都有
$$
f(xy + xf(x)) = f(x)(f(x) + f(y)).
$$
Let $\mathbb{R}$ denote the set of all real numbers. Determine all functions $f : \mathbb{R} \to \mathbb{R}$ such that, for all real numbers $x$ and $y$,
$$
f(xy + xf(x)) = f(x)(f(... | [
"定義 $P(a, b)$ 為將 $x = a, y = b$ 代入函數方程。\n\n(1) $P(0, y)$ : $f(0) = f(0)(f(0) + f(y))$，當 $f(0) \\neq 0$ 有 $f$ 是常數函數，此時 $f(x) = \\frac{1}{2}, \\forall x \\in \\mathbb{R}$ 是一個解。\n\n(2) 若 $f(a) = 0$，$P(a, y)$ : $f(ay) = 0$，得到 $a = 0$ 或是 $f$ 全為 $0$，其中 $f$ 全為 $0$ 確實是一解。\n\n(3) $P(x, -f(x))$ : $0 = f(x)(f(x) + f(-f(x)))$，... | Taiwan | 2020 Taiwan IMO 2J | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x for all real x; f(x) = 0 for all real x; f(x) = 1/2 for all real x | |
0729 | Problem:
Let $R$ denote the circumradius of a triangle $ABC$; $a, b, c$ its sides $BC, CA, AB$; and $r_{a}, r_{b}, r_{c}$ its exradii opposite $A, B, C$. If $2R \leq r_{a}$, prove that
(i) $a > b$ and $a > c$;
(ii) $2R > r_{b}$ and $2R > r_{c}$. | [
"Solution:\nWe know that $2R = \\frac{abc}{2\\triangle}$ and $r_{a} = \\frac{\\triangle}{s-a}$, where $a, b, c$ are the sides of the triangle $ABC$, $s = \\frac{a+b+c}{2}$ and $\\triangle$ is the area of $ABC$. Thus the given condition $2R \\leq r_{a}$ translates to\n$$\nabc \\leq \\frac{2\\triangle^{2}}{s-a}\n$$\n... | India | INMO 2004 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
032c | Problem:
Solve the inequality
$$
\sqrt{x^{2}-1}+\sqrt{2 x^{2}-3}+x \sqrt{3}>0
$$ | [
"Solution:\nThe inequality makes sense for $x \\in \\left(-\\infty,-\\frac{\\sqrt{6}}{2}\\right] \\cup \\left[\\frac{\\sqrt{6}}{2},+\\infty\\right)$. All $x$ in the second interval are solutions of the inequality.\n\nLet $x \\in \\left(-\\infty,-\\frac{\\sqrt{6}}{2}\\right]$. Then the inequality is equivalent to\n$... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (-∞, -\frac{\sqrt{5+\sqrt{33}}}{2}) \cup [\frac{\sqrt{6}}{2}, +∞) | |
053w | A positive integer $n$ is interesting, if for some positive integer $m$ and positive integers $a$, $b$ that are smaller than $m$, $\frac{m^2}{ab} = n$. For example, $10$ is interesting because $\frac{20^2}{4 \cdot 10} = 10$. Find the smallest interesting integer. | [
"For $n = 2$ we can take $m = 12$, $a = 8$ and $b = 9$, because $\\frac{12^2}{8 \\cdot 9} = \\frac{144}{72} = 2$.\n\nOn the other hand, $1$ is not interesting, because if $\\frac{m^2}{ab} = 1$, or $m^2 = ab$, then $a$ and $b$ cannot both be less than $m$ at the same time."
] | Estonia | Estonian Math Competitions | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 2 | |
0gn2 | If $a_1, a_2, \dots, a_n$ are positive real numbers such that
$$
t = a_1 + a_2 + \dots + a_n = a_1^2 + a_2^2 + \dots + a_n^2
$$
and $n \ge 2$, then show that
$$
\sum_{i \neq j} \frac{a_i}{a_j} \geq \frac{(n-1)^2 t}{t-1}.
$$ | [
"By the Cauchy-Schwarz inequality,\n$$\n\\sum_{i \\neq j} \\frac{a_i}{a_j} \\cdot \\sum_{i \\neq j} a_i a_j \\geq \\left( \\sum_{i \\neq j} a_i \\right)^2 = \\left( (n-1) \\sum_{i=1}^{n} a_i \\right)^2 = (n-1)^2 t^2.\n$$\nOn the other hand,\n$$\n\\sum_{i \\neq j} a_i a_j = \\left( \\sum_{i=1}^{n} a_i \\right)^2 - \... | Turkey | 14th Turkish Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0j8w | Problem:
Unit circle $\Omega$ has points $X$, $Y$, $Z$ on its circumference so that $XYZ$ is an equilateral triangle. Let $W$ be a point other than $X$ in the plane such that triangle $WYZ$ is also equilateral. Determine the area of the region inside triangle $WYZ$ that lies outside circle $\Omega$. | [
"Solution:\n\nAnswer: $\\frac{3 \\sqrt{3}-\\pi}{3}$\n\nLet $O$ be the center of the circle. Then, we note that since $\\angle WYZ = 60^{\\circ} = \\angle YXZ$, that $YW$ is tangent to $\\Omega$. Similarly, $WZ$ is tangent to $\\Omega$.\n\nNow, we note that the circular segment corresponding to $YZ$ is equal to $\\f... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | (3√3 − π)/3 | |
0097 | Given a finite sequence with terms belonging to the set $A = \{0, 1, \ldots, 121\}$, an allowed operation consists in replacing each term by a number of the set $A$ so that equal terms are replaced by equal numbers, and different terms are replaced by different numbers. (Some terms may remain unreplaced.) The goal is t... | [] | Argentina | XXIX Olimpíada Matemática Argentina National Round | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
0k6f | Problem:
Compute the sum of all positive integers $n$ for which
$$
9 \sqrt{n} + 4 \sqrt{n+2} - 3 \sqrt{n+16}
$$
is an integer. | [
"Solution:\nFor the expression to be an integer at least one of $n$ and $n+2$ must be a perfect square. We also note that at most one of $n$ and $n+2$ can be a square, so exactly one of them is a square.\n\nCase 1: $n$ is a perfect square. By our previous observation, it must be that $4 \\sqrt{n+2} = 3 \\sqrt{n+16}... | United States | HMMT November 2019 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 18 | |
06e5 | Let $p$ be a prime number and $n$ a positive integer such that $n \ge p$. Show that $\binom{n}{p} - \lfloor \frac{n}{p} \rfloor$ is divisible by $p$. (Here the combinatorial symbol $\binom{n}{p} = \frac{n!}{p!(n-p)!}$ and $\lfloor x \rfloor$ stands for the largest integer less than or equal to $x$.) | [
"This is an immediate consequence of Lucas' theorem. We give an elementary proof as follows.\n\nLet $n = ap + r$ where $0 \\le r \\le p - 1$. Then we have\n$$\n\\binom{n}{p} = \\frac{n(n-1)\\cdots(n-p+1)}{p!} \\\\ = \\frac{(ap+r)(ap+r-1)\\cdots(ap+1)(a)(ap-1)(ap-2)\\cdots(ap+r-p+1)}{(p-1)!}\n$$\nNote that the numer... | Hong Kong | IMO HK TST | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0g6a | 設圓 $O$ 為 $\Delta ABC$ 的外接圓, $\angle A$ 的角平分線交圓 $O$ 於第二點 $P$, 並交 $BC$ 於 $D$。由 $D$ 對 $AB$ 作垂線, 交 $AB$ 於 $E$, 並交圓 $O$ 於 $Q$, 使得 $E$ 在 $DQ$ 線段上。再連 $PQ$ 交 $BC$ 於 $F$, 連 $AF$ 交 $CQ$ 於 $G$。試證: $EG$ 平行 $FC$. | [
"解：首先，由於\n$$\n\\angle PDB = \\frac{1}{2}(\\widehat{CA} + \\widehat{BP}) = \\frac{1}{2}(\\widehat{CA} + \\widehat{CP}) = \\angle AQP,\n$$\n因此 $A$, $D$, $F$, $Q$ 四點共圓。故, $\\angle AQE = \\angle AQD = \\angle AFD = \\angle GFC$.\n$$\n\\angle EAQ = \\angle BAQ = \\angle BCQ = \\angle FCG,\n$$\n故 $\\angle CGF = \\angle A... | Taiwan | 二〇一二數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0iw8 | Problem:
If $\tan x + \tan y = 4$ and $\cot x + \cot y = 5$, compute $\tan(x+y)$. | [
"Solution:\n\nWe have $\\cot x + \\cot y = \\frac{\\tan x + \\tan y}{\\tan x \\tan y}$, so $\\tan x \\tan y = \\frac{4}{5}$. Thus, by the tan sum formula,\n\n$$\n\\tan(x+y) = \\frac{\\tan x + \\tan y}{1 - \\tan x \\tan y} = \\frac{4}{1 - \\frac{4}{5}} = \\frac{4}{\\frac{1}{5}} = 20.\n$$"
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 20 | |
0bkj | Let $ABC$ be a triangle and let $X, Y, Z$ be interior points on the sides $BC, CA, AB$, respectively. Show that the magnified image of the triangle $XYZ$ under a homothety of factor $4$ from its centroid covers at least one of the vertices $A, B, C$. | [
"Since the problem is of an affine nature, we may (and will) assume that the triangle $XYZ$ is equilateral. The triangle $ABC$ has at least one vertex angle, say at $A$, greater than or equal to $60^\\circ$, so $A$ is covered by the closed circumdisc $OYZ$, where $O$ is the center of the triangle $XYZ$. Since the l... | Romania | 65th NMO Selection Tests for BMO and IMO | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
09ki | Let $p \ge 5$ be a prime number and $f(x) = (x-1)(x^3-1)\dots(x^{p-4}-1)(x^{p-2}-1)$. Show that $f(n^k) - f(n)$ is divisible by $p$ for any positive integer $k$ relatively prime to $p-1$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof only | null | |
0czu | Let $f_{n} = 2^{2^{n}} + 1$, $n = 1, 2, 3, \ldots$, be the Fermat's numbers. Find the least real number $C$ such that
$$
\frac{1}{f_{1}} + \frac{2}{f_{2}} + \frac{2^{2}}{f_{3}} + \ldots + \frac{2^{n-1}}{f_{n}} < C
$$
for all positive integers $n$. | [
"We have for $k = 1, 2, \\ldots$\n$$\n2^{2^{k+1}} - 1 = \\left(2^{2^{k}}\\right)^{2} - 1 = \\left(2^{2^{k}} - 1\\right)\\left(2^{2^{k}} + 1\\right),\n$$\nhence we get\n$$\n\\frac{1}{2^{2^{k+1}} - 1} = \\frac{1}{2} \\left( \\frac{1}{2^{2^{k}} - 1} - \\frac{1}{2^{2^{k}} + 1} \\right)\n$$\nIt follows\n$$\n\\begin{equa... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | 1/3 | |
0319 | Problem:
Three of $n$ equal balls are radioactive. A detector measures radioactivity. Any measurement of a set of balls gives as a result whether 0, 1 or more than 1 balls are radioactive. Denote by $L(n)$ the least number of measurements that one needs to find the three radioactive balls.
a) Find $L(6)$.
b) Prove t... | [
"Solution:\n\na. We shall prove that 4 measurements are enough. Denote the balls by $1,2,3,4,5,6$ and measure consecutively $\\{1,2\\}$, $\\{1,3\\}$, $\\{1,4\\}$ and $\\{1,5\\}$.\n\nCase 1. If all the measurements show radioactivity, then $1$ is a radioactive ball. If $\\{1, a\\}$, $a=2,3,4,5$, contains two radioac... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | L(6) = 4 and for all n, L(n) ≤ (n+5)/2 | |
0jvd | Problem:
Let $ABCD$ be a convex quadrilateral. Assume that the incircle of triangle $ABD$ is tangent to $\overline{AB}$, $\overline{AD}$, $\overline{BD}$ at points $W$, $Z$, $K$. Also assume that the incircle of triangle $CBD$ is tangent to $\overline{CB}$, $\overline{CD}$, $\overline{BD}$ at points $X$, $Y$, $K$. Pro... | [
"Solution:\n\nFrom the concurrence of the Gergonne point, it follows that lines $WZ$, $XY$, and $BD$ concur at the harmonic conjugate $T$ of $K$ with respect to $\\overline{BC}$. (One can also see the concurrence directly by applying Ceva and Menelaus.) Then $TK^2 = TW \\cdot TZ = TX \\cdot TY$, as desired."
] | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Ad... | null | proof only | null | |
0dn9 | Problem:
Одредити све $n \in \mathbb{N}$ за које је могуће поделити скуп $\{1,2, \ldots, 3 n\}$ на $n$ дисјунктних трочланих подскупова облика $\{a, b, c\}$ у којима су $b-a$ и $c-b$ различити бројеви из скупа $\{n-1, n, n+1\}$. | [
"Solution:\n\nТражена партиција скупа $\\{1,2, \\ldots, 3 n\\}$ одговара партицији темена правилног $3 n$-угла $P_{1} P_{2} \\ldots P_{3 n}$ на тројке $\\left\\{A_{i}, B_{i}, C_{i}\\right\\}$ такве да су углови сваког од троуглова $A_{i} B_{i} C_{i}$ једнаки $\\frac{n-1}{3 n} \\pi, \\frac{n}{3 n} \\pi$ и $\\frac{n+... | Serbia | Serbian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | All even n | |
0l6w | Let $m$ and $n$ be positive integers with $m \ge n$. There are $m$ cupcakes of different flavors arranged around a circle and $n$ people who like cupcakes. Each person assigns a nonnegative real number score to each cupcake, depending on how much they like the cupcake. Suppose that for each person $P$, it is possible t... | [] | United States | USAMO | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Graph Theory > Menger's theorem / max-flow, min-cut",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0a3p | Let $ABCD$ be a parallelogram and let $M$ be the intersection of its diagonals. The circumcircle of $\triangle ABM$ intersects the line segment $AD$ in $E \ne A$ and the circumcircle of $\triangle EMD$ intersects the line segment $BE$ in the point $F \ne E$.
Prove that $\angle ACB = \angle DCF$. | [
"We first show that $CBFD$ is a cyclic quadrilateral. Note that\n$$\n\\angle BCD = \\angle BAD \\quad \\text{(parallelogram)}\n= \\angle BAE = 180^\\circ - \\angle EMB \\quad \\text{(EABM is cyclic)}\n= \\angle EMD = \\angle EFD \\quad \\text{(inscribed angle theorem in EFMD)}\n= 180^\\circ - \\angle BFD\n$$\nThere... | Netherlands | IMO Team Selection Test 3 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
04dy | Determine the complex number $z$ such that
$$
\text{Re} \frac{1}{1-z} = 2 \quad \text{and} \quad \text{Im} \frac{1}{1-z} = -1
$$ | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | final answer only | (3 - i)/5 | |
016h | Problem:
A unit square is cut into $m$ quadrilaterals $Q_{1}, \ldots, Q_{m}$. For each $i=1, \ldots, m$ let $S_{i}$ be the sum of the squares of the four sides of $Q_{i}$. Prove that
$$
S_{1}+\ldots+S_{m} \geq 4
$$ | [] | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0aka | Let $k$ be a semicircle with center $O$ and diameter $AB$. Let $C$ be a point on $k$ such that $CO \perp AB$. The symmetrical of $\angle ABC$ intersects $k$ at the point $D$. Let $E$ be the point of $AB$ such that $DE \perp AB$ and let $F$ be the midpoint of $CB$. Prove that the quadrilateral $EFCD$ is cyclic. | [
"\n\nNote that the triangle $ABC$ is isosceles right triangle. Let $CD \\cap AB = \\{H\\}$. From\n$$\n\\angle AED = \\angle ADB = 90^\\circ \\text{ and } \\angle DAE = \\angle DAB\n$$\nfollows that $\\triangle ADE \\sim \\triangle ABD$. Since $ABCD$ is cyclic, it follows\n$$\n\\angle ADC = ... | North Macedonia | Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates"
] | English | proof only | null | |
0l9w | Solve the equation system
$$
\begin{cases} 1 - \frac{12}{y+3x} = \frac{2}{\sqrt{x}} \\ 1 + \frac{12}{y+3x} = \frac{6}{\sqrt{x}} \end{cases}
$$ | [
"The necessary conditions for the given equation system are $x, y > 0; y+3x \\neq 0$. The given equation system is equivalent to the following equation systems\n$$\n\\begin{cases} 1 - \\frac{12}{y+3x} = \\frac{2}{\\sqrt{x}} \\\\ 1 + \\frac{12}{y+3x} = \\frac{6}{\\sqrt{x}} \\end{cases} \\Rightarrow \\begin{cases} \\... | Vietnam | Vijetnam 2007 | [
"Algebra > Intermediate Algebra > Other"
] | English | proof and answer | (16, -24) | |
083s | Problem:
Date nel piano due rette parallele $r, s$ e due punti $P, Q$ con $P \in r$ e $Q \in s$, si considerino coppie di circonferenze $\left(C_{P}, C_{Q}\right)$, la prima tangente a $r$ in $P$ e la seconda tangente a $s$ in $Q$, che siano anche tangenti esternamente tra loro, in un punto che chiamiamo $T$. Determin... | [
"Solution:\n\nIl luogo cercato è l'unione del segmento aperto $P Q$ e della parte della circonferenza di diametro $P Q$ che si trova al di fuori della striscia delimitata da $r$ ed $s$.\nCiascuna circonferenza può essere tangente alla rispettiva retta in due modi: in un caso interseca la striscia $S$ compresa tra l... | Italy | Olimpiade Italiana di Matematica | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02fe | $ABCD$ is a quadrilateral with a circumcircle center $O$ and an inscribed circle center $I$. The diagonals intersect at $S$. Show that if two of $O$, $I$, $S$ coincide, then it must be a square. | [
"If $S = O$ then $AC = BD = 2R$, where $R$ is the circumradius of $ABCD$ and $S$ is the midpoint of both $AC$ and $BD$. This means that $ABCD$ is an inscribed parallelogram. The sum of its opposite angles, which are congruent, is $180^\\circ$, so all angles of $ABCD$ are right, that is, $ABCD$ is a rectangle. But t... | Brazil | XVII OBM | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0182 | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$$
f(x + f(y)) - f(x) = (x + f(y))^3 - x^3
$$
for all $x, y \in \mathbb{R}$. | [
"The function $f$ is either $f(x) = 0$ or $f(x) = x^3 + c$ with an arbitrary $c \\in \\mathbb{R}$.\n\nProof: Obviously $f(x) = 0$ is a solution, so let us assume that a real number $a \\neq 0$ belongs to the range of $f$. Let us first assume $a > 0$. Taking $x = -f(y)$ in the given equation we get\n$$\nf(-f(y)) = f... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = 0 for all x, or f(x) = x^3 + c for some real constant c | |
00fg | Determine all positive integers $n$ for which the equation
$$
x^{n}+(2+x)^{n}+(2-x)^{n}=0
$$
has an integer as a solution. | [
"If $n$ is even, $x^{n}+(2+x)^{n}+(2-x)^{n}>0$, so $n$ is odd.\n\nFor $n=1$, the equation reduces to $x+(2+x)+(2-x)=0$, which has the unique solution $x=-4$.\n\nFor $n>1$, notice that $x$ is even, because $x$, $2-x$, and $2+x$ have all the same parity. Let $x=2y$, so the equation reduces to\n$$\ny^{n}+(1+y)^{n}+(1-... | Asia Pacific Mathematics Olympiad (APMO) | APMO 1993 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 1 | |
0i66 | Problem:
What is the minimum vertical distance between the graphs of $2+\sin (x)$ and $\cos (x)$? | [
"Solution:\n\nThe derivative of $2+\\sin (x)-\\cos (x)$ is $\\cos x+\\sin x$, which in the interval $0 \\leq x<2 \\pi$ is zero at $x=\\frac{3 \\pi}{4}, \\frac{7 \\pi}{4}$. At $\\frac{7 \\pi}{4}$, when $\\sin (x)$ is negative and $\\cos (x)$ is positive, the distance reaches its minimal value of $2-\\sqrt{2}$."
] | United States | Harvard-MIT Math Tournament | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications",
"Precalculus > Trigonometric functions"
] | null | proof and answer | 2 - sqrt(2) | |
00ve | Determine all natural numbers $n$ such that the inequality
$$
x^n + 2x + 1 \ge 4x^2
$$
holds for every $x > 0$. | [
"Suppose that $n$ is a solution to the problem. The polynomial\n$$\nP(x) = x^n - 4x^2 + 2x + 1\n$$\nclearly has a root at $1$. Therefore we may write $P(x) = (x - 1)Q(x)$ for some polynomial $Q$. Since $P(x) \\ge 0$ for $x > 0$, the polynomial $Q$ changes sign at $1$ and so $Q(1) = 0$. Calculating\n$$\nQ(x) = \\fra... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlisted Problems | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 6 | |
03fe | Points $A$, $B$, $Y$ and $C$ lie in this order on circle $k$ with center $O$, such that $BC = 2$ cm, $\angle BAY = 42^\circ$ and $\angle CAY = 78^\circ$. It is known that the circle $\omega$ through the points $A$, $O$ and $B$ is tangent to the line $BY$. The circle through the points $A$ and $C$, tangent to the line $... | [
"a) Clearly $\\angle BAC = \\angle BAY + \\angle CAY = 120^\\circ$, respectively $\\angle BOC = 360^\\circ - 2\\angle BAC = 120^\\circ$. Thus, if $M$ is the midpoint of $BC$, then $OM \\perp BC$ (because $BO = OC$), $\\angle BOM = 60^\\circ$ and $BM = \\frac{BC}{2} = 1$. Let $BO = x$ and from triangle $BOM$ we have... | Bulgaria | 3 Bulgarian Spring Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof and answer | BO = 2√3/3 cm; ∠YAN = 36° | |
0h2s | $$
\max\{x^2+3x+3, x^{2011}+x^4+x^2+x+1\} \le \min\{1-x-x^2, x^{2011}+x^4+x^2+x+1\},
$$
where $\max\{a,b\} = \begin{cases} a, & \text{if } a \ge b \\ b, & \text{if } a < b \end{cases}$, and $\min\{a,b\} = \begin{cases} b, & \text{if } a \ge b \\ a, & \text{if } a < b \end{cases}$. | [
"For our convenience we denote $P(x) = x^{2011} + x^4 + x^2 + x + 1$. It is clear that\n$$\n\\max\\{x^2+3x+3, P(x)\\} \\ge P(x) \\ge \\min\\{1-x-x^2, P(x)\\},\n$$\nso the inequality from the problem condition can be satisfied only if\n$$\n\\max\\{x^2+3x+3, P(x)\\} = P(x) = \\min\\{1-x-x^2, P(x)\\},\n$$\nand this, i... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials"
] | English | proof and answer | -1 | |
0d4w | Let $ABC$ be a triangle and $D$ a point on the side $BC$. Point $E$ is the symmetric of $D$ with respect to $AB$. Point $F$ is the symmetric of $E$ with respect to $AC$. Point $P$ is the intersection of line $DF$ with line $AC$. Prove that the quadrilateral $AEDP$ is cyclic. | [
"Let $\\alpha = \\angle BAC$ and $\\theta = \\angle BAD$. Because $E$ is the symmetric of $D$ with respect to $AB$, we have $AD = AE$ and $DE$ is perpendicular to $AB$. We deduce that $\\angle EAD = 2\\theta$ and $\\angle DEA = 90^\\circ - \\theta$.\n\nBecause $F$ is the symmetric of $E$ with respect to $AC$, we ha... | Saudi Arabia | SAMC 2015 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English, Arabic | proof only | null | |
0fub | Problem:
Gegeben sind $m \geq 3$ Punkte in der Ebene. Beweise, dass man stets drei dieser Punkte $A, B, C$ auswählen kann, sodass gilt
$$
\Varangle A B C \leq \frac{180^{\circ}}{m} .
$$ | [
"Solution:\n\nDie konvexe Hülle der $m$ Punkte sei ein $k$-Eck. Die Innenwinkelsumme in diesem $k$-Eck beträgt $(k-2) \\cdot 180^{\\circ}$. Für den kleinsten Innenwinkel $\\alpha$ gilt daher die Abschätzung\n$$\n\\alpha \\leq \\frac{k-2}{k} \\cdot 180^{\\circ} \\leq \\frac{m-2}{m} \\cdot 180^{\\circ}\n$$\nSei $P$ d... | Switzerland | SMO Finalrunde | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
023n | Problem:
Rogério coloca seis bolinhas sobre a mesa de modo a formar dois quadrados, como na figura. Ele percebe que havia esquecido de colocar mais uma bolinha. Complete a figura formada pelas bolinhas com essa bolinha a mais, de modo a formar 3 quadrados.
 | [
"Solution:\n\n"
] | Brazil | Desafios | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | final answer only | Place the extra ball at the position that serves as the fourth corner of a square formed with three already placed balls, completing an additional square overlapping the two given ones. | |
0lam | In the plane given a circle $(O)$ and two fixed points $B$, $C$ on the circle, such that $BC$ is not a diameter. Consider a point $A$ moving on $(O)$ in such a way that $AB = AC$ and $A$ is not coincident with $B$, $C$. Denote by $D$ and $E$ the intersections of line $BC$ with the internal bisector and the external bis... | [
"a/ First, we show $MN \\parallel OA$.\nIndeed, without loss of generality assume that $\\overline{ABC} > \\overline{ACB}$. Then\n$$\n\\overline{OAD} = \\overline{OAB} - \\overline{DAB} = \\frac{180^\\circ - \\overline{AOB}}{2} - \\frac{\\overline{BAC}}{2} = 90^\\circ - \\left( \\hat{C} + \\frac{\\overline{BAC}}{2}... | Vietnam | Vietnamese Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalitie... | null | proof and answer | a) The line MN always passes through the fixed point O′, the image of the circle’s center O under translation by the vector 2·OK, where K is the midpoint of BC.
b) The maximal area occurs when A is at the two endpoints of the diameter parallel to BC. | |
05dj | Problem:
Find all positive integers $a$ and $b$ for which there are three consecutive integers at which the polynomial
$$
P(n)=\frac{n^{5}+a}{b}
$$
takes integer values. | [
"Solution:\nDenote the three consecutive integers by $x-1, x$, and $x+1$, so that\n$$\n(x-1)^{5}+a \\equiv 0 \\quad(\\bmod b), \\quad x^{5}+a \\equiv 0 \\quad(\\bmod b), \\quad(x+1)^{5}+a \\equiv 0 \\quad(\\bmod b) .\n$$\nBy computing the differences of the equations in (1) we get\n$$\n\\begin{aligned}\nA & :=(x+1)... | European Girls' Mathematical Olympiad (EGMO) | EGMO | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | b = 1 with any positive integer a; or b = 11 with a congruent to plus or minus 1 modulo 11 | |
0gog | Let $A$ be the set of points in the plane whose coordinates are integers and let
$F$ be the collection of all functions from $A$ to $\{1, -1\}$. We call a function $f$ in $F$ perfect if every function $g$ in $F$ that differs from $f$ at finitely many points satisfies the condition
$$
\sum_{0 < d(P,Q) < 2010} \frac{f(P... | [
"Let $L$ be a line in the plane, and let $\\pi_1$ and $\\pi_2$ be the corresponding open half-planes. We set\n$$\nf_L(P) = \\begin{cases} 1 & \\text{if } P \\in \\pi_1 \\cup L, \\\\ -1 & \\text{if } P \\in \\pi_2. \\end{cases}\n$$\nWe will show that $f_L$ is a perfect function. Then the family $\\{f_L : (0,0) \\in ... | Turkey | Team Selection Test for IMO 2010 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0kph | Problem:
Sofiya and Marquis play a game by taking turns. They form a circle with $2023$ other people, and on each turn Sofiya or Marquis can remove one of their neighbors to the left or to the right from the circle. The person who removes the other player wins. If Sofiya starts, who has the winning strategy? | [
"Solution:\n\nNote that there are an odd number of people in the circle beside Sofiya and Marquis, so Sofiya and Marquis divide the circle into two arcs, one with an even number of people and the other with an odd number of people. Sofiya's winning strategy will be to always remove a neighbor from the even side. Th... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Sofiya | |
00ek | In each cell of a $3 \times 3$ board, there is a lamp and a button. Pressing the button in a cell changes the state of the lamps in its neighboring cells (those that are on turn off and vice versa). The lamp in the same cell as the button does not change its state. Initially, all lamps are *off*.
a. Is it possible, af... | [
"First we make some useful observations that will simplify the problem:\n\n* The order in which the buttons are pressed is not relevant to the final state. We only care about the number of times each button was pressed.\n* Since pressing the same button twice makes no changes, we can assume that each button was pre... | Argentina | Rioplatense Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | a: no; b: 64 | |
0ce3 | Let $ABC$ be a triangle. On the lines $BC$, $CA$, respectively $AB$ the points $M$ and $N$, $P$ and $Q$, respectively $R$ and $S$ are taken, outside of the triangle, so that $MB = BS = AC$, $CP = CN = AB$ and $AR = AQ = BC$. Prove that the points $M, N, P, Q, R$ and $S$ are concyclic. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | null | proof only | null | |
08uq | Suppose there are 3 islands each for 3 different colors, red, blue and yellow. Try to build several bridges so as to satisfy the following conditions:
* For any pair of islands, they are either connected by a bridge or not connected by any bridge, and every bridge connects two distinct islands.
* For any pair of island... | [
"If the conditions specified in the problem for building bridges are satisfied, then clearly the following modified condition is satisfied:\n(†) For any choice of two colors out of the given three, both of the requirements stated for the problem are satisfied if we restrict our consideration only to the islands of ... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 39304 | |
0cxr | Let $a_{0}$ be a positive integer and $a_{n+1} = \sqrt{a_{n}^{2} + 1}$, for all $n \geq 0$.
1) Prove that for all $a_{0}$ the sequence contains infinitely many integers and infinitely many irrational numbers.
2) Is there an $a_{0}$ for which $a_{2010}$ is an integer? | [] | Saudi Arabia | SAMC | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | No | |
02sc | Problem:
Os números $x$, $y$, $z$ e $w$ na figura são números inteiros todos diferentes entre si, maiores do que $1$, e foram colocados nas casas abaixo de modo que cada número (a partir de $y$) é divisor do número na casa da esquerda.
| $x$ | $y$ | $z$ | $w$ |
| :--- | :--- | :--- | :--- |
Descubra todas as soluçõe... | [
"Solution:\n\nPelo enunciado do problema sabemos que\n$$\n1 < w < z < y < x < 329\n$$\nComo $w$ é divisor de $z$ e $z$ é divisor de $y$, então $w$ é divisor de $y$. Mas $y$ é divisor de $x$ então, $w$ é também divisor de $x$ e, portanto, de cada número colocado nas casas.\n\nO número $w$ é portanto divisor da soma,... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | x = 280, y = 28, z = 14, w = 7 | |
0f5x | Problem:
The real sequence $\{x_n\}$ is defined by $x_1 = 1$, $x_2 = 1$, $x_{n+2} = x_{n+1}^2 - x_n / 2$. Show that the sequence converges and find the limit. | [] | Soviet Union | 18th ASU | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 0 | |
0if0 | Problem:
Let $x$, $y$, and $z$ be positive real numbers such that $(x \cdot y) + z = (x + z) \cdot (y + z)$. What is the maximum possible value of $x y z$? | [
"Solution:\nThe condition is equivalent to $z^{2} + (x + y - 1) z = 0$. Since $z$ is positive, $z = 1 - x - y$, so $x + y + z = 1$. By the AM-GM inequality,\n$$\nx y z \\leq \\left(\\frac{x + y + z}{3}\\right)^{3} = \\frac{1}{27}\n$$\nwith equality when $x = y = z = \\frac{1}{3}$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 1/27 | |
036j | Problem:
Consider the function
$$
f(x) = \frac{x^{2} - 2006x + 1}{x^{2} + 1}
$$
a. Solve the inequality $f'(x) \geq 0$.
b. Prove that $|f(x) - f(y)| \leq 2006$ for all real numbers $x$ and $y$. | [] | Bulgaria | Bulgarian Mathematical Competitions | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications",
"Precalculus > Functions"
] | null | proof and answer | f'(x) ≥ 0 for x ≤ −1 or x ≥ 1; moreover, |f(x) − f(y)| ≤ 2006 for all real x, y. | |
02pf | Problem:
Quantos números naturais de cinco algarismos têm o produto de seus algarismos igual a 2000? | [
"Solution:\nInicialmente, observe que $2000 = 2^{4} \\times 5^{3}$. Como os algarismos do número são menores que $10$, cada fator $5$ deve ser um algarismo desse número. Além disso, o produto dos outros algarismos deve ser $2^{4} = 16$. Assim, temos dois casos:\n\n- Os algarismos que faltam são $2$ e $8$. Nesse cas... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 30 | |
0imy | Problem:
$ABCD$ is a cyclic quadrilateral in which $AB = 3$, $BC = 5$, $CD = 6$, and $AD = 10$. $M$, $I$, and $T$ are the feet of the perpendiculars from $D$ to lines $AB$, $AC$, and $BC$ respectively. Determine the value of $MI / IT$.
 | [
"Solution:\n\nAnswer: $\\frac{25}{9}$. Quadrilaterals $AMID$ and $DICT$ are cyclic, having right angles $\\angle AMD$, $\\angle AID$, and $\\angle CID$, $\\angle CTD$ respectively. We see that $M$, $I$, and $T$ are collinear. For, $m \\angle MID = \\pi - m \\angle DAM = \\pi - m \\angle DAB = m \\angle BCD = \\pi -... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 25/9 | |
00zx | Problem:
In a forest each of $n$ animals ($n \geqslant 3$) lives in its own cave, and there is exactly one separate path between any two of these caves. Before the election for King of the Forest some of the animals make an election campaign. Each campaign-making animal visits each of the other caves exactly once, use... | [
"Solution:\n\na) As each campaign-making animal uses exactly $n$ paths and the total number of paths is $\\frac{n(n-1)}{2}$, the number of campaign-making animals cannot exceed $\\frac{n-1}{2}$. Labeling the caves by integers $0,1,2, \\ldots, n-1$, we can construct $\\frac{n-1}{2}$ non-intersecting campaign routes ... | Baltic Way | Baltic Way 1997 | [
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | a) (n-1)/2; b) 4 | |
01bo | Points $X$, $Y$, $Z$ lie on a line $k$ in this order. Let $\omega_1$, $\omega_2$, $\omega_3$ be three circles of diameters $XZ$, $XY$, $YZ$, respectively. Line $l$ passing through point $Y$ intersects $\omega_1$ at points $A$ and $D$, $\omega_2$ at $B$ and $\omega_3$ at $C$ in such manner that points $A$, $B$, $Y$, $C$... | [
"Let $E$ be the second intersection of line $XB$ with circle $\\omega_1$ and $F$ be the second intersection of line $ZC$ with $\\omega_1$. Note that $\\angle XEZ = \\angle ZFX = 90^\\circ$. What is more, $XE \\parallel ZF$ as $XE \\perp BC \\perp ZE$. Hence $XEZF$ is a rectangle and $AD \\perp XE$, so $AB = CD$."
] | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
05ka | Problem:
Soit $ABC$ un triangle. Soient $D \in [AC]$ et $E \in [AB]$ tels que $BE = CD$. Soit $P$ le point d'intersection de $(BD)$ et $(CE)$. Les cercles circonscrits à $BEP$ et $CDP$ se recoupent en $Q$. Soient $K$ et $L$ les milieux respectifs de $[BE]$ et $[CD]$. Soit $R$ l'intersection entre la perpendiculaire à ... | [
"Solution:\n\nRappelons d'abord la méthode de construction du centre d'une similitude directe qui envoie deux points distincts $A$ et $B$ sur deux points distincts $A'$ et $B'$ respectivement.\n\n\n\nSupposons que $A \\neq A'$, $B \\neq B'$, et que les droites $(AA')$ et $(BB')$ ne sont pas... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
059p | We call a prime number $p$ cute if there exists a prime $q$ so that both $pq - 2$ and $pq + 2$ are also primes. We call $p$ wonderful if both $p$ and $p+2$ are cute primes. Find all wonderful numbers. | [
"Let $p$ be a cute prime. The numbers $pq - 2$, $pq$ and $pq + 2$ give all the possible remainders modulo $3$, so one of them must be divisible by $3$. If $3 \\mid pq - 2$, then $pq - 2 = 3$ and $pq = 5$, which is impossible, as $p$, $q$ are primes. Similarly if $3 \\mid pq + 2$, then $pq + 2 = 3$ and $pq = 1$, whi... | Estonia | Estonian Math Competitions | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 3, 5 | |
044u | In a plane rectangular coordinate system $xOy$, $\Gamma_1$ is a unit circle centred at $(2, 1)$ and $\Gamma_2$ is a unit circle centred at $(10, 11)$. Make a line $l$ through the origin $O$ such that $l$ has two intersections with each of $\Gamma_1$ and $\Gamma_2$, dividing $\Gamma_1$ and $\Gamma_2$ into four arcs, and... | [
"Denote the centres $(2, 1)$, $(10, 11)$ of the two circles $\\Gamma_1, \\Gamma_2$ as $T_1, T_2$, respectively.\n\nIf $l$ passes through $T_1$ or $T_2$, then $l$ bisects the circumference of $\\Gamma_1$ or that of $\\Gamma_2$, which yields two equal arcs. The possible slopes of $l$ at this point are $k_1 = k_{OT_1}... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 67/20 | |
04ps | Find all positive integers $n$ such that some three consecutive coefficients in the power series representation of $(1+x)^n$ are in the $3:4:5$ ratio. (AIME 1992) | [] | Croatia | Croatian Mathematical Society Competitions | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | final answer only | 62 | |
0hx8 | Problem:
Given a fixed triangle $\triangle ABC$ and a point $P$, find the maximum value of
$$
\frac{AB^{2} + BC^{2} + CA^{2}}{PA^{2} + PB^{2} + PC^{2}}
$$ | [
"Solution:\nWe use the following lemma.\n\nLemma. Given $a, b, c, p \\in \\mathbb{R}$ such that not all of $a, b, c$ are equal,\n$$\n(a-b)^{2} + (b-c)^{2} + (c-a)^{2} \\leq 3\\left((p-a)^{2} + (p-b)^{2} + (p-c)^{2}\\right)\n$$\nwith equality if and only if $p = (a+b+c)/3$.\n\nProof. For fixed $a, b, c$, the right h... | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algeb... | null | proof and answer | 3 | |
0fsk | Problem:
Bestimme alle ganzen Zahlen $n$, für die es positive reelle Zahlen $0 < a_{1} \leq a_{2} \leq \ldots \leq a_{n}$ gibt mit
$$
\sum_{k=1}^{n} a_{k}=96, \quad \sum_{k=1}^{n} a_{k}^{2}=144, \quad \sum_{k=1}^{n} a_{k}^{3}=216
$$ | [
"Solution:\nFür positive Zahlen $a_{k}$ gilt nach CS\n$$\n\\left(\\sum_{k=1}^{n} a_{k}\\right)\\left(\\sum_{k=1}^{n} a_{k}^{3}\\right) \\geq\\left(\\sum_{k=1}^{n} a_{k}^{2}\\right)^{2}\n$$\nSetzt man hier die gegeben Werte für die drei Summen ein, dann gilt Gleichheit. Nach den allgemeinen Gleichheitsbedingungen fü... | Switzerland | IMO - Selektion | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | n = 64 | |
07db | Let $n > 1$ be an integer and $a_1, a_2, \dots, a_{n+1}$ be distinct positive integers. Does there exist a polynomial $p(x)$ with integer coefficients of degree at most $n$ that satisfies the following conditions?
* $\forall 1 \le i < j \le n+1 : \gcd(p(a_i), p(a_j)) > 1$
* $\forall 1 \le i < j < k \le n+1 : \gcd(p(a_i... | [
"Let $b_{(i,j)}$ for $1 \\le i, j \\le n+1$ be positive integers in which for every $i, j, k, l$ where $\\{i, j\\} \\ne \\{k, l\\}$, we have $(b_{(i,j)}, b_{(k,l)}) = 1$ and $b_{(i,j)} = b_{(j,i)}$. According to Lagrange interpolation formula, there exists a unique polynomial $P(x)$ with rational coefficients of de... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0epn | For how many positive values of $n$ are both $\frac{1}{2}n$ and $2n$ two-digit integers? | [
"15\n\nThe smallest value of $n$ is 20 (for which $\\frac{1}{2}n$ is 10) and the largest value of $n$ is 49 (for which $2n$ is 98). From 20 to 49 inclusive represents 30 integers, half of which are even (so that $\\frac{1}{2}n$ is an integer)."
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | final answer only | 15 | |
0dwh | Problem:
Igralca imata vsak po $2004$ žetonov. Izmenoma mečeta neobičajno igralno kocko, na kateri je napisanih prvih $6$ praštevil. Igralec, ki je na potezi, vrže kocko, drugi pa mu da toliko žetonov, kolikor je ostanek pri deljenju števila $2004$ s številom, ki je padlo pri metu kocke. Ali je mogoče, da bi imel eden... | [
"Solution:\n\nČe bi imel eden izmed igralcev $7$-krat toliko žetonov kot drugi, bi jih imel eden $3507$, drugi pa $501$, kajti žetonov je ves čas igre $4008$. Prvih $6$ praštevil je $2$, $3$, $5$, $7$, $11$ in $13$, število $2004$ pa da pri deljenju s temi praštevili zaporedoma ostanke $0, 0, 4, 2, 2, 2$. Pri vsake... | Slovenia | 48. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | No | |
03nu | Problem:
Let $n$ be a positive integer. A set of $n$ distinct lines divides the plane into various (possibly unbounded) regions. The set of lines is called "nice" if no three lines intersect at a single point. A "colouring" is an assignment of two colours to each region such that the first colour is from the set $\{A_... | [
"Solution:\n\nThe answer is $n \\geq 5$. If $n \\leq 4$, consider $n$ parallel lines. There are 6 total colour combinations required, and only $n+1 \\leq 5$ total regions, hence the colouring is not possible.\n\nNow, assume $n \\geq 5$. Rotate the picture so that no line is horizontal, and orient each line so that ... | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | n ≥ 5 | |
04ib | Prove the following statement: if $z$ is a complex number such that $\text{Re}\left(\frac{z-i}{z+i}\right) = 0$, then $|z| = 1$. (Ilko Brnetić) | [
"Let $z = x + iy$, where $x, y \\in \\mathbb{R}$.\n\nWe have:\n$$\n\\frac{z - i}{z + i} = \\frac{(x + iy) - i}{(x + iy) + i} = \\frac{x + i(y - 1)}{x + i(y + 1)}\n$$\n\nMultiply numerator and denominator by the conjugate of the denominator:\n$$\n\\frac{x + i(y - 1)}{x + i(y + 1)} \\cdot \\frac{x - i(y + 1)}{x - i(y... | Croatia | Croatia Mathematical Competitions | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
09tr | Problem:
Een verzameling lijnen in het vlak noemen we mooi indien elke lijn in de verzameling een oneven aantal van de andere lijnen in de verzameling snijdt.
Bepaal het kleinste gehele getal $k \geq 0$ met de volgende eigenschap: voor iedere 2018 verschillende lijnen $\ell_{1}, \ell_{2}, \ldots, \ell_{2018}$ in het vl... | [
"Solution:\nWe bewijzen eerst dat in een mooie verzameling het aantal lijnen even moet zijn. Stel namelijk dat het aantal lijnen oneven zou zijn. Dan ligt op elk van het oneven aantal lijnen een oneven aantal snijpunten, dus het totale aantal snijpunten is dan ook oneven. Elk snijpunt wordt hier echter twee keer ge... | Netherlands | IMO-selectietoets III | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 1010 | |
09l9 | Consider a sphere and two of its tangent planes in space. Prove that the center of a sphere tangent to all three lies on a fixed ellipse. | [
"The problem is clear if the two planes are parallel, thus we assume that they intersect on line $l$. Let $r$ denote the radius of the sphere and let $O$ denote the center of the sphere. Let $a > r$ denote the distance from $O$ to $l$.\n\nA sphere with center $P$ and radius $s$ satisfies the condition of the proble... | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
040s | Suppose there are beetles on a chessboard consisting of $2012 \times 2012$ unit squares. Each unit square can accommodate at most one beetle. At a moment, all beetles fly and land on the chessboard again. For a beetle, we call the vector from its flying unit to its landing unit the beetle's "displacement vector". We ca... | [
"Set up a coordinate system with origin at the center of chessboard $O$ and the grid line as the coordinate line. Denote the set of the centers of squares by $S$, and the set where the beetles initially stand on by $M_1 \\subseteq S$, and the set that the beetles land on by $M_2 \\subseteq S$. Let $f: M_1 \\to M_2$... | China | China National Team Selection Test | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | 2*1006^3 | |
04mg | Let $p$ be a prime number, and let $n \ge p-1$ be an integer. If $np+1$ is a perfect square, prove that $n+1$ can be represented as a sum of squares of exactly $p$ positive integers. | [
"Let $a$ be a positive integer such that $np + 1 = a^2$. Note that $a > 1$.\n$$\n\\text{We have } a^2 - 1 = np, \\text{ i.e. } (a-1)(a+1) = np.\n$$\nSince $p$ is prime, it follows that $p \\mid a-1$ or $p \\mid a+1$. We treat these two cases separately:\n\ni) Let $p \\mid a-1$, i.e. let $a = kp + 1$ for some intege... | Croatia | Croatia_2018 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
02nq | Problem:
Ada e Luisa treinam todos os dias, cada uma delas sempre com a mesma velocidade, para a grande corrida que vai acontecer no final do ano na escola. O treino começa num ponto $A$ e termina no ponto $B$, distantes $3000~\mathrm{m}$. Elas partem no mesmo instante, mas quando Luisa termina a corrida, ainda faltam... | [
"Solution:\n\nSejam $v_A$ e $v_L$ as velocidades de Ada e Luisa, respectivamente. Ambas são constantes.\n\nQuando Luisa termina o percurso de $3000~\\mathrm{m}$, Ada ainda tem $120~\\mathrm{m}$ para percorrer, ou seja, Ada percorreu $3000 - 120 = 2880~\\mathrm{m}$ no mesmo tempo.\n\nAssim, o tempo que ambas levaram... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 125 | |
0ino | Problem:
The curves $y = x^{2}(x-3)^{2}$ and $y = (x^{2}-1)(x-2)$ intersect at a number of points in the real plane. Determine the sum of the $x$-coordinates of these points of intersection. | [
"Solution:\n\nAnswer: 7. Because the first curve touches the $x$-axis at $x=0$ and $x=3$ while the second curve crosses the $x$-axis at $x= \\pm 1$ and $x=2$, there are four points of intersection. In particular, the points of intersection have $x$-coordinates determined by the difference of the two curves:\n$$\n0 ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 7 | |
0bnj | Denote by $p(a)$ the first digit of the natural number $a$. Show that each of the sets
$$
A = \{n \in \mathbb{N} \mid p(5^n) - p(2^n) > 0\}, \quad B = \{n \in \mathbb{N} \mid p(5^n) - p(2^n) < 0\}
$$
has infinitely many elements. | [
"For $k \\in \\mathbb{N}^*$ there exists $n_k \\in \\mathbb{N}^*$ such that $2^{n_k} < 10^k < 2^{n_k+1}$ ($n_k + 1$ is the smallest element of the set $\\{m \\in \\mathbb{N}^* \\mid 10^k < 2^m\\}$). As $2^{n_k} < 10^k < 2^{n_k+1}$, we get $10^k < 2^{n_k+1} < 2 \\cdot 10^k$, so $p(2^{n_k+1}) = 1$.\n\nMultiplying by ... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Other",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
04so | Let $ABC$ be a right-angled triangle with a hypotenuse $AB$ and longer leg $BC$. Let $D$ be a foot of an altitude from the vertex $C$. Circle $k$ with the center $D$ and the radius $CD$ intersects the leg $BC$ in a point $Q$ and line $AB$ in points $E$ and $F$ ($E \neq F$), where $F$ is a point on the hypotenuse $AB$. ... | [
"The circle $k$ is the Thales' circle with the diameter $EF$ and the center $D$. A triangle $EFC$ is the isosceles right-angled triangle, so $EC = EF$. We will show that triangles $EPC$ and $FQC$ are congruent, which will prove the statement of the problem.\n\n\nFig. 2\n\nAngles $CEQ$ and $... | Czech Republic | 65th Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0eml | Let $\omega$ be a circle, and let $A$ and $B$ be points on a line not intersecting $\omega$. Given a point $X_0$ on $\omega$, define a sequences $X_0, X_1, X_2, \ldots$ and $Y_0, Y_1, Y_2, \ldots$ as follows: $Y_n$ is the second intersection of the line $AX_n$ with $\omega$ and $X_{n+1}$ is the second intersection of t... | [
"Consider circles $\\omega_1$ and $\\omega_2$ centred at $A$ and $B$ respectively such that both $\\omega_1$ and $\\omega_2$ are orthogonal to $\\omega$. Let $I$ be one of the intersection points of $\\omega_1$ and $\\omega_2$. Note that $I$ must exist because $AB$ is outside $\\omega$. Then invert the diagram thro... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
0j1w | Problem:
To set up for a Fourth of July party, David is making a string of red, white, and blue balloons. He places them according to the following rules:
- No red balloon is adjacent to another red balloon.
- White balloons appear in groups of exactly two, and groups of white balloons are separated by at least two no... | [
"Solution:\n\nAnswer: 99 It is possible to achieve 99 red balloons with the arrangement\n$$\n\\text{WWBBBWW} \\underbrace{\\text{RBBBWWRBBBWW \\ldots RBBBWW}}_{99\\ \\text{RBBBWW's}},\n$$\nwhich contains $99 \\cdot 6 + 7 = 601$ balloons.\n\nNow assume that one can construct a chain with 98 or fewer red balloons. Th... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 99 | |
0df0 | Let $O$ be the circumcenter of triangle $ABC$. Points $X$ and $Y$ on side $BC$ are such that $AX = BX$ and $AY = CY$. Prove that the circumcircle of triangle $AXY$ passes through the circumcenters of triangles $AOB$ and $AOC$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
053r | Find all pairs $(a, b)$ of integers satisfying the equality
$$
3(a^2 + b^2) - 7(a + b) = -4.
$$ | [
"The given equation is equivalent to $(6a - 7)^2 + (6b - 7)^2 = 50$. Number $50$ can be represented as the sum of two squares as $25 + 25$ or $1 + 49$. Hence both $6a - 7$ and $6b - 7$ must be among the numbers $7, 5, 1, -1, -5$ and $-7$. As both $a$ and $b$ are integers, only $5, -1$ and $-7$ fit. We obtain the fo... | Estonia | Estonian Math Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | [(2, 2), (1, 0), (0, 1)] | |
0gao | 設實數 $x_i \ge 0$ ($i = 1, 2, \dots, m$), $n \ge 2$, $\sum_{i=1}^{m} x_i = S$。試證：
$$
\sum_{i=1}^{m} \sqrt[n]{\frac{x_i}{S - x_i}} \ge 2,
$$
若且唯若當 $x_i$ 中有兩個相等且不為 $0$，其他的都為 $0$ 時，等號成立。 | [
"先以數學歸納法證明引理 1.\n\n引理 1：當 $x, y \\ge 0$ 時，$n \\ge 2$，其中 $n$ 為正整數時，\n$$\n(x^n + y^n)^2 \\le (x^2 + y^2)^n. \\quad (1)\n$$\n證明：當 $n = 2$ 時，顯然成立。\n假設 $n = k$ ($k \\ge 2$) 時，結論成立，即\n$$\n(x^k + y^k)^2 \\le (x^2 + y^2)^k.\n$$\n當 $n = k + 1$ 時，\n$$\n\\begin{aligned}\n(1) \\text{ 式右邊} &= (x^2 + y^2)^{k+1} \\\\\n&= (x^2 + y... | Taiwan | 二〇一七數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | The sum is at least 2, with equality if and only if exactly two variables are equal and positive and all others are zero. | |
04k0 | For a positive integer $k$, let $a_k = \lfloor \frac{2^k}{k} \rfloor$. Prove that the sequence $(a_k)$ contains infinitely many odd numbers.
($\lfloor x \rfloor$ denotes the largest integer not greater than $x$.) | [
"Let $k = 3 \\cdot 4^l$ for an arbitrary positive integer $l$. Then we have\n$$\na_k = \\left\\lfloor \\frac{2^k}{k} \\right\\rfloor = \\left\\lfloor \\frac{2^{3 \\cdot 4^l}}{3 \\cdot 4^l} \\right\\rfloor = \\left\\lfloor \\frac{2^{3 \\cdot 4^l - 2l}}{3} \\right\\rfloor = \\left\\lfloor \\frac{2^{3 \\cdot 4^l - 2l}... | Croatia | Croatian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof only | null | |
01e2 | Find all functions $f: \mathbb{R} \to \mathbb{R}$ that for all real $x$ and $y$ satisfy the equation
$$
f(y^2 - f(x)) = yf(x)^2 + f(x^2y + y).
$$ | [
"Answer: The only such a function is $f(x) = 0$.\nAt first, assume that $f(x) > 0$ for some $x \\in \\mathbb{R}$. It means that we can choose $y$ such that\n$$\ny^2 - f(x) = x^2y + y\n$$\n(because for $f(x) > 0$ this equation has two solutions with respect to $y$), and if we insert it into the given equation we obt... | Baltic Way | Baltic Way shortlist | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 0 for all real x | |
0hrr | Problem:
Let $ABCD$ be a quadrilateral whose diagonals are perpendicular and intersect at $P$. Let $h_{1}, h_{2}, h_{3}, h_{4}$ be the lengths of the altitudes from $P$ to $AB, BC, CD, DA$. Show that
$$
\frac{1}{h_{1}^{2}}+\frac{1}{h_{3}^{2}}=\frac{1}{h_{2}^{2}}+\frac{1}{h_{4}^{2}}
$$ | [
"Solution:\n\nThe area of triangle $ABP$ is equal to $h_{1} \\cdot AB / 2$ and also to $AP \\cdot BP / 2$. Hence\n$$\n\\frac{1}{h_{1}^{2}}=\\frac{AB^{2}}{AP^{2} \\cdot BP^{2}}=\\frac{AP^{2}+BP^{2}}{AP^{2} \\cdot BP^{2}}=\\frac{1}{BP^{2}}+\\frac{1}{AP^{2}}\n$$\nApplying the same transformation to all the terms of (1... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0gw1 | a) Find all real numbers $x$ such that
$$
x^2 + \frac{1}{x^2 + 2x} = 1
$$
b) Find all pairs of positive real numbers $x$ and $y$ such that
$$
x^2 + y^2 + \frac{1}{x^2 + 2x} + \frac{1}{y^2 + 2y} = 2
$$ | [
"a. Відповідь: $x = \\frac{-1 \\pm \\sqrt{5}}{2}$.\nДане рівняння можна записати у вигляді\n$$\nx^2 + 2x + \\frac{1}{x^2 + 2x} = 2(x + 1) - 1.\n$$\n$$\n\\frac{1}{x^2 + 2x} \\left( (x^2 + 2x) - (x + 1) \\right)^2 = 0.\n$$\n\nb. Відповідь: $x = y = \\frac{-1 + \\sqrt{5}}{2}$.\nМаємо:\n$$\nx^2 + 2x + y^2 + 2y + \\frac... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | a) x = (-1 ± √5)/2; b) x = y = (-1 + √5)/2 | |
04fb | A group of people of different height decided to dance the Hungarian traditional dance at the opening ceremony of MEMO 2013 in Veszprém. We say that a person is *average* if he is taller than one of her neighbours and shorter than the other. (People stand in a circle and every person has exactly two neighbours.)
If the... | [
"We first observe how the height of people changes as we go around the circle in the clockwise direction. For every pair of neighbours $A$ and $B$ (where $B$ is after $A$ in the clockwise direction) we put the symbol ♣ between them if $B$ is taller than $A$ and the symbol ♠ if $B$ is shorter than $A$. In that way w... | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | All integers strictly less than N and having the same parity as N | |
045b | Given an oblique triangle $ABC$ with $BC > AC > AB$. Let $P_1 \neq P_2$ be two points on the plane such that, for $i = 1, 2$, if $AP_i, BP_i$, and $CP_i$ intersect the circumcircle of $\triangle ABC$ at $D_i, E_i$, and $F_i$, respectively, then $D_iE_i \perp D_iF_i$ and $D_iE_i = D_iF_i \neq 0$. Let the line $P_1P_2$ i... | [
"(1) This problem can be solved by computing the angles, but it will (seriously) depend on the relative positions of points. To avoid a case-by-case discussion, we use complex numbers. Assume that the circumcircle of $\\triangle ABC$ is the unit circle on the complex plane. We use the corresponding lowercase letter... | China | 2022 China Team Selection Test for IMO | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > ... | English | proof only | null | |
0iv1 | Problem:
The differentiable function $F: \mathbb{R} \rightarrow \mathbb{R}$ satisfies $F(0) = -1$ and
$$
\frac{d}{d x} F(x) = \sin (\sin (\sin (\sin (x)))) \cdot \cos (\sin (\sin (x))) \cdot \cos (\sin (x)) \cdot \cos (x)
$$
Find $F(x)$ as a function of $x$. | [
"Solution:\n\nSubstituting $u = \\sin (\\sin (\\sin (x)))$, we find\n$$\nF(x) = \\int \\sin (u) \\, d u = -\\cos (u) + C\n$$\nfor some $C$. Since $F(0) = -1$ we find $C = 0$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | proof and answer | -cos(sin(sin(sin(x)))) | |
04oc | Let $\triangle ABC$ be an acute-angled triangle such that $|AB| < |AC|$. Point $D$ is the midpoint of the shorter arc $\widehat{BC}$ of the circumcircle of the triangle $ABC$. Point $I$ is the incentre of the triangle $ABC$, and point $J$ is the reflection of $I$ across the line $BC$. Line $DJ$ intersects the circumcir... | [
"Let $O$ be the circumcentre of the triangle $ABC$.\n\n\n\nPoint $D$ lies on the angle bisector of $\\angle BAC$. Therefore $A$, $I$ and $D$ are collinear. Lines $IJ$ and $OD$ are parallel, since they are both perpendicular to $BC$.\n\nWe will prove the claim by showing that the triangles $... | Croatia | Croatian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
05vm | Problem:
Soit $ABC$ un triangle et $\Omega$ son cercle circonscrit. Soit $D$ le pied de la hauteur issue du sommet $A$. La bissectrice issue du sommet $A$ coupe le segment $[BC]$ au point $P$ et recoupe le cercle $\Omega$ au point $S$. Soit $A'$ le point diamétralement opposé au sommet $A$ dans le cercle $\Omega$. Dém... | [
"Solution:\n\nSoit $X$ le point où la droite $(A'P)$ recoupe le cercle $\\Omega$. Nous allons démontrer que les points $D$, $S$ et $X$ sont alignés.\n\nSoit $O$ le centre de $\\Omega$. Le point $S$ n'est autre que le pôle Sud de $ABC$ issu de $A$, donc il appartient à la médiatrice de $[BC]$, qui n'est autre que la... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
0hho | Let $ABC$ be a triangle and let its incircle, centred at $I$, touch the side $BC$ at $D$. A line through $A$ intersects the lines $BC$, $BI$ and $CI$ at $X$, $Y$ and $Z$, respectively. The circle $(ABC)$ intersects the circles $(AIY)$ and $(AIZ)$ again at $U$ and $V$, respectively. Prove that the points $D$, $U$, $V$ a... | [
"Let $ABC$ be a triangle with orthocenter $I$, $D$ is the antipode of $I$ in $(BIC)$ (we'll call this circle as $\\omega$). So, $ABCD$ is a parallelogram. A circle, which passes through the points $A$ and $I$, intersects $\\omega$ at $X$ and lines $BI$ and $CI$ at points $Y$ and $Z$ respectively ($X,Y,Z \\neq I$). ... | Ukraine | Problems from Ukrainian Authors | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
... | English | proof only | null | |
07s0 | Suppose $a_i > 0$, $i = 1, \dots, n$. Prove that the polynomial
$$
p(x) = -1 + \sum_{i=1}^{n} a_i x^i
$$
has precisely one positive root. | [
"Since\n$$\n\\lim_{x \\to \\infty} \\frac{p(x)}{x^n} = a_n > 0,\n$$\n$p(x)$ is positive for all sufficiently large positive $x$. But, in any event, $p(0) = -1$. Hence, by the Intermediate Value theorem, $p$ has at least one positive root. Suppose $a, b$ are two positive roots of $p$, so that $p(a) = p(b) = 0$. Then... | Ireland | Irish | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
056u | Some knights are on a playground. Each knight has three properties: speed, smartness and sightliness. For every knight, each property takes a certain integral value $x$ such that $1 \le x \le n$. A knight $A$ can win a knight $B$ if the speed, smartness and sightliness of $A$ are all greater than those of $B$. It is kn... | [
"For each knight, we can define a unique non-negative integer $x$ such that the properties of the knight are $a+x$, $b+x$ and $c+x$ and $\\min(a,b,c) = 1$. Call the vector $(a,b,c)$ the *base triple* of the knight.\nIf two knights had the same base triple then either their all properties would be equal or one of th... | Estonia | Final Round of National Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 3n^2 - 3n + 1 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.