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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0j9k | Problem:
Jacob starts with some complex number $x_{0}$ other than $0$ or $1$. He repeatedly flips a fair coin. If the $n^{\text{th}}$ flip lands heads, he lets $x_{n} = 1 - x_{n-1}$, and if it lands tails he lets $x_{n} = \frac{1}{x_{n-1}}$. Over all possible choices of $x_{0}$, what are all possible values of the pro... | [
"Solution:\n\n$1, \\frac{2^{2011}+1}{3 \\cdot 2^{2011}}$\n\nLet $f(x) = 1 - x$, $g(x) = \\frac{1}{x}$. Then for any $x$, $f(f(x)) = x$ and $g(g(x)) = x$. Furthermore,\n\n$f(g(x)) = 1 - \\frac{1}{x}$,\n\ng(f(g(x))) = \\frac{x}{x-1}$,\n\n$f(g(f(g(x)))) = \\frac{1}{1-x}$,\n\ng(f(g(f(g(x))))) = 1 - x = f(x)$,\n\nso for... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 1, (2^{2011}+1)/(3*2^{2011}) | |
0a2l | We have an $n \times n$-board with $n \ge 3$ of which each square can be coloured white or black separately. Each move we change the colour of five chosen squares in the possibly rotated version of the following pentomino pattern.
At the beginning, all the squares are white. Determine for whic... | [
"Answer: this can be done exactly if $n$ is divisible by $2$ or $3$, except for $3$ itself.\n\nWe first show that for $n = 3$, recolouration is not possible. We consider the boxes $A$, $B$ and $M$ in the following figure.\n\nThen each pentomino box contains $M$ and each pentomino contains $... | Netherlands | BxMO/EGMO Team Selection Test | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | All n that are even, and all multiples of 3 except n = 3. | |
0d0s | Prove that for every positive integer $n$ the number
$$
N = \underbrace{44\dots4}_{n} \underbrace{88\dots8}_{n} - 1 \underbrace{33\dots3}_{n-1} 2,
$$
is a perfect square. | [
"We have\n$$\n\\begin{aligned}\nN &= \\underbrace{44\\dots4}_{n-1} \\underbrace{355\\dots56}_{n-1} = \\underbrace{44\\dots4}_{2n} - \\underbrace{88\\dots8}_{n} \\\\\n&= 4(10^{2n-1} + \\dots + 10 + 1) - 8(10^{n-1} + \\dots + 10 + 1) \\\\\n&= 4 \\cdot \\frac{10^{2n}-1}{9} - 8 \\cdot \\frac{10^n-1}{9} = \\frac{1}{9}(4... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0d7c | Let $u$ and $v$ be positive rational numbers with $u \neq v$. Assume that there are infinitely many positive integers $n$ with the property that $u^{n}-v^{n}$ are integers. Prove that $u$ and $v$ are integers. | [
"Let $u = x / z$ and $v = y / z$ with $x$, $y$, $z$ positive integers, and $(x, y, z) = 1$. The statement is equivalent to saying that $x^{n} \\equiv y^{n} \\pmod{z^{n}}$.\n\nIf $z > 1$, assume that there is an odd prime divisor $p \\mid z$. Let $r$ be the least positive integer such that $x^{r} \\equiv y^{r} \\pmo... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
... | English | proof only | null | |
04i0 | There are $100$ circles in the plane, all having the same centre with radii $1$, $2$, $\ldots$, $100$. The smallest circle is coloured in red and each of the $99$ annuli bounded by two circles is coloured in red or green, such that adjacent regions are of different colour. Determine the total area of green regions. | [] | Croatia | Croatia Mathematical Competitions | [
"Geometry > Plane Geometry > Circles",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 5050π | |
04a3 | Let $k$ be an arbitrary nonnegative integer. Prove that we can find $4 \cdot 2^k$ distinct positive integers not greater than $5 \cdot 3^k$, such that no three of them are consecutive terms of an arithmetic progression. | [
"We will prove the statement using induction on $k$.\n\nFor $k = 0$ we need to choose four numbers from $\\{1, 2, 3, 4, 5\\}$ so that among them don't exist three consecutive members of an arithmetic progression. The only such choice is $\\{1, 2, 4, 5\\}$.\n\nAssume that for some $k \\ge 0$ we found a set $S_k$ tha... | Croatia | CroatianCompetitions2011 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
0d4o | Let $a_{1} \geq a_{2} \geq \cdots \geq a_{n}>0$ be real numbers. Prove that
$$
a_{1} a_{2}\left(a_{1}-a_{2}\right)+a_{2} a_{3}\left(a_{2}-a_{3}\right)+\cdots+a_{n-1} a_{n}\left(a_{n-1}-a_{n}\right) \geq a_{1} a_{n}\left(a_{1}-a_{n}\right) .
$$ | [
"First solution. Assume that there exists $1 \\leq i_{0}<n$ such that $a_{i_{0}}= a_{i_{0}+1}$. The sum of the two terms in the left hand side $a_{i_{0}-1} a_{i_{0}}\\left(a_{i_{0}-1}-a_{i_{0}}\\right)+ a_{i_{0}} a_{i_{0}+1}\\left(a_{i_{0}}-a_{i_{0}+1}\\right)$ is equal to $a_{i_{0}-1} a_{i_{0}+1}\\left(a_{i_{0}-1}... | Saudi Arabia | SAMC | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English, Arabic | proof only | null | |
0gqo | Show that for all positive real numbers $x$, $y$, $z$ satisfying $x^2 + y^2 + z^2 \le x + y + z$ the following inequality is held:
$$
\frac{x^2+3}{x^3+1} + \frac{y^2+3}{y^3+1} + \frac{z^2+3}{z^3+1} \ge 6
$$ | [
"Let $f(t) = \\frac{t^2 + 3}{t^3 + 1}$. Since $t^2 + 1 \\ge 2t$ we get $f(t) \\ge \\frac{2t + 2}{t^3 + 1} = \\frac{2}{t^2 - t + 1}$. Let $x^2 - x + 1 = a$, $y^2 - y + 1 = b$, $z^2 - z + 1 = c$. Then the condition $x^2 + y^2 + z^2 \\le x + y + z$ becomes $a + b + c \\le 3$. Therefore,\n$$\nf(x) + f(y) + f(z) \\ge \\... | Turkey | Team Selection Test | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0fwn | Problem:
Finde alle Paare $(\alpha, \beta)$ von positiven reellen Zahlen mit folgenden Eigenschaften:
a. Für alle positiven reellen Zahlen $x, y, z, w$ gilt
$$
x + y^{2} + z^{3} + w^{6} \geq \alpha (x y z w)^{\beta}
$$
b. Es gibt ein Quadrupel $(x, y, z, w)$ von positiven reellen Zahlen, sodass in (a) Gleichheit gilt... | [
"Solution:\nSei $(\\alpha, \\beta)$ ein Paar wie in der Aufgabenstellung. Wir bemerken zuerst, dass für festes $\\beta$ höchstens ein $\\alpha$ wie in (b) existiert, denn $\\alpha$ ist dann die grösste positive reelle Zahl, für die (a) gilt. Setze nun $x = t^{6}, y = t^{3}, z = t^{2}$ und $w = t$ für $t > 0$, dann ... | Switzerland | SMO Finalrunde | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | (alpha, beta) = (2^{4/3} * 3^{1/4}, 1/2) | |
02h7 | A square is contained in a cube when all of its points are in the faces or in the interior of the cube. Determine the biggest $l$ such that there exists a square of side $l$ contained in a cube with edge $1$. | [
"The answer is $\\frac{3\\sqrt{2}}{4}$. Just consider the points $(\\frac{3}{4}, 0, 0)$, $(0, \\frac{3}{4}, 0)$, $(1, \\frac{1}{4}, 1)$ and $(\\frac{1}{4}, 1, 1)$.\n\nNow suppose that there is a square with side $\\ell > \\frac{3\\sqrt{2}}{4}$ inside the cube. First, it's not hard to prove that we can suppose witho... | Brazil | Brazil | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Translation"
] | English | proof and answer | 3√2/4 | |
0iqa | Problem:
Given right triangle $ABC$, with $AB = 4$, $BC = 3$, and $CA = 5$. Circle $\omega$ passes through $A$ and is tangent to $BC$ at $C$. What is the radius of $\omega$? | [
"Solution:\n\nThe radius is $\\frac{25}{8}$.\n\nLet $O$ be the center of $\\omega$, and let $M$ be the midpoint of $AC$. Since $OA = OC$, $OM \\perp AC$. Also, $\\angle OCM = \\angle BAC$, and so triangles $ABC$ and $CMO$ are similar. Then, $CO / CM = AC / AB$, from which we obtain that the radius of $\\omega$ is $... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 25/8 | |
0cww | The chess King was placed on a cell of the board $8 \times 8$, and then he made $64$ moves so that he visited all the cells and returned back to the initial cell. At each moment, we calculate the distance from the center of the cell occupied by the King to the center of the board. We call the move *pleasant* if after t... | [
"Ответ. $44$ moves.\n\nLet us prove that there must have been at least $20$ unpleasant moves (and thus the number of pleasant moves cannot exceed $44$). Let's place numbers in the cells as shown in рис. 3; cells with the same numbers are equidistant from the center, and cells with smaller numbers are closer to the ... | Russia | LI Всероссийская математическая олимпиада школьников | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Russian | proof and answer | 44 | |
072g | Problem:
Let $x_{1}$ be a given positive integer. A sequence $\langle x_{n}\rangle_{n=1}^{\infty} = \langle x_{1}, x_{2}, x_{3}, \cdots \rangle$ of positive integers is such that $x_{n}$, for $n \geq 2$, is obtained from $x_{n-1}$ by adding some nonzero digit of $x_{n-1}$. Prove that
(a) the sequence has an even number... | [
"Solution:\n(a) Let us assume that there are no even numbers in the sequence. This means that $x_{n+1}$ is obtained from $x_{n}$, by adding a nonzero even digit of $x_{n}$ to $x_{n}$, for each $n \\geq 1$.\nLet $E$ be the left most even digit in $x_{1}$ which may be taken in the form\n$$\nx_{1} = O_{1} O_{2} \\cdot... | India | INMO | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0e5q | Prove that there is no injective function $f: \mathbb{R} \to \mathbb{R}$ such that
$$
f(f(x) + y) = f(x + y) + f(2012) \quad \text{for all } x, y \in \mathbb{R}.
$$ | [
"If we substitute $y = -x$ in the equation, we get $f(f(x)-x) = f(0)+f(2012)$. Since the right side of the equation is a constant, and $f$ is an injective function, $f(x)-x$ must also be a constant. Hence $f(x) = x + c$ for a real number $c$. If we substitute this in the initial equation, we get $x + y + 2c = x + y... | Slovenia | National Math Olympiad 2012 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
08a3 | Problem:
Sapendo che il polinomio $p$ è tale che, per ogni intero $n$, $p\left(5^{n}-1\right)=5^{5 n}-1$, quanto varrà $p(3)$?
(A) 1023
(B) 999
(C) 874
(D) 242
(E) 0 | [
"Solution:\n\nLa risposta è **(A)**. Consideriamo il polinomio $q(x) = (x+1)^5 - 1$. Per ogni intero $n$ abbiamo\n$$\nq\\left(5^{n}-1\\right) = \\left(5^{n}-1+1\\right)^5 - 1 = 5^{5 n} - 1 = p\\left(5^{n}-1\\right)\n$$\nquindi il polinomio differenza $p(x)-q(x)$ si annulla per infiniti valori di $x$ (tutti quelli d... | Italy | Progetto Olimpiadi della Matematica | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | MCQ | A | |
0103 | Problem:
Find all triples $(a, b, c)$ of non-negative integers satisfying $a \geqslant b \geqslant c$ and $1 \cdot a^{3} + 9 \cdot b^{2} + 9 \cdot c + 7 = 1997$. | [
"Solution:\n$(10, 10, 10)$ is the only such triple.\n\nThe equality immediately implies $a^{3} + 9 b^{2} + 9 c = 1990 \\equiv 1 \\pmod{9}$. Hence $a^{3} \\equiv 1 \\pmod{9}$ and $a \\equiv 1 \\pmod{3}$. Since $13^{3} = 2197 > 1990$ then the possible values for $a$ are $1, 4, 7, 10$.\n\nOn the other hand, if $a \\le... | Baltic Way | Baltic Way 1997 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (10, 10, 10) | |
0jxp | Problem:
Triangle $ABC$ has side lengths $AB = 15$, $BC = 18$, $CA = 20$. Extend $CA$ and $CB$ to points $D$ and $E$ respectively such that $DA = AB = BE$. Line $AB$ intersects the circumcircle of $CDE$ at $P$ and $Q$. Find the length of $PQ$. | [
"Solution:\n\nWLOG suppose that $P$ is closer to $A$ than to $B$. Let $DA = AB = BE = c = 15$, $BC = a = 18$, $CA = b = 20$, $PA = x$, and $QB = y$. By Power of a Point on $B$ and $A$, we get $ac = (x + c) y$ and $bc = (y + c) x$, respectively. Subtracting the two equations gives $c y - c x = ac - bc \\Rightarrow y... | United States | HMMT November | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 37 | |
0fre | Problem:
$ABCD$ es un cuadrilátero convexo verificando $AB > BC$, $CD = DA$ y $\angle ABD = \angle DBC$. Sea $E$ el punto de la recta $AB$ tal que $\angle DEB = 90^{\circ}$. Probar que $AE = \frac{AB - BC}{2}$. | [
"Solution:\n\nSea $C'$ el punto simétrico de $C$ con respecto a la recta $BD$, lo incluimos en la figura.\n\n\n\nPor simetría se tiene que $\\angle C'BD = \\angle CBD$, $BC' = BC$ y $DC' = DC$. En particular $C'$ está en el segmento $AB$ y cumple que $AC' = AB - BC$.\n\nAdemás, el triángulo... | Spain | Spain | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
09hn | Is the sum $1^1 + 2^2 + 3^3 + \dots + 2021^{2021}$ a perfect square? | [] | Mongolia | Round 2 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | No | |
01x5 | Positive real numbers $a$ and $b$ satisfy the following conditions: the function $f(x) = x^3 + a x^2 + 2b x - 1$ has three different real roots, while the function $g(x) = 2x^2 + 2b x + a$ does not have real roots.
Prove that $a - b > 1$. | [
"Since the polynomial $f(x) = x^3 + a x^2 + 2b x - 1$ has three different roots, the equation\n$$\nx^3 + a x^2 + 2b x - 1 = 0 \\quad (1)\n$$\nhas three different solutions, denote them by $x_1$, $x_2$ and $x_3$. Vieta's formulas implies the equalities $x_1 + x_2 + x_3 = -a$, $x_1 x_2 + x_1 x_3 + x_2 x_3 = 2b$ and $... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof only | null | |
06jk | Let $k$ be a positive integer. Find the number of nonnegative integers $n$ less than or equal to $10^k$ satisfying the following conditions:
(i) $n$ is divisible by $3$;
(ii) Each decimal digit of $n$ is one of the digits $2$, $0$, $1$ or $7$. | [
"The answer is $\\frac{4^k - 1}{3}$ if $3 \\nmid k$, and $\\frac{4^k + 2}{3}$ if $3 \\mid k$.\n\nNote that $n$ has exactly $k$ digits, possibly starting with $0$. We need to count the number $N$ of integers of the form $\\overline{a_1a_2\\cdots a_k}$ such that each $a_j$ is one of $0$, $1$, $2$, $7$ and $3 \\mid a_... | Hong Kong | CHKMO | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | proof and answer | If k is divisible by 3: (4^k + 2)/3; otherwise: (4^k - 1)/3. | |
0j09 | Problem:
A circle of radius $6$ is drawn centered at the origin. How many squares of side length $1$ and integer coordinate vertices intersect the interior of this circle? | [
"Solution:\nAnswer: $132$\n\nBy symmetry, the answer is four times the number of squares in the first quadrant. Let's identify each square by its coordinates at the bottom-left corner, $(x, y)$. When $x=0$, we can have $y=0 \\ldots 5$, so there are $6$ squares. (Letting $y=6$ is not allowed because that square inte... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 132 | |
0bl4 | Prove that if $a, b, c \in [\frac{1}{\sqrt{6}}, 6]$, then
$$
\frac{4}{a+3b} + \frac{4}{b+3c} + \frac{4}{c+3a} \ge \frac{3}{a+2b} + \frac{3}{b+2c} + \frac{3}{c+2a}.
$$ | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0efy | Problem:
Najmanj kolikokrat moramo hkrati vreči dve pošteni igralni kocki, da bo verjetnost, da bomo vsaj enkrat na obeh kockah hkrati vrgli enako število pik, večja od $\frac{1}{2}$ ? | [
"Solution:\n\nMoč algebre dogodkov pri metu dveh kock je $6 \\cdot 6 = 36$. Hkrati lahko na obeh kockah pade isto število pik na 6 načinov. Verjetnost, da na obeh kockah pade isto število pik, če kocki vržemo enkrat, je torej enaka $p = \\frac{6}{36} = \\frac{1}{6}$.\n\nDenimo, da kocki vržemo $n$-krat. Naj bo $A$ ... | Slovenia | 61. matematično tekmovanje srednješolcev Slovenije, Odbirno tekmovanje | [
"Statistics > Probability > Counting Methods > Other",
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 4 | |
0coi | Given a square grid $n \times n$ ($n \ge 4$). Initially each of $n$ cells of one diagonal contains a sign “+” while each other cell contains a sign “-”. By one move it is allowed to choose a row or a column and replace each sign “-” by “+”, and vice versa. Prove that at any moment the number of signs “+” is not less th... | [
"Первое решение. Пронумеруем строки числами $1, \\dots, n$ сверху вниз, а столбцы — теми же числами слева направо. Клетку будем обозначать парой номеров её строки и столбца; при этом будем считать, что клетки диагонали из плюсов имеют координаты $(i, i)$ $(i = 1, \\dots, n)$.\nЗаметим, что если четыре клетки лежат ... | Russia | Final round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof only | null | |
01dv | A polynomial $f(x)$ with real coefficients is called generating, if for each polynomial $\varphi(x)$ with real coefficients there exists positive integer $k$ and polynomials $g_1(x), \dots, g_k(x)$ such that
$$
\varphi(x) = f(g_1(x)) + \dots + f(g_k(x)).
$$
Find all generating polynomials. | [
"Answer: the generating polynomials are exactly the polynomials of odd degree.\nTake an arbitrary polynomial $f$. We call a polynomial *good* if it can be represented as $\\sum f(g_i(x))$ for some polynomials $g_i$. It is clear that the sum of good polynomials is good, and if $\\varphi$ is a good polynomial then ea... | Baltic Way | Baltic Way shortlist | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | All real-coefficient polynomials of odd degree. | |
00ug | Joe and Penny play a game. Initially there are 5000 stones in a pile, and the two players remove stones from the pile by making a sequence of moves. On the $k$th move, any number of stones between 1 and $k$ inclusive may be removed. Joe makes the odd-numbered moves and Penny makes the even-numbered moves. The player wh... | [
"If on move $2y - 1$, Joe removes $1 \\le a \\le 2y - 1$ stones, then on move $2y$, Penny can remove either $2y - a$ or $2y + 1 - a$ stones (both of which are in the valid range). Thus Penny can ensure that the two moves remove either $2y$ or $2y + 1$ stones.\n\nIf on $k$ occasions, Penny chooses to ensure $2y + 1$... | Balkan Mathematical Olympiad | BMO 2023 Short List | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Penny | |
0iew | Problem:
If $a$, $b$, and $c$ are random real numbers from $0$ to $1$, independently and uniformly chosen, what is the average (expected) value of the smallest of $a$, $b$, and $c$? | [
"Solution:\nLet $d$ be a fourth random variable, also chosen uniformly from $[0,1]$. For fixed $a$, $b$, and $c$, the probability that $d < \\min \\{a, b, c\\}$ is evidently equal to $\\min \\{a, b, c\\}$. Hence,\nif we average over all choices of $a$, $b$, $c$, the average value of $\\min \\{a, b, c\\}$ is equal t... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | final answer only | 1/4 | |
0jet | Problem:
Suppose $A$, $B$, $C$, and $D$ are four circles of radius $r>0$ centered about the points $(0, r)$, $(r, 0)$, $(0,-r)$, and $(-r, 0)$ in the plane. Let $O$ be a circle centered at $(0,0)$ with radius $2 r$. In terms of $r$, what is the area of the union of circles $A$, $B$, $C$, and $D$ subtracted by the area... | [
"Solution:\n\nAnswer: $8 r^{2}$\n\nLet $U$ denote the union of the four circles, so we seek\n$$\nU-([O]-U)=2U-[O]=2\\left[(2r)^{2}+4 \\cdot \\frac{1}{2} \\pi r^{2}\\right]-\\pi(2r)^{2}=8 r^{2}\n$$\n(Here we decompose $U$ into the square $S$ with vertices at $(\\pm r, \\pm r)$ and the four semicircular regions of ra... | United States | HMMT November 2013 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | final answer only | 8r^2 | |
03di | Factorize the polynomial $P = (x^4 - 40x^2 + 144)(x^3 - 16x)$ and compute $\gcd(P(10), P(91))$. | [
"$P = (x^4 - 36x^2 - 4x^2 + 144)(x^3 - 16x) = (x^2 - 36)(x^2 - 4)x(x^2 - 16) = (x-6)(x+6)(x-2)(x+2)x(x-4)(x+4)$.\n\nDenote by $d = \\gcd(P(10), P(91))$. $P(10)$ equals $4 \\cdot 6 \\cdot 8 \\cdot 10 \\cdot 12 \\cdot 14 \\cdot 16 = 2^{14} \\cdot 3^2 \\cdot 5 \\cdot 7$, while $P(91)$ is odd, thus $d \\le 3^2 \\cdot 5... | Bulgaria | Bulgaria 2022 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 315 | |
0gda | Alice 和 Bob 組隊玩一場遊戲。在遊戲開始時,他們兩人會被丟到一台列車上的兩個位置,列車總長為 $1$ 公里。列車是封閉且全黑的,所以除非他們在車頭或車尾,否則他們不會知道自己的位置,他們也無從知道隊友的位置。遊戲配給兩人各一台儀器,儀器的左半邊會顯示:
- 持有人目前面向什麼方向(所以他/她可以選擇往車頭或車尾移動);
- 持有人目前已經移動的總距離;
- 持有人目前是否碰到車頭;
- 持有人目前是否碰到車尾。
儀器的右半邊,則顯示隊友的儀器左半邊所呈現的資訊。遊戲會在兩人遇到彼此(雙方位置重合)的瞬間結束。
假設 Alice 和 Bob 在遊戲開始前先被充分告知列車的狀態以及儀器的所有功能,且可以先共同討論他們的策略。試... | [
"答. $x = 1.5$。\n\n- 構造: 首先 Alice 往車頭走 $0.5$ 公里。如果 Alice 碰到車頭,那麼 Bob 就開始往車頭走,於是在兩人共走 $1.5$ 公里前可以保證重合。假設 Alice 沒有碰到車頭,那就換 Bob 往車頭走,直到碰到車頭或是遇見 Alice 為止,碰到車頭就折返。那麼在兩人共走 $1.5$ 公里前可以保證重合。\n\n- 下界：定義一個人走到的最右界跟走到的最左界的相隔距離是他的探索值。如果兩人的探索值的和不到 $1$ 公里，就可以把兩個人探索的區間擺在列車上互斥的位置。不失一般性，假設 Alice 是第一個讓探索值抵達 $0.5$ 公里的人，那就讓他在探索值達到 $0.5$... | Taiwan | 二〇一九數學奧林匹亞競賽第二階段選訓營,模擬競賽(一) | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1.5 | |
0jv5 | Problem:
The numbers $1, 2, \cdots, 11$ are arranged in a line from left to right in a random order. It is observed that the middle number is larger than exactly one number to its left. Find the probability that it is larger than exactly one number to its right. | [
"Solution:\n\nSuppose the middle number is $k$. Then there are $k-1$ ways to pick the number smaller than $k$ to its left and $\\binom{11-k}{4}$ ways to pick the 4 numbers larger than $k$ to its right. Hence there is a total of $\\sum_{k=2}^{7} (k-1) \\cdot \\binom{11-k}{4}$ ways for there to be exactly one number ... | United States | HMMT November 2016 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 10/33 | |
0fhe | Problem:
Sean $A'$, $B'$ y $C'$ los puntos de tangencia de los lados $BC$, $CA$ y $AB$ de un triángulo con su circunferencia inscrita. Sea $D$ el punto de intersección de $C'A'$ con la bisectriz del ángulo del vértice $A$. Calcular el valor del ángulo $\widehat{ADC}$. | [
"Solution:\n\n\n\nEl triángulo $BC'A'$ es isósceles, luego $2\\beta + B = 180^\\circ$, de donde $\\beta = 90^\\circ - \\frac{B}{2}$. El ángulo $\\gamma$ es suplementario del $\\beta$ y vale $\\gamma = 180^\\circ - \\beta = 90^\\circ + \\frac{B}{2}$. El triángulo $DC'A$ nos permite calcular ... | Spain | OME 27 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof and answer | 90° | |
0iqf | Problem:
Let $ABC$ be a triangle, and let $M$ be the midpoint of side $AB$. If $AB$ is $17$ units long and $CM$ is $8$ units long, find the maximum possible value of the area of $ABC$. | [
"Solution:\n\nAnswer: $68$\n\nLet $h$ be the length of the altitude from $C$ to $AB$. Observe that $K = \\frac{1}{2} \\cdot h \\cdot AB \\leq \\frac{1}{2} \\cdot CM \\cdot AB = 68$ and that equality is achieved when $CM \\perp AB$."
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 68 | |
0ay0 | Problem:
Find the largest positive integer $x$ such that $2017!$ is divisible by $19^{x}$. | [
"Solution:\n\nThere are $\\left\\lfloor\\frac{2017}{19}\\right\\rfloor = 106$ numbers from $1$ to $2017$ which are divisible by $19$.\n\nAmong these, five numbers, $19^{2}, 2 \\cdot 19^{2}, 3 \\cdot 19^{2}, 4 \\cdot 19^{2}, 5 \\cdot 19^{2}$, are divisible by $19^{2}$.\n\nTherefore, $19^{111} \\mid 2017!$ and so the... | Philippines | Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 111 | |
02cu | Problem:
Uma distração - Em vez de multiplicar certo número por $6$, Julia se distraiu e dividiu o número por $6$. O erro cometido por Julia foi de aproximadamente
(a) $100\%$
(b) $97\%$
(c) $83\%$
(d) $17\%$
(e) $3\%$ | [] | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | b | |
0ft3 | Problem:
Seien $a$, $b$ und $c$ die Seiten eines Dreiecks. Beweise die Ungleichung
$$
\sqrt{a+b-c} + \sqrt{c+a-b} + \sqrt{b+c-a} \leq \sqrt{a} + \sqrt{b} + \sqrt{c}
$$
Wann gilt das Gleichheitszeichen? | [
"Solution:\nDa $a$, $b$, $c$ Seiten eines Dreiecks sind, existieren positive Zahlen $x$, $y$, $z$ mit $a = x + y$, $b = y + z$ und $c = z + x$. Damit wird die Ungleichung zu\n$$\n\\sqrt{2x} + \\sqrt{2y} + \\sqrt{2z} \\leq \\sqrt{x+y} + \\sqrt{y+z} + \\sqrt{z+x}\n$$\nNach AM-GM gilt nun\n$$\n\\sqrt{x+y} = \\sqrt{2} ... | Switzerland | IMO - Selektion | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | Equality holds if and only if the triangle is equilateral. | |
08gj | Problem:
Diciamo che una coppia $(a, b)$ di interi positivi è bella se $a$ divide $b$ e $a+1$ divide $b+1$.
a. Dimostrare che, fissato un intero positivo $a$, esistono infiniti interi positivi $b$ tali che la coppia $(a, b)$ sia bella.
b. Fissato un intero positivo $a$, determinare il minimo intero $b>a$ tale che $(... | [
"Solution:\n\nPer definizione, una coppia di interi positivi $(a, b)$ è bella se esistono $k, h$ interi positivi tali che $b=k a$ e $b+1=h(a+1)$. In particolare, si ha $k a+1=h(a+1)$ e quindi\n$$\nh=\\frac{k a+1}{a+1} .\n$$\nPoiché $h$ è intero, si ha che $k a+1=k(a+1)-(k-1)$ è divisibile per $a+1$, da cui $k-1$ è ... | Italy | Olimpiadi di Matematica | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | a) Infinitely many such b exist. b) The minimal b is a^2 + 2a. c) 3438. | |
03n2 | Let $\triangle ABC$ be an acute-angled triangle with altitudes $AD$ and $BE$ meeting at $H$. Let $M$ be the midpoint of segment $AB$, and suppose that the circumcircles of $\triangle DEM$ and $\triangle ABH$ meet at points $P$ and $Q$ with $P$ on the same side of $CH$ as $A$. Prove that the lines $ED$, $PH$, and $MQ$ a... | [
"Let $R$ denote the intersection of lines $ED$ and $PH$. Since quadrilaterals $ECDH$ and $APHB$ are cyclic, we have $\\angle RDA = 180^\\circ - \\angle EDA = 180^\\circ - \\angle EDH = 180^\\circ - \\angle ECH = 90^\\circ + A$, and $\\angle RPA = \\angle HPA = 180^\\circ - \\angle HBA = 90^\\circ + A$. Therefore, $... | Canada | Kanada | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Colline... | null | proof only | null | |
0cx2 | Using each of the first eight primes exactly once and several algebraic operations, obtain the result $2010$. | [] | Saudi Arabia | SAMC | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | 2×3×5×(7+11+13+17+19) = 2010 | |
02lc | Problem:
Um jardim retangular - O retângulo $ABCD$ representa um terreno retangular cuja largura é $3/5$ do comprimento. O retângulo $ABEF$ representa um jardim retangular cuja largura é também $3/5$ do comprimento. Qual a razão entre a área do jardim e a área total do terreno?
(a) $30\%$
(... | [
"Solution:\n\nPelos dados do problema sabemos que\n$$\nAD = \\frac{5}{3} AB \\quad \\text{e} \\quad AB = \\frac{5}{3} AF\n$$\nLogo,\n$$\nAD = \\left(\\frac{5}{3}\\right)^2 AF = \\frac{25}{9} AF\n$$\nA área do terreno é $AB \\times AD$ e a área do jardim é $AB \\times AF$, portanto a razão entre as áreas é\n$$\n\\fr... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | b | |
0kje | Inside a right circular cone with base radius $5$ and height $12$ are three congruent spheres each with radius $r$. Each sphere is tangent to the other two spheres and also tangent to the base and side of the cone. What is $r$?
(A) $\frac{3}{2}$ (B) $\frac{90 - 40\sqrt{3}}{11}$ (C) $2$ (D) $\frac{144 - 25\sqrt{3}}{44}... | [
"Let $C$ be the center of the base of the cone, $A$ be the center of one of the spheres, $B$ be the point where that sphere is tangent to the base of the cone, $D$ be the point such that $\\overline{CD}$ is a radius of the base of the cone containing $B$, $V$ be the vertex of the cone, and $E$ be the point on $\\ov... | United States | AMC 10 A | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | B | |
0i0w | Problem:
Circles $k_{1}$, $k_{2}$, $k_{3}$ intersect as follows: $k_{1} \cap k_{2} = \{A, D\}$, $k_{1} \cap k_{3} = \{B, E\}$, $k_{2} \cap k_{3} = \{C, F\}$. Also, $A B C D E F$ is a non-self-intersecting hexagon. Prove that
$$
A B \cdot C D \cdot E F = B C \cdot D E \cdot F A.
$$
(Hint: first prove that $A D$, $B E$,... | [
"Solution:\n\nWe take the hint. This statement is actually the \"Radical Axis Theorem.\" To prove it, we let $O_{1}$, $O_{2}$, $O_{3}$ be the centers of the circles and $r_{1}$, $r_{2}$, $r_{3}$ their respective radii. We shall consider the set of points $P$ satisfying $P O_{1}^{2} - P O_{2}^{2} = r_{1}^{2} - r_{2}... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
05qu | Problem:
Est-il possible de ranger 601 disques de rayon $1$ dans une boîte rectangulaire de taille $4 \times 600$ sans que deux disques ne se chevauchent (ils peuvent toutefois se toucher)? | [
"Solution:\n\nLa réponse est oui! Pour le montrer, on considère la disposition en triangle sur la figure ci-dessous.\nCalculons $d$. On note $(x_{A}, y_{A})$ les coordonnées du centre du disque $A$, et ainsi de suite. On a donc directement\n$$\nx_{A}=1, \\quad x_{B}=2, \\quad x_{C}=3, \\quad \\text{ et } \\quad y_{... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 5 : Pot-POURRI | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | Yes | |
05n9 | Problem:
Sur chaque sommet d'un $n$-gone régulier on place un signe $+$ ou $-$. À chaque étape, on a le droit de changer les signes de trois sommets consécutifs du $n$-gone.
Quels sont les $n$ pour lesquels, quelle que soit la configuration de départ, on peut obtenir en un nombre fini d'étapes des $+$ sur tous les so... | [
"Solution:\n\nOn va montrer que ce sont exactement les $n$ qui ne sont pas divisibles par $3$.\n\nSi $n$ est divisible par $3$, on note $A_{1}, \\ldots, A_{n}$ les sommets du polygone, et on note $P$ le nombre de $k$ qui ne sont pas divisibles par $3$ et tels qu'il y a un signe $-$ sur $A_{k}$. On montre qu'à chaqu... | France | OCympiades Françaises de Mathématiques - Envoi Numéro 4 - Combinatoire | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | All positive integers n that are not divisible by 3 | |
0d92 | How many ways are there to insert plus signs $+$ between the digits of number $111111 \ldots 111$ which includes thirty digits $1$ so that the result will be a multiple of $30$? | [
"Let $S$ be the final sum, then the necessary conditions are $3|S$, $10|S$.\nOne can check that for every positive integer $n$ then $S(n) \\equiv n \\pmod{9}$ which implies that for all ways to insert the plus signs, we have\n$$\nS \\equiv 1+1+1+\\cdots+1 \\equiv 0 \\pmod{3} .\n$$\nThen the condition that divisible... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | binom(30,10) + 1 | |
04iq | Every team at a basketball tournament is playing with every other team exactly twice. Each victory is worth 2 points, defeat 0 points and there are no draws. Determine all positive integers $n$ such that there exists a basketball tournament with $n$ teams with the overall winner having 26 points and exactly two teams h... | [] | Croatia | Croatia Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 12 | |
0a4y | Problem:
Find the smallest positive integer $N$ satisfying the following three properties.
- $N$ leaves a remainder of $5$ when divided by $7$.
- $N$ leaves a remainder of $6$ when divided by $8$.
- $N$ leaves a remainder of $7$ when divided by $9$. | [
"Solution:\nWe notice that $\\{5,6,7\\}$ are each $2$ less than $\\{7,8,9\\}$ respectively. Therefore $N + 2$ must be a multiple of $7$, $8$ and $9$. Since $7$, $8$ and $9$ are pairwise coprime, this means that\n$$(N + 2) \\text{ is a multiple of } 7 \\times 8 \\times 9 = 504.$$ \nTherefore the smallest positive po... | New Zealand | New Zealand Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | final answer only | 502 | |
0kn4 | The knights in a certain kingdom come in two colors: $\frac{2}{7}$ of them are red, and the rest are blue. Furthermore, $\frac{1}{6}$ of the knights are magical, and the fraction of red knights who are magical is 2 times the fraction of blue knights who are magical. What fraction of red knights are magical?
(A) $\frac{... | [] | United States | 2021 AMC 10 B Fall | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | C | |
06un | Find all pairs $(p, q)$ of prime numbers with $p > q$ for which the number
$$
\frac{(p+q)^{p+q}(p-q)^{p-q}-1}{(p+q)^{p-q}(p-q)^{p+q}-1}
$$
is an integer. | [
"Let $M = (p+q)^{p-q}(p-q)^{p+q} - 1$, which is relatively prime with both $p+q$ and $p-q$. Denote by $(p-q)^{-1}$ the multiplicative inverse of $(p-q)$ modulo $M$.\nBy eliminating the term $-1$ in the numerator,\n$$\n\\begin{align*}\n(p+q)^{p+q}(p-q)^{p-q}-1 &\\equiv (p+q)^{p-q}(p-q)^{p+q}-1 \\quad (\\bmod M) \\\\... | IMO | International Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization... | English | proof and answer | (3, 2) | |
0gx3 | There are five piles made of $42$, $70$, $105$, $462$, and $2009$ stones respectively. You are allowed to do the following: instead of two piles consisting of $a$ and $b$ stones you can make
* two piles of $a$ and $(a+b)$ stones;
* two piles of $a$ and $|a-b|$ stones.
You can also choose $k$ different piles, in which t... | [
"a) We can easily find that the GCD of all the numbers $42$, $70$, $105$, $462$, and $2009$ is $7$. For every operation you are allowed to perform the number of stones in piles stays divisible by $7$. Therefore the answer is \"No\" as the number $2008$ is not divisible by $7$.\n\nb) In the second case the answer is... | Ukraine | Ukrajina 2008 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | a) No; b) Yes | |
0a5l | Problem:
Is it possible to pair up the numbers $0, 1, 2, 3, \ldots , 61$ in such a way that when we sum each pair, the product of the 31 numbers we get is a perfect fifth power? | [
"Solution:\nClaim: It can be achieved. One way to achieve it is:\nto pair $0$ with $1$, and\nto pair $k$ with $(63 - k)$ for $k = 2,3,4,\\ldots ,31$\nThis would result in a product equal to:\n$$(0 + 1)\\prod_{k = 2}^{31}\\left(k + (63 - k)\\right) = 1\\times 63^{30} = \\left(63^{6}\\right)^{5}$$\nwhich is a perfect... | New Zealand | New Zealand Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
09zj | Ayman writes down the numbers $1$ through $10$ in a sequence in some order, writes down the nine (positive) differences between adjacent numbers and computes the sum of these differences. The result is called the *dynamic* of the sequence. For example, the dynamic of the sequence $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$ is $9$,... | [] | Netherlands | Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | MCQ | C | |
0ehs | Problem:
Maja je v dva enaka kozarca nalila limonin sok, tako da je bil prvi kozarec napolnjen do $\frac{1}{3}$, drugi pa do $\frac{2}{5}$. Nato je v oba kozarca dolivala vodo, dokler nista bila polna. Nazadnje je oba kozarca izpraznila v večjo skledo, pri čemer se tekočina ni prelila čez rob sklede. Kolikšen del teko... | [
"Solution:\n\nOznačimo količino tekočine, ki jo drži en kozarec z $e$. Tedaj je v skledi $\\frac{1}{3} e + \\frac{2}{5} e = \\frac{11}{15} e$ limoninega soka od skupno $2e$ tekočine. Limonin sok torej predstavlja $\\frac{\\frac{11}{15} e}{2e} = \\frac{11}{30}$ tekočine v skledi."
] | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | C | |
09hf | Let $D$ and $E$ be points on sides $AB$ and $AC$ of triangle $ABC$, respectively, such that $BD = CE$. Let $M$ and $N$ be midpoints of $BC$ and $DE$, respectively. Prove that $MN$ is parallel to the bisector of $\angle BAC$. | [
"Let $P$ and $Q$ be midpoints of the sides $BE$ and $CD$ respectively. Then $PN$ and $MQ$ are midlines of the triangles $BDE$ and $BDC$. Thus\n$$\nPN = QM = \\frac{BD}{2}\n$$\nand similarly\n$$\nNQ = PM = \\frac{EC}{2}.\n$$\nFrom the hypothesis of the problem we have $BD = EC$. So $PNQM$ is a rhombus. Thus $MN$ is ... | Mongolia | Mongolian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0khx | An organization has 30 employees, 20 of whom have a brand A computer while the other 10 have a brand B computer. For security, the computers can only be connected to each other and only by cables. The cables can only connect a brand A computer to a brand B computer. Employees can communicate with each other if their co... | [] | United States | AMC 12 A | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | MCQ | B | |
0dz5 | Find all positive real numbers $x$ and $y$ such that
$$
x^{x+y} = y^{x-y} \quad \text{and} \quad x^2y = 1
$$ | [
"The second equation implies $y = x^{-2}$, so\n$$\nx^{x + x^{-2}} = x^{-2(x - x^{-2})}.\n$$\nTaking the logarithm on both sides we get\n$$\n(x + x^{-2}) \\log x = -2(x - x^{-2}) \\log x.\n$$\nIf $\\log x = 0$, then $x = 1$ and $y = 1$. Otherwise,\n$$\nx + x^{-2} = -2x + 2x^{-2},\n$$\nso $3x^3 = 1$. This implies $x ... | Slovenia | Slovenija 2008 | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof and answer | Two solutions: (x, y) = (1, 1) and (x, y) = (3^{-1/3}, 3^{2/3}). | |
0542 | Is there a two-digit number $n$ that does not end with zero such that
a) all numbers that can be formed by adding one or more zeros between the two-digit number's digits are its multiples?
b) none of the numbers that can be formed by adding one or more zeros between the two-digit number's digits are its multiples?
c... | [
"a) One such number is $n = 15$. All numbers formed by adding zeros between its digits are divisible by $3$ and $5$.\n\nb) One such number is $n = 12$. Adding zeros between the digits, the last two digits will always be $02$. Hence no such number will be divisible by $4$.\n\nc) One such number is $n = 11$, because ... | Estonia | Estonian Math Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | null | proof and answer | a) 15; b) 12; c) 11 | |
09u9 | The numbers $abcd$ and $dcba$ consist of the same four digits $a$, $b$, $c$, and $d$, but in opposite orders. When we add the two numbers, we get $13552$.
Determine $a + b + c + d$. | [] | Netherlands | Junior Mathematical Olympiad, September 2019 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 26 | |
0b2t | Problem:
How many subsets of the set $\{1,2,3, \ldots, 9\}$ do not contain consecutive odd integers? | [
"Solution:\n\nLet $a_{n}$ be the number of subsets of $\\{1,2, \\ldots, n\\}$ that do not contain consecutive odd integers. We work on the following cases depending on whether such a subset contains $1$ or not:\n\n- If such a subset does not contain $1$, then its elements must consist of elements from $\\{2,3, \\ld... | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 208 | |
0fam | Problem:
Circles $C$ and $C'$ intersect at $O$ and $X$. A circle center $O$ meets $C$ at $Q$ and $R$ and meets $C'$ at $P$ and $S$. $PR$ and $QS$ meet at $Y$ distinct from $X$. Show that $\angle YXO = 90^{\circ}$. | [
"Solution:\n\nWe show first that $YRSX$ is cyclic.\n\n\n\nIt is sufficient to show that $\\angle RYS = \\angle RXS$. We have $\\angle RYS = \\angle PRQ - \\angle RQY = \\angle PSQ - \\angle RQY = \\angle OSQ - \\angle OSP - \\angle RQY = \\angle OQS - \\angle RQY - \\angle OSP = \\angle OQR... | Soviet Union | 1st CIS | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0g45 | Problem:
Let $n$ be a positive integer. Find all polynomials $P$ with real coefficients such that
$$
P\left(x^{2}+x-n^{2}\right)=P(x)^{2}+P(x)
$$
for all real numbers $x$. | [
"Solution:\n\nWe quickly check that $P \\equiv 0$ is indeed a solution. So from here on out let's assume $P \\not \\equiv 0$. Plugging in $x=n$, we get that $P(n)=0$, from which it follows that $P(x)=(x-n)^{r} Q(x)$, for some polynomial $Q \\in \\mathbb{R}[x]$ with $Q(n) \\neq 0$. We now have\n$$\n\\left(x^{2}+x-n-... | Switzerland | IMO Selection | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | P(x) ≡ 0 | |
0hn5 | Problem:
Let $ABC$ be an isosceles right triangle with $\angle C = 90^{\circ}$. Let $X$ be a point on the line segment $AC$, and let $Y$ and $Z$ be, respectively, the feet of the perpendiculars from $A$ and $C$ to the line $BX$. Prove that
$$
BZ = YZ + AY
$$ | [
"Solution:\nDraw the perpendicular $CW$ from $C$ to line $AY$. We get $\\triangle CBZ \\cong \\triangle CAW$ by AAS since:\n- $\\angle BCZ = \\angle ACW$ (this comes from the right angles $BCA$ and $ZCW$).\n- $\\angle BZC = \\angle AWC$ (both $90^{\\circ}$)\n- $BC = AC$.\n\n\n\nQuadrilatera... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0gxz | Find the minimum positive integer number which has a product of its digits equals $5120$. | [
"Since $5120 = 2^{10} \\cdot 5$, then the main question is – how many digits have its number? Obviously among these digits it hasn't $0$ and has $5$ because only $5$ from these digits divides by $5$. From the ten factors $2$ we can produce at the least four digits – $2$, $8$, $8$, $8$ or $4$, $4$, $8$, $8$. Therefo... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 25888 | |
09z8 | In triangle $ABC$, the point $D$ lies on segment $AB$ such that $CD$ is the angle bisector of angle $C$. The perpendicular bisector of segment $CD$ intersects the line $AB$ in $E$. Suppose that $|BE| = 4$ and $|AB| = 5$.
a. Prove that $\angle BAC = \angle BCE$.
b. Prove that $2|AD| = |ED|$. | [
"a.\nIn triangle $\\triangle ADC$, the sum of the angles is $180^\\circ$, hence\n$$\n\\angle BAC = \\angle DAC = 180^\\circ - \\angle ADC - \\angle ACD.\n$$\nBecause $CD$ is the angle bisector of $\\angle ACB$, we have $\\angle ACD = \\angle DCB$ and hence the equality above can be rewritten as\n$$\n\\angle BAC = 1... | Netherlands | Final Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
02zm | Problem:
Seja $x$ o menor número inteiro positivo que satisfaz simultaneamente as seguintes condições: $2x$ é o quadrado de um número inteiro, $3x$ é o cubo de um número inteiro e $5x$ é a quinta potência de um número inteiro. Encontre a fatoração em primos de $x$. | [
"Solution:\nDo enunciado, podemos escrever\n$$\n2x = a^{2}, \\quad 3x = b^{3}, \\quad 5x = c^{5}\n$$\nem que $a, b, c$ são inteiros. Daí podemos concluir que $x$ possui fatores primos $2, 3$ e $5$. Ou seja, a fatoração em primos pode ser escrita como $x = 2^{p} 3^{q} 5^{r} \\cdot m$, em que $m$ é o produto de primo... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 2^15 3^20 5^24 | |
0az2 | Problem:
In how many ways can the letters of the word $C H I E F$ be arranged such that $I$ appears at some position after $E$? | [
"Solution:\n\nStart with an empty string that should have 5 letters. We first choose where to put $I$ and $E$, and there are $\\binom{5}{2}$ ways of doing this. Each way has $3!$ ways to rearrange the remaining letters, so the number of ways is $10 \\times 6 = 60$ ways.\n\nAlternatively, note that the letters can b... | Philippines | 20th Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 60 | |
0002 | Tenemos un tablero cuadrado de $2002 \times 2002$ con casillas pintadas de negro y blanco, como un tablero de ajedrez. Las filas han sido numeradas del $1$ al $2002$ y las columnas del $1$ al $2002$ (la casilla en la fila $1$ y la columna $1$ es negra). En cada casilla escribimos el producto del número de la fila por e... | [] | Argentina | XI Olimpiada Matemática Rioplatense | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | español | proof only | null | |
09t7 | Problem:
Zij $n \geq 4$ een geheel getal. Bekijk een regelmatige $2n$-hoek waarbij aan elk hoekpunt een geheel getal is toegekend, wat we de waarde van dat hoekpunt noemen. Als vier verschillende hoekpunten van deze $2n$-hoek een rechthoek vormen, dan noemen we de som van de waarden van deze hoekpunten een rechthoekss... | [
"Solution:\n\nNummer de hoekpunten van de $2n$-hoek met de klok mee $1$ tot en met $2n$ en zij $a_{i}$ de waarde van hoekpunt $i$. Omdat het aantal hoekpunten van de veelhoek even is, kunnen we elk hoekpunt koppelen aan het punt er precies tegenover. We tellen de waarden van elk tweetal tegenoverliggende hoekpunten... | Netherlands | MO-selectietoets | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | m = -n | |
0kkw | Problem:
Suppose two distinct competitors of the HMMT 2021 November contest are chosen uniformly at random. Let $p$ be the probability that they can be labelled $A$ and $B$ so that $A$'s score on the General round is strictly greater than $B$'s, and $B$'s score on the Theme round is strictly greater than $A$'s. Estimat... | [
"Solution:\nIf competitors' scores on the General and Theme rounds were completely uncorrelated, we would expect the answer to be approximately $\\frac{1}{2}$. If they were maximally correlated, we would expect the answer to be exactly $0$. It turns out that guessing $\\frac{1}{4} \\rightarrow 2500$ achieves almost... | United States | HMMT November 2021 | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 2500 | |
017x | Given an acute-angled triangle, describe all interior points whose orthogonal projections on the sides form a triangle similar to the original one. | [
"Let the triangle be $ABC$. Pick an interior point $P$, and call its orthogonal projections on the sides $A'$, $B'$, $C'$, respectively. Denote $A_1 = \\angle CAP$, $A_2 = \\angle PAB$, $B_1 = \\angle ABP$, etc. The quadrilateral $AC'PB'$ is cyclic, whence $\\angle PC'B' = A_1$ and $\\angle PB'C' = A_2$. Applying t... | Baltic Way | BALTIC WAY | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Eul... | null | proof and answer | Exactly six interior points, one of which is the circumcenter; for each permutation of the triangle’s vertices there is a unique interior point whose perpendicular projections onto the sides form a triangle similar to the original. | |
0cmp | A Kingdom consists of $N$ cities, some pairs of cities are connected with pairwise non-intersecting roads (two cities connected by a road are called *neighboring* cities). One can start from each city and get to any other city passing through roads, but it is not possible to get back to the initial city using each road... | [
"Применим индукцию по $N$. Утверждение задачи очевидно при $N=1$ и $N=2$. Пусть $\\Gamma_1$ — множество городов, из которых исходит одна дорога, а $\\Gamma$ — множество остальных городов. Начав движение по различным дорогам из некоторого города, согласно условию, мы не сможем попасть дважды в один и тот же город, п... | Russia | Russian mathematical olympiad | [
"Discrete Mathematics > Other",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English; Russian | proof only | null | |
05m2 | Problem:
Soit $P$ un point à l'extérieur d'un cercle $\mathcal{C}$. Les tangentes à $\mathcal{C}$ passant par $P$ touchent $\mathcal{C}$ en $A$ et $B$. Une droite passant par $P$ intersecte $\mathcal{C}$ aux points $Q$ et $R$. Soit $S$ un point de $\mathcal{C}$ tel que $(B S) \parallel (Q R)$. Montrer que $(S A)$ pass... | [
"Solution:\n\nLa puissance du point $P$ par rapport au cercle $\\mathcal{C}$ est égale à $P B^{2} = P Q \\cdot P R$. Ainsi, on a égalité des rapports de longueurs $\\frac{P R}{P B} = \\frac{P B}{P Q}$, ce qui montre que les triangles $P R B$ et $P B Q$ sont indirectement semblables. En particulier, on en déduit que... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
02uo | Problem:
O mágico Magimático chama três pessoas da plateia: Ana, Beto e Caio, para ajudarem em sua matemágica. Ele diz para cada um pensar em um número de 1 a 50, sem revelá-lo ao mágico, e contá-lo para cada um dos outros dois participantes. Em seguida, cada um deles deve simultaneamente trocar o seu número pela soma... | [
"Solution:\n\nVamos chamar o número de Ana de $A$, o de Beto de $B$ e o de Caio de $C$. Na primeira troca, Ana passou a ter $B+C$, Beto $A+C$ e Caio $A+B$. Após a segunda troca, Ana passou a ter $A+C+A+B=2A+B+C$, Beto passou a ter $A+2B+C$ e Caio passou a ter $A+B+2C$. A partir das respostas finais que eles deram, ... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | Ana 13, Beto 32, Caio 46 | |
0anq | Problem:
Find the solution set of the inequality
$$\left(\frac{\pi}{2}\right)^{(x-1)^{2}} \leq \left(\frac{2}{\pi}\right)^{x^{2}-5x-5}.$$ | [] | Philippines | QUALIFYING STAGE | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | [-1/2, 4] | |
0ai8 | A special number is a positive integer $n$ for which there exist positive integer $a$, $b$, $c$ and $d$ with
$$
n = \frac{a^3 + 2b^3}{c^3 + 2d^3}
$$
Prove that
(a) There are infinitely many special numbers;
(b) 2014 is not a special number. | [
"(a) Every perfect cube $k^3$ of a positive integer is special because we can write\n$$\nk^3 = k^3 \\frac{a^3 + 2b^3}{a^3 + 2b^3} = \\frac{(ka)^3 + 2(kb)^3}{a^3 + 2b^3},\n$$\nfor some positive integers $a$, $b$.\n\n(b) Observe that $2014 = 2 \\cdot 19 \\cdot 53$. If $2014$ is special, then we have,\n$$\nx^3 + 2y^3 ... | North Macedonia | Balkan Mathematical Olympiads | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
04x6 | Find all triples $(k, m, n)$ of positive integers with the following property: The square with the side length $m$ can be cut into some number of rectangles of dimensions $1 \times k$ and exactly one square of the side length $n$. | [
"Answer: The triples $(k, m, n)$ must satisfy $n \\leq m$ and at least one of the two conditions:\n$$\n1^{\\circ}\\ k \\mid m-n,\n$$\n$$\n2^{\\circ}\\ k \\mid m+n \\text{ and } r+n \\leq m, \\text{ where } r \\text{ is the remainder of } m \\text{ modulo } k.\n$$\nWe start from proving that if $(k, m, n)$ are as ab... | Czech-Polish-Slovak Mathematical Match | Czech-Slovak-Polish Match | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | All triples of positive integers satisfy n ≤ m and either k divides m − n, or k divides m + n and r + n ≤ m, where r is the remainder when m is divided by k. | |
07gw | Prove there are coprime polynomials $P(x)$ and $Q(x)$ with integer coefficients and a real number $u > 0$ such that if
$$
\left| \frac{a}{c} - 1 \right|^{2021} \leq \frac{u}{d c^{1010}}, \quad \left| \left( \frac{a}{c} \right)^{2020} - \frac{b}{d} \right| \leq \frac{u}{d c^{1010}},
$$
for some positive integers $a, b, ... | [
"We shall prove a more general statement in the following lines.\n**Proposition.** Let $k$ be a positive integer. Prove that for all pairs $(c, d)$ of integers such that $|c|$ and $|d|$ are prime numbers there is a positive integer $N$ such that the number of pairs $(a, b)$ of integers satisfying the following ineq... | Iran | 38th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Abstract Algebra > Ring Theory",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0dad | $$
f(x)=\left(x-F_{1}\right)\left(x-F_{2}\right) \ldots\left(x-F_{3030}\right)
$$
with $\left(F_{n}\right)$ is the Fibonacci sequence, which defined as $F_{1}=1, F_{2}=2, F_{n+2}= F_{n+1}+F_{n}, n \geq 1$. Suppose that on the range $\left(F_{1}, F_{3030}\right)$, the function $|f(x)|$ takes on the maximum value at $x=x... | [
"We will prove that $x_{0} \\in\\left(F_{3029}, F_{3030}\\right)$ by showing that for all $x^{*} \\in\\left(F_{1}, F_{3029}\\right]$, there is some $x^{**} \\in\\left(F_{3029}, F_{3030}\\right)$ for which $\\left|f\\left(x^{**}\\right)\\right|>\\left|f\\left(x^{*}\\right)\\right|$.\n\nIndeed, if\n$$\nx^{*} \\in\\le... | Saudi Arabia | Team selection tests for IMO 2018 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
00qi | Let $I$ be the incenter of $\triangle ABC$ and let $H_a$, $H_b$, and $H_c$ be the orthocenters of $\triangle BIC$, $\triangle CIA$, and $\triangle AIB$, respectively. The line $H_aH_b$ meets $AB$ at $X$ and the line $H_aH_c$ meets $AC$ at $Y$. If the midpoint $T$ of the median $AM$ of $\triangle ABC$ lies on $XY$, prov... | [
"Let the lines through $B$ and $C$ parallel to $XY$ meet $AM$ at $P$ and $Q$, respectively. Since $BM = MC$, we have $PM = MQ$ and\n$$\n\\frac{BX}{XA} + \\frac{CY}{YA} = \\frac{PT}{TA} + \\frac{QT}{TA} = \\frac{PT + QT}{TA} = \\frac{2MT}{TA} = 2.\n$$\n\nLet $I_a$, $I_b$, and $I_c$ be the ex... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
06fc | Let $f(x) = c_m x^m + c_{m-1} x^{m-1} + \cdots + c_1 x + c_0$, where each $c_i$ is a nonzero integer. Define a sequence $\{a_n\}$ by $a_1 = 0$ and $a_{n+1} = f(a_n)$ for all positive integers $n$.
a. Let $i$ and $j$ be positive integers with $i < j$. Show that $a_{j+1} - a_j$ is a multiple of $a_{i+1} - a_i$.
b. Show... | [
"a.\nRecall the fact that $b - c \\mid f(b) - f(c)$ for any integers $b$ and $c$. Putting $b = a_{i+1}$ and $c = a_i$, we immediately obtain\n$$\na_{i+1} - a_i \\mid f(a_{i+1}) - f(a_i) = a_{i+2} - a_{i+1}.\n$$\nBy induction on $j$, $a_{i+1} - a_i \\mid a_{j+1} - a_j$ for any $j > i$.\n\nb.\nSuppose on the contrary... | Hong Kong | CHKMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
08x7 | Let for $k$, $0 \le k \le 9$, $a_k$ be the coefficient of the $x^k$-term in the polynomial $(x+1)^3(x+2)^3(x+3)^3$. Find the value of $a_2 + a_4 + a_6 + a_8$. | [
"Using the coefficients $a_0, a_1, a_2, \\dots, a_9$, we can write\n$$\n(x+1)^3(x+2)^3(x+3)^3 = a_0 + a_1x + a_2x^2 + \\dots + a_8x^8 + a_9x^9.\n$$\nSubstitution of $x = 1$ and $x = -1$ in the equation above yields, respectively\n$$\na_0 + a_1 + a_2 + \\dots + a_8 + a_9 = 2^3 \\cdot 3^3 \\cdot 4^3\n$$\n$$\na_0 - a_... | Japan | Japan 2013 Initial Round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | final answer only | 6696 | |
0157 | Problem:
How many pairs $(m, n)$ of positive integers with $m < n$ fulfill the equation
$$
\frac{3}{2008} = \frac{1}{m} + \frac{1}{n} ?
$$ | [
"Solution:\nLet $d$ be the greatest common divisor of $m$ and $n$, and let $m = d x$ and $n = d y$. Then the equation is equivalent to\n$$\n3 d x y = 2008(x + y).\n$$\nThe numbers $x$ and $y$ are relatively prime and have no common divisors with $x + y$ and hence they are both divisors of $2008$. Notice that $2008 ... | Baltic Way | Baltic Way 2008 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 5 | |
01qk | Given positive real numbers $a$, $b$, $c$, with $ab + bc + ca \ge a + b + c$, prove that
$$(a+b+c)(ab+bc+ca) + 3abc \ge 4(ab+bc+ca).$$
(I. Gorodnin) | [
"By the Cauchy-Buniakowski inequality,\n$$\n(\\sqrt{a}^2 + \\sqrt{b}^2 + \\sqrt{c}^2)((\\sqrt{a}(b+c))^2 + (\\sqrt{b}(c+a))^2 + (\\sqrt{c}(a+b))^2) \\geq (a(b+c) + b(c+a) + c(a+b))^2 = 4(ab + bc + ca)^2.\n$$\nHence,\n$$\n(a+b+c)(a(b+c)^2+b(c+a)^2+c(a+b)^2) \\geq 4(ab+bc+ca)^2 \\geq 4(ab+bc+ca)(a+b+c) \\Leftrightarr... | Belarus | Selection and Training Session | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0lad | Let $x$, $y$, $z$ be pairwise distinct non-negative real numbers. Prove that
$$
(xy + yz + zx) \left( \frac{1}{(x-y)^2} + \frac{1}{(y-z)^2} + \frac{1}{(z-x)^2} \right) \ge 4.
$$
When does equality hold? | [] | Vietnam | Vijetnam 2008 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | Equality holds iff one of the numbers is zero and the other two are positive with their ratio equal to (3 + sqrt(5)) / 2 or its reciprocal (3 − sqrt(5)) / 2, up to permutation. | |
0fru | Sea $ABC$ un triángulo acutángulo y escaleno con incentro $I$ y ortocentro $H$. Sea $M$ el punto medio de $AB$. Sobre la recta $AH$ se consideran puntos $D$ y $E$ tales que la recta $MD$ es paralela a $CI$ y $ME$ es perpendicular a $CI$. Prueba que $AE = DH$. | [
"Demostraremos que los segmentos $AH$ y $DE$ tienen el mismo punto medio, lo cual probará que $AE = DH$. Sea $F = AH \\cap BC$. Sea $N$ el punto\n\n\n\nEsquema para resolver el problema 2.\nmedio de $DE$, que es el circuncentro del triángulo rectángulo $DEM$. Vemos que las rectas $DE$ y $EM... | Spain | LIX Olimpiada Matemática Española | [
"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"
] | Spanish | proof only | null | |
08kv | Problem:
Let $M$ be an interior point of the triangle $A B C$ with angles $\varangle B A C = 70^{\circ}$ and $\varangle A B C = 80^{\circ}$. If $\varangle A C M = 10^{\circ}$ and $\varangle C B M = 20^{\circ}$, prove that $A B = M C$. | [
"Solution:\n\nLet $O$ be the circumcenter of the triangle $A B C$. Because the triangle $A B C$ is acute, $O$ is in the interior of $\\triangle A B C$. Now we have that $\\varangle A O C = 2 \\varangle A B C = 160^{\\circ}$, so $\\varangle A C O = 10^{\\circ}$ and $\\varangle B O C = 2 \\varangle B A C = 140^{\\cir... | JBMO | 2007 Shortlist JBMO | [
"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 | |
00va | Let $k$ be a positive integer. Find all sequences $(a_n)_{n \ge 1}$ of positive integers such that
$$
a_{n+2}(a_{n+1} - k) = a_n(a_{n+1} + k)
$$
for all $n \ge 1$. | [
"Denote the given equation by (1). Note that if $a_n \\le k$, for $n \\ge 2$, we get a contradiction, so $a_n > k$ for all $n \\ge 2$. Then (1) becomes\n$$\na_{n+2} = \\frac{a_n(a_{n+1} + k)}{a_{n+1} - k}.\n$$\nWe deduce that $a_{n+2} \\ge a_n + 1$.\nNow $a_{n+2} = \\frac{a_n(a_{n+1}+k)}{a_{n+1}-k}$ so $a_{n+1} - k... | Balkan Mathematical Olympiad | 41st Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | All arithmetic progressions with common difference k: a_n = a_1 + (n−1)k for any positive integer a_1. | |
0axe | Problem:
Evaluate: $6+\frac{16}{6+\frac{16}{6+\frac{16}{6+\cdots}}}$ | [
"Solution:\nLet $x$ be equal to the continued fraction to be evaluated. Assuming the continued fraction converges, we must have $x=6+\\frac{16}{x}$, which has $8$ and $-2$ as solutions. Since $x$ must be positive, $x=8$."
] | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | 8 | |
0a95 | Problem:
The number $10^{2007}$ is written on a blackboard. Anne and Berit play a game where the player in turn makes one of two operations:
(i) Replace a number $x$ on the blackboard by two integer numbers $a$ and $b$ greater than $1$ such that $x = ab$;
(ii) Erase one or both of two equal numbers on the blackboard... | [
"Solution:\n\nWe describe a winning strategy for Anne. Her first move is\n$$\n10^{2007} \\rightarrow 2^{2007}, 5^{2007}\n$$\nWe want to show that Anne can act in such a way that the numbers on the blackboard after each of her moves are of the form\n$$\n2^{\\alpha_{1}}, \\ldots, 2^{\\alpha_{k}}, 5^{\\alpha_{1}}, \\l... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 21 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | Anne | |
0h1p | There are candies on the table. At each step Petro can take away some of them. At the first step he takes away one candy and on each next step he can take away either the same amount or twice the amount that he has taken at the previous step. What is the minimal number of steps does Petro need in order to take from the... | [
"**Answer:** $17$.\n\nNote, that on each step the number of candies that Petro takes from the table is a power of $2$. Moreover, if at some step Petro takes away $2^n$ candies, then there were the moves where he was taking away $2, 4, \\ldots, 2^{n-1}$. Thus, the maximum number that Petro could take can not exceed ... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Other"
] | English | proof and answer | 17 | |
035k | Problem:
Does there exist a strictly increasing sequence of positive integers $\{a_{n}\}_{n=1}^{\infty}$ such that $a_{n} \leq n^{3}$ for every $n$ and every positive integer can be written in a unique way as a difference of two terms of the sequence? | [
"Solution:\n\nAnswer: there is such a sequence. We shall define the required sequence inductively. We set $a_{1}=1$, $a_{2}=2$ and assume that $a_{1}, a_{2}, \\ldots, a_{2k}$ are already determined. Denote by $m$ the smallest positive integer which cannot be represented as $a_{j}-a_{i}$, $1 \\leq i < j \\leq 2k$. S... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | null | proof only | null | |
0dv1 | Problem:
Pri deljenju števila $a$ s $7$ dobimo ostanek $3$, pri deljenju števila $b$ s $7$ pa ostanek $4$. Kolikšen je ostanek pri deljenju kvadrata vsote števil $a$ in $b$ s $7$? Odgovor utemelji. | [
"Solution:\n\nZapis: $a = 7x + 3$\n\nZapis $b = 7y + 4$\n\n$$(a + b)^2 = (7x + 3 + 7y + 4)^2 = (7x + 7y + 7)^2 = (7(x + y + 1))^2 = 7 \\cdot (7(x + y + 1)^2)$$\n\nZapisana ugotovitev: Ostanek je $0$, saj je $(a + b)^2$ večkratnik števila $7$."
] | Slovenia | 2. matematično tekmovanje dijakov srednjih tehniških in strokovnih šol | [
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 0 | |
047k | Given positive integers $m, n$ and non-negative real numbers $a_0, a_1, \dots, a_m, b_0, b_1, \dots, b_n$. Define
$$
c_k = \max_{i+j=k} a_i b_j \quad (k = 0, 1, \dots, m+n).
$$
Prove that:
$$
\frac{c_0 + c_1 + \dots + c_{m+n}}{m+n+1} \ge \frac{a_0 + a_1 + \dots + a_m}{m+1} \cdot \frac{b_0 + b_1 + \dots + b_n}{n+1}.
$$ | [
"*Proof.* Consider the set of points $\\mathcal{P} = \\{(i_0, j_0), (i_1, j_1), \\dots, (i_{m+n}, j_{m+n})\\}$ as a \"monotone path\" from $(0, 0)$ to $(m, n)$ if $(i_0, j_0) = (0, 0)$, $(i_{m+n}, j_{m+n}) = (m, n)$, and for every index $0 \\le k < m + n$, we have\n$$\n(i_{k+1}, j_{k+1}) \\in \\{(i_k + 1, j_k), (i_... | China | The 65th IMO China National Team Selection Test | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0amu | Problem:
A camera's aperture determines the size of the circular opening in the lens that allows light in. If we want to allow twice as much light in, what should be the ratio of the new radius to the current radius? | [] | Philippines | 18th PMO Area Stage | [
"Geometry > Plane Geometry > Circles"
] | null | final answer only | sqrt(2):1 | |
0da8 | Let $ABCD$ be a convex quadrilateral inscribed in circle $(O)$ such that $DB = DA + DC$. The point $P$ lies on the ray $AC$ such that $AP = BC$. The point $E$ is on $(O)$ such that $BE \perp AD$. Prove that $DP$ is parallel to the angle bisector of $\angle BEC$. | [
"Denote $R, F$ as the intersections of $AC$ with the angle bisector of $\\angle BEC$ and $BE$, respectively. Let $Q$ be a point on $BD$ such that $DQ = DC$, hence $QB = DB - DQ = DB - DC = AD$.\n\n\n\nThus two triangles $CQB$ and $PDA$ are congruent (s.a.s). Note that $AD \\perp BE$ then $\... | Saudi Arabia | Team selection tests for IMO 2018 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0302 | Problem:
Seja $\overline{a b}$ um número inteiro de dois dígitos. Um inteiro positivo é um parente de $\overline{a b}$ se:
i) o dígito das unidades de $n$ também é $b$.
ii) os outros dígitos de $n$ são distintos de zero e somam $a$.
Por exemplo, os parentes de 31 são 31, 121, 211 e 1111. Encontre todos os números de d... | [
"Solution:\n\nVamos dividir o problema em casos:\n\n1) Se $a=1$ e $n$ é parente de $\\overline{a b}$, então a única possibilidade é $n=\\overline{1 b}$. Esse número é sempre solução, pois $\\overline{1 b}$ divide $\\overline{1 b}$.\n\n2) Se $a=2$, como $\\overline{2 b}$ deve dividir $\\overline{11 b}$, então $\\ove... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 30, 45, 90 | |
0ldf | There are $44$ distinct holes on a line and $2017$ ants. Each ant crawled up from a hole, then moved to another hole and crawled down. Denote $T$ as the set of time points that the ants crawled up or crawled down from some hole. Suppose that the speed of the ants are pairwise distinct and they did not change their own ... | [
"We call a time point \"special\" if at the time, there some ant crawled up/down from the hole. It is easy to see that we just need to solve the problem in case $|T| = 45$ (if $|T| < 45$, we consider some more special time points and that will make the below argument stronger).\n\nConsider the system of coordinatio... | Vietnam | Vietnamese Team Selection Test for IMO | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0if1 | Problem:
How many nonempty subsets of $\{1,2,3, \ldots, 12\}$ have the property that the sum of the largest element and the smallest element is $13$? | [
"Solution:\nIf $a$ is the smallest element of such a set, then $13-a$ is the largest element, and for the remaining elements we may choose any (or none) of the $12-2a$ elements $a+1, a+2, \\ldots, (13-a)-1$. Thus there are $2^{12-2a}$ such sets whose smallest element is $a$. Also, $13-a \\geq a$ clearly implies $a ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1365 | |
02ff | Find all real-valued functions on the positive integers such that $f(x+1019) = f(x)$ for all integer $x$, and $f(xy) = f(x)f(y)$ for all integers $x, y$. | [
"The condition $f(x + 1019) = f(x)$ allows to make all computations modulo $1019$, which is a prime number. Hence the problem asks the number of multiplicative functions in $\\mathbb{Z}/1019\\mathbb{Z}$.\n\nTwo such functions are $f(x) \\equiv 1 \\pmod{1019}$ and $f(x) \\equiv 0 \\pmod{1019}$. From now on suppose $... | Brazil | XVII OBM | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof and answer | All solutions f: N -> R are exactly the following four functions (with p = 1019):
1) f(n) = 0 for all n.
2) f(n) = 1 for all n.
3) The principal character modulo p: f(n) = 0 if p divides n, and f(n) = 1 otherwise.
4) The quadratic character modulo p: f(n) = 0 if p divides n; if not, then f(n) = 1 when n is a quadratic ... |
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