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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0j3w | Problem:
A function $f(x, y, z)$ is linear in $x$, $y$, and $z$ such that $f(x, y, z) = \frac{1}{x y z}$ for $x, y, z \in \{3,4\}$. What is $f(5,5,5)$? | [
"Solution:\nAnswer: $\\frac{1}{216}$ We use a similar method to the previous problem. Notice that $f(x, y, 5) = 2 f(x, y, 4) - f(x, y, 3)$. Let $f_{2}$ denote the function from the previous problem and $f_{3}$ the function from this problem.\nSince $3 f_{3}(x, y, 3)$ is linear in $x$ and $y$, and $3 f_{3}(x, y, 3) ... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | final answer only | 1/216 | |
04vx | Consider a half-disc with diameter $AB$ and centre $O$ and let $X$ be a point in the interior of the half disc. Denote the centroid of $XOB$ by $G$ and the second intersection of the line $AX$ with the boundary of the half-disc by $Y$. Prove that $YG = GB$. (Jiří Blažek, Josef Tkadlec) | [
"\n\nSince $B$ and $Y$ lie on the same circle centered at $O$, we have $OY = OB$. Moreover, by Thales' Theorem, we see that $\\angle AYB = 90^\\circ$. This means that $M$ is the midpoint of the hypotenuse of a right triangle $XYB$, so by Thales' Theorem, it is the circumcentre of $XYB$, hen... | Czech Republic | First Round of the 73rd Czech and Slovak Mathematical Olympiad (December 12th, 2023) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0ez5 | Problem:
Every city in a certain state is directly connected by air with at most three other cities in the state, but one can get from any city to any other city with at most one change of plane. What is the maximum possible number of cities? | [
"Solution:\nAnswer: 10.\nTake a particular city $X$. At most $3$ cities are directly connected to $X$. Each of those is directly connected to at most $2$ other cities (apart from $X$). So $X$ is connected with at most one change to at most $9$ other cities. Thus the maximum number is at most $10$.\n\nWe can achieve... | Soviet Union | 3rd ASU | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 10 | |
0fct | Problem:
Se considera un triángulo equilátero de lado $1$ y centro $O$, como el de la figura.
Un rayo parte de $O$ y se refleja en los tres lados, $\overline{AB}$, $\overline{AC}$ y $\overline{BC}$, (en el orden dado), hasta alcanzar el vértice $A$.
Determina la longitud mínima del recorrido ... | [
"Solution:\n\nComo el rayo se refleja en los lados indicados, basta con desarrollar el camino recorrido por el rayo, para ello desdoblamos el triángulo según la siguiente figura.\n\nEsta figura nos indica que existe un único camino para ir del punto $O$ al punto $A$ reflejándose en los lado... | Spain | null | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(39)/3 | |
07lc | Suppose $a$, $b$, $c$ are real numbers such that $a + b + c = 0$ and $a^2 + b^2 + c^2 = 1$. Prove that
$$
a^2 b^2 c^2 \leq \frac{1}{54}
$$
and determine the cases of equality. | [
"First of all, $0 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$, i.e.,\n$$\nab + bc + ca = -\\frac{1}{2},\n$$\nand so $a$, $b$, $c$ are the roots of the cubic $x^3 - \\frac{1}{2}x - abc = 0$, and, by hypothesis, these are real. Hence the product of the local extrema of this cubic is non-positive. But these e... | Ireland | Irska | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | Maximum: a^2 b^2 c^2 = 1/54. Equality iff {a, b, c} is a permutation of {1/√6, 1/√6, −2/√6} or {−1/√6, −1/√6, 2/√6}. | |
05b5 | Call a positive integer $n$ supereven, if its largest odd factor $d$ is less than $\frac{n}{d}$. How many positive integers less than 1000 are supereven? | [
"Let $n$ be supereven and $d$ its greatest odd divisor. If $\\frac{n}{d}$ were divisible by some odd $p > 1$, then $pd > d$ would also be a factor of $n$, contradiction. Thus $\\frac{n}{d}$ is a power of 2. Hence supereven numbers are exactly those that can be expressed as a product of a power of two and an odd num... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | 46 | |
0fjv | Problem:
Diremos que un triángulo es multiplicativo si el producto de las longitudes de dos de sus lados es igual a la longitud del tercer lado.
Sea $A B C \ldots X Y Z$ un polígono regular de $n$ lados con todos sus lados de longitud $1$. Las $n-3$ diagonales que salen del vértice $A$ dividen al triángulo $Z A B$ en $... | [
"Solution:\nEl ángulo formado por dos diagonales consecutivas con un extremo en $A$ es el mismo por estar inscrito en el mismo arco $\\alpha$.\n\nSean $P Q$ y $Q R$ dos segmentos adyacentes sobre el segmento $Z B$, determinados por tres diagonales consecutivas. Entonces\n$$\n\\begin{aligned}\n& S_{A P Q} = \\frac{1... | Spain | Spanish National Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
030l | Problem:
Se $6 x y - \sqrt{3} x^{2} = \sqrt{3} y^{2}$, calcule
$$
\left(\frac{x}{y}\right)^{4} + \left(\frac{y}{x}\right)^{4}
$$ | [
"Solution:\nTemos:\n$$\n\\begin{aligned}\n6 x y - \\sqrt{3} x^{2} & = \\sqrt{3} y^{2} \\\\\n6 x y & = \\sqrt{3} (x^{2} + y^{2}) \\\\\n\\frac{x y}{x^{2} + y^{2}} & = \\frac{\\sqrt{3}}{6} \\\\\n\\frac{x^{2} + y^{2}}{x y} & = \\frac{6}{\\sqrt{3}} \\\\\n\\frac{x}{y} + \\frac{y}{x} & = 2 \\sqrt{3}\n\\end{aligned}\n$$\nE... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 98 | |
0dk9 | Let the alphabet has only two letters: $A$ and $B$. Is it possible to choose one word of length 5, one word of length 6, ..., one word of length 30 such that any word of length 300 contains one of the chosen words as substring? | [
"The answer is NO.\nConsider all words of length 13 and continue them periodically in both directions to create the sequences. For details, take $x_1x_2\\dots x_{13}$ as some original word of length 13 then add $x_1, x_2, x_3, \\dots$ to the right and add $x_{13}, x_{12}, x_{11}, \\dots$ to the left will result:\n$... | Saudi Arabia | Saudi Arabia booklet 2024 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | No | |
0823 | Problem:
È noto che i Marziani maschi dicono sempre la verità, mentre le Marziane mentono sempre; al contrario i Venusiani maschi mentono e le Venusiane dicono sempre il vero. Atterra un'astronave piena di Marziani e Venusiani; all'ufficio immigrazione due degli occupanti, Ark e Bark, fanno le seguenti dichiarazioni:
... | [
"Solution:\n\nLa risposta è (E). Se Ark dice il vero, allora Bark è un venusiano maschio, quindi mente, ma allora Ark deve essere un venusiano maschio, assurdo. Dunque Ark mente, quindi Bark è una femmina marziana, che mente, e dunque Ark è effettivamente un venusiano maschio (che mente). Quindi tutto è determinato... | Italy | Progetto Olimpiadi di Matematica | [
"Discrete Mathematics > Logic"
] | null | MCQ | E | |
02qe | Problem:
João vai pintar figuras compostas por quadrados e triângulos. Cada quadrado pode ser pintado de azul, vermelho ou verde e cada triângulo de azul, vermelho ou amarelo, de modo que polígonos com um lado comum não tenham a mesma cor. Em cada um dos itens a seguir, determine de quantas maneiras João pode pintar a ... | [
"Solution:\na) Se João pintar o quadrado de azul, ele terá as escolhas vermelho e amarelo para o triângulo. Se ele pintar o quadrado de vermelho, ele terá as escolhas azul e amarelo para pintar o triângulo. Finalmente, se ele pintar o quadrado de verde, ele terá as escolhas azul, vermelho e amarelo para pintar o tr... | Brazil | Nível 2 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a = 7; b = 43; c = 667 | |
08av | Problem:
Sia $ABCD$ un quadrilatero tale che $AB=24$, $BC=20$, $CD=15$, $DA=7$, $BD=25$. Quanto è lungo $AC$?
(A) 18
(B) $14\sqrt{2}$
(C) 20
(D) 21
(E) 24 | [
"Solution:\n\nDato che $AB^{2} + AD^{2} = BD^{2}$ e $BC^{2} + CD^{2} = BD^{2}$ il quadrilatero $ABCD$ si può inscrivere in una circonferenza di diametro $BD$. Sia $DH$ l'altezza del triangolo $DAC$. Allora il triangolo $AHD$ è simile al triangolo $BCD$ perché sono entrambi triangoli rettangoli e $\\widehat{CBD} = \... | Italy | Progetto Olimpiadi della Matematica | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | C | |
05w1 | Problem:
Un chasseur et un lapin invisible jouent sur une grille infinie à maille carrée, c'est-à-dire où chaque case a quatre voisines : à gauche, à droite, en haut et en bas. Tout d'abord, le chasseur colorie chaque case de la grille, mais ne peut utiliser qu'un nombre fini de couleurs. Le lapin choisit ensuite une ... | [
"Solution:\n\nLe chasseur dispose en effet d'une stratégie gagnante, par exemple celle que nous donnons ci-dessous. Tout d'abord, il identifie la grille à l'ensemble $\\mathbb{Z}^{2}$, puis il associe à chaque case $\\mathbf{c}=(x, y)$ le triplet $\\varphi(\\mathbf{c})=(x, y, x+y)$.\n\nChaque fois que le lapin se d... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Yes. The hunter has a winning strategy via a finite coloring scheme that allows reconstruction of the rabbit’s starting position (for example, using triple modular colorings or a five-color linear coloring). | |
0j26 | Problem:
Suppose that there exist nonzero complex numbers $a$, $b$, $c$, and $d$ such that $k$ is a root of both the equations $a x^{3} + b x^{2} + c x + d = 0$ and $b x^{3} + c x^{2} + d x + a = 0$. Find all possible values of $k$ (including complex values). | [
"Solution:\n$1, -1, i, -i$\n\nLet $k$ be a root of both polynomials. Multiplying the first polynomial by $k$ and subtracting the second, we have $a k^{4} - a = 0$, which means that $k$ is either $1, -1, i$, or $-i$. If $a = b = c = d = 1$, then $-1, i$, and $-i$ are roots of both polynomials. If $a = b = c = 1$ and... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 1, -1, i, -i | |
0jx7 | Problem:
How many ways are there to insert +'s between the digits of $111111111111111$ (fifteen 1's) so that the result will be a multiple of $30$? | [
"Solution:\n\nAnswer: $2002$\n\nNote that because there are $15$ 1's, no matter how we insert +'s, the result will always be a multiple of $3$. Therefore, it suffices to consider adding +'s to get a multiple of $10$. By looking at the units digit, we need the number of summands to be a multiple of $10$. Because the... | United States | February 2017 | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 2002 | |
0hiw | Problem:
Are there natural numbers $a$ and $b$ that make the equation
$$
2 a^{2} + 1 = 4 b^{2}
$$
true? (The natural numbers are the counting numbers, $1, 2, 3, \ldots$ ) | [
"Solution:\n\nThe answer is no. Suppose that $a$ and $b$ are any natural numbers. Then $2 a^{2}$ is even, and so is $4 b^{2}$. But an even number plus $1$ will be odd, not even, so the equation can never be made to work, i.e. there are no natural numbers that satisfy the equation."
] | United States | Berkeley Math Circle Monthly Contest 1 | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | No | |
0hhp | The problem gives us a right triangle $ABC$ with a right angle at $ACB$. Let $W_A$ and $W_B$ be the midpoints of the smaller arcs $BC$ and $AC$ of the circumcircle of $\triangle ABC$, and $N_A$ and $N_B$ be the midpoints of the larger arcs $BC$ and $AC$. Let $P$ and $Q$ be the intersection points of segment $AB$ with l... | [
"Let $M$ be the midpoint of the hypotenuse $AB$ of triangle $ABC$ (see figure 7). It is clear that $N_A W_B W_A N_B$ is a rectangle with center $M$. Therefore, its sides $N_A W_B$ and $N_B W_A$ are symmetric with respect to $M$. This means that $AP = BQ$."
] | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals"
] | English | proof only | null | |
0enw | Let $\triangle ABC$ be a triangle such that $AB \neq AC$. We denote its orthocentre by $H$, its circumcentre by $O$ and the midpoint of $BC$ by $D$. The extensions of $HD$ and $AO$ meet in $P$. Prove that triangles $AHP$ and $ABC$ have the same centroid.
 | [
"Let $AO$ intersect the circumcircle at $X$. Since $AX$ is a diameter, $ABX$ is a right angle, hence $BX \\parallel CH$ (both are perpendicular to $AB$). For the same reason, $CX \\parallel BH$, so $BHCX$ is a parallelogram. This means that $XH$ passes through $D$, the midpoint of $BC$. Hence $X$ coincides with $P$... | South Africa | South African Mathematical Olympiad Third Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0j8f | Problem:
How many ordered triples of positive integers $(a, b, c)$ are there for which $a^{4} b^{2} c = 54000$? | [
"Solution:\n\nWe note that $54000 = 2^{4} \\times 3^{3} \\times 5^{3}$. Hence, we must have $a = 2^{a_{1}} 3^{a_{2}} 5^{a_{3}}$, $b = 2^{b_{1}} 3^{b_{2}} 5^{b_{3}}$, $c = 2^{c_{1}} 3^{c_{2}} 5^{c_{3}}$. We look at each prime factor individually:\n\n- $4 a_{1} + 2 b_{1} + c_{1} = 4$ gives 4 solutions: $(1,0,0), (0,2... | United States | Harvard-MIT November Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 16 | |
0dnk | Problem:
Стражар предлаже затвореницима следећу игру. Сви ће бити изведени у двориште, где ће свакоме од њих бити стављен на главу шешир у једној од 5 могућих боја. Стражар ће их потом поређати у врсту тако да сваки затвореник види све шешире осим сопственог и питати првог затвореника у врсти да ли зна боју свог шешир... | [
"Solution:\n\nДоделимо бојама вредности $0,1,2,3,4$ и означимо са $B$ боју шешира другог затвореника, а са $S$ збир боја шешира од трећег до последњег затвореника по модулу 5. Описаћемо стратегију прва два затвореника након које ће сви остали затвореници знати $S$, те ће, знајући боје осталих шешира, моћи да одреде... | Serbia | 9. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 2013 | |
0jby | Problem:
A rectangular piece of paper with vertices $A$, $B$, $C$, $D$ is being cut by a pair of scissors. The pair of scissors starts at vertex $A$, and then cuts along the angle bisector of $DAB$ until it reaches another edge of the paper. One of the two resulting pieces of paper has 4 times the area of the other pi... | [
"Solution:\n\nAnswer: $\\sqrt{\\frac{5}{2}}$\n\nWithout loss of generality, let $AB > AD$, and let $x = AD$, $y = AB$. Let the cut along the angle bisector of $\\angle DAB$ meet $CD$ at $E$. Note that $ADE$ is a $45$-$45$-$90$ triangle, so $DE = AD = x$, and $EC = y - x$. Now, $[ADE] = \\frac{x^2}{2}$, and $[AECB] ... | United States | HMMT November 2012 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 5/2 | |
07xu | Let $O$ denote the circumcentre of $\triangle ABC$, and $M$ the midpoint of $BC$. The line perpendicular to $AC$ passing through the point $C$ intersects line $AB$ (extended) at $T$, and intersects the circumcircle of $\triangle ABC$ for the second time at $D$. Let $L$ denote a point on the line $AD$ (extended) such th... | [
"Because $ABDC$ is cyclic,\n$$\n\\angle TAL = \\angle BAD = \\angle BCD = \\angle DCL = \\angle TCL\n$$\nhence quadrilateral *ACLT* is cyclic. This implies\n$$\n\\angle OLP = \\angle ALT = \\angle ACT = 90^{\\circ} \\quad \\text{and}\n$$\n$$\n\\angle CLP = \\angle CLT = 180^{\\circ} - \\angle A = 180^{\\circ} - \\a... | Ireland | IRL_ABooklet_2025 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations... | null | proof only | null | |
01fw | Let $ABC$ be a triangle and $H$ its orthocenter. Let $P$ be an arbitrary point on segment $BC$. Prove that $H$ lies on the line passing through reflections of $P$ with respect to $AB$ and $AC$ iff triangle $ABC$ is right angled. | [
"Denote by $P'$ and $P''$ reflections of $P$ with respect to $AB$ and $AC$. Easy to see that $P$ is an anti-Steiner point of the line $P'P''$, so $P \\equiv B$ or $P \\equiv C$— therefore $\\angle BAC = 90^\\circ$.",
"The implication $\\iff$ is very easy. We shall prove $\\implies$.\nSuppose that triangle $ABC$ i... | Baltic Way | Baltic Way 2019 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
02n4 | Problem:
No triângulo $\triangle ABC$, $M$ é o ponto médio do lado $AC$, $D$ é um ponto do lado $BC$, tal que $AD$ é a bissetriz do ângulo $B\widehat{A}C$, e $P$ é o ponto de interseção de $AD$ e $BM$. Sabendo que $AB = 10\ \mathrm{cm}$, $AC = 30\ \mathrm{cm}$ e a área do triângulo $\triangle ABC$ mede $100\ \mathrm{c... | [
"Solution:\n\n$$\n\\frac{\\text{área}(\\triangle ABM)}{\\text{área}(\\triangle ABC)} = \\frac{\\frac{1}{2} AM \\cdot d}{\\frac{1}{2} AC \\cdot d} = \\frac{AM}{AC} = \\frac{1}{2}\n$$\nde modo que\n$$\n\\text{área}(\\triangle ABM) = \\frac{1}{2} \\text{área}(\\triangle ABC) = \\frac{1}{2} \\times 100 = 50\\ \\mathrm{... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles"
] | null | final answer only | 20 | |
0kvd | Problem:
Compute the smallest positive integer $k$ such that $49$ divides $\binom{2k}{k}$. | [
"Solution:\n\nThe largest $a$ such that $7^{a}$ divides $\\binom{2k}{k}$ is equal to the number of carries when you add $k + k$ in base $7$, by Kummer's Theorem. Thus, we need two carries, so $2k$ must have at least $3$ digits in base $7$. Hence, $2k \\geq 49$, so $k \\geq 25$. We know $k = 25$ works because $25 + ... | United States | HMMT November 2023 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 25 | |
05ya | Problem:
Soit $n$ un entier strictement positif et $x \geqslant n$ un réel. Montrer que $x+\frac{n}{x} \geqslant n+1$ et donner les cas d'égalité. | [
"Solution:\n\nLe réel $x$ est strictement positif donc on peut multiplier l'équation par $x$ pour avoir que l'équation est équivalente à\n$$\nx^{2}+n \\geqslant (n+1)x\n$$\nce qui donne, en réarrangeant les termes,\n$$\nx^{2}-(n+1)x+n \\geqslant 0\n$$\nOn reconnait alors une factorisation\n$$\n(x-n)(x-1) \\geqslant... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOi 2 : Algèbre | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Equality holds if and only if the real number equals the integer. | |
06js | Mable and Nora play a game according to the following steps in order.
(1) Mable writes down any $2015$ distinct prime numbers in ascending order in a row. The product of these primes is Mable's score.
(2) Nora writes down a positive integer.
(3) Mable draws a vertical line between two adjacent primes she has written in... | [
"Nora has a winning strategy.\nLet Mable write down $p_1 < p_2 < \\dots < p_{2015}$ in step (1). Then $P = p_1p_2\\dots p_{2015}$ is Mable's score.\nLet $q_k = p_1p_2\\dots p_k$ for each $k$. Nora needs to write down a positive integer $Q$ in step (2) such that $(P, Q + q_j) = 1$ for all $j$. Indeed, for each $1 \\... | Hong Kong | Year 2016 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Nora has a winning strategy. | |
0f5n | Problem:
Show that we can find $n$ integers whose sum is $0$ and whose product is $n$ if and only if $n$ is divisible by $4$. | [] | Soviet Union | 18th ASU | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
06x5 | Let $ABCD$ be a cyclic quadrilateral. Assume that the points $Q$, $A$, $B$, $P$ are collinear in this order, in such a way that the line $AC$ is tangent to the circle $ADQ$, and the line $BD$ is tangent to the circle $BCP$. Let $M$ and $N$ be the midpoints of $BC$ and $AD$, respectively. Prove that the following three ... | [
"We first prove that triangles $ADQ$ and $CDB$ are similar. Since $ABCD$ is cyclic, we have $\\angle DAQ = \\angle DCB$. By the tangency of $AC$ to the circle $AQD$ we also have $\\angle CBD = \\angle CAD = \\angle AQD$. The claimed similarity is proven.\n\nLet $R$ be the midpoint of $CD$. Points $N$ and $R$ corres... | IMO | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | English | proof only | null | |
07fk | Find all functions $f : \mathbb{N} \to \mathbb{N}$ such that for any distinct positive integers $x$, $y$, $z$, $x + y + z$ is a perfect square if and only if $f(x) + f(y) + f(z)$ is a perfect square. | [
"**Claim i.** $f(x)$ is injective.\n**Proof.** Assume there exist two natural numbers $x_1$ and $x_2$ such that $f(x_1) = f(x_2)$ and $x_1 \\neq x_2$. Now choose $y$ and $z$ such that $y + z = t^2 - x_1$ then we have\n$$\nf(y) + f(z) + f(x_1) = a^2\n$$\nfor some natural number $a$. Therefore,\n$$\nf(y) + f(z) + f(x... | Iran | 37th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | f(n) = k^2 n for some positive integer k | |
08km | Problem:
Find all the three digit numbers $\overline{a b c}$ such that
$$
\overline{a b c}=a b c(a+b+c)
$$ | [
"Solution:\nWe will show that the only solutions are 135 and 144.\nWe have $a>0, b>0, c>0$ and\n$$\n9(11 a+b)=(a+b+c)(a b c-1)\n$$\n- If $a+b+c \\equiv 0\\pmod{3}$ and $a b c-1 \\equiv 0\\pmod{3}$, then $a \\equiv b \\equiv c \\equiv 1\\pmod{3}$ and $11 a+b \\equiv 0\\pmod{3}$. It follows now that\n$$\na+b+c \\equi... | JBMO | OJBM | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 135 and 144 | |
01ob | Determine the greatest possible value of the area of a quadrilateral $ABCD$ if the length of broken line $ABCD$ is equal to $L$.
(I. Gorodnin) | [
"**Answer:** $S(ABCD) = \\frac{L^2}{8}$.\nLet the area of $ABCD$ be a maximum for some $AB = x$, $BD = y$, $CD = z$, $x + y + z = L$. Since $S(ABCD) = S(ABD) + S(DBC) = \\frac{1}{2} AB \\cdot BD \\sin \\angle ABD + \\frac{1}{2} BD \\cdot BC \\sin \\angle BDC$, we see that the area of the quadrilateral with fixed va... | Belarus | 62nd Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | L^2/8 | |
0366 | Problem:
Find all four-digit positive integers $m$ less than $2005$ for which there exists a positive integer $n < m$, such that $mn$ is a perfect square and $m - n$ has at most three distinct positive divisors. | [
"Solution:\nThe number $m - n$ has at most three different positive divisors if and only if $m - n = p^k$, where $p$ is a prime and $k \\in \\{0, 1, 2\\}$. If $k = 0$ then $m = n + 1$ and $n(n + 1)$ is a perfect square, which is impossible.\n\nLet $m - n = p^k$, $mn = t^2$, where $k \\in \\{1, 2\\}$ and $t$ is a po... | Bulgaria | Bulgarian Mathematical Competitions | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 1156, 1296, 1369, 1377, 1600, 1764, 1900 | |
0ki6 | Problem:
Let $ABCDEFGH$ be an equilateral octagon with $\angle A \cong \angle C \cong \angle E \cong \angle G$ and $\angle B \cong \angle D \cong \angle F \cong \angle H$. If the area of $ABCDEFGH$ is three times the area of $ACEG$, then $\sin B$ can be written as $\frac{m}{n}$, where $m, n$ are positive integers and ... | [
"Solution:\n\nAssume $AC=1$. Note that from symmetry, it can be seen that all angles in $ACEG$ must be equal. Further, by similar logic all sides must be equal which means that $ACEG$ is a square. Additionally, as $AB=BC$, $ABC$ is an isosceles triangle, which means the octagon consists of a unit square with four i... | United States | HMMT November 2021 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 405 | |
07ks | A triangle $ABC$ has an obtuse angle at $B$. The perpendicular at $B$ to $AB$ meets $AC$ at $D$, and $|CD| = |AB|$. Prove that
$$
|AD|^2 = |AB| \cdot |BC| \text{ if and only if } \angle CBD = 30^\circ.
$$ | [
"Let $|BC| = a$, $|AD| = b$, $|AB| = |CD| = c$. We are given $b^2 = a c$ and need to prove $\\angle CBD = 30^\\circ$.\n$$\n\\frac{a^2 + c^2 - (b + c)^2}{2 a c} = \\cos(90^\\circ + \\angle CBD) = -\\sin(\\angle CBD). \\text{ Hence } \\sin(\\angle BCD) = \\frac{b^2 + 2 b c - a^2}{2 a c}.\n$$\n$$\n\\text{Also } \\sin(... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
04rr | Let $I$ be the incenter of a triangle $ABC$. The circle passing through the vertex $B$ and touches the line $AI$ at $I$ intersects the sides $AB$ and $BC$ at points $P$ and $Q$, respectively. Let $R$ be the intersection point of the line $QI$ and the side $AC$. Prove that
$$
|AR| \cdot |BQ| = |PI|^2.
$$ | [
"Let $\\alpha, \\beta, \\gamma$ denote the measures of the interior angles at the vertices $A, B, C$, respectively, of the triangle $ABC$, and let $J$ be the intersection point of the line $AI$ and the side $BC$. (Fig. 2). The inscribed angle $PBI$ corresponds to the chord $PI$, while the inscribed angle $QBI$ corr... | Czech Republic | 62nd Czech and Slovak 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"
] | English | proof only | null | |
08e5 | Problem:
Lucia, dopo aver disegnato il quadrato $ABCD$ di lato unitario, traccia una circonferenza di centro $C$ e raggio uguale al lato del quadrato. Indica poi con $X$ l'intersezione tra la diagonale $AC$ e la circonferenza, e con $Y$ l'intersezione della retta $DX$ con il lato $AB$. Quanto vale la lunghezza del seg... | [
"Solution:\n\nLa risposta è $\\mathbf{(B)}$. Tracciamo la retta $r$ perpendicolare al lato $AD$ e passante per $X$. Sia $H$ l'intersezione fra $r$ ed $AD$, e sia $K$ l'intersezione fra $CB$ ed $r$. Sia $J$ la proiezione di $X$ sul lato $CD$. Mostriamo intanto che i triangoli $DHX$ e $DAY$ sono simili. Infatti, $\\a... | Italy | Olimpiadi della Matematica | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Qu... | null | MCQ | B | |
0kdc | Problem:
Harvard has recently built a new house for its students consisting of $n$ levels, where the $k$th level from the top can be modeled as a 1-meter-tall cylinder with radius $k$ meters. Given that the area of all the lateral surfaces (i.e. the surfaces of the external vertical walls) of the building is 35 percen... | [
"Solution:\n\nThe $k$th layer contributes a lateral surface area of $2k\\pi$, so the total lateral surface area is\n$$\n2(1+2+\\cdots+n)\\pi = n(n+1)\\pi\n$$\nOn the other hand, the vertical surface area is $2n^2\\pi$ (no need to sum layers, just look at the building from above and from below). Therefore,\n$$\nn+1 ... | United States | HMMO 2020 | [
"Geometry > Solid Geometry > Surface Area",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 13 | |
03z9 | $n$ ($n \ge 3$) table tennis players have a round-robin tournament — each player will play all the others exactly once, and there is no draw game. Suppose, after the tournament, all the players can be arranged in a circle such that, for any three players $A$, $B$, $C$, if $A$, $B$ are adjacent, then at least one of the... | [
"We will prove that $n$ can be any odd number not less than $3$.\n\nSuppose $n = 2k + 1$, an odd number greater than $3$, and $n$ players are represented by $A_1, A_2, \\dots, A_{2k+1}$. Let us arrange the competition result as follows: $A_i$ ($1 \\le i \\le 2k+1$) defeated $A_{i+2}, A_{i+4}, \\dots, A_{i+2k}$ (we ... | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof and answer | All odd integers n ≥ 3 | |
0j93 | Find all real-valued functions $f$ defined on pairs of real numbers, having the following property: for all real numbers $a, b, c$, the median of $f(a, b)$, $f(b, c)$, $f(c, a)$ equals the median of $a, b, c$. (The median of three real numbers, not necessarily distinct, is the number that is in the middle when the thre... | [
"There are two solutions:\n* $f(a, b) = a$ for all $a, b$, and\n* $f(a, b) = b$ for all $a, b$.\nClearly these functions meet the condition. We must show there are no others.\n\nBy setting $a = b = c$ we get $f(a, a) = a$ for all $a$. Next, for all $a, b$, the median of $f(a, a)$, $f(a, b)$, $f(b, a)$ must equal th... | United States | Team Selection Test Selection Test | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(a, b) = a for all real a, b; and f(a, b) = b for all real a, b. | |
0iku | Problem:
Suppose we have an octagon with all angles of $135^{\circ}$, and consecutive sides of alternating length $1$ and $\sqrt{2}$. We draw all its sides and diagonals. Into how many regions do the segments divide the octagon? (No proof is necessary.) | [
"Solution:\n\nThe easiest way to see the answer is to view the octagon as five unit squares in a cross arrangement, with four half-squares wedged at the corners. The center square is divided into $8$ regions. The other $4$ squares are each divided into $15$ regions. The $4$ half-squares are each divided into $4$ re... | United States | Harvard-MIT Mathematics Tournament, Team Round B | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | final answer only | 84 | |
03iw | Problem:
For each real number $r$ let $T_{r}$ be the transformation of the plane that takes the point $(x, y)$ into the point $(2^{r} x, r 2^{r} x + 2^{r} y)$. Let $F$ be the family of all such transformations i.e. $F = \{ T_{r} : r \text{ a real number} \}$. Find all curves $y = f(x)$ whose graphs remain unchanged by ... | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Homothety",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All functions of the form f(0) = 0 and, for x ≠ 0,
f(x) = x log_2(|x|) + c_+ x for x > 0,
f(x) = x log_2(|x|) + c_- x for x < 0,
where c_+ and c_- are arbitrary real constants. | |
0ank | Problem:
The points $(3, m)$, $(x_1, y_1)$ and $(x_2, y_2)$ are on the graph of the function $f(x) = \log_{a} x$. If $y_1 + y_2 = 2m$, find the value of $x_1 x_2$. | [
"Solution:\n\nSince $(3, m)$ is on the graph of $f(x) = \\log_{a} x$, we have:\n$$\nm = \\log_{a} 3\n$$\n\nSimilarly, $(x_1, y_1)$ and $(x_2, y_2)$ are on the graph, so:\n$$\ny_1 = \\log_{a} x_1 \\\\\ny_2 = \\log_{a} x_2\n$$\nGiven $y_1 + y_2 = 2m$, so:\n$$\n\\log_{a} x_1 + \\log_{a} x_2 = 2 \\log_{a} 3\n$$\nUsing ... | Philippines | 18th PMO Area Stage | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | 9 | |
06an | The positive real numbers $a$, $b$, $c$ satisfy: $a^2 + b^2 + c^2 = 3$. Prove that
$$
\frac{a^2 + b^2}{2ab} + \frac{b^2 + c^2}{2bc} + \frac{c^2 + a^2}{2ca} + \frac{2(ab + bc + ca)}{3} \geq 5.
$$
When does equality hold? | [
"$$\nB\\hat{A}C = B\\hat{D}C = D\\hat{B}C = D\\hat{A}C.\n$$\nHence $AC$ is the bisector of the angle $B\\hat{A}D$, and hence it is the perpendicular bisector of the base $BE$ od the isosceles $\\triangle BAE$.\nNow we can complete the proof in two ways:\n**First way:** Since the points $E, B, F$ belong to the circl... | Greece | Selection Examination | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Equality holds when a = b = c = 1. | |
06q4 | Given trapezoid $A B C D$ with parallel sides $A B$ and $C D$, assume that there exist points $E$ on line $B C$ outside segment $B C$, and $F$ inside segment $A D$, such that $\angle D A E=\angle C B F$. Denote by $I$ the point of intersection of $C D$ and $E F$, and by $J$ the point of intersection of $A B$ and $E F$.... | [
"Assume that the disposition of points is as in the diagram.\nSince $\\angle E B F=180^{\\circ}-\\angle C B F=180^{\\circ}-\\angle E A F$ by hypothesis, the quadrilateral $A E B F$ is cyclic. Hence $A J \\cdot J B=F J \\cdot J E$. In view of this equality, $I$ belongs to the circumcircle of $A B K$ if and only if $... | IMO | 49th International Mathematical Olympiad Spain | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
00oa | Let $x$ and $y$ be positive real numbers with $x + y = 1$. Prove that
$$
\frac{x+1}{y} + \frac{y+1}{x} \geq 6.
$$
When does equality hold? | [
"We have\n$$\n\\frac{x+1}{y} + \\frac{y+1}{x} = \\frac{x+x+y}{y} + \\frac{y+x+y}{x} = 2\\left(\\frac{x}{y} + \\frac{y}{x}\\right) + 2.\n$$\nFor $x, y > 0$, the AM-GM inequality gives\n$$\n\\frac{\\frac{x}{y} + \\frac{y}{x}}{2} \\geq \\sqrt{\\frac{x}{y} \\cdot \\frac{y}{x}} = 1,\n$$\nwhich immediately implies the de... | Austria | Austrian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Equality holds when x = y = 1/2. | |
0fw1 | Problem:
Finde alle Paare $(a, b)$ natürlicher Zahlen, sodass
$$
\frac{a^{3}+1}{2 a b^{2}+1}
$$
eine ganze Zahl ist. | [
"Solution:\n\nNach Voraussetzung ist $2 a b^{2}+1$ ein Teiler von $a^{3}+1$, also auch von $a^{2}\\left(2 a b^{2}+1\\right)-2 b^{2}\\left(a^{3}+1\\right)=a^{2}-2 b^{2}$. Weiter gilt sicher $2 b^{2} \\leq a^{2}$, denn sonst wäre der Nenner grösser als der Zähler. Der Fall $2 b^{2}=a^{2}$ ist nicht möglich, denn sons... | Switzerland | SMO Finalrunde | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic ... | null | proof and answer | (a, b) = (2 n^2 + 1, n) for n in natural numbers | |
0br5 | How many three-digit primes can be turned into perfect cubes with a permutation of their digits? | [
"The three-digit perfect cubes are $125$, $216$, $343$, $512$ and $729$. The primes must end with an odd digit, different from $5$, and must be not divisible by $3$. Therefore we must consider only $251$, $521$, $433$. All of them are primes."
] | Romania | 67th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | final answer only | 3 | |
0dl8 | Let $D$ be an interior point on the side $BC$ of an acute-angled triangle $ABC$. Let the circumcircle of triangle $ADB$ intersect the side $AC$ again at $E$ ($\neq A$) and the circumcircle of triangle $ADC$ intersect the side $AB$ again at $F$ ($\neq A$). Let $AD$, $BE$ and $CF$ intersect the circumcircle of triangle $... | [] | Saudi Arabia | Saudi Booklet | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscel... | null | proof only | null | |
0gx1 | Point $M$ is placed on the side $BC$ of the triangle $ABC$ so that $BM = AC$. $H$ is the foot of the perpendicular dropped on $AM$ from point $B$. We know that $BH = CM$, and $\angle MAC = 30^\circ$. Find degree measure of the $\angle ACB$.
**Answer:** $15^\circ$ or $105^\circ$. | [
"Let $AC = BM = b$, $BH = MC = h$. We can easily show that point $H$ can not be on the segment $AM$. Let's consider two cases.\n\n1) Point $H$ is on the ray $MA$ (fig.1).\nLet's draw perpendicular $LC \\perp AM$, point $L$ is on the line $AM$. As one of the angles of the right-angled triangle $ALC$ is $30^\\circ$, ... | Ukraine | Ukrajina 2008 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 15° or 105° | |
0fr8 | Problem:
Sea $n$ un entero positivo. Calcular la siguiente suma:
$$
\frac{3}{1 \cdot 2 \cdot 4 \cdot 5}+\frac{4}{2 \cdot 3 \cdot 5 \cdot 6}+\frac{5}{3 \cdot 4 \cdot 6 \cdot 7}+\ldots+\frac{n+2}{n \cdot(n+1) \cdot(n+3) \cdot(n+4)}
$$ | [
"Solution:\n\nDenotemos por $S$ la suma buscada. En lugar de $S$ vamos a calcular $2S$, que se escribe como\n$$\n\\begin{gathered}\n2S=\\sum_{k=1}^{n} \\frac{2k+4}{k(k+1)(k+3)(k+4)}=\\sum_{k=1}^{n}\\left(\\frac{1}{k(k+3)}-\\frac{1}{(k+1)(k+4)}\\right) \\\\\n=\\left(\\frac{1}{4}-\\frac{1}{10}\\right)+\\left(\\frac{1... | Spain | FASE LOCAL DE LA OLIMPIADA MATEMÁTICA ESPAÑOLA. | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | n(n+5)/(8(n+1)(n+4)) | |
09x6 | Let $ABCD$ be a cyclic quadrilateral with $|AB| = |BC|$. Point $E$ lies on the arc $CD$ which does not contain $A$ and $B$. The intersection of $BE$ and $CD$ is denoted by $P$, the intersection of $AE$ and $BD$ is denoted by $Q$. Prove that $PQ \parallel AC$. | [
"Because $|AB| = |BC|$, we have $\\angle AEB = \\angle BDC$, hence $\\angle QEP = \\angle AEB = \\angle BDC = \\angle QDP$, which yields that $QPED$ is a cyclic quadrilateral. Therefore, $\\angle QPD = \\angle QED = \\angle AED = \\angle ACD$. From this, we get that $QP$ and $AC$ are parallel. $\\square$"
] | Netherlands | BxMO Team Selection Test | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0fdl | Problem:
Sean $x$, $y$, $z$ números reales positivos.
1. Si $x + y + z \geq 3$, ¿Se verifica necesariamente que $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \leq 3$?
2. Si $x + y + z \leq 3$, ¿se verifica necesariamente que $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq 3$? | [
"Solution:\n1. La respuesta es no.\nContraejemplo con los números $1$, $2$ y $0.1$\n\n2. La respuesta es sí.\nEn efecto,\n$$\n3\\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right) \\geq (x + y + z)\\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right) \\geq 9\n$$\nde donde $\\frac{1}{x} + \\frac{1}{y} + \\f... | Spain | null | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 1: No. 2: Yes. | |
0l5k | Problem:
Find all polynomials $P$ with real coefficients for which there exists a polynomial $Q$ with real coefficients such that for all real $t$,
$$\cos (P(t)) = Q(\cos t).$$ | [
"Solution:\nIt is well-known that these polynomials work by taking $Q$ to be a Chebyshev polynomial (if $P$ is linear) or a constant (if $P$ is constant).\n\nSuppose that $\\deg P \\geq 2$. Now consider the density of the roots of $\\cos (P(t))$, i.e.\n$$\n\\lim_{n \\to \\infty} \\frac{\\text{number of roots in the... | United States | HMIC | [
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials"
] | null | proof and answer | All such P are of degree at most one. Precisely: either P is constant, or P(t) = a t + kπ with a ∈ ℤ and k ∈ ℤ. In the nonconstant case, one can take Q(x) = cos(kπ) T_a(x), where T_a is the Chebyshev polynomial of the first kind. | |
080h | Problem:
Sia $n$ il più piccolo intero positivo $>200$ che si può scrivere sia come somma di 5 interi consecutivi che come somma di 6 interi consecutivi e di 7 interi consecutivi. Quanto vale $n$ ? | [] | Italy | Progetto Olimpiadi di Matematica 2000 GARA di SECONDO LIVELLO | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | final answer only | 315 | |
04vl | Suppose that we fill a $3 \times 3$ table with natural numbers from $1$ to $9$ using each of them precisely once. After that, we write down the sums in each of the four $2 \times 2$ squares in the ascending order. Determine whether it is possible to obtain the following quadruples of natural numbers:
a) $24$, $24$, $2... | [
"a) Yes, the following table gives us the desired quadruple of sums.\n\n| 1 | 8 | 2 |\n|---|---|---|\n| 6 | 9 | 5 |\n| 3 | 7 | 4 |\n\nb) No, it is impossible to get this quadruple. Denote the sum of the four numbers as $S$, we shall proceed by proving that $S \\le 98$ and then describing all the cases where we get ... | Czech Republic | Second Round of the 73rd Czech and Slovak Mathematical Olympiad (January 16th, 2024) | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) Yes; b) No. | |
03ke | Problem:
Let $AB$ be a diameter of a circle $\Omega$ and $P$ be any point not on the line through $A$ and $B$. Suppose the line through $P$ and $A$ cuts $\Omega$ again in $U$, and the line through $P$ and $B$ cuts $\Omega$ again in $V$. (Note that in case of tangency $U$ may coincide with $A$ or $V$ may coincide with $... | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | cos(APB) = sqrt(s t) if P is outside the circle; cos(APB) = -sqrt(s t) if P is inside the circle; and cos(APB) = 0 if P lies on the circle | |
0bib | Let $a$, $b$, $c$, $d$ be positive real numbers so that $abc + bcd + cda + dab = 4$. Prove that
$$
a^2 + b^2 + c^2 + d^2 \ge 4.
$$ | [
"We successively have:\n$$\n\\begin{align*}\n4 = abc + bcd + cda + dab &= ab(c+d) + cd(a+b) \\\\ &\\le \\frac{a^2+b^2}{2} \\cdot \\sqrt{2(c^2+d^2)} + \\frac{c^2+d^2}{2} \\cdot \\sqrt{2(a^2+b^2)} \\\\ &= \\sqrt{(a^2+b^2)(c^2+d^2)} \\cdot \\left( \\sqrt{\\frac{a^2+b^2}{2}} + \\sqrt{\\frac{c^2+d^2}{2}} \\right) \\\\ &... | Romania | 65th NMO Selection Tests for JBMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0gzj | Find all nonnegative integers $k$, $n$ that satisfy the following equality:
$$
2^{2k+1} + 9 \cdot 2^k + 5 = n^2.
$$ | [
"Consider our equation modulo $8$. For $k \\geq 3$ the left hand side is divisible by $8$, but the right hand side is not $8$. So we have to consider $k \\in \\{0, 1, 2\\}$.\n\nIf $k = 0$ we have $n^2 = 16$ or $n = 4$.\n\nFor $k = 1$ we have $n^2 = 31$, that gives no solution."
] | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 23, 2010) | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | k = 0, n = 4 | |
0k5a | Problem:
Po picks 100 points $P_{1}, P_{2}, \ldots, P_{100}$ on a circle independently and uniformly at random. He then draws the line segments connecting $P_{1} P_{2}, P_{2} P_{3}, \ldots, P_{100} P_{1}$. When all of the line segments are drawn, the circle is divided into a number of regions. Find the expected number... | [
"Solution:\n\nIf the 100 segments do not intersect on the interior, then the circle will be cut into 101 regions. By Euler's formula, each additional intersection cuts two edges into two each, and adds one more vertex, so since $V-E+F$ is constant, there will be one more region as well. It then suffices to compute ... | United States | HMMT February | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F"
] | null | proof and answer | 4853/3 | |
0hki | Problem:
For an arrangement of the digits $0$ through $9$ around a circle, a number is called a neighbor sum if it is the sum of some two adjacent digits in the arrangement. For example, the arrangement
has five neighbor sums: $4, 7, 8, 11$, and $14$. What is the minimal possible number of ne... | [
"Solution:\n\nThe answer is $3$. It can be achieved using the arrangement\n\nin which the only neighbor sums are $5, 9$, and $10$.\n\nTo prove that no smaller number is possible, first consider the digit $0$. It is adjacent to two different numbers between $1$ and $9$ inclusive, yielding tw... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 3 | |
09u7 | Problem:
Zij $ABC$ een driehoek en zij $I$ het middelpunt van de ingeschreven cirkel van deze driehoek. De lijn door $I$ loodrecht op $AI$ snijdt de omgeschreven cirkel van $\triangle ABC$ in de punten $P$ en $Q$, waarbij $P$ aan dezelfde kant van $AI$ ligt als $B$. Zij $S$ het tweede snijpunt van de omgeschreven cirke... | [
"Solution:\nOplossing I. Laat $M$ en $N$ de snijpunten zijn van respectievelijk $BI$ en $CI$ met de omgeschreven cirkel van $\\triangle ABC$. Schrijf verder $\\angle CAB=2\\alpha$, $\\angle ABC=2\\beta$ en $\\angle BCA=2\\gamma$. Dan geldt vanwege de omtrekshoekstelling\n$$\n\\angle AMN=\\angle ACN=\\angle ACI=\\ga... | Netherlands | Selectietoets | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08yp | Let $n > 1$ be an integer. Alice and Bob play the following game using an $n \times 2021$ grid. First, Alice colors each square either black or white. Bob places a piece in one of the squares in the top row and designates one square in the bottom row as the goal. Then, Alice repeatedly does the following operation $n-1... | [
"The answer is $n = 2022$.\n\nFirst, we will prove that $n \\ge 2022$. We denote the square in the $i^{\\text{th}}$ row and the $j^{\\text{th}}$ column as $(i, j)$. When the piece in $(i, j)$ is moved to $(i', j')$ in a single operation, it holds that $i' = i+1$ and $|j' - j| \\le 1$. Thus, in order to move the pie... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2022 | |
0juu | Problem:
Nine pairwise noncongruent circles are drawn in the plane such that any two circles intersect twice. For each pair of circles, we draw the line through these two points, for a total of $\binom{9}{2}=36$ lines. Assume that all 36 lines drawn are distinct. What is the maximum possible number of points which lie... | [
"Solution:\n\nThe lines in question are the radical axes of the 9 circles. Three circles with noncollinear centers have a radical center where their three pairwise radical axes concur, but all other intersections between two of the $\\binom{9}{2}$ lines can be made to be distinct. So the answer is\n$$\n\\left(\\beg... | United States | HMMT February 2016 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 462 | |
0k1n | Problem:
A positive integer is called primer if it has a prime number of distinct prime factors. A positive integer is called primest if it has a primer number of distinct primer factors. Find the smallest primest number. | [
"Solution:\n\nWe claim the answer is $72$, as it has $6$ primer factors: $6, 12, 24, 18, 36, 72$, and $6$ is a primer.\n\nWe now prove that there is no smaller primest number. Suppose there were a smaller primest number $r < 72$. We do casework on the number of distinct prime factors of $r$.\n\n- $r$ has $\\geq 4$ ... | United States | HMMT November 2018 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 72 | |
06f0 | Let $ABCD$ be a cyclic quadrilateral such that $AD + BC = AB$. Show that the intersection of the bisectors of $\angle ADC$ and $\angle BCD$ is on line $AB$. | [
"Let $P$ be the point on the side $AB$ such that $AP = AD$. Then we have\n$$\nBP = AB - AP = AD + BC - AD = BC.\n$$\nLet the internal angle bisector of $\\angle ADC$ meet $AB$ at $Q$. We find that\n$$\n\\angle QPC = \\angle BPC = 90^\\circ - \\frac{1}{2}\\angle CBP = \\frac{1}{2}(180^\\circ - \\angle CBA) = \\frac{... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0eq2 | It is said that you can only fold a piece of paper in half 7 times. Harold folds a sheet of paper in half 5 times and then makes a hole in the folded paper. How many holes does the sheet of paper have after it is unfolded again?
(A) 32 (B) 36 (C) 81 (D) 50 (E) 64 | [
"Each fold doubles the number of layers that will be pierced. There will be $2^5$ layers and therefore $2^5 = 32$ holes."
] | South Africa | South African Mathematics Olympiad | [
"Math Word Problems"
] | English | MCQ | A | |
08q1 | Problem:
In a certain city there are $n$ straight streets, such that every two streets intersect, and no three streets pass through the same intersection. The City Council wants to organize the city by designating the main and the side street on every intersection. Prove that this can be done in such way that if one g... | [
"Solution:\n\nPick any street $s$ and organize the intersections along $s$ such that the intersections of the two types alternate, as in the statement of the problem.\n\nOn every other street $s_{1}$, exactly one intersection has been organized, namely the one where $s_{1}$ intersects $s$. Call this intersection $I... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
03na | Problem:
Two positive integers $a$ and $b$ are prime-related if $a = p b$ or $b = p a$ for some prime $p$. Find all positive integers $n$, such that $n$ has at least three divisors, and all the divisors can be arranged without repetition in a circle so that any two adjacent divisors are prime-related.
Note that $1$ a... | [
"Solution:\n\nWe say that a positive integer is good if it has the given property. Let $n$ be a good number, and let $d_{1}, d_{2}, \\ldots, d_{k}$ be the divisors of $n$ in the circle, in that order. Then for all $1 \\leq i \\leq k$, $d_{i+1} / d_{i}$ (taking the indices modulo $k$) is equal to either $p_{i}$ or $... | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Exactly those positive integers that are neither perfect squares nor powers of a single prime. | |
0j8p | Problem:
Let $G, A_{1}, A_{2}, A_{3}, A_{4}, B_{1}, B_{2}, B_{3}, B_{4}, B_{5}$ be ten points on a circle such that $G A_{1} A_{2} A_{3} A_{4}$ is a regular pentagon and $G B_{1} B_{2} B_{3} B_{4} B_{5}$ is a regular hexagon, and $B_{1}$ lies on minor arc $G A_{1}$. Let $B_{5} B_{3}$ intersect $B_{1} A_{2}$ at $G_{1}$,... | [
"Solution:\nAnswer: $12^{\\circ}$\n\nNote that $G B_{3}$ is a diameter of the circle. As a result, $A_{2}, A_{3}$ are symmetric with respect to $G B_{3}$, as are $B_{1}, B_{5}$. Therefore, $B_{1} A_{2}$ and $B_{5} A_{3}$ intersect along line $G B_{3}$, so in fact, $B_{1}, A_{2}, G_{1}, G_{2... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 12° | |
06kp | There are three piles of coins, with $a$, $b$ and $c$ coins respectively, where $a$, $b$, $c \ge 2015$ are positive integers. The following operations are allowed:
(1) Choose a pile with an even number of coins and remove all coins from this pile. Add coins to each of the remaining two piles with amount equal to half o... | [
"$(a, b, c)$ can be any integer triple with $a$, $b$, $c \\ge 2015$ except $(2015, 2015, 2015)$.\nIt is obvious that no operation can be carried out in the exceptional case. For the other cases, we claim that it is always possible to increase the total number of coins in some operations. If this is true, then the t... | Hong Kong | null | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | All integer triples (a, b, c) with a, b, c ≥ 2015 except (2015, 2015, 2015). | |
0gtm | Let $ABC$ be a triangle with incenter $I$ such that $|AB| < |BC| < |CA|$. Let $H_A, H_B, H_C$ be the orthocenters of the triangles $IBC, IAC, IAB$, respectively. Let $K_A$ and $L_A$ be points on $BC$ such that $K_A \in H_B H_C$ and $IL_A \perp H_B H_C$. Define $K_B, L_B, K_C, L_C$, similarly. Prove that $|K_A L_A| = |K... | [
"We will show that $L_A, L_B, L_C$ are the midpoints of the sides, and $K_A, K_B, K_C$ are the tangent points of the incircle to the sides in the triangle $ABC$. Note that the desired equality holds in this case. Indeed, if this is the case, we obtain\n\n$$\n|L_A K_A| = |BL_A| - |BK_A| = \\frac{|AC| - |AB|}{2}\n$$\... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Po... | null | proof only | null | |
03w6 | Let $D$ be a point on side $BC$ of triangle $ABC$ such that $\angle CAD = \angle CBA$. A circle with center $O$ passes through $B$, $D$, and meets segments $AB$, $AD$ at $E$, $F$, respectively. Lines $BF$ and $DE$ meet at point $G$. $M$ is the midpoint of $AG$. Prove that $CM \perp AO$. (Posed by Xiong Bin) | [
"As shown in Fig. 1, extend $EF$ and meet $BC$ at point $P$, and join and extend $GP$, which meets $AD$ at $K$ and the extension of $AC$ at $L$.\n\n\nFig. 1\n\nAs shown in Fig. 2, let $Q$ be a point on $AP$ such that\n$$\n\\angle PQF = \\angle AEF = \\angle ADB.\n$$\nIt is easy to see that ... | China | China National Team Selection Test | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Circles > Radical axis theorem... | English | proof only | null | |
01qr | Prime numbers $p$, $q$, $r$ ($p + q < 111$) satisfy the equality
$$
\frac{p + q}{r} = p - q + r.
$$
Find the largest possible value of the product $pqr$. | [
"We have\n$$\n\\frac{p + q}{r} = p - q + r \\Leftrightarrow q(r + 1) - p(r - 1) = r^2.\n$$\nIf $r > 2$, then $r$ is odd (since $r$ is prime). Thus the left-hand side of the latter equation is an even number while its right-hand side is an odd number, a contradiction. Therefore, $r = 2$. Then from the initial equali... | Belarus | Final Round | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 2014 | |
03ml | Problem:
Let $n$ be a positive integer. For any positive integer $j$ and positive real number $r$, define
$$
f_{j}(r)=\min (j r, n)+\min \left(\frac{j}{r}, n\right), \quad \text{and} \quad g_{j}(r)=\min (\lceil j r\rceil, n)+\min \left(\left\lceil\frac{j}{r}\right\rceil, n\right),
$$
where $\lceil x\rceil$ denotes the... | [
"Solution:\n\nWe first prove the left hand side inequality. We begin by drawing an $n \\times n$ board, with corners at $(0,0),(n, 0),(0, n)$ and $(n, n)$ on the Cartesian plane.\nConsider the line $\\ell$ with slope $r$ passing through $(0,0)$. For each $j \\in\\{1, \\ldots, n\\}$, consider the point $(j, \\min (j... | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0arg | Problem:
$x^{2} + 4x + 8 = 0$ has roots $a$ and $b$. Find a quadratic polynomial with integer coefficients whose roots are $\frac{1}{a}$ and $\frac{1}{b}$. | [
"Solution:\n\n(ans. $8x^{2} + 4x + 1$.\n$a b = 8$, $a + b = -4 \\Rightarrow \\frac{1}{a} \\frac{1}{b} = \\frac{1}{8}$, $\\frac{a + b}{a b} = \\frac{1}{a} + \\frac{1}{b} = -\\frac{1}{2}$. This means that the reciprocals are roots of the polynomial $x^{2} + \\frac{1}{2}x + \\frac{1}{8}$ and hence of $8x^{2} + 4x + 1$... | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | 8x^2 + 4x + 1 | |
0hxj | Problem:
Let $ABC$ be a triangle with incenter $I$. The incircle of $ABC$ is tangent to sides $BC$, $CA$, $AB$ at $D$, $E$, $F$. Let $H$ denote the orthocenter of triangle $BIC$, and let $P$ denote the midpoint of the altitude from $D$ to $EF$. Prove that $HP$ bisects $EF$. | [
"Solution:\n\nWithout loss of generality, $AB \\leq AC$. Let $B_1$ and $C_1$ be the projections of $C$ and $B$ onto lines $BI$ and $CI$, respectively. Then $\\angle IEC = \\angle IB_1C = 90^{\\circ}$, so quadrilateral $IEB_1C$ is cyclic. But then\n$$\n\\angle B_1EC = \\angle B_1IC = 180^{\\circ} - \\angle BIC = 180... | United States | Berkeley Math Circle Monthly Contest 3 | [
"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 > Transformations > Homot... | null | proof only | null | |
0azv | Problem:
Let $N$ be the smallest positive integer such that $N / 15$ is a perfect square, $N / 10$ is a perfect cube, and $N / 6$ is a perfect fifth power. Find the number of positive divisors of $N / 30$. | [
"Solution:\n\n$N$ must be of the form $N = 2^{m} 3^{n} 5^{p}$ for some nonnegative integers $m, n, p$.\n\nSince $N / 15 = 2^{m} 3^{n-1} 5^{p-1}$ is a perfect square, we have $m \\equiv 0 \\bmod 2$ and $n \\equiv p \\equiv 1 \\bmod 2$.\n\nSince $N / 10 = 2^{m-1} 3^{n} 5^{p-1}$ is a perfect cube, we have $n \\equiv 0... | Philippines | 20th Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | 8400 | |
0ivp | Problem:
In how many ways can the positive integers from $1$ to $100$ be arranged in a circle such that the sum of every two integers placed opposite each other is the same? (Arrangements that are rotations of each other count as the same.) Express your answer in the form $a!\cdot b^{c}$. | [
"Solution:\n\nAnswer: $49!\\cdot 2^{49}$\n\nSplit the integers up into pairs of the form $(x, 101-x)$. In the top half of the circle, exactly one element from each pair occurs, and there are thus $50!$ ways to arrange them, and also $2^{50}$ ways to decide whether the larger or smaller number in each pair occurs in... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 49!·2^49 | |
06m9 | There are 50 rods of lengths $1, 3, 5, 7, \ldots, 99$. How many ways are there to pick three of these rods to form a triangle? | [
"The answer is $\\binom{50}{3}$ minus the number of choices of three rods that violate the triangle inequality. The number to be subtracted is the size of the set\n$$\nS = \\{(a, b, c) : 1 \\le a < b < c \\le 99,\\ a + b < c,\\ a, b, c \\text{ are odd numbers}\\}.\n$$\nNote that for $(a, b, c) \\in S$, $a+b$ is an ... | Hong Kong | HongKong 2022-23 IMO Selection Tests | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | English | proof and answer | 9500 | |
0kej | Problem:
Let $A B C D E F$ be a regular hexagon with side length $2$. A circle with radius $3$ and center at $A$ is drawn. Find the area inside quadrilateral $B C D E$ but outside the circle. | [
"Solution:\n\nRotate the region $6$ times about $A$ to form a bigger hexagon with a circular hole. The larger hexagon has side length $4$ and area $24 \\sqrt{3}$, so the area of the region is $\\frac{1}{6}(24 \\sqrt{3}-9 \\pi)=4 \\sqrt{3}-\\frac{3}{2} \\pi$.\n\n"
] | United States | HMMT February | [
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | 4√3 − 3/2·π | |
0ich | Problem:
For a string of $P$ 's and $Q$ 's, the value is defined to be the product of the positions of the $P$ 's. For example, the string $P P Q P Q Q$ has value $1 \cdot 2 \cdot 4=8$.
Also, a string is called antipalindromic if writing it backwards, then turning all the $P$ 's into $Q$ 's and vice versa, produces th... | [
"Solution:\n$2005^{1002} / 2004$ !\nConsider the product\n$$\n\\left(\\frac{1}{1}+\\frac{1}{2004}\\right)\\left(\\frac{1}{2}+\\frac{1}{2003}\\right)\\left(\\frac{1}{3}+\\frac{1}{2002}\\right) \\cdots\\left(\\frac{1}{1002}+\\frac{1}{1003}\\right) .\n$$\nThis product expands to $2^{1002}$ terms, and each term gives t... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2005^1002/2004! | |
0dm5 | Suppose $G$ is a simple planar graph with minimum degree at least $5$. Prove that $G$ contains an edge $\{x, y\}$ such that $\deg(x) + \deg(y) \le 11$. | [
"By adding edges we can assume that $G$ is a triangulation (note that if we add edges the minimum degree only increases and if we find an edge $\\{x, y\\}$ such that $\\deg(x) + \\deg(y) \\le 11$ in the new graph, this edge will also be present in the original graph, since otherwise one of its endpoints would have ... | Saudi Arabia | Saudi Booklet | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
07u1 | Find the smallest number of colours needed for colouring all positive integers in such a way that any two integers that differ by a prime number have different colours. | [
"Assume such a colouring exists and consider the numbers $1, 3, 6, 8$. Any two of these numbers differ by a prime number: $3 - 1 = 2$, $6 - 1 = 5$, $8 - 1 = 7$, $6 - 3 = 3$, $8 - 3 = 5$ and $8 - 6 = 2$. Thus, at least four colours are needed.\n\nTo prove that four colours are sufficient, we pick four colours and nu... | Ireland | IRL_ABooklet | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 4 | |
0iwo | Problem:
Write down a pair of integers $(a, b)$, where $-100000 < a < b < 100000$. You will get $\max (25, k)$ points, where $k$ is the number of other teams' pairs that you interleave. (Two pairs $(a, b)$ and $(c, d)$ of integers interleave each other if $a < c < b < d$ or $c < a < d < b$.) | [] | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | final answer only | null | |
0h9g | We know that a certain number $A$ has exactly 2018 positive integer factors (including 1 and number $A$ itself), and it is divisible by 2018. Prove that $A$ is not divisible by $2018^2$. | [
"Let $A = p_1^{k_1} p_2^{k_2} \\dots p_n^{k_n}$ be a canonical prime factorization of $A$, where $p_i$ are different prime factors, and $k_i$ are positive integers, $i = 1, n$. Knowing the total number of positive integer factors of $A$ we can conclude that\n$$\n(k_1 + 1)(k_2 + 1)\\dots(k_n + 1) = 2018.\n$$\n\nSinc... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0bmz | A quadratic function $f$ sends any interval $I$ of length 1 to an interval $f(I)$ of length at least 1.
Prove that for any interval $J$ of length 2, the length of the interval $f(J)$ is at least 4. | [
"If $f(x) = ax^2 + bx + c$ and $v = -b/2a$ is the abscissa of parabola's vertex, then, taking $I = [v - 1/2, v + 1/2]$, the interval $f(I)$ has length $|a|/4$, hence $|a| \\ge 4$.\n\nNow, taking an interval $J$ of length 2, one can find $x, y \\in J$ such that $x - y = 1$ and $v \\notin (y, x)$.\nWe have $|f(x) - f... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
07on | Suppose $(a_n)$ is a sequence of integers such that $a_1$ and $a_2$ are two distinct digits of the number $2014$, and $a_n = a_{n-1}^{10} + a_{n-2}^{10}$ for all $n > 2$.
Let $m = 20142014\ldots2014$ be the number given by the string $2014$ repeated $2014$ times. Show that $m$ cannot be $a_n$ for any positive integer $... | [
"Note first that\n$$\n\\begin{align*}\na_3 &= a_1^{10} + a_2^{10} \\le 2^{10} + 4^{10} < 2 \\cdot 4^{10} = 2^{21} < 10^{21} < m \\quad \\text{and} \\\\\na_4 &= a_2^{10} + a_3^{10} < 4^{10} + 10^{20} < 2 \\cdot 10^{20} < m.\n\\end{align*}\n$$\nHence, if $m = a_n$, then $n \\ge 5$.\nGiven the possible values for $a_1... | Ireland | Irska 2014 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0hbl | Let **ABC** be a right triangle such that its legs satisfy the following: $BC = \sqrt{2}AC$. Show that the medians **AN** and **CM** are perpendicular to each other.
(Danylo Khilko)
**Fig. 6** | [
"Since **AB** is a hypothenuse, then $\\angle MCB = \\angle MBC$ (Fig. 6).\n\nOn the other hand, $AC^2 = CN \\cdot CB$, since $BC = \\sqrt{2}AC$ and $CN = \\frac{1}{2}BC = \\frac{\\sqrt{2}}{2}AC$.\n\nThen $\\triangle ACN \\sim \\triangle ABC$. Hence $\\angle CAN = \\angle ABC$.\n\nThus, $\\angle AKC = \\angle ACK +... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
02d0 | Problem:
Um número surpreendente é um número divisível por $9$, de nove algarismos diferentes, nenhum deles igual a $0$ tal que:
a. O número formado pelos $2$ primeiros algarismos é divisível por $2$;
b. O número formado pelos $3$ primeiros algarismos é divisível por $3$;
c. O número formado pelos $4$ primeiros algar... | [
"Solution:\n\n381654729"
] | Brazil | Desafios | [
"Number Theory > Divisibility / Factorization"
] | null | final answer only | 381654729 | |
0apk | Problem:
Solve for $x$: $16^{1 / 8} + x^{1 / 4} = \frac{23}{5 - \sqrt{2}}$. | [
"Solution:\nAfter rationalizing the denominator, we get\n$$\n16^{1 / 8} + x^{1 / 4} = 5 + \\sqrt{2}\n$$\nIt follows that $x^{1 / 4} = 5$ or $x = 625$."
] | Philippines | Tenth Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | 625 | |
05eg | Problem:
Déterminer tous les entiers $a$ et $b$ tels que $(a+1)(b-1)=a^{2} b^{2}$. | [
"Solution:\nSoit $a, b$ des entiers tels que $(a+1)(b-1)=a^{2} b^{2}$.\nPuisque d'une part $a$ et $a+1$ sont premiers entre eux, et d'autre part $b$ et $b-1$ sont premiers entre eux, c'est donc que $b-1= \\pm a^{2}$ et que $a+1= \\pm b^{2}$, où les signes sont les mêmes dans les deux relations.\n\n- Si $b-1=a^{2}$ ... | France | Olympiades Françaises de Mathématiques, Envoi No. 6 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (0,1) and (-1,0) | |
03so | Suppose that positive integers $a_1, a_2, \dots, a_{2\,006}$ (some of them may be equal) satisfy the condition: any two of $\frac{a_1}{a_2}, \frac{a_2}{a_3}, \dots, \frac{a_{2\,005}}{a_{2\,006}}$ are unequal. At least how many different numbers are there in $\{a_1, a_2, \dots, a_{2\,006}\}$? (posed by Chen Yonggao) | [
"With 45 different positive integers we can only get $45 \\times 44 + 1 = 1\\,981$ fractions. So there are more than 45 different numbers in $\\{a_1, a_2, \\dots, a_{2\\,006}\\}$.\n\nOn the other hand, let $p_1, p_2, \\dots, p_{46}$ be 46 different prime. Set $a_1, a_2, \\dots, a_{2\\,006}$ to be:\n$$\n\\begin{alig... | China | China Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 46 | |
04jn | If $n$ is a positive integer, let $R_n$ denote the integer that in the decimal representation consists of $n$ digits 1. Prove the following statement: if $R_n$ is a prime number, then $n$ is a prime number, too. | [
"Let $R_n$ be the integer consisting of $n$ digits 1. That is,\n$$\nR_n = \\underbrace{111\\ldots1}_{n\\ \\text{digits}} = \\frac{10^n - 1}{9}.\n$$\n\nSuppose $R_n$ is a prime number. We want to show that $n$ is also a prime number.\n\nAssume, for contradiction, that $n$ is composite. Then $n = ab$ for integers $a,... | Croatia | Croatia Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0eve | Let $H$ be the orthocenter of an acute triangle $ABC$, and a circle $\omega$ that passes through all of the points $H$, $A$, $B$ meet the line segment $BC$ at the point $D (\neq B)$. Let $P$ be the intersecting points of the line $DH$ and the line segment $AC$, and $Q$ be the circumcenter of the triangle $ADP$. Show th... | [
"Let $R$ be the center of the circle $\\omega$, and $E$ be the intersecting points of $BH$ and $AC$. Then we have\n$$\n\\begin{aligned}\n\\angle RBD &= 90^\\circ - \\frac{1}{2}\\angle DRB && \\text{since } RB = RD \\\\\n&= 90^\\circ - \\angle DHB \\\\\n&= 90^\\circ - \\angle PHE \\\\\n&= \\angle EPH \\\\\n&= 180^\\... | South Korea | Korean Mathematical Olympiad Final Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0c9o | Problem:
Fie $n \in \mathbb{N}, n \geq 2$ şi numerele reale $a_{1}, a_{2}, \ldots, a_{n} \in [0,1]$. Aflaţi valoarea maximă a celui mai mic dintre numerele:
$$
a_{1} - a_{1} a_{2},\ a_{2} - a_{2} a_{3},\ \ldots,\ a_{n} - a_{n} a_{1}
$$ | [] | Romania | Primul test de selecţie pentru OBMJ | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1/4 | |
0afd | а) На табла $5 \times 5$ се поставени $21$ жетони со белата страна нагоре така да секој жетон лежи врз посебно $1 \times 1$ квадратче (секој жетон е двобоен, има една бела и една црна страна). Во секој потег, Марта зема од таблата еден “бел” жетон, го превртува и го враќа врз некое слободно $1 \times 1$ квадратче. Нејз... | [
"а)\nОд принципот на Дирихле, во секој момент кога на таблата се поставени $21$ жетони, постои редица целосно исполнета со жетони и постои колона целосно исполнета со жетони. Да претпоставиме дека Марта успеала да ја реализира поставената цел. Тогаш на почетокот, на таблата има “бел” крст жетони, а на крајот, на та... | North Macedonia | Републички натпревар по математика за средно образование | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Macedonian, English | proof and answer | a) Impossible for any starting arrangement. b) Yes; for example, start with one entire column empty and sweep tokens into the empty column while flipping. | |
03gx | Problem:
Given three non-collinear points $A$, $B$, $C$, construct a circle with centre $C$ such that the tangents from $A$ and $B$ to the circle are parallel. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0a8n | Problem:
Stones are placed on the squares of a chessboard having $10$ rows and $14$ columns. There is an odd number of stones on each row and each column. The squares are coloured black and white in the usual fashion. Show that the number of stones on black squares is even. Note that there can be more than one stone o... | [
"Solution:\n\nChanging the order of rows or columns does not influence the number of stones on a row, on a column or on black squares. Thus we can order the rows and columns in such a way that the $5 \\times 7$ rectangles in the upper left and lower right corner are black and the other two $5 \\times 7$ rectangles ... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 17 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0kdu | Problem:
Let $p_{i}$ be the $i$\textsuperscript{th} prime. Let
$$
f(x)=\sum_{i=1}^{50} p_{i} x^{i-1}=2+3 x+\cdots+229 x^{49}
$$
If $a$ is the unique positive real number with $f(a)=100$, estimate $A=\lfloor 100000 a\rfloor$. An estimate of $E$ will earn $\max (0,\lfloor 20-|A-E| / 250\rfloor)$ points. | [
"Solution:\nNote $f(x)$ is increasing. Since $f(0)=2$ and $f(1) \\approx 50000$, we have $0<a<1$.\nSince we know that $p_{50}=229$, we can crudely bound\n$$\nf(x) \\lesssim \\sum_{i=1}^{\\infty} 5 i x^{i-1}=\\frac{5}{(1-x)^{2}}\n$$\nSetting this equal to 100 yields $x=1-20^{-1 / 2} \\approx 0.78$, so this is a good... | United States | HMMO 2020 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 84000 |
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