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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
06pf | Let $A_{0} = (a_{1}, \ldots, a_{n})$ be a finite sequence of real numbers. For each $k \geq 0$, from the sequence $A_{k} = (x_{1}, \ldots, x_{n})$ we construct a new sequence $A_{k+1}$ in the following way.
1. We choose a partition $\{1, \ldots, n\} = I \cup J$, where $I$ and $J$ are two disjoint sets, such that the ex... | [
"Lemma. Suppose that all terms of the sequence $(x_{1}, \\ldots, x_{n})$ satisfy the inequality $|x_{i}| < a$. Then there exists a partition $\\{1,2, \\ldots, n\\} = I \\cup J$ into two disjoint sets such that\n$$\n|\\sum_{i \\in I} x_{i} - \\sum_{j \\in J} x_{j}| < a. \\tag{1}\n$$\nProof. Apply an induction on $n$... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
0iug | Problem:
Let $\omega$ be a circle of radius $1$ centered at $O$. Let $B$ be a point on $\omega$, and let $l$ be the line tangent to $\omega$ at $B$. Let $A$ be on $l$ such that $\angle A O B = 60^\circ$. Let $C$ be the foot of the perpendicular from $B$ to $O A$. Find the length of line segment $O C$. | [
"Solution:\n$O C / O B = \\cos(60^\\circ)$. Since $O B = 1$, $O C = \\frac{1}{2}$."
] | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | final answer only | 1/2 | |
0hlf | Problem:
Determine, with proof, whether or not there exist positive integers $a$, $b$, and $c$ such that
$$
a b + b c = a c \quad \text{and} \quad a b c = 10! .
$$ | [
"Solution:\n\nThe answer is no. Note that $10!$ has exactly one prime factor of $7$. Therefore, exactly one of $a$, $b$, and $c$ is divisible by $7$. If $7$ divides $b$ (we write this as $7 \\mid b$), then $7 \\mid a b + b c$ but $7 \\nmid a c$, so the equation cannot hold. Likewise, if $7 \\mid a$, then $7 \\mid a... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | No | |
0l9i | Let $f$ be a function defined on the set of real numbers $\mathbb{R}$, taking values in $\mathbb{R}$ and satisfying the condition
$$
f(\cot x) = \sin 2x + \cos 2x
$$
for every $x$ belonging to the open interval $(0; \pi)$.
Find the least and the greatest values of the function $g(x) = f(x) \cdot f(1-x)$ on the closed i... | [
"We have:\n$$\nf(\\cot x) = \\sin 2x + \\cos 2x \\quad \\forall x \\in (0; \\pi)\n$$\n$$\n\\Leftrightarrow f(\\cot x) = \\frac{2 \\cot x}{\\cot^2 x + 1} + \\frac{\\cot^2 x - 1}{\\cot^2 x + 1} = \\frac{\\cot^2 x + 2 \\cot x - 1}{\\cot^2 x + 1} \\quad \\forall x \\in (0; \\pi).\n$$\nTherefore, remarking that for ever... | Vietnam | 2003 Vietnamese Mathematical Olympiad | [
"Precalculus > Trigonometric functions",
"Precalculus > Functions",
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | English | proof and answer | minimum = 4 - sqrt(34), maximum = 1/25 | |
093x | Problem:
Let $n$, $b$ and $c$ be positive integers. A group of $n$ pirates wants to fairly split their treasure. The treasure consists of $c \cdot n$ identical coins distributed over $b \cdot n$ bags, of which at least $n-1$ bags are initially empty. Captain Jack inspects the contents of each bag and then performs a s... | [
"Solution:\n\nWe proceed by induction on $n$. The case $n=1$ is trivial. Below we show that using one move we can always create a $b$-tuple of non-empty bags with precisely $c$ coins in total. This finishes the proof as we can put that $b$-tuple of bags aside for one pirate and finish by induction.\n\nSort the non-... | Middle European Mathematical Olympiad (MEMO) | 15th Middle European Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0afe | Димитар има два квадратни картони кои имаат страни $3$ cm и $4$ cm.
Дали може од нив, со сечење, да формира квадрат без да отфрли материјал? Во случај на потврден одговор, колку е страната на тој квадрат? | [
"Димитар може да состави квадрат со должина на страната од $5$ см. Квадратот со страна $3$ см ќе го раздели на правоаголници со страни $1$ см и $3$ см, еден правоаголник со страна $1$ см и $2$ см и еден квадрат со страна $1$ см. Нив ќе ги додаде на квадратот со страна $4$ см како на цртежот.\n\n![](attached_image_1... | North Macedonia | Регионален натпревар по математика за основно образование | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Macedonian, English | proof and answer | Yes; the side length is 5 cm. | |
0cpg | 2011 real numbers are written on a blackboard. It appears that for each three written numbers $a$, $b$, $c$, the sum $a+b+c$ is also written. Find the least possible number of zeroes among the written numbers.
На доску выписаны 2011 чисел. Оказалось, что сумма любых трёх выписанных чисел также является выписанным числ... | [
"Ответ. 2009.\n\nПример из 2009 нулей и чисел 1, -1 удовлетворяет условию, поэтому количество нулей может быть ровно 2009.\n\nПредположим, что количество нулей меньше 2008. Нетрудно видеть, что тогда на доске либо найдутся три неотрицательных числа, среди которых хотя бы два строго положительных, либо найдутся три ... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English, Russian | proof and answer | 2009 | |
09ps | Problem:
Julian en Johan spelen een spel met een even aantal, zeg $2n$, kaarten ($n \in \mathbb{Z}_{>0}$). Op elke kaart staat een positief geheel getal. De kaarten worden geschud en in een rij op tafel gelegd met de getallen zichtbaar. Een speler die aan de beurt is, mag ofwel de meest linker kaart ofwel de meest rec... | [] | Netherlands | TOETS TRAININGSKAMP | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0hrc | Problem:
Let $f(x)$ be a quadratic polynomial. Prove that there exist quadratic polynomials $g(x)$ and $h(x)$ such that
$$
f(x) f(x+1)=g(h(x)) .
$$ | [
"Solution:\nWrite $f(x)=a x^{2}+b x+c$, $a \\neq 0$. It is a familiar fact that the graph of a quadratic function always has an axis of symmetry, specifically the line $x=-\\frac{b}{2 a}$. By substituting\n$$\nu=x+\\frac{b}{2 a}+\\frac{1}{2}$$\nwe can make $f(x)=a \\nu^{2}-a \\nu+d$ symmetric about the line $\\nu=1... | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
0izt | Problem:
Pick a random integer between $0$ and $4095$, inclusive. Write it in base $2$ (without any leading zeroes). What is the expected number of consecutive digits that are not the same (that is, the expected number of occurrences of either $01$ or $10$ in the base $2$ representation)? | [
"Solution:\n\nAnswer: $\\frac{20481}{4096}$\n\nNote that every number in the range can be written as a $12$-digit binary string. For $i=1,2, \\ldots, 11$, let $R_{i}$ be a random variable which is $1$ if the $i$th and $(i+1)$st digits differ in a randomly chosen number in the range. By linearity of expectation, $E\... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | final answer only | 20481/4096 | |
0g15 | Problem:
Trouver tous les polynômes $P$ à coefficients entiers tels que $P(2017 n)$ est un nombre premier pour tout nombre naturel $n$. | [
"Solution:\n\nPuisque $P$ est un polynôme à coefficients entiers, on a que $a-b \\mid P(a)-P(b)$ pour tous $a, b \\in \\mathbb{Z}$. Ainsi en particulier nous avons que, pour $q = P(2017)$, $P(2017 k q + 2017) \\equiv P(2017) \\bmod q$, mais par hypothèse ces deux valeurs sont des nombres premiers, donc on a $P(2017... | Switzerland | IMO-Selektion | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | P is a constant polynomial equal to a prime number. | |
0bu9 | Problem:
Fie $a \geq 2$ un număr natural. Arătaţi că afirmaţiile următoare sunt echivalente:
a) Există numerele naturale nenule $b, c$, astfel încât $a^{2}=b^{2}+c^{2}$;
b) Există un număr natural nenul $d$, astfel încât ecuaţiile $x^{2}-a x+d=0$ şi $x^{2}-a x-d=0$ au rădăcinile întregi. | [
"Solution:\n\nSă presupunem că $a^{2}=b^{2}+c^{2}$. Numerele $b$ şi $c$ nu pot fi ambele impare (suma a două numere impare e de forma $4k+2$ şi nu poate fi pătrat), deci cel puţin unul dintre ele este par, adică produsul $bc$ este par.\n\nDiscriminanţii celor două ecuaţii sunt $\\Delta_{1}=a^{2}-4d$ şi $\\Delta_{2}... | Romania | Olimpiada Naţională de Matematică, Etapa Judeţeană şi a Municipiului Bucureşti | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
07hr | Find all functions $f : \mathbb{R}^{+} \to \mathbb{R}^{+}$ such that for all $x, y, z \in \mathbb{R}^{+}$
$$
f(x + f(y) + f(f(z))) = z + f(y + f(x)).
$$ | [
"We shall firstly prove that $f$ is injective. Note that if $f(z) = f(z')$ for some $z \\neq z'$, by substituting $(x, y, z)$ and $(x, y, z')$ in the given equation, we have:\n$$\n\\begin{align*}\nz + f(y + f(x)) &= f(x + f(y) + f(f(z))) \\\\\n&= f(x + f(y) + f(f(z'))) \\\\\n&= z' + f(y + f(x)) \\\\\n\\implies z &=... | Iran | 40th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x for all x > 0 | |
0bir | Let $n$ be a positive integer and $x_1, x_2, \dots, x_n > 0$ be real numbers so that
$$
x_1 + x_2 + \dots + x_n = \frac{1}{x_1^2} + \frac{1}{x_2^2} + \dots + \frac{1}{x_n^2}.
$$
Show that for each positive integer $k \le n$, there are $k$ numbers among $x_1, x_2, \dots, x_n$ whose sum is at least $k$. | [
"Arguing by contradiction, suppose that every sum of $k$ numbers from $x_1, x_2, \\dots, x_n$ is strictly less than $k$. Then the numbers\n$$\na_j = x_j + x_{j+1} + \\dots + x_{j+k-1}, \\quad j = 1, 2, \\dots, n\n$$\nare also less than $k$ (where the indices from the sums $a_j$ are considered to be taken modulo $n$... | Romania | 65th NMO Selection Tests for JBMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
09jn | Let $\overline{abcd}$ be a four-digit number, where $ab$ and $cd$ are two-digit numbers. If the sum of $ab$ and $cd$ is equal to $bc$, then $\overline{abcd}$ is called an *interesting number*. For example, $13+18=31$, but $1208$ is not an interesting number because $\overline{08}$ is not a two-digit number. How many in... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Other"
] | English | proof and answer | 28 | |
04qp | Let $ABC$ be a triangle, $k$ its incircle and $k_a, k_b, k_c$ three circles orthogonal to $k$ passing through $B$ and $C$, $A$ and $C$, and $A$ and $B$ respectively. The circles $k_a, k_b$ meet again in $C'$; in the same way we obtain the points $B'$ and $A'$. Prove that the radius of the circumcircle of $A'B'C'$ is ha... | [
"\nLet $I$ and $r$ denote the center and the radius of circle $k$. Let $D, E$, and $F$ denote the points where $k$ touches $BC, AC$, and $AB$, respectively. Let $P, Q$, and $R$ denote the midpoints of $EF, DF$, and $DE$ respectively. We will use the well known lemma:\n**LEMMA.** The circles... | Czech Republic | 6-th Czech-Slovak Match | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | English | proof only | null | |
0cw3 | A circle $\Omega$ centered at $O$ is circumscribed about an acute-angled triangle $ABC$ with $AB < BC$. Let $H$ be the orthocenter of $ABC$. A point $D$ is chosen on the extension of $BO$ beyond $O$ so that $\angle ADC = \angle ABC$. A line through $H$ parallel to $BO$ meets the smaller arc $AC$ of $\Omega$ at $E$. Pro... | [
"Пусть $P$ — вторая точка пересечения $BO$ с окружностью $\\Omega$ (см. рис. 15). Тогда $BP$ — диаметр $\\Omega$, и $\\angle BCP = 90^\\circ = \\angle BAP$. Значит, $CP \\parallel AN$ и $AP \\parallel CH$. Следовательно, четырёхугольник $ANCP$ — параллелограмм. Обозначим через $M$ точку пересечения его диагоналей. ... | Russia | 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 > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous >... | English; Russian | proof only | null | |
07qt | The four digit number $ABCD$ has the property that
$$
ABCD = A \times BCD + ABC \times D.
$$
What is the smallest possible value of $ABCD$? | [
"To find the smallest number, we try $A = 1$. The given equation becomes $1BCD = BCD + 1BC \\times D$, hence $1000 = 1BC \\times D$. This means that $D$ is a divisor of $1000 = 2^3 \\times 5^3$ and so $D \\in \\{1, 2, 4, 5, 8\\}$, because $D$ is a digit. On the other hand, $1BC < 200$ implies that $1000 < 200 \\tim... | Ireland | Ireland_2017 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 1258 | |
0dcv | It is given a graph whose vertices are positive integers and an edge between numbers $a$ and $b$ exists if and only if
$$
a+b+1 \mid a^{2}+b^{2}+1 .
$$
Is this graph connected? | [
"If $x, y \\in \\mathbb{Z}^{+}$, define $x \\leftrightarrow y$ if and only if $x, y$ are connected by some edge. We have for all $a \\in \\mathbb{Z}^{+}$,\n$$\na^{2}+a+1 \\mid (a^{2}-a+1)(a^{2}+a+1) = a^{4}+a^{2}+1 .\n$$\nThus $a \\leftrightarrow a^{2}$ for all $a$, then also true for $a+1 \\leftrightarrow (a+1)^{2... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Graph Theory",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | Yes, the graph is connected. | |
0fzt | Problem:
Im Teil-Land gibt es $n$ Städte. Je zwei Städte sind durch eine Einbahnstrasse verbunden, die entweder nur mit dem Töff oder nur mit dem Auto befahrbar ist. Zeige, dass es eine Stadt gibt, von der aus jede andere Stadt entweder mit dem Töff oder mit dem Auto erreicht werden kann.
Bemerkung: Es muss nicht jed... | [
"Solution:\n\nWir benützen starke Induktion nach $n$. $n=1$ ist trivial. Betrachte nun $n+1$ Städte $A_{0}, A_{1}, \\ldots, A_{n}$, welche nach Induktionsvoraussetzung wie folgt geordnet werden können: $A_{i}$ besitzt die gewünschte Eigenschaft für den Subgraphen induziert durch $A_{i}, A_{i+1}, \\ldots, A_{n}$ für... | Switzerland | IMO-Selektion | [
"Discrete Mathematics > Other",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0gcn | 令 $a$, $b$, $c$, $d$ 為非負實數且滿足 $a + b + c + d = 100$。試證：
$$
\sqrt[3]{\frac{a}{b+7}} + \sqrt[3]{\frac{b}{c+7}} + \sqrt[3]{\frac{c}{d+7}} + \sqrt[3]{\frac{d}{a+7}} \le \frac{8}{\sqrt[3]{7}}
$$ | [
"Let\n$$\nS = \\sqrt[3]{\\frac{a}{b+7}} + \\sqrt[3]{\\frac{b}{c+7}} + \\sqrt[3]{\\frac{c}{d+7}} + \\sqrt[3]{\\frac{d}{a+7}}\n$$\nAssume that $x$, $y$, $z$, $t$ is a permutation of the variables, with\n$x \\le y \\le z \\le t$. Then, by the rearrangement inequality,\n$$\nS \\le \\left( \\sqrt[3]{\\frac{x}{t+7}} + \\... | Taiwan | 二〇一九數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
09w6 | Given is a parallelogram $ABCD$ with $\angle A < 90^\circ$ and $|AB| < |BC|$. The angular bisector of angle $A$ intersects side $BC$ in $M$ and intersects the extension of $DC$ in $N$. Point $O$ is the centre of the circle through $M$, $C$, and $N$.
Prove that $\angle OBC = \angle ODC$.
 | [
"As an intermediate step, we first show that triangles *OCM* and *OCN* are congruent. Since $AD$ and $BC$ are parallel, we have (F angles): $\\angle CMN = \\angle DAM = \\frac{1}{2} \\angle DAB$. Since $DN$ and $AB$ are parallel, we have (Z angles):\n$\\angle CNM = \\angle NAB = \\frac{1}{2} \\angle DAB$. It follow... | Netherlands | Final Round | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0kgf | Problem:
A polygon is regular if all its sides and angles are the same. Find the measure of each angle in a regular dodecagon (12-sided polygon). | [
"Solution:\n\nA regular dodecagon can be divided into $10$ triangles, as shown above. The sum of the angles in each triangle is $180^{\\circ}$, so the sum of the angles in the dodecagon is $10 \\cdot 180^{\\circ} = 1800^{\\circ}$. Since it is regular, all the angles are the same, so we divi... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 150° | |
0d3x | Let $\mathbb{N}$ denote the set of positive integers, and let $S$ be a set. There exists a function $f: \mathbb{N} \rightarrow S$ such that if $x$ and $y$ are a pair of positive integers with their difference being a prime number, then $f(x) \neq f(y)$. Determine the minimum number of elements in $S$. | [
"Let $f: \\mathbb{N} \\rightarrow S$ be such a function. Because the difference of any two numbers in $\\{1, 3, 6, 8\\}$ is a prime number, the cardinality of $\\{f(1), f(3), f(6), f(8)\\}$ is $4$. Hence, the minimum number of elements in $S$ is greater than or equal to $4$.\n\nNow, consider the function $f: \\math... | Saudi Arabia | SAMC | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Modular Arithmetic"
] | English, Arabic | proof and answer | 4 | |
0axy | Problem:
In triangle $A B C$, the medians $A D$ and $B E$ meet at the centroid $G$. Determine the ratio of the area of quadrilateral $C D G E$ to the area of triangle $A B C$. | [
"Solution:\n\nRefer to the figure on the right.\n$$\n\\begin{aligned}\n\\{[C D G E] \\} & =[C D E]+[G E D] \\\\\n& =\\frac{1}{4}[A B C]+\\frac{1}{3}[B E D] \\\\\n& =\\frac{1}{4}[A B C]+\\frac{1}{3}\\left(\\frac{1}{4}[A B C]\\right) \\\\\n& =\\frac{1}{4}[A B C]+\\frac{1}{12}[A B C] \\\\\n& =\\frac{1}{3}[A B C]\n\\en... | Philippines | Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | 1/3 | |
0apo | Problem:
Find the smallest positive integer $x$ such that the sum of $x, x+3, x+6, x+9$, and $x+12$ is a perfect cube. | [
"Solution:\n19\n$$\nx + (x+3) + (x+6) + (x+9) + (x+12) = 5x + 30 = 5(x+6)\n$$\nTo make up the least possible cube, we must have $x+6 = 5^{2}$ or $x = 19$."
] | Philippines | Tenth Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 19 | |
0j81 | Problem:
Three circles $k_{1}$, $k_{2}$, and $k_{3}$ intersect in point $O$. Let $A$, $B$, and $C$ be the second intersection points (other than $O$) of $k_{2}$ and $k_{3}$, $k_{1}$ and $k_{3}$, and $k_{1}$ and $k_{2}$, respectively. Assume that $O$ lies inside of the triangle $ABC$. Let lines $AO$, $BO$, and $CO$ int... | [
"Solution:\n\nIn this solution we will use a method called Inversion in the Plane.\nWe invert with respect to point $O$ with an arbitrary radius $r$. We will label the images of objects (points, circles, lines, segments) under this inversion by putting a bar over them. By properties of inversion, the three given ci... | United States | 13th Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0isi | Problem:
Kermit the frog enjoys hopping around the infinite square grid in his backyard. It takes him 1 Joule of energy to hop one step north or one step south, and 1 Joule of energy to hop one step east or one step west. He wakes up one morning on the grid with 100 Joules of energy, and hops till he falls asleep with... | [
"Solution:\n\nAnswer: 10201 Same as Combinatorics Test problem 4."
] | United States | 11th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 10201 | |
08pu | Problem:
Find all the integers pairs $(x, y)$ which satisfy the equation
$$
x^{5}-y^{5}=16 x y
$$ | [
"Solution:\nIf one of $x, y$ is $0$, the other has to be $0$ too, and $(x, y) = (0, 0)$ is one solution.\n\nIf $x y \\neq 0$, let $d = \\gcd(x, y)$ and we write $x = d a$, $y = d b$, $a, b \\in \\mathbb{Z}$ with $(a, b) = 1$. Then, the given equation is transformed into\n$$\nd^{3} a^{5} - d^{3} b^{5} = 16 a b\n$$\n... | JBMO | Junior Balkan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | (0, 0) and (-2, 2) | |
07ek | a $>$ k are two positive integers and two strictly increasing sequences $r_1 < r_2 < \dots < r_n$ and $s_1 < s_2 < \dots < s_n$ of positive integers have the following property,
$$
(a^{r_1} + k)(a^{r_2} + k)\dots(a^{r_n} + k) = (a^{s_1} + k)(a^{s_2} + k)\dots(a^{s_n} + k).
$$
Prove that these two sequences are equals, ... | [
"Without loss of generality, suppose $s_1 \\ge r_1 + 1$. Let $d = \\gcd(a, k)$, $a = da_1$ and $k = dk_1$. By using this equation and rewriting the problem's equality it is obtained that\n$$\n\\begin{align*}\n\\prod_{i=1}^{n} (d^{r_i} a_1^{r_i} + dk_1) &= \\prod_{i=1}^{n} (d^{s_i} a_1^{s_i} + dk_1) \\\\\n\\implies ... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | English | proof only | null | |
0agj | Let $A' \in (BC)$, $B' \in (AC)$, $C' \in (AB)$ be the points of tangency of the excribed circles of the triangle $ABC$ with the sides of $ABC$. Let $R'$ the circumradius of $A'B'C'$. Show that
$$
R' = \frac{1}{2r} \sqrt{2R(2R - h_a)(2R - h_b)(2R - h_c)},
$$
where, as usual, $R$ is the circumradius of $ABC$, $r$ is the... | [
"The triangle $A'B'C'$ is the pedal triangle of the symmetrical point of the incenter $I$ of $ABC$ with respect to the circumcenter of $ABC$. So, the relation between the areas $S = [ABC]$ and $S' = [A'B'C']$ is given by\n$$\nS' = S \\cdot \\frac{R^2 - \\overline{OI}^2}{4R^2} = \\frac{r}{2R}.\n$$\nIn the triangle $... | North Macedonia | Mediterranean Mathematics Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Method... | English | proof only | null | |
0buo | Problem:
Determinaţi funcţia $f: \mathbb{R} \rightarrow \mathbb{R}$ care admite primitive şi verifică pentru orice număr real $x$, egalitatea: $f(x)-F(x)=|x-1|$, unde $F$ este o primitivă a lui $f$. | [
"Solution:\n\nFie $F$ o primitivă a funcţiei $f$, adică $F'(x) = f(x)$ pentru orice $x \\in \\mathbb{R}$.\n\nDin ipoteză avem:\n$$\nf(x) - F(x) = |x-1|, \\quad \\forall x \\in \\mathbb{R}.\n$$\n\nRezultă:\n$$\nf(x) = F(x) + |x-1|.\n$$\n\nDerivăm ambele părţi după $x$:\n$$\nf'(x) = F'(x) + \\frac{d}{dx}|x-1|.\n$$\nD... | Romania | Olimpiada de Matematică Etapa Locală | [
"Calculus > Differential Equations > ODEs",
"Calculus > Differential Calculus > Derivatives"
] | null | proof and answer | All solutions are given, for an arbitrary real parameter a, by
f(x) = (a − 1/e) e^x + 1 for x < 1;
f(1) = a e;
f(x) = (a + 1/e) e^x − 1 for x > 1.
For this f, a valid primitive is F(x) = f(x) − |x − 1|. | |
0i0i | Problem:
Two sets of points in the coordinate plane are given: $\{(-1,1),(-1,2), \ldots,(-1,2000)\}$ and $\{(1,1),(1,2), \ldots,(1,2000)\}$. $2000$ line segments are drawn connecting these points so that each point in the first set is connected to exactly one point in the second set, and vice versa. Find, with proof, ... | [
"Solution:\n\nNote that, for any real numbers $a$ and $b$, the segment connecting $(-1, a)$ and $(1, b)$ has midpoint $\\left(0, \\frac{a+b}{2}\\right)$, so its $y$-intercept is $\\frac{a+b}{2}$. Now suppose that our given segments connect $(-1,1)$ to $\\left(1, y_{1}\\right)$, $(-1,2)$ to $\\left(1, y_{2}\\right)$... | United States | Berkeley Math Circle Monthly Contest #1 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2001000 | |
04o8 | Let $a \ge 2018$ be a real number. There are 2018 bowls, each containing a finite number of balls. It is known that the weight of each ball is of the form $a^k$, where $k$ is an integer, and that the total weight of balls in any bowl is the same. Let $B$ denote the total number of occurrences of the most frequently use... | [
"The smallest value $B$ can attain is 2018.\n\nWithout loss of generality we can assume that the weight of the lightest ball is equal to 1. If this is not the case, we can divide all the weights by the weight of the lightest ball.\n\nLet us assume that there are at most 2017 balls of each weight appearing in all th... | Croatia | Croatian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 2018 | |
0586 | The bisector of the internal angle on vertex $A$ of a triangle $ABC$ intersects the side $BC$ at point $D$. The line tangent to the circumcircle of the triangle $ABC$ at point $A$ intersects the line $BC$ at point $K$. Prove that $KA = KD$. | [
"Assume w.l.o.g. that $\\angle ABC > \\angle ACB$ (Fig. 31; otherwise change the roles of points $B$ and $C$). Note that\n$$\n\\angle ADK = 180^{\\circ} - \\angle CDA = \\angle DAC + \\angle ACD = \\angle BAD + \\angle ACB.\n$$\nBy inscribed angle property, $\\angle KAB = \\angle ACB$, whence\n$$\n\\angle BAD + \\a... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0jwi | Problem:
Let $L B C$ be a fixed triangle with $L B = L C$, and let $A$ be a variable point on arc $L B$ of its circumcircle. Let $I$ be the incenter of $\triangle A B C$ and $\overline{A K}$ the altitude from $A$. The circumcircle of $\triangle I K L$ intersects lines $K A$ and $B C$ again at $U \neq K$ and $V \neq K$.... | [
"Solution:\nLet $M$ be the midpoint of arc $B C$ not containing $L$ and let $D$ be the point where the incircle of triangle $A B C$ touches $B C$. Also let $N$ be the projection from $I$ to $A K$. We claim that $M$ is the desired fixed point.\nBy Simson's Theorem on triangle $K U V$ and point $I$ we have that point... | United States | February 2017 | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcen... | null | proof only | null | |
00tk | Let $ABC$ be an acute triangle such that $AB < AC$. Let $\omega$ be the circumcircle of $ABC$ and assume that the tangent to $\omega$ at $A$ intersects the line $BC$ at $D$. Let $\Omega$ be the circle with center $D$ and radius $AD$. Denote by $E$ the second intersection point of $\omega$ and $\Omega$. Let $M$ be the m... | [
"$$\n\\angle BAS = \\angle DAS - \\angle DAB = \\angle DSA - \\angle DCA = \\angle CAS.\n$$\nThis means that the line $AS$ is the angle bisector of $\\angle BAC$.\n\n\n\nNotice that $DE$ is also tangent to $\\omega$, because it is the second intersection point of $\\omega$ and $\\Omega$. Fr... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"... | null | proof only | null | |
0bqm | Find all functions $f : \mathbb{R} \to \mathbb{R}$ with the property
$$
|f(x+y) - f(x-y) - 2y| \le y^2, \quad \forall x, y \in \mathbb{R}.
$$ | [] | Romania | 67th NMO Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof and answer | f(x) = x + c for any real constant c | |
004x | Para $n = 1, 2, \ldots$ sea $1 + \frac{1}{2} + \ldots + \frac{1}{n} = \frac{u}{v}$, donde $u$ y $v$ son enteros positivos primos entre sí. Halle todos los $n$ para los cuales $u$ es divisible por $5$. | [] | Argentina | XVI Olimpiada Matemática Rioplatense | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | Spanish | proof and answer | All n such that 4·5^k ≤ n ≤ 5^{k+1} − 1 for some integer k ≥ 0 (equivalently, n whose base-five representation begins with the digit 4). | |
0eo2 | Let $f: \mathbb{Q}^{+} \to \mathbb{R}^{+}$ be a function that satisfies
$$
f(x + y) - f(x - y) = 4\sqrt{f(x)f(y)}
$$
for all $0 < y < x$.
(a) Prove that $f(2x) = 4f(x)$ for all $x \in \mathbb{Q}^{+}$.
(b) Find all such functions. | [
"Choose any positive rational $p, q$ and let $x = \\frac{p+q}{2}$ and $y = \\frac{p-q}{2}$. Then\n$$\nf(p) - f(q) = 4\\sqrt{f(x)f(y)} > 0,\n$$\nso $f(p) > f(q)$ and hence $f$ is strictly increasing.\n\nNow we show that $f$ has values arbitrarily close to $0$, i.e., for any $\\epsilon \\in \\mathbb{Q}^+$, we can fin... | South Africa | South-Afrika 2011-2013 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All functions f: Q^+ -> R^+ of the form f(x) = c x^2 with c > 0; in particular, f(2x) = 4 f(x). | |
0fwf | Problem:
Seien $a$, $b$, $c$ positive reelle Zahlen mit $a+b+c \geq a b c$. Beweise, dass von den folgenden drei Ungleichungen mindestens zwei richtig sind:
$$
\frac{2}{a}+\frac{3}{b}+\frac{6}{c} \geq 6, \quad \frac{2}{b}+\frac{3}{c}+\frac{6}{a} \geq 6, \quad \frac{2}{c}+\frac{3}{a}+\frac{6}{b} \geq 6
$$ | [
"Solution:\n\nSetze $x=\\frac{1}{a}$, $y=\\frac{1}{b}$ und $z=\\frac{1}{c}$. Die Nebenbedingung lautet dann $x y+y z+z x \\geq 1$ und die drei Ungleichungen werden zu\n$$\n\\begin{aligned}\n& 2 x+3 y+6 z \\geq 6 \\\\\n& 2 y+3 z+6 x \\geq 6 \\\\\n& 2 z+3 x+6 y \\geq 6\n\\end{aligned}\n$$\nEs genügt nun zu zeigen, da... | Switzerland | IMO Selektion | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0h56 | Numbers $a$, $b$ fulfill both equalities simultaneously:
$$
a^2 + b^2 = 1 \text{ and } a^3 + b^3 = -1.
$$
What is the possible value of the expression $a^3 + b^2$? | [
"From the first equation $-1 \\leq a \\leq 1$ and $-1 \\leq b \\leq 1$, therefore $0 \\leq 1+a \\leq 2$ and $0 \\leq 1+b \\leq 2$. Add both equations and get\n$$\na^2(1+a) + b^2(1+b) = 0.\n$$\nAs both items are non-negative, their sum equals zero if and only if every item equals $0$. So $a, b \\in \\{-1, 0\\}$. Fro... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round (Second Tour) | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | {-1, 1} | |
0f5c | Problem:
Interior points $D$, $E$, $F$ are chosen on the sides $BC$, $CA$, $AB$ (not at the vertices). Let $k$ be the length of the longest side of $DEF$. Let $a$, $b$, $c$ be the lengths of the longest sides of $AFE$, $BDF$, $CDE$ respectively. Show that $k \geq \dfrac{\sqrt{3}}{2} \min(a, b, c)$. When do we have equ... | [] | Soviet Union | 17th ASU | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0515 | Let $ABC$ be an acute triangle and $D$ an interior point of its side $AC$. We call a side of the triangle $ABD$ friendly, if the excircle of $ABD$ tangent to that side has its center on the circumcircle of $ABC$. Prove that there are exactly two friendly sides of $ABD$ if and only if $|BD| = |DC|$. | [
"Let $E$, $F$ and $G$ be the centers of excircles touching $BD$, $AD$ and $AB$ respectively, and let $\\omega$ be the circumcircle of $ABC$ (see Fig. 5). To prove the assertion of the problem, we will show that $F$ and $G$ cannot both lie on $\\omega$ and that $E \\in \\omega \\iff |BD| = |DC| \\iff F \\in \\omega$... | Estonia | Estonian Math Competitions | [
"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"
] | null | proof only | null | |
05ig | Problem:
$ABC$ est un triangle dont tous les angles sont aigus. Soit respectivement $H$ le pied de la hauteur de ce triangle issue de $C$ et $K$ le milieu du côté $AC$. On suppose que $BK = CH$ et que les angles $\widehat{KBC}$ et $\widehat{HCB}$ sont égaux. Montrer que le triangle $ABC$ est équilatéral. | [
"Solution:\n\nVoici la figure :\n\n\nMaintenant considérons la médiatrice de $[BC]$ (dessinée en pointillés sur la figure) et la symétrie par rapport à cet axe. Il est évident que par cette symétrie $B$ devient $C$ et vice-versa. Appelons $A'$ l'image de $A$ par la symétrie. Les hypothèses ... | France | Olympiades Françaises de Mathématiques - Envoi Numéro 1 - Corrigé | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
08cd | Problem:
Abelarda, Brunilda e Callisto, tre vecchi conoscenti, vogliono comprare una casa a testa tra le 10 casette in fila sulla via principale della città. Siccome non si sopportano, vogliono assolutamente evitare di essere vicini di casa: desiderano perciò che le case che acquistano siano due a due non adiacenti. I... | [
"Solution:\n\nLa risposta è $\\mathbf{( C )}$. Consideriamo dapprima un problema leggermente diverso: lasciamo indeterminato il numero delle case, per ora, e trascuriamo l'ipotesi che le case non possano essere adiacenti. In questo caso, detto $n$ il numero di case sulla via, avremmo $\\binom{n}{3}$ modi di sceglie... | Italy | Gara di Febbraio | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | MCQ | C | |
0jl8 | Problem:
How many two-digit prime numbers have the property that both digits are also primes? | [
"Solution:\n\nAnswer: $4$\n\nWhen considering the $16$ two-digit numbers with $2, 3, 5$, and $7$ as digits, we find that only $23, 37, 53$, and $73$ have this property."
] | United States | HMMT November 2014 | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | final answer only | 4 | |
06vn | Let $I$ be the incentre of acute-angled triangle $A B C$. Let the incircle meet $B C$, $C A$, and $A B$ at $D$, $E$, and $F$, respectively. Let line $E F$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\angle D P A + \angle A Q D = \angle Q I P$.
(Slovaki... | [
"Let $N$ and $M$ be the midpoints of the $\\operatorname{arcs}\\ \\overparen{B C}$ of the circumcircle, containing and opposite vertex $A$, respectively. By $\\angle F A E = \\angle B A C = \\angle B N C$, the right-angled kites $A F I E$ and $N B M C$ are similar. Consider the spiral similarity $\\varphi$ (dilatio... | IMO | IMO 2019 Shortlisted Problems | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configur... | English | proof only | null | |
05wq | 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$$\nOn peut chercher les racines de ce polynôme de degré 2 ou alors... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOi 2 : Algèbre | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | Equality holds if and only if x = n. | |
0akm | $ABC$ is acute triangle. $AE$ and $AF$ are isogonal cevians, where $E \in BC$ and $F \in BC$. The straight lines $AE$ and $AF$ intersect again the circumcircle of $ABC$ at points $M$ and $N$, respectively. In the rays $AB$ and $AC$ we get points $P$ and $R$ such that $\angle PEA = \angle B$ and $\angle AER = \angle C$.... | [] | North Macedonia | Mediterranean Mathematical Olympiad | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Meth... | English | proof only | null | |
04b5 | Let $a$ be a real number. Determine the sum of all three solutions of the equation
$$
x^3 - a^2x + ax - x + a^2 - a = 0.
$$ | [
"Let us write the equation as:\n$$\nx^3 - a^2x + ax - x + a^2 - a = 0.\n$$\nGroup the $x$ terms:\n$$\nx^3 + (a - a^2 - 1)x + (a^2 - a) = 0.\n$$\nThis is a cubic equation of the form:\n$$\nx^3 + px + q = 0,\n$$\nwhere $p = a - a^2 - 1$ and $q = a^2 - a$.\n\nBy Vieta's formulas, the sum of the roots of the cubic equa... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | 0 | |
07si | Find the smallest number with exactly $2020$ distinct positive divisors. You should include $1$ and the number as divisors. | [
"The divisors of an integer $n$ with prime factorisation\n$$\nn = p_1^{e_1} \\cdot p_2^{e_2} \\cdots p_k^{e_k}\n$$\nare all of the form $p_1^{f_1} \\cdot p_2^{f_2} \\cdots p_k^{f_k}$, where $0 \\le f_i \\le e_i$. Hence, the number of positive divisors of $n$ is equal to $(e_1+1)(e_2+1)\\cdots(e_k+1)$ and for each p... | Ireland | IRL_ABooklet_2020 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 2^100 * 3^4 * 5 * 7 | |
0c31 | Let $x$, $y$, $z$ be positive real numbers satisfying $2x^2 + 3y^2 + 6z^2 + 12(x + y + z) = 108$. Find the maximum value of $x^3y^2z$. | [
"Guessing that the maximum is obtained when $x = 3$, $y = 2$, $z = 1$, we multiply the given equality by $6$, and apply AM-GM:\n\n$$6 \\cdot 108 = 4x^2 + 4x^2 + 4x^2 + 9y^2 + 9y^2 + 36z^2 + 12x + 12x + 12x + 12x + 12x + 12x + 18y + 18y + 18y + 18y + 36z + 36z \\ge 18 \\sqrt{2^{28} \\cdot 3^{24} \\cdot x^{12} \\cdot... | Romania | 69th NMO Selection Tests for JBMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 108 | |
0k3x | Problem:
In the quadrilateral $M A R E$ inscribed in a unit circle $\omega$, $A M$ is a diameter of $\omega$, and $E$ lies on the angle bisector of $\angle R A M$. Given that triangles $R A M$ and $R E M$ have the same area, find the area of quadrilateral $M A R E$. | [
"Solution:\n\nSince $A E$ bisects $\\angle R A M$, we have $R E = E M$, and $E, A$ lie on different sides of $R M$. Since $A M$ is a diameter, $\\angle A R M = 90^{\\circ}$. If the midpoint of $R M$ is $N$, then from $[R A M] = [R E M]$ and $\\angle A R M = 90^{\\circ}$, we find $A R = N E$. Note that $O$, the cent... | United States | HMMT February | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 8√2/9 | |
005z | Determine todos los enteros $k \ge 2$ para los cuales, para todo entero $n \ge 2$, $n$ no divide al mayor divisor impar de $k^n + 1$. | [] | Argentina | XVII Olimpiada Matemática Rioplatense | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | Spanish | proof and answer | All integers of the form 2^t − 1 with t ≥ 2 | |
0aj4 | Let $m, n$ be integers greater than $1$, and let $a_1, a_2, \dots, a_m$ be positive integers not greater than $n^m$. Prove that there exist positive integers $b_1, b_2, \dots, b_m$ not greater than $n$ such that
$$
\text{gcd}(a_1 + b_1, a_2 + b_2, a_3 + b_3, \dots, a_m + b_m) < n,
$$
where $\text{gcd}(x_1, x_2, \dots, ... | [
"Suppose without loss of generality that $a_1$ is the smallest of the $a_i$. If $a_1 \\ge n^m - 1$, then the problem is simple: either all the $a_i$ are equal, or $a_1 = n^m - 1$ and $a_j = n^m$ for some $j$. In the first case we can take (say) $b_1 = 1$, $b_2 = 2$, and the rest of the $b_i$ can be arbitrary, and w... | North Macedonia | Girls European Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0khf | Let $c = \frac{2\pi}{11}$. What is the value of
$$
\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c} ?
$$
(A) $-1$ (B) $-\frac{\sqrt{11}}{5}$ (C) $\frac{\sqrt{11}}{5}$ (D) $\frac{10}{11}$ (E) $1$ | [
"Because the sine function has period $2\\pi$, it follows that $\\sin mc = \\sin nc$ if $m \\equiv n \\pmod{11}$. Because the sine function is also an odd function, it follows also that $\\sin mc = -\\sin nc$ if $m \\equiv -n \\pmod{11}$ and thus $\\sin 6c = -\\sin 5c$, $\\sin 9c = -\\sin 2c$, $\\sin 12c = \\sin c$... | United States | AMC 12 B | [
"Precalculus > Trigonometric functions"
] | null | MCQ | E | |
0dw7 | Problem:
Na vsaki ploskvi kocke je napisano naravno število, v vsakem oglišču pa je napisan zmnožek števil na 3 ploskvah, ki se stikajo v tem oglišču. Vsota števil v ogliščih kocke je 70. Kolikšna je vsota števil na ploskvah kocke? | [
"Solution:\n\nOznačimo števila na ploskvah kocke z $a_{1}$, $a_{2}$, $a_{3}$, $a_{4}$, $a_{5}$ in $a_{6}$. V ogliščih kocke so zapisana števila $a_{1} a_{2} a_{5}$, $a_{2} a_{3} a_{5}$, $a_{3} a_{4} a_{5}$, $a_{4} a_{1} a_{5}$, $a_{1} a_{2} a_{6}$, $a_{2} a_{3} a_{6}$, $a_{3} a_{4} a_{6}$ in $a_{4} a_{1} a_{6}$, za... | Slovenia | 48. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 14 | |
0j41 | Problem:
Let $a$, $b$, and $c$ be positive real numbers. Determine the largest total number of real roots that the following three polynomials may have among them: $a x^{2} + b x + c$, $b x^{2} + c x + a$, and $c x^{2} + a x + b$. | [
"Solution:\n\nAnswer: 4\n\nIf all the polynomials had real roots, their discriminants would all be nonnegative: $a^{2} \\geq 4 b c$, $b^{2} \\geq 4 c a$, and $c^{2} \\geq 4 a b$. Multiplying these inequalities gives $(a b c)^{2} \\geq 64(a b c)^{2}$, a contradiction. Hence one of the quadratics has no real roots.\n... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 4 | |
0akt | Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that for any $x, y \in \mathbb{R}$ it holds that
$$
x f(x+y) + y f(y-x) = f(x^2 + y^2).
$$ | [] | North Macedonia | Macedonian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | All functions of the form f(x) = c x for real constants c. | |
0iru | Problem:
Below is a list of famous mathematicians. Your task is to list a subset of them in the chronological order of their birth dates. Your submission should be a sequence of letters. If your sequence is not in the correct order, then you get 0 points. Otherwise your score will be $\min \{\max \{5(N-4), 0\}, 25\}$,... | [
"Solution:\n\nAnswer: any subsequence of $EJKFIHADCGBNL$\n\nThe corresponding birth dates are listed below:\n\n(A) Niels Abel (1802-1829)\n(B) Arthur Cayley (1821-1895)\n(C) Augustus De Morgan (1806-1871)\n(D) Gustav Dirichlet (1805-1859)\n(E) Leonhard Euler (1707-1783)\n(F) Joseph Fourier (1768-1830)\n(G) Évariste... | United States | Harvard-MIT Mathematics Tournament | [
"Math Word Problems"
] | null | final answer only | any subsequence of EJKFIHADCGBNL | |
0j0t | Problem:
Let $f(x) = -x^{2} + 10x - 20$. Find the sum of all $2^{2010}$ solutions to
$$
\underbrace{f(f(\ldots(x) \ldots))}_{2010\ f\text{~s}} = 2.
$$ | [
"Solution:\nAnswer: $5 \\cdot 2^{2010}$\n\nDefine $g(x) = f(f(\\ldots(x) \\ldots))$.\n\nWe calculate:\n$$\nf(10-x) = -(10-x)^{2} + 10(10-x) - 20 = -100 + 20x - x^{2} + 100 - 10x - 20 = -x^{2} + 10x - 20 = f(x).\n$$\nThis implies that $g(10-x) = g(x)$. So if $g(x) = 2$, then $g(10-x) = 2$.\n\nMoreover, we can calcul... | United States | Harvard-MIT November Tournament | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 5 * 2^2010 | |
0cst | Имеются 2013 карточек, на которых написана цифра 1, и 2013 карточек, на которых написана цифра 2. Вася складывает из этих карточек 4026-значное число. За один ход Петя может поменять местами некоторые две карточки и заплатить Васе 1 рубль. Процесс заканчивается, когда у Пети получается число, делящееся на 11. Какую наи... | [
"Рассмотрим 4026-значное число $A$, состоящее из 2013 единиц и 2013 двоек. Пусть в этом числе в нечётных разрядах стоит $k$ единиц и $\\ell = 2013 - k$ двоек, тогда в чётных разрядах будет $k$ двоек и $\\ell$ единиц (здесь $k$ может принимать любое целое значение от 0 до 2013). Разность сумм цифр в нечётных разряда... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 5 | |
0dws | Problem:
Dano je geometrijsko zaporedje $\left(a_{n}\right)$ s pozitivnimi členi. Označimo
$$
S_{n}=\log a_{1}+\log a_{2}+\ldots+\log a_{n-1}+\log a_{n}
$$
Dokaži: če za nek $m \neq n$ velja $S_{n}=S_{m}$, je $S_{n+m}=0$. | [
"Solution:\n\nČlene geometrijskega zaporedja lahko zapišemo v obliki $a_{n}=a_{1} q^{n-1}$. Ker so členi pozitivni, je $a_{1}>0$ in $q>0$. Potem je\n$$\nS_{n}=\\log a_{1}+\\log a_{1} q+\\ldots+\\log a_{1} q^{n-1}=\\log a_{1}^{n} q^{\\frac{n(n-1)}{2}}\n$$\nČe je $S_{n}=S_{m}$, sledi\n$$\na_{1}^{n} q^{\\frac{n(n-1)}{... | Slovenia | 49. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof only | null | |
02m6 | Compute
$$
\int_{0}^{\pi/4} \frac{x}{(\sin x + \cos x) \cos x} \, dx.
$$ | [
"First notice the identity $\\sin x + \\cos x = \\sqrt{2}(\\cos x \\cos \\frac{\\pi}{4} + \\sin x \\sin \\frac{\\pi}{4}) = \\sqrt{2} \\cos(\\frac{\\pi}{4} - x)$. So\n$$\nI = \\int_{0}^{\\pi/4} \\frac{x}{(\\sin x + \\cos x) \\cos x} \\, dx = \\int_{0}^{\\pi/4} \\frac{x}{\\sqrt{2} \\cos(\\frac{\\pi}{4} - x) \\cos x} ... | Brazil | Brazilian Math Olympiad | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Trigonometric functions"
] | null | proof and answer | π ln 2 / 8 | |
04o2 | Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$$
f(xf(y)) = (1-y)f(xy) + x^2y^2f(y)
$$
holds for all real numbers $x$ and $y$. | [
"Let $P(x,y)$ denote plugging $x$ and $y$ into the original equation. We have\n$$\n\\begin{aligned}\nP(0,1) &: f(0) = 0, \\\\\nP(1,1) &: f(f(1)) = f(1), \\\\\nP(1,f(1)) &: f(f(f(1))) = (1-f(1))f(f(1)) + f(1)^2 f(f(1)).\n\\end{aligned}\n$$\nBy combining the second and the third equality we get\n$$\nf(1) = (1 - f(1))... | Croatia | Croatian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = 0 for all real x; f(x) = x - x^2 for all real x | |
0fle | Problem:
Sean $a$, $b$, $c$ números reales positivos. Demuestra que
$$
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\sqrt{\frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}}} \geq \frac{5}{2}
$$
¿Cuándo se alcanza la igualdad? | [
"Solution:\n\nNótese en primer lugar que, en virtud de la desigualdad entre medias aritmética y geométrica, se tiene\n$$\n\\frac{1}{2} \\frac{a^{2}+b^{2}+c^{2}}{a b+b c+c a}+\\sqrt{\\frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}}} \\geq \\frac{3}{2} \\sqrt[3]{\\frac{a^{2}+b^{2}+c^{2}}{a b+b c+c a}\\left(\\sqrt{\\frac{a b+b c... | Spain | XLVII Olimpiada Matemática Española, Fase nacional (Pamplona) | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | Equality holds if and only if a = b = c. | |
04oi | Determine all positive integers $n \ge 2$ which satisfy the following condition: For all integers $a_1, a_2, \dots, a_n$ such that their sum is not divisible by $n$, there exists an integer $i \in \{1, 2, \dots, n\}$ such that none of the $n$ numbers
$$
a_i,\ a_i + a_{i+1},\ \dots,\ a_i + a_{i+1} + \dots + a_{i+n-1}
$... | [
"All such numbers are primes.\n\nLet $n$ be a composite number, i.e. $n = ab$, where $a \\ge 2$ and $b \\ge 2$ are positive integers.\nObserve the sequence\n$0, b, b, \\dots, b,$\nin which the number $b$ is appearing $ab-1$ times. The sum of all numbers in that sequence is $ab^2-b$, which is not divisible by $n=ab$... | Croatia | Croatian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | All prime numbers | |
024b | Problem:
Joana escreveu uma sequência em 10 linhas usando os algarismos de 0 a 9, seguindo o padrão seguinte.
$$
\begin{array}{lllllllllll}
0 & & & & & & & & & \\
1 & 1 & 0 & & & & & & & \\
2 & 2 & 2 & 1 & 1 & 0 & & & & \\
3 & 3 & 3 & 3 & 2 & 2 & 2 & 1 & 1 & 0
\end{array}
$$
Qual foi o algarismo mais usado? Quantas ve... | [
"Solution:\n\nDe acordo com o padrão da sequência, temos\n\n\n\nLogo,\num algarismo $0$ em cada linha dá $1 \\times 10 = 10$ algarismos $0$ no total; dois algarismos $1$ em nove linhas dá $2 \\times 9 = 18$ algarismos $1$ no total; três algarismos $2$ em oito linhas dá $3 \\times 8 = 24$ al... | Brazil | Nível 2 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Digits 4 and 5, each used 30 times. | |
0hxq | Problem:
G.H. Hardy once went to visit Srinivasa Ramanujan in the hospital, and he started the conversation with: "I came here in taxi-cab number 1729. That number seems dull to me, which I hope isn't a bad omen." "Nonsense," said Ramanujan. "The number isn't dull at all. It's quite interesting. It's the smallest numb... | [
"Solution:\n\nLet this smallest positive integer be represented as $a^{3}+b^{3}+c^{3}=d^{3}+e^{3}+f^{3}$. By inspection, a solution is not possible with the first 4 cubes. We prove that it is impossible to write the same number as two different sums of the first 5 cubes. Because we necessarily need to use the 5th c... | United States | HMMT 1998 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 251 | |
0665 | A regular tetrahedron of height $h$ has a tetrahedron of height $xh$ cut off by a plane parallel to the base. When the remaining frustrum is placed on one of its slant faces on a horizontal plane, it is just on the point of falling over, that is the projection of the center of gravity $G$ of the frustrum is a point of ... | [] | Greece | Mediterranean Mathematical Competition | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Algebraic Expressions > Polynomials"
] | English | proof only | x^3 + x^2 + x = 2 | |
0h3d | Can a product of four consecutive odd positive integers be a cube of an integer? | [
"Нехай для деякого парного числа $k$ і натурального числа $l$ виконується рівність $(k-3)(k-1)(k+1)(k+3) = l^3$, тобто $(k^2-9)(k^2-1) = l^3$. Якщо числа $k^2-9$ і $k^2-1$ мають спільний простий дільник, то він є дільником їхньої різниці, а тому дорівнює 2. Але числа $k^2-9$ і $k^2-1$ непарні, отже вони взаємно про... | Ukraine | Ukrainian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | No | |
07uc | Solve, for all real $x$ and $y$,
$$
x^3 + y^3 = 19, \\
x^2 + y^2 + 5x + 5y + xy = 12.
$$ | [
"Let $s = x + y$ and $p = xy$, then $x$ and $y$ are the roots of $T^2 - sT + p = 0$,\n$$\nx^2 + y^2 = s^2 - 2p \\quad (3)\n$$\nand\n$$\nx^2 + y^2 + 5x + 5y + xy = s^2 + 5s - p = 12,\n$$\nthus\n$$\np = s^2 + 5s - 12. \\quad (4)\n$$\nUsing (3) and (4), we get\n$$\n\\begin{align*}\nx^3 + y^3 &= (x + y)(x^2 + y^2 - xy)... | Ireland | IRL_ABooklet | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | (-2, 3) and (3, -2) | |
0836 | Problem:
Quanti sono i numeri di due cifre $AB$ tali che $(AB)^2 = CAAB$, con $C = B-1$ (in notazione decimale)?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 9. | [
"Solution:\n\nLa risposta è (B).\n\nInnanzi tutto sappiamo che il numero $AB$ e il suo quadrato hanno entrambi la stessa cifra finale $B$. Quindi $B$ deve assumere uno dei quattro valori $0, 1, 5, 6$. I casi 0 e 1 li escludiamo visto che $C = B-1$ è una cifra positiva. Negli altri casi la relazione $(AB)^2 = CAAB$ ... | Italy | Progetto Olimpiadi di Matematica 2003 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | B | |
0lg7 | Problem:
The Crocodile thought of four unit squares of a $2018 \times 2018$ forming a rectangle with sides $1$ and $4$. The Bear can choose any square formed by $9$ unit squares and ask whether it contains at least one of the four Crocodile's squares. What minimum number of questions should he ask to be sure of at lea... | [
"Solution:\n\nWe call checked any square chosen by the Bear, and all its unit squares. The position of a unit square in the table can be defined by the numbers of its row and column, that is, the square $(x, y)$ is in the $x$-th row and $y$-th column.\n\nFirst we prove that $\\frac{673^{2}-1}{2}$ questions is enoug... | Zhautykov Olympiad | IZhO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 226464 | |
07xd | A circle $C$ of radius $r$ is tangent to $L_1$ and $L_2$, two mutually perpendicular lines in the plane. Circles $C_1$ and $C_2$ have radius $s$. Circle $C_1$ is tangent to $L_1$ and externally tangent to the circles $C$ and $C_2$. Circle $C_2$ is tangent to $L_2$ and externally tangent to the circles $C$ and $C_1$. Fi... | [
"There are two possible configurations as shown in the diagrams below.\n\n\nLet $O$ be the intersection point of $L_1$ and $L_2$. The centre $P$ of circle $C$ lies on the bisector of the angle between $L_1$ and $L_2$. Because $C_1$ and $C_2$ have equal radius, the l... | Ireland | IRL_ABooklet_2024 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | s = r(√2 − 1)(2√2 − 1 ± 2√(2 − √2)) | |
0ft9 | Problem:
Finde die zwei kleinsten natürlichen Zahlen $n$, sodass die Brüche
$$
\frac{68}{n+70}, \frac{69}{n+71}, \frac{70}{n+72}, \ldots, \frac{133}{n+135}
$$
alle irreduzibel sind. | [
"Solution:\nDie Brüche haben die Form $k/(k+n+2)$ für $68 \\leq k \\leq 133$. Wegen $\\operatorname{ggT}(k, k+n+2) = \\operatorname{ggT}(k, n+2)$ sind sie genau dann alle irreduzibel, wenn $n+2$ teilerfremd ist zu $68, \\ldots, 133$. Also darf $n+2$ durch keine Primzahl $p$ teilbar sein, die ein Teiler ist von mind... | Switzerland | IMO - Selektion | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 65, 135 | |
0azc | Problem:
Determine $a$ and $b$ in the following:
$$
(5!)^{8} + (5!)^{7} = 4a,356,487,b80,000,000
$$ | [
"Solution:\nObserve that $(5!)^{8} + (5!)^{7} = (5!)^{7} \\times 121$. Hence it is divisible by $9$ and $11$.\n\nTherefore,\n$$\n4 + a + 3 + 5 + 6 + 4 + 8 + 7 + b + 8 = a + b + 45\n$$\nis divisible by $9$ and\n$$\n(4 + 3 + 6 + 8 + b) - (a + 5 + 4 + 7 + 8) = b - a - 3\n$$\nis divisible by $11$.\nHence $a + b = 0, 9,... | Philippines | 20th Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a = 3, b = 6 | |
01ng | Find all pairs $(n, m)$ of integers $n$ and $m$ satisfying the equality $n^2 + m = m^2 + 2n - 9$. | [
"Answer: $(-10, -11), (-10, 9), (-3, -5), (-3, 3), (2, -5), (2, 3), (9, -11), (9, 9)$.\n\nMultiplying the given equality by $4$, we obtain\n$$\n4n^2 + 4n = 4m^2 + 8m - 36 \\Leftrightarrow 4n^2 + 4n + 1 = 4m^2 + 8m + 4 - 39 \\Leftrightarrow\n$$\n$$\n(2n+1)^2 = (2m+2)^2 - 39 \\Leftrightarrow (2m+2)^2 - (2n+1)^2 = 39 ... | Belarus | Belorusija 2012 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | [(-10, -11), (-10, 9), (-3, -5), (-3, 3), (2, -5), (2, 3), (9, -11), (9, 9)] | |
097x | Problem:
Să se afle toate funcțiile derivabile $f:(0,+\infty) \rightarrow (0,+\infty)$ ce verifică relațiile $f\left(\frac{1}{2}\right)=1$ și $f'(x) = -f^3(x)$. | [
"Solution:\nAvem $\\left(f^{-2}(x)\\right)' = -2 f^{-3}(x) f'(x) = 2 f^{-3}(x) f^{3}(x) = 2$, adică derivata funcției $f^{-2}(x)$ este o constantă. Rezultă că funcția $f^{-2}(x)$ reprezintă o funcție liniară, ce poate fi scrisă în forma $f^{-2}(x) = a x + b$.\n\nDerivând, obținem $2 = \\left(f^{-2}(x)\\right)' = a$... | Moldova | Olimpiada Republicană la Matematică | [
"Calculus > Differential Equations > ODEs"
] | null | proof and answer | 1/sqrt(2x) | |
0emt | Does there exist a natural number $N$ which is a power of $2$ such that the digits of $N$ can be permuted to form a power of $2$ different from $N$? | [
"Suppose that the digits of $2^a$ can be rearranged to form $2^b$, with $a > b$. Then, since the two numbers have the same digit set, it follows that they're congruent modulo $9$. Hence $9 \\mid 2^a - 2^b = 2^b(2^{a-b} - 1)$ and so $2^{a-b} \\equiv_9 1$. However, the smallest positive power of $2$ with this propert... | South Africa | South-Afrika 2011-2013 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | No | |
00c3 | All diagonals of a convex 10-gon are drawn. They divide the angles into 80 parts. It is known that 59 of these parts are equal. Determine the maximum of different values among the 80 angles of division. How many times does each of these values occur? | [
"The sides of each of the 80 angles pass through the endpoints of a side of the 10-gon $P$. We say that such an angle and such a side are adjacent; each side is adjacent to exactly 8 angles. Call *black* the 59 angles that are known to be equal and $\\alpha$ the measure of a black angle. Let $a$ be a side of $P$. T... | Argentina | Argentina_2018 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | Maximum number of distinct values is 3; the occurrences are 64, 8, and 8. | |
0865 | Problem:
Sia $ABC$ un triangolo rettangolo in $A$, con $\angle ABC = 15^\circ$. Sia $H$ il piede dell'altezza da $A$ e siano $J, K$ le proiezioni di $H$ su $AB$ e su $AC$. Sapendo che l'area di $AJHK$ è $45~\mathrm{cm}^2$, quanti $\mathrm{cm}^2$ vale il prodotto $BJ \cdot CK$? | [
"Solution:\n\nLa risposta è $45$. Ponendo $x = AJ$, $y = AK$, $x' = JB$ e $y' = KC$, si ha, usando il secondo teorema di Euclide, $x \\cdot x' = y^2$ e $y \\cdot y' = x^2$. Moltiplicando membro a membro si ha $\\left(x \\cdot x'\\right)\\left(y \\cdot y'\\right) = \\left(x' \\cdot y'\\right)(x \\cdot y) = x^2 \\cdo... | Italy | Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 45 | |
0frt | Tenemos un cubo de lado $3$ formado por $27$ piezas cúbicas de lado $1$. Dentro de cada pieza hay una bombilla que puede estar encendida o apagada. Cada vez que se pulsa una pieza (no es posible pulsar la del centro del cubo), cambia el estado de su bombilla y el de las que comparten una cara con la pulsada. Inicialmen... | [
"Llamamos $N$ a la pieza de lado $1$ que está en el interior, $C$ a cada pieza que es centro de una cara, $V$ a cada pieza que es vértice del cubo y $A$ a cada una de las que comparten cara con un vértice. Es evidente que el resultado obtenido tras pulsar varias piezas es independiente del orden en que las pulsemos... | Spain | LIX Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Spanish | proof and answer | 1: no; 2: yes; 3: no | |
06b2 | Let $A B \Gamma$ be an acute angled triangle with circumcircle $\omega$. A circle $\gamma$ with center $A$ intersects the arc $AB$ of the circle $\omega$, not containing $\Gamma$, at point $\Delta$ and the arc $A\Gamma$, not containing $B$, at point $E$. We suppose that the point of intersection $K$ of the lines $BE$ a... | [
"From the relationship of a central angle and an inscribed angle that go on the same arc $\\Delta K$ of the circle $\\gamma$, we have:\n$$\n\\angle A\\Delta K = 2 \\cdot \\angle E\\Delta K \\quad (1)\n$$\nAlso we have the equality of inscribed angles\n$$\n\\angle E\\Delta K = \\angle AB\\Delta \\quad (2)\n$$\nFrom ... | Greece | Hellenic 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 | |
09xd | Prickle and Sting are playing a game on an $m \times n$-board, where $m$ and $n$ are positive integers. They alternatingly take turns, and Prickle goes first. Prickle must, during his turn, place a pawn on a square which doesn't contain a pawn yet. Sting must, during his turn, also place a pawn on a square which doesn'... | [
"We use the convention that $m$ is the number of rows, and that $n$ is the number of columns. If $m$ is even, then we pair the squares of the board as follows: in every column we pair the top two squares, then squares 3 and 4, etc. As the number of rows is even, this pairs the squares of every column completely. St... | Netherlands | IMO Team Selection Test 2 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof and answer | Sting wins if at least one of m or n is even, or if (m, n) is (1, 1), (1, 3), or (3, 1). Prickle wins if both m and n are odd with at least one of them at least 5, or if (m, n) = (3, 3). | |
02lq | Let $f: [0, 1] \to [0, 1]$ be an increasing and derivable function that has an inverse $f^{-1}$.
If $\int_0^1 f(x) dx = \int_0^1 f^{-1}(x) dx$, prove that there exist two distinct real numbers $a$ and $b$, $0 \le a < b \le 1$, such that $f'(a) = f'(b) = 1$. | [
"First notice that, since $f$ is increasing and has an inverse, then it is bijective and hence $f(0) = 0$ and $f(1) = 1$ (if $f(0) > 0$ then $f$ would never be equal to $0$; the same applies if $f(1) < 1$). So, by applying the substitution $x = f(y)$, for which $\\frac{dx}{dy} = f'(y)$, and integration by parts,\n$... | Brazil | XXXI Brazilian Math Olympiad | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Differential Calculus > Applications"
] | English | proof only | null | |
0k7g | Problem:
Dan is walking down the left side of a street in New York City and must cross to the right side at one of 10 crosswalks he will pass. Each time he arrives at a crosswalk, however, he must wait $t$ seconds, where $t$ is selected uniformly at random from the real interval $[0,60]$ ($t$ can be different at diffe... | [
"Solution:\n\nWith probability $\\left(1-\\frac{k}{60}\\right)^{9}$, Dan reaches the last crosswalk without crossing at any previous site, in which case the expected value of his wait time is 30 seconds. Otherwise, with probability $1-\\left(1-\\frac{k}{60}\\right)^{9}$, Dan crosses at an earlier crosswalk, in whic... | United States | HMMT November 2019 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 60(1 - (1/10)^{1/9}) | |
0ihy | Problem:
Determine the maximum value attained by
$$
\frac{x^{4}-x^{2}}{x^{6}+2 x^{3}-1}
$$
over real numbers $x>1$. | [
"Solution:\nWe have the following algebra:\n$$\n\\begin{aligned}\n\\frac{x^{4}-x^{2}}{x^{6}+2 x^{3}-1} & = \\frac{x-\\frac{1}{x}}{x^{3}+2-\\frac{1}{x^{3}}} \\\\\n& = \\frac{x-\\frac{1}{x}}{\\left(x-\\frac{1}{x}\\right)^{3}+2+3\\left(x-\\frac{1}{x}\\right)} \\\\\n& \\leq \\frac{x-\\frac{1}{x}}{3\\left(x-\\frac{1}{x}... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 1/6 | |
0kaf | Problem:
How many pairs of integers $(x, y)$ are there such that $|x^{2}-2 y^{2}| \leq 1$ and $|3 x-4 y| \leq 1$? | [
"Solution:\nNote that if $(x, y)$ is a solution, so is $(-x,-y)$. Thus, we consider $x \\geq 0$.\n\nWhen $x \\equiv 0 \\pmod{4}$, $y = 3x/4$ by inequality 2. Inequality 1 gives $|x^{2}/9| \\leq 1$, so $x \\leq 3$, so $x = 0$.\n\nWhen $x \\equiv 1 \\pmod{4}$, $y = (3x+1)/4$ by inequality 2. Beyond $x = 1$, $2y^{2} -... | United States | HMMT February 2019 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 7 | |
0dbb | Prove the inequality for non-negative $a, b, c$
$$
a \sqrt{3 a^{2}+6 b^{2}}+b \sqrt{3 b^{2}+6 c^{2}}+c \sqrt{3 c^{2}+6 a^{2}} \geq (a+b+c)^{2}.
$$ | [
"Note that $3 a^{2}+6 b^{2} \\geq (a+2 b)^{2}$ is true for all real numbers $a, b$. Indeed, after expanding and grouping we get $2 a^{2}-4 a b+2 b^{2} \\geq 0$ which is equivalent to $2(a-b)^{2} \\geq 0$. So with non-negative numbers $a, b, c$ we have\n$$\n\\begin{aligned}\na \\sqrt{3 a^{2}+6 b^{2}}+b \\sqrt{3 b^{2... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
03ew | Given a natural number $n$. We have $n + 1$ balls numbered $1$, $1$, $2$, $3$, ..., $n$ (only the first two are the same). We need to color these balls in $n$ given colors so that every ball is a single color and every color is used at least once. We denote by $a_n$ the number of possible colorings. Find the smallest $... | [
"Exactly one of the colors will be used for two of the balls; let their numbers be $a$ and $b$, such that $a \\le b$.\n\nIf $a > 1$, then we have $(n-1)(n-2)/2$ choices for $a$ and $b$, and $n$ choices for their color. The remaining $n-1$ balls (two of which are the same) must be colored in the remaining $n-1$ colo... | Bulgaria | Bulgarian Winter Tournament | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 13 | |
05gi | Problem:
Un grand carré de côté $n$ est découpé en $n^{2}$ petits carrés de côté $1$. On veut colorier en rouge ou bleu chacun des $(n+1)^{2}$ sommets des petits carrés de telle manière que chacun des petits carrés a exactement $2$ sommets rouges. Combien y a-t-il de coloriages possibles? | [
"Solution:\n\nLa réponse est $2^{n+2} - 2$. Pour le montrer, on commence par colorier les $n+1$ sommets les plus hauts. Il y a $2^{n+1}$ manières de le faire.\n\nSi il existe deux sommets consécutifs de même couleur sur la rangée supérieure ($2^{n+1} - 2$ coloriages de la rangée supérieure), alors ils fixent les co... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 2^{n+2} - 2 | |
072m | Let $A_1, A_2, A_3, \dots, A_n$ be $n$ subsets of a finite set $S$ such that $|A_j| = 8$ for each $j$, $1 \le j \le n$. For a subset $B$ of $S$, let $F(B) = \{j : 1 \le j \le n \text{ and } A_j \subset B\}$. Suppose for each subset $B$ of $S$, at least one of the following conditions holds:
(i) $|B| > 25$;
(ii) $F(B) =... | [
"We use induction on $n$. The case $n = 1$ is immediate. Assume the result for $n$ and consider the case of $n+1$. Thus $A_1, A_2, \\dots, A_{n+1}$ are subsets of $S$ with $|A_j| = 8$.\nNow if we consider the sets $A_1, A_2, \\dots, A_{j-1}, A_{j+1}, \\dots, A_{n+1}$, these satisfy induction hypothesis for each $j ... | India | Indija TS 2006 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof only | null | |
03x9 | Every day at a railway station, there is just one train arriving between $8{:}00$ am and $9{:}00$ am and between $9{:}00$ am and $10{:}00$ am, respectively. The arrival times and their probabilities for the two trains are shown in the following table:
| Arrival time | Train A | $8{:}10$ | $8{:}30$ | $8{:}50$ |
|------... | [
"The distribution table for the waiting times of the traveler is shown below.\n\n| Waiting time/min | $10$ | $30$ | $50$ | $70$ | $90$ |\n|------------------|------|------|------|------|------|\n| Probability | $\\frac{1}{2}$ | $\\frac{1}{3}$ | $\\frac{1}{6} \\times \\frac{1}{6}$ | $\\frac{1}{2} \\times \\frac... | China | China Mathematical Competition | [
"Statistics > Probability > Counting Methods > Other",
"Math Word Problems"
] | English | final answer only | 27 | |
08lf | Problem:
Two perpendicular chords of a circle, $A M$, $B N$, which intersect at point $K$, define on the circle four arcs with pairwise different length, with $A B$ being the smallest of them. We draw the chords $A D$, $B C$ with $A D \parallel B C$ and $C, D$ different from $N, M$. If $L$ is the point of intersection... | [
"Solution:\n\nFirst we prove that $N L \\perp M C$. The arguments depend slightly on the position of $D$. The other cases are similar.\nFrom the cyclic quadrilaterals $A D C M$ and $D N B C$ we have:\n$$\n\\varangle D C L = \\varangle D A M \\text{ and } \\varangle C D L = \\varangle C B N.\n$$\nSo we obtain\n$$\n\... | JBMO | 2008 Shortlist JBMO | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0cs7 | Two players play the following card game. They have a deck of $n$ cards. For every two cards it is known which one of them takes the other (it may happen that $A$ takes $B$, $B$ takes $C$, and $C$ takes $A$). Initially the deck is distributed between the players in an arbitrary way. On each move, the players open the t... | [
"Выпишем все возможные ситуации, которые могут встречаться в игре (т. е. все возможные пары колод у участников). Назовём ситуацию *фундаментальной*, если все карты у одного игрока; *критической*, если у одного из игроков ровно одна карта; и регулярной, если у обоих игроков хотя бы по две карты. Проведём стрелку от ... | Russia | XL Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
01zl | Olya and Tolya have paints of two opposite colors – white and black. They play the following game on the segment $[0, 1]$. Each round of the game takes place in two stages: one of the players chooses a number $l \in [0, 1]$, and then the other player chooses some segment $J \subseteq [0, 1]$ of length $l$ and recolors ... | [
"Here is a winning strategy for Olya. Denote by $L_n$ the total length of the white segments after the $n$-th round ($L_0$ is considered to be equal to 1).\n**Statement.** Olya can play in such a way that for $k = 0, 1, \\dots, 1012$ after the $(2k)$-th round the total length of the white segments is greater than $... | Belarus | SELECTION TESTS OF THE BELARUSIAN TEAM TO THE IMO | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Olya | |
002e | En el planeta Alfa usan un alfabeto de $100$ letras. Una palabra es una secuencia de letras que satisface las siguientes condiciones:
* No hay dos letras consecutivas iguales. Por ejemplo, BELLEZA no es una palabra, porque tiene LL, y COOPERAR no es una palabra porque tiene OO.
* No hay dos letras distintas $U$ y $V$ q... | [] | Argentina | XX OLIMPIADA MATEMÁTICA ARGENTINA | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | español | proof and answer | 199 | |
014y | Problem:
Let $ABCD$ be a parallelogram. The circle with diameter $AC$ intersects the line $BD$ at points $P$ and $Q$. The perpendicular to the line $AC$ passing through the point $C$ intersects the lines $AB$ and $AD$ at points $X$ and $Y$, respectively. Prove that the points $P$, $Q$, $X$ and $Y$ lie on the same circl... | [
"Solution:\nIf the lines $BD$ and $XY$ are parallel the statement is trivial. Let $M$ be the intersection point of $BD$ and $XY$.\nBy Intercept Theorem $\\dfrac{MB}{MD} = \\dfrac{MC}{MY}$ and $\\dfrac{MB}{MD} = \\dfrac{MX}{MC}$, hence $MC^2 = MX \\cdot MY$. By the circle property $MC^2 = MP \\cdot MQ$ (line $MC$ is... | Baltic Way | Baltic Way 2008 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | null | proof only | null |
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