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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
00bd | Given are 51 natural numbers written in a row. Their sum is 100. An integer is *representable* if it can be expressed as the sum of several consecutive numbers in the given row. Prove that for each $k \in \{1, 2, \dots, 100\}$ one of the numbers $k$ and $100-k$ is representable. | [
"Let the given row be $a_1, \\dots, a_{51}$. Take a circle $\\gamma$ of length 100. Mark 51 blue points on it so that they determine 51 consecutive arcs of lengths $a_1, \\dots, a_{51}$. One may imagine each number $a_i$ written next to an arc of $\\gamma$ with length $a_i$. So, starting from a certain blue point $... | Argentina | Argentina_2018 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0f28 | Problem:
Given 1000 square plates in the plane with their sides parallel to the coordinate axes (but possibly overlapping and possibly of different sizes). Let $S$ be the set of points covered by the plates. Show that you can choose a subset $T$ of plates such that every point of $S$ is covered by at least one and at ... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0e2k | Problem:
Prijatelja Miha in Blaž za nedeljsko potepanje najameta vsak svoje motorno kolo pri različnih ponudnikih. Miha mora plačati na začetku 100 evrov, ko pa motorno kolo vrne, še 4 evre za vsak prevoženi kilometer. Blaž na začetku plača 200 evrov, potem pa 3 evre za vsak prevoženi kilometer. Najmanj koliko kilomet... | [
"Solution:\n\nNaj bo $x$ število prevoženih kilometrov in $m(x)$ ter $b(x)$ zneska, ki sta odvisna od $x$. Znesek za Mihovo kolo je $m(x) = 4x + 100$, znesek za Blaževo kolo pa $b(x) = 3x + 200$. Miha bo plačal več kot Blaž, ko bo veljalo $4x + 100 > 3x + 200$. Rešimo neenačbo in dobimo rešitev $x > 100$. Prevoziti... | Slovenia | 10. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | 101 | |
0928 | Problem:
Find all surjective functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for all positive integers $a$ and $b$, exactly one of the following equations is true:
$$
\begin{aligned}
f(a) &= f(b) \\
f(a+b) &= \min \{f(a), f(b)\}
\end{aligned}
$$
Remarks: $\mathbb{N}$ denotes the set of all positive integer... | [
"Solution:\n\nEach positive integer can be uniquely written as $n=2^{k} l$ where $k \\geqslant 0$ and $l$ is odd. We will show that the only function satisfying the conditions is $f\\left(2^{k} l\\right)=k+1$ for all $k \\geqslant 0$ and all odd $l$.\n\nAssume that $f(1) \\neq 1$. Since $f$ is surjective, there exi... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | For every positive integer written uniquely as 2^k times an odd number l, f(2^k l) = k + 1. | |
0j4m | Problem:
Let $A B C D E F$ be a convex equilateral hexagon such that lines $B C$, $A D$, and $E F$ are parallel. Let $H$ be the orthocenter of triangle $A B D$. If the smallest interior angle of the hexagon is $4$ degrees, determine the smallest angle of the triangle $H A D$ in degrees. | [
"Solution:\n\nAnswer: $3$\n\nNote that $A B C D$ and $D E F A$ are isosceles trapezoids, so $\\angle B A D = \\angle C D A$ and $\\angle F A D = \\angle E D A$. In order for the hexagon to be convex, the angles at $B$, $C$, $E$, and $F$ have to be obtuse, so $\\angle A = \\angle D = 4^{\\circ}$. Letting $s$ be a si... | United States | Harvard-MIT Mathematics Tournament | [
"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",
"Geometry > Plane Geometry > Analytic / C... | null | proof and answer | 3 | |
0ba2 | Show that there are infinitely many positive integer numbers $n$ such that $n^2 + 1$ has two positive divisors whose difference is $n$. | [
"Define the sequence $(a_k)_{k \\ge 0}$ by $a_0 = 1$, $a_1 = 2$ and $a_{k+2} a_k = a_{k+1}^2 + 1$, $k = 0, 1, 2, \\dots$, and check inductively that the $a_k$ are all positive integer numbers, the $n_k = a_{k+1} - a_k$ form a strictly increasing sequence of positive integer numbers, and $a_k$ and $a_{k+1}$ both div... | Romania | 62nd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
06w3 | Let $\mathbb{R}^{+}$ be the set of positive real numbers. Determine all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that, for all positive real numbers $x$ and $y$,
$$
f(x+f(xy))+y=f(x)f(y)+1
$$ | [
"A straightforward check shows that $f(x)=x+1$ satisfies (*). We divide the proof of the converse statement into a sequence of steps.\n\nStep 1: $f$ is injective.\nPut $x=1$ in (*) and rearrange the terms to get\n$$\ny=f(1)f(y)+1-f(1+f(y))\n$$\nTherefore, if $f(y_1)=f(y_2)$, then $y_1=y_2$.\n\nStep 2: $f$ is (stric... | IMO | IMO 2020 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x + 1 | |
05gs | Problem:
Trouver tous les triplets de réels $\left(a, b, c\right)$ vérifiant le système d'égalités :
$$
\left\{
\begin{array}{l}
a\left(b^{2}+c\right)=c(c+a b) \\
b\left(c^{2}+a\right)=a(a+b c) \\
c\left(a^{2}+b\right)=b(b+a c)
\end{array}
\right.
$$ | [
"Solution:\n\nLes égalités se réécrivent $a b(b-c)=c(c-a)$, $b c(c-a)=a(a-b)$ et $c a(a-b)=b(b-c)$. En multipliant ces trois égalités, on obtient $(a b c)^{2}(c-a)(b-c)(a-b)=a b c(c-a)(b-c)(a-b)$. En particulier deux cas se présentent :\n\n- Soit $a b c=0$. Les égalités étant cycliques, supposons $a=0$. Par la prem... | France | ENVOI 2 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | All real triples with a = b = c (any real value). | |
00zn | Problem:
Let $n$ and $k$ be integers, $1 < k \leq n$. Find an integer $b$ and a set $A$ of $n$ integers satisfying the following conditions:
(i) No product of $k-1$ distinct elements of $A$ is divisible by $b$.
(ii) Every product of $k$ distinct elements of $A$ is divisible by $b$.
(iii) For all distinct $a, a'$ in $A$... | [
"Solution:\nLet $p_1, \\ldots, p_n$ be the first $n$ odd primes. Then we can take $A = \\{2 p_1, 2 p_2, \\ldots, 2 p_n\\}$ and $b = 2^k$. It is easily seen that the conditions are satisfied."
] | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | Let p1, ..., pn be the first n odd primes. Take A = {2*p1, 2*p2, ..., 2*pn} and b = 2^k. | |
0ahz | Prove that
$$
\frac{1}{1 \cdot 2013} + \frac{1}{2 \cdot 2012} + \frac{1}{3 \cdot 2011} + \dots + \frac{1}{2012 \cdot 2} + \frac{1}{2013 \cdot 1} < 1.
$$
Докажи дека
$$
\frac{1}{1 \cdot 2013} + \frac{1}{2 \cdot 2012} + \frac{1}{3 \cdot 2011} + \dots + \frac{1}{2012 \cdot 2} + \frac{1}{2013 \cdot 1} < 1.
$$ | [
"For arbitrary natural numbers $n \\neq 1 \\neq m$ the inequality $nm \\geq n+m$ holds, since $(n-1)(m-1) \\geq 1 \\Rightarrow nm-n-m+1 \\geq 1 \\Rightarrow nm-n-m \\geq 0 \\Rightarrow nm \\geq n+m$, with equality only when $n=m=2$. Then for $n \\geq 2$, we have\n$$\n\\frac{1}{n(2014-n)} < \\frac{1}{n+2014-n} = \\f... | North Macedonia | Macedonian Junior Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof only | null | |
061r | Problem:
Eine Funktion $f$ ist gegeben durch $f(x)+f\left(1-\frac{1}{x}\right)=1+x$ für $x \in \mathbb{R} \backslash\{0,1\}$.
Man ermittle eine Formel für $f$. | [
"Solution:\n\nSei $x \\in \\mathbb{R} \\backslash\\{0,1\\}$ und $y=1-\\frac{1}{x}$ und $z=\\frac{1}{1-x}$. Es ist leicht einzusehen, dass zusammen mit $x$ auch $y$ und damit auch $z$ zu $\\mathbb{R} \\backslash\\{0,1\\}$ gehören. Einsetzen von $y$ und $z$ in die Ausgangsgleichung führt zu:\n$$\nf\\left(1-\\frac{1}{... | Germany | Auswahlwettbewerb zur IMO | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 1/2 * (1/(1 - x) + 1/x + x) | |
0ef6 | Problem:
Za realno število $x$, kjer $x \notin\left\{0, \frac{1}{2}, \frac{2}{3}, 1,2\right\}$, poenostavi izraz
$$
\left(\frac{3 x^{-\frac{1}{3}}}{x^{\frac{2}{3}}-2 x^{-\frac{1}{3}}}-\frac{x^{\frac{1}{3}}}{x^{\frac{4}{3}}-x^{\frac{1}{3}}}\right)^{-1}-\left(\frac{1-2 x}{3 x-2}\right)^{-1}
$$
in izračunaj njegovo vredn... | [
"Solution:\n\nV imenovalcu ulomka izpostavimo potenco $z$ najmanjšim eksponentom:\n$$\n\\left(\\frac{3 x^{-\\frac{1}{3}}}{x^{-\\frac{1}{3}}(x-2)}-\\frac{x^{\\frac{1}{3}}}{x^{\\frac{1}{3}}(x-1)}\\right)^{-1}-\\left(\\frac{1-2 x}{3 x-2}\\right)^{-1}\n$$\nKrajšamo ulomka, razširimo ulomek v oklepaju na skupni imenoval... | Slovenia | 17. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Odbirno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Other",
"Algebra > Intermediate Algebra > Other"
] | null | final answer only | (6*sqrt(3)+3)/11 | |
053p | Determine the largest possible number of primes among 100 consecutive natural numbers. | [
"There are 25 primes among the numbers from 1 to 100. The number of primes in the next a few number intervals with length 100 are shown in the table:\n\n| Interval | out | in | Number of primes |\n|--------------|-----|-----|------------------|\n| 1,...,100 | | | 25 |\n| 2,...,101 ... | Estonia | Estonian Math Competitions | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 26 | |
0hd2 | Natural numbers $a < b$ are written on the board. At each step, two numbers written on the board are wiped out, and their sum and the modulus of difference are written down instead. At some point the number 2019 appeared on the board. What is the smallest possible value of $b$? | [
"It is easy to write down all the pairs of the numbers, that will successively appear on the board:\n$$\na, b \\rightarrow b+a, b-a \\rightarrow 2a, 2b \\rightarrow 2(b+a), 2(b-a) \\rightarrow 2a^2, 2b^2 \\dots\n$$\n\nAs we see, after the appearance of the first four numbers: $a, b, b+a, b-a$ – all the other number... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 1010 | |
04cz | Let $x$, $y$ and $z$ be positive real numbers such that $x + y + z = 18xyz$. Prove the inequality
$$
\frac{x}{\sqrt{x^2 + 2yz + 1}} + \frac{y}{\sqrt{y^2 + 2xz + 1}} + \frac{z}{\sqrt{z^2 + 2xy + 1}} \ge 1.
$$ | [
"The A-H inequality together with the condition of the problem gives\n$$\nxy + yz + zx = xyz \\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right) \\ge xyz \\cdot \\frac{9}{x + y + z} = \\frac{9xyz}{18xyz} = \\frac{1}{2}.\n$$\nUsing $1 \\le 2xy + 2yz + 2zx$ we get\n$$\nx^2 + 2yz + 1 \\le x^2 + 2xy + 2zx + 4yz =... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
00zr | Problem:
Two circles $\mathcal{C}_1$ and $\mathcal{C}_2$ intersect in $P$ and $Q$. A line through $P$ intersects $\mathcal{C}_1$ and $\mathcal{C}_2$ again in $A$ and $B$, respectively, and $X$ is the midpoint of $AB$. The line through $Q$ and $X$ intersects $\mathcal{C}_1$ and $\mathcal{C}_2$ again in $Y$ and $Z$, res... | [
"Solution:\n\nDepending on the radii of the circles, the distance between their centres and the choice of the line through $P$ we have several possible arrangements of the points $A$, $B$, $P$ and $Y$, $Z$, $Q$. We shall show that in each case the triangles $A X Y$ and $B X Z$ are congruent, whence $|Y X| = |X Z|$.... | Baltic Way | Baltic Way 1997 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
091c | Problem:
Three strictly increasing sequences
$$
a_{1}, a_{2}, a_{3}, \ldots, \quad b_{1}, b_{2}, b_{3}, \ldots, \quad c_{1}, c_{2}, c_{3}, \ldots
$$
of positive integers are given. Every positive integer belongs to exactly one of the three sequences. For every positive integer $n$, the following conditions hold:
(i) $c... | [
"Solution:\nSince $\\{c_{n}\\}$ is a strictly increasing sequence of positive integers, it is clear that $c_{n} \\geq n, n \\in \\mathbb{N}$. Hence, $c_{a_{n}} \\geq a_{n}, n \\in \\mathbb{N}$. However, the given sequences do not contain equal terms, so $c_{a_{n}}>a_{n}$ and $b_{n}=c_{a_{n}}-1>a_{n}, n \\in \\mathb... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | a2010 = 2010^2, b2010 = 2011^2 - 2, c2010 = 2099 | |
0gn6 | Prove that there exists no triangle whose side lengths, area and angles (measured in degrees) are rational numbers. | [
"By the law of sines and the law of cosines, the sines and cosines of the angles of such a triangle are also rational. Therefore it suffices to prove that if $\\theta$ is an angle whose degree measure is a rational number that is not an integer multiple of $90$, then both $\\sin\\theta$ and $\\cos\\theta$ cannot be... | Turkey | 14th Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
0khp | In square $ABCD$, points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$, respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$, with $BR = 6$ and $PR = 7$. What is the area of the square?
(A) 85 (B) 93 (C) 100 (D) 117 (E) 125 | [
"Because $\\angle RBC$ is complementary to both $\\angle RCB$ and $\\angle PBA$, those two angles are congruent. Therefore $\\triangle BAP \\cong \\triangle CBQ$ by ASA, so $CQ = BP = 13$. Let $d = CR$; then $QR = 13 - d$, so the Altitude-to-Hypotenuse Theorem yields $6^2 = d(13 - d)$, which has solutions $d = 4$ a... | United States | Fall 2021 AMC 10 B | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | MCQ | D | |
0dsq | Determine all positive integers $n$ with at least 4 factors such that $n$ is the sum of the squares of its 4 smallest factors. | [
"Let $a < b < c < d$ be the 4 smallest factors of $n$. Then $n = a^2 + b^2 + c^2 + d^2$.\n\nIf $n$ is odd, then $a, b, c, d$ are all odd, impossible. Thus $n$ is even. Hence $a = 1, b = 2$.\n\nIf $4 \\mid n$, then one of $c, d$ is $4$ and the other is odd. By taking mod $4$, we see the LHS is $0$ while the RHS is $... | Singapore | Singapore Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 130 | |
001w | Maxi eligió 3 dígitos y haciendo todas las permutaciones posibles obtuvo 6 números distintos de 3 dígitos cada uno. Si exactamente uno de los números que obtuvo Maxi es un cuadrado perfecto y exactamente tres son primos, hallar los 3 dígitos que eligió Maxi. | [] | Argentina | 15ª Olimpiada Matemática del Cono Sur | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | español | proof and answer | 1, 3, 6 | |
04a0 | Determine all integers $a$ such that $\log_2(1 + a + a^2 + a^3)$ is also an integer. | [] | Croatia | CroatianCompetitions2011 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | [0, 1] | |
08eb | Problem:
Aisha scrive su un foglio tutti i numeri da $1$ a $2020$. Quanto vale la differenza tra il numero di cifre "1" e il numero di cifre "0" che ha scritto?
(A) $78$
(B) $1010$
(C) $1089$
(D) $2020$
(E) $5005$ | [
"Solution:\n\nLa risposta è (C). Osserviamo che, dato un numero con scrittura decimale $abcd$, se fissiamo $a \\neq 0$, abbiamo $10$ scelte per ciascuna delle cifre $b, c, d$: tra queste $10$ scelte, esattamente una è \"1\", come anche esattamente una è \"0\". Notiamo inoltre che lo stesso ragionamento può essere f... | Italy | Olimpiadi della Matematica | [
"Discrete Mathematics > Other"
] | null | MCQ | C | |
00sa | Find all functions $f: \mathbb{Z}_{>0} \to \mathbb{Z}_{>0}$ such that the number $x f(x) + f^2(y) + 2x f(y)$ is a perfect square for all positive integers $x, y$. | [
"Let $p$ be a prime number. Then for $x = y = p$ the given condition gives us that the number $f^2(p) + 3p f(p)$ is a perfect square. Then, $f^2(p) + 3p f(p) = k^2$ for some positive integer $k$.\nCompleting the square gives us that $(2f(p) + 3p)^2 - 9p^2 = 4k^2$, or\n$$\n(2f(p) + 3p - 2k)(2f(p) + 3p + 2k) = 9p^2. ... | Balkan Mathematical Olympiad | BMO 2017 | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | f(x) = x | |
06qy | A sequence $x_{1}, x_{2}, \ldots$ is defined by $x_{1}=1$ and $x_{2k}=-x_{k}$, $x_{2k-1}=(-1)^{k+1} x_{k}$ for all $k \geq 1$. Prove that $x_{1}+x_{2}+\cdots+x_{n} \geq 0$ for all $n \geq 1$. | [
"We start with some observations. First, from the definition of $x_{i}$ it follows that for each positive integer $k$ we have\n$$\nx_{4k-3}=x_{2k-1}=-x_{4k-2} \\quad \\text{ and } \\quad x_{4k-1}=x_{4k}=-x_{2k}=x_{k} .\n$$\nHence, denoting $S_{n}=\\sum_{i=1}^{n} x_{i}$, we have\n$$\nS_{4k}=\\sum_{i=1}^{k}\\left(\\l... | IMO | 51st IMO Shortlisted Problems | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0h05 | Given natural numbers $a$, $b$, $c$, $d$ for which $ab^2 + ad^2 + cb^2 = ba^2 + bd^2 + ca^2$ and the number $a^2 + b^2 + c^2 + d^2$ is prime. Prove that $a = b$. | [
"Assume the contrary. Let $a \\ne b$. The condition $ab^2 + ad^2 + cb^2 = ba^2 + bd^2 + ca^2$ we can rewrite in the form: $(a-b)(d^2 - ab - ac - bc) = 0$. Since $a \\ne b$ it implies that $d^2 = ab + ac + bc$. Then $a^2 + b^2 + c^2 + d^2 = a^2 + b^2 + c^2 + ab + ac + bc = (a+b+c)^2 - d^2 = (a+b+c+d)(a+b+c-d)$. The ... | Ukraine | 50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010) | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
02h8 | A positive integer is **bold** iff it has $8$ positive divisors that sum up to $3240$. For example, $2006$ is bold because its $8$ positive divisors, $1$, $2$, $17$, $34$, $59$, $118$, $1003$ and $2006$, sum up to $3240$. Find the smallest positive bold number. | [
"Let $n = p_1^{\\alpha_1} \\cdots p_k^{\\alpha_k}$. Then $8 = (\\alpha_1 + 1) \\cdots (\\alpha_k + 1)$, and $3240 = \\frac{p_1^{\\alpha_1+1}-1}{p_1-1} \\cdots \\frac{p_k^{\\alpha_k+1}-1}{p_k-1}$. Hence there are three cases:\n(a) $8 = \\alpha_1 + 1 \\implies n = p^7$.\n(b) $8 = (\\alpha_1 + 1)(\\alpha_2 + 1) \\impl... | Brazil | Brazilian Math Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Resi... | English | proof and answer | 1614 | |
0kmf | Problem:
Let $S=\{1,2, \ldots, 2021\}$, and let $\mathcal{F}$ denote the set of functions $f: S \rightarrow S$. For a function $f \in \mathcal{F}$, let
$$
T_{f}=\left\{f^{2021}(s): s \in S\right\}
$$
where $f^{2021}(s)$ denotes $f(f(\cdots(f(s)) \cdots))$ with 2021 copies of $f$. Compute the remainder when
$$
\sum_{f \... | [
"Solution:\nThe key idea is that $t \\in T_{f}$ if and only if $f^{k}(t)=t$ for some $k>0$. To see this, let $s \\in S$ and consider\n$$\ns, f(s), f(f(s)), \\ldots, f^{2021}(s)\n$$\nThis sequence has 2022 terms that are all in $S$, so we must have a repeat. Suppose $f^{m}(s)=f^{n}(s)$ with $0 \\leq n<m \\leq 2021$.... | United States | HMMT Spring 2021 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 255 | |
03rn | A circle intersects sides $BC$, $CA$, $AB$ of $ABC$ at two points for each side in the following order: \{$D_1, D_2$\}, \{$E_1, E_2$\} and \{$F_1, F_2$\}. Line segments $D_1E_1$ and $D_2E_2$ intersect at point $L$, $E_1F_1$ and $E_2D_2$ intersect at point $M$, $F_1D_1$ and $F_2E_2$ intersect at point $N$. Prove that $A... | [
"**Proof** Through point $L$ draw perpendicular lines to $AB$ and to $AC$, the feet are $L'$ and $L''$ respectively. Let $\\angle LAB = \\alpha_1$, $\\angle LAC = \\alpha_2$, $\\angle LF_2A = \\alpha_3$, and $\\angle LE_1A = \\alpha_4$. We have\n$$\n\\frac{\\sin \\alpha_1}{\\sin \\alpha_2} = \\frac{LL'}{LL''} = \\f... | China | China Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0bv6 | Problem:
a) Să se arate că $\forall x>1$, are loc inegalitatea: $2^{x} \cdot \sqrt{4^{x}-1}+4^{x} \cdot \sqrt{2^{x}-1}<8^{x}$.
b) Să se determine funcțiile bijective $f: \mathbb{R} \rightarrow \mathbb{R}$ astfel încât $\forall x, y \in \mathbb{R}$ are loc relația: $f(2 x+f(x)+3 f(y))=f(3 x)+f(3 y)$. | [] | Romania | Olimpiada de Matematică Etapa Locală | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x | |
0dh7 | Let $ABC$ be a triangle with midpoints $K$, $M$, $N$ of $BC$, $CA$, $AB$ respectively. Let $AD$, $BE$, $CF$ be the altitudes of the triangle $ABC$ and let $U$, $V$, $W$ be the midpoints of $FD$, $DE$, $EF$ respectively. Prove that $KW$, $MV$, $NU$ intersect at one point. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscell... | English | proof only | null | |
0a83 | Problem:
Determine all functions $f$ defined in the set of rational numbers and taking their values in the same set such that the equation $f(x+y)+f(x-y)=2 f(x)+2 f(y)$ holds for all rational numbers $x$ and $y$. | [
"Solution:\nInsert $x=y=0$ in the equation to obtain $2 f(0)=4 f(0)$, which implies $f(0)=0$.\n\nSetting $x=0$, one obtains $f(y)+f(-y)=2 f(y)$ or $f(-y)=f(y)$.\n\nThen assume $y=n x$, where $n$ is a positive integer. We obtain\n$$\nf((n+1) x)=2 f(x)+2 f(n x)-f((n-1) x)\n$$\nIn particular, $f(2 x)=2 f(x)+2 f(x)-f(0... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 12 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | All and only functions of the form f(x) = a x^2 for all rational x, where a is a rational constant. | |
0cgh | Find all positive integers $x, y, z$ so that $2^x - 2^y - 2^z = 1023$. | [
"If none of $x, y, z$ is nil, then the left term of the equation is even, so it cannot be equal to $1023$. Hence, at least one of $x, y, z$ is nil.\nFrom $2^x = 2^y + 2^z + 1023$ follows that $x \\ge 11$, therefore $y = 0$ or $z = 0$.\nThe case $z = 0$ yields $2^x - 2^y = 1024$, so $2^y \\cdot (2^{x-y} - 1) = 2^{10... | Romania | 74th Romanian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | (x, y, z) = (11, 10, 0) or (11, 0, 10) | |
08i7 | Problem:
Let $[AB]$ be a segment and $\sigma$ be one of the halfplanes determined by the straight line $AB$. The segments $[AP]$ and $[BQ]$ with integer lengths are situated in $\sigma$ and are perpendicular to the straight line $AB$. The intersection point $M$ of the straight lines $AQ$ and $BP$ is distanced at 8 uni... | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | AP = 9, BQ = 72 | |
07ye | Problem:
Sia $n$ un intero maggiore o uguale a 2. Ci sono $n$ persone in fila indiana, ognuna delle quali è o un furfante (e mente sempre) oppure un cavaliere (e dice sempre la verità). Ogni persona, eccetto la prima, indica una delle persone davanti a lei e dichiara "Questa persona è un furfante" oppure "Questa perso... | [
"Solution:\n\nDiremo che due persone sono dello stesso tipo se sono entrambe cavalieri o entrambe furfanti, che sono di tipo diverso altrimenti.\nNotiamo che, se la persona $A$ indica la persona $B$ e la dichiara cavaliere, allora $A$ e $B$ sono dello stesso tipo: entrambi cavalieri (se $A$ dice la verità) o entram... | Italy | null | [
"Discrete Mathematics > Logic"
] | null | proof only | null | |
0fa5 | Problem:
A sequence of positive integers is constructed as follows. If the last digit of $a_n$ is greater than $5$, then $a_{n+1}$ is $9a_n$. If the last digit of $a_n$ is $5$ or less and $a_n$ has more than one digit, then $a_{n+1}$ is obtained from $a_n$ by deleting the last digit. If $a_n$ has only one digit, which... | [] | Soviet Union | 25th ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | No; every such sequence terminates. | |
04ro | There is written a number $N$ (in the decimal representation) on the board. In a step we erase the last digit $c$ and instead of the number $m$, which is now left on the board, we write number $|m - 3c|$ (for example, if $N = 1204$ was written on the board, then after the step there will be $120 - 3 \cdot 4 = 108$). We... | [
"Let us find $N$, which lead to zero on the board after only one step. Obviously $|m - 3c| = 0$ iff $m = 3c$, which is $N = 10m + c = 31c$. All such $N$ are of the form $N = 31c$, $c \\in \\{1, 2, \\dots, 9\\}$.\n\nWe show that the solution of the problem are exactly all multiples of $31$. Since $c = N - 10m$, ther... | Czech Republic | 62nd Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | All positive multiples of 31 | |
0g0a | Problem:
Sei $ABCD$ ein Trapez, wobei $AB$ und $CD$ parallel sind. $P$ sei ein Punkt auf der Seite $BC$. Zeige, dass sich die Parallelen zu $AP$ und $PD$ durch $C$ respektive $B$ auf $DA$ schneiden. | [
"Solution:\n\nSei $X$ der Schnittpunkt von $DA$ und der Parallelen zu $AP$ durch $C$ und sei $Y$ der Schnittpunkt von $DA$ und der Parallelen zu $PD$ durch $B$. Wir betrachten zuerst den Fall, dass $BC$ und $DA$ parallel sind. Dann ist $ABCD$ ein Parallelogramm und eine Punktspiegelung am Diagonalenschnittpunkt füh... | Switzerland | SMO - Finalrunde | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05u8 | Problem:
Soit $\left(F_{k}\right)_{k \geqslant 0}$ la suite définie par $F_{0}=0, F_{1}=1$, et $F_{k+2}=F_{k}+F_{k+1}$ pour tout entier $k \geqslant 0$. Soit ensuite $n \geqslant 1$ un entier. Démontrer qu'il existe exactement $F_{n+1}$ façons d'ordonner les nombres $1,2, \ldots, n$ de manière à obtenir un $n$-uplet $\... | [
"Solution:\nOn dit qu'une permutation $\\mathbf{a}=\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ des entiers $1,2, \\ldots, n$ est jolie si elle satisfait les inégalités $a_{1} \\leqslant 2 a_{2} \\leqslant \\ldots \\leqslant n a_{n}$.\n\nTout d'abord, soit $a$ une jolie permutation. S'il existe un entier $k$ pour l... | France | Préparation Olympique Française de Mathématiques - Test du 14 et du 21 Février 2021 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | F_{n+1} | |
02lw | Given positive integers $n$, $a_1$, $a_2$, $\dots$, $a_n$, define $q_0 = 1$, $q_1 = a_1$ and $q_{k+1} = a_{k+1}q_k + q_{k-1}$, for $1 \le k \le n-1$.
Prove that, given $c > 1$, there exists $K > 0$ such that, for all $M > K$, there exist a positive integer $n$ and $a_1, a_2, \dots, a_n \in \{1, 2\}$ such that $M \le q_... | [
"We will choose two large positive integers $r$, $s$, and take $m = r + s$, $a_j = 1$ for $1 \\le j \\le r$ and $a_j = 2$ for $r + 1 \\le j \\le r + s = m$.\nWe have $q_{k+1} = q_k + q_{k-1}$, for $1 \\le k \\le r-1$, and so $q_j = F_{j+1}$, for $0 \\le j \\le r$, where\n$$\nF_j = \\frac{1}{\\sqrt{5}} \\left( \\lef... | Brazil | XXXI Brazilian Math Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other"
] | English | proof only | null | |
0b3d | Problem:
The numbers $2, b, c, d, 72$ are listed in increasing order so that $2, b, c$ form an arithmetic sequence, $b, c, d$ form a geometric sequence, and $c, d, 72$ form a harmonic sequence (that is, a sequence whose reciprocals of its terms form an arithmetic sequence). What is the value of $b+c$ ?
(a) 7
(b) 13
(... | [] | Philippines | 24th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | MCQ | 19 | |
0dgg | Let $S(n)$ be the sum of divisors of $n$ (for example $S(6) = 1+2+3+6 = 12$). Find all $n$ for which $S(2n) = 3S(n)$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)"
] | English | proof and answer | All odd positive integers | |
00yy | Problem:
The positive integers $a$, $b$, $c$ are pairwise relatively prime, $a$ and $c$ are odd and the numbers satisfy the equation $a^{2} + b^{2} = c^{2}$. Prove that $b + c$ is a square of an integer. | [
"Solution:\n\nSince $a$ and $c$ are odd, $b$ must be even. We have $a^{2} = c^{2} - b^{2} = (c + b)(c - b)$. Let $d = \\operatorname{gcd}(c + b, c - b)$. Then $d$ divides $(c + b) + (c - b) = 2c$ and $(c + b) - (c - b) = 2b$. Since $c + b$ and $c - b$ are odd, $d$ is odd, and hence $d$ divides both $b$ and $c$. But... | Baltic Way | Baltic Way | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0kbq | Problem:
Jarris is a weighted tetrahedral die with faces $F_{1}$, $F_{2}$, $F_{3}$, $F_{4}$. He tosses himself onto a table, so that the probability he lands on a given face is proportional to the area of that face (i.e. the probability he lands on face $F_{i}$ is $\frac{[F_{i}]}{[F_{1}]+[F_{2}]+[F_{3}]+[F_{4}]}$ where... | [
"Solution:\nSince the maximum distance to the table is just the height, the expected value is equal to $\\frac{\\sum_{i=1}^{4} h_{i}[F_{i}]}{\\sum_{i=1}^{4}[F_{i}]}$. Let $V$ be the volume of Jarris. Recall that $V=\\frac{1}{3} h_{i}[F_{i}]$ for any $i$, but also $V=\\frac{r}{3}\\left(\\sum_{i=1}^{4}[F_{i}]\\right)... | United States | HMMT February 2020 | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Surface Area"
] | null | proof and answer | 12 | |
010k | Problem:
Let $P_{k}(x) = 1 + x + x^{2} + \cdots + x^{k-1}$. Show that
$$
\sum_{k=1}^{n} \binom{n}{k} P_{k}(x) = 2^{n-1} P_{n}\left(\frac{1+x}{2}\right)
$$
for every real number $x$ and every positive integer $n$. | [
"Solution:\n\nLet $A$ and $B$ be the left- and right-hand side of the claimed formula, respectively. Since\n$$\n(1-x) P_{k}(x) = 1 - x^{k},\n$$\nwe get\n$$\n(1-x) \\cdot A = \\sum_{k=1}^{n} \\binom{n}{k} (1 - x^{k}) = \\sum_{k=0}^{n} \\binom{n}{k} (1 - x^{k}) = 2^{n} - (1+x)^{n}\n$$\nand\n$$\n\\begin{aligned}\n(1-x... | Baltic Way | Baltic Way 1998 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
03gd | Problem:
Show that there are no integers $a, b, c$ for which $a^{2} + b^{2} - 8c = 6$. | [
"Solution:\nSuppose there exist integers $a, b, c$ such that $a^{2} + b^{2} - 8c = 6$.\n\nThen $a^{2} + b^{2} = 8c + 6$.\n\nConsider $a^{2} + b^{2}$ modulo $8$.\n\nThe possible quadratic residues modulo $8$ are $0, 1, 4$ (since $0^{2} \\equiv 0$, $1^{2} \\equiv 1$, $2^{2} \\equiv 4$, $3^{2} \\equiv 1$, $4^{2} \\equ... | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0cgw | Let $M$ be the midpoint of the side $AD$ of the square $ABCD$. Consider the equilateral triangles $DFM$ and $BFE$, such that $F$ lies in the interior of $ABCD$ and the lines $EF$ and $BC$ are concurrent. Denote by $P$ the midpoint of $ME$. Prove that:
a) $P$ lies on the line $AC$;
b) the halfline $PM$ is the bisector o... | [
"a) Construct the equilateral triangle $BDQ$, such that $C$ lies in its interior. $Q$ is situated on the perpendicular bisector of the diagonal $BD$, therefore $Q$, $C$ and $A$ are collinear. Since $\\angle EBQ = \\angle FBD = 60^\\circ - \\angle FBQ$, $BQ = BD$ and $BE = BF$, triangles $BEQ$ and $BFD$ are congruen... | Romania | 74th NMO Selection Tests for JBMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
099r | Let $\{1, 2, \ldots, 2009\} \supseteq A = \{a_1, \ldots, a_k\}$. Find all $A$ sets such that $2009 \mid \sum_{i=1}^{k} a_i$.
(proposed by B. Bayasgalan) | [
"Those $z_i$ are odd numbered $2n$th root of unity. In the figure regular $n$-gon.\nConsider following function's decomposition. $f(x) = (1+x)(1+x^2)\\ldots(1+x^n)$. Then $x^k$'s coefficient is a set whose sum of elements. We need to find $x^k$'s sum of coefficients, which is denoted by $A_n$. Here $n = 2009$. Now ... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | English | proof and answer | (2^2009 + 6*2^287 + 40*2^49 + 42*2^41 + 240*2^7 + 3360) / 2009 | |
01cj | The numbers
$$
\frac{1}{2016}, \frac{2}{2016}, \frac{3}{2016}, \dots, \frac{2015}{2016}
$$
are written on a blackboard. With each move, one may erase any two numbers $a$ and $b$ and replace them with
$$
3ab - 2a - 2b + 2.
$$
What will be the single remaining number after 2014 moves? | [
"Note that if $a = \\frac{1344}{2016} = \\frac{2}{3}$, then\n$$\n3ab - 2a - 2b + 2 = \\frac{2}{3},\n$$\nirrespective of the value of $b$. Hence, $\\frac{2}{3}$ will always remain on the blackboard. $\\square$"
] | Baltic Way | Baltic Way 2015 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 2/3 | |
0aev | Let $\triangle ABC$ be an equilateral triangle. Let $C_1$ and $C_2$ be on $AB$, $B_1$ and $B_2$ on $AC$ and $A_1$ and $A_2$ on $BC$ such that $\overline{A_1A_2} = \overline{B_1B_2} = \overline{C_1C_2}$. Let $A_2B_1$ and $B_2C_1$, $B_2C_1$ and $C_2A_1$, $C_2A_1$ and $A_2B_1$ intersect at $E$, $F$, $G$ correspondently. P... | [
"Let us denote the triangle formed by the segments $B_1A_2$, $A_1C_2$ and $C_1B_2$ with $\\triangle A_3B_3C_3$. Let $P$ be a point of the interior of the triangle $\\triangle EFG$ such that $C_1C_2PB_2$ is a parallelogram. Then $\\triangle B_2PB_1$ is equilateral, hence $PA_1A_2B_1$ is a parallelogram. From the abo... | North Macedonia | Junior Macedonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Translation"
] | Macedonian, English | proof only | null | |
077f | Problem:
Suppose $r \geq 2$ is an integer, and let $m_{1}, n_{1}, m_{2}, n_{2}, \cdots, m_{r}, n_{r}$ be $2 r$ integers such that
$$
\left|m_{i} n_{j}-m_{j} n_{i}\right|=1
$$
for any two integers $i$ and $j$ satisfying $1 \leq i<j \leq r$. Determine the maximum possible value of $r$. | [
"Solution:\nLet $m_{1}, n_{1}, m_{2}, n_{2}$ be integers satisfying $m_{1} n_{2}-m_{2} n_{1}= \\pm 1$. By changing the signs of $m_{2}, n_{2}$ if need be, we may assume that\n$$\nm_{1} n_{2}-m_{2} n_{1}=1\n$$\nIf $m_{3}, n_{3}$ are integers satisfying $m_{1} n_{3}-m_{3} n_{1}= \\pm 1$, again we may assume (by chang... | India | INMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Linear Algebra > Determinants",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | 3 | |
0khk | Una rolls 6 standard 6-sided dice simultaneously and calculates the product of the 6 numbers obtained. What is the probability that the product is divisible by 4?
(A) $\frac{3}{4}$ (B) $\frac{57}{64}$ (C) $\frac{59}{64}$ (D) $\frac{187}{192}$ (E) $\frac{63}{64}$ | [
"The product will not be divisible by 4 precisely when all 6 rolls are odd, or exactly one of them is equal to either 2 or 6 and the rest are odd. The probability of this complementary event is\n$$\n\\left(\\frac{1}{2}\\right)^6 + 6 \\cdot \\left(\\frac{1}{3}\\right) \\cdot \\left(\\frac{1}{2}\\right)^5 = \\frac{5}... | United States | Fall 2021 AMC 10 B | [
"Statistics > Probability > Counting Methods > Other",
"Statistics > Probability > Counting Methods > Other"
] | English | MCQ | (C) | |
039r | Find all positive integers $x$ and $y$ such that the number $(x^2 + y)(y^2 + x)$ is the fifth power of a prime. | [
"Let $(x^2 + y)(y^2 + x) = p^5$, where $p$ is a prime. Then $x^2 + y = p^s$, $y^2 + x = p^t$, where $\\{s, t\\} = \\{1, 4\\}$ or $\\{2, 3\\}$. In the first case we can assume without loss of generality that $x < y$, $x^2 + y = p$ and $y^2 + x = p^4$. Then $p^2 = (x^2 + y)^2 > x + y^2 = p^4$, a contradiction.\nLet $... | Bulgaria | Winter Mathematical Competition | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (2, 5) and (5, 2) | |
0ddq | Let $ABC$ be a non-isosceles triangle with altitudes $AD$, $BE$, $CF$ with orthocenter $H$. Suppose that $DF \cap HB = M$, $DE \cap HC = N$ and $T$ is the circumcenter of triangle $HBC$. Prove that $AT \perp MN$. | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Polar... | null | proof only | null | |
04xw | For each rational number $r$ consider the statement: If $x$ is a real number such that $x^2 - rx$ and $x^3 - rx$ are rational numbers, then $x$ is rational as well.
a) Prove the statement for $r \ge \frac{4}{3}$ and for $r \le 0$.
b) Let $p, q$ be different odd primes such that $3p < 4q$. Show the statement is false ... | [
"a) Let $s = x^2 - rx$ and $t = x^3 - rx$ be rational. Then\n$$\nx^2 = s + rx,\n$$\n$$\nx^3 = x^2 \\cdot x = (s + rx)x = sx + rx^2 = sx + r(s + rx) = (r^2 + s)x + rs,\n$$\nconsequently\n$$\nt = x^3 - rx = ((r^2 + s)x + rs) - rx = (r^2 - r + s)x + rs.\n$$\nif $(r^2 - r + s) \\neq 0$ (which is $x^2 - rx + r^2 - r \\n... | Czech-Polish-Slovak Mathematical Match | Cesko-Slovacko-Poljsko 2013 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0awo | Problem:
Let $\{a_n\}$ be a sequence such that $a_1 = 20$, $a_2 = 17$ and $a_n + 2 a_{n-2} = 3 a_{n-1}$. Determine the value of $a_{2017} - a_{2016}$. | [
"Solution:\nWe have $a_n = 3 a_{n-1} - 2 a_{n-2} \\Longrightarrow a_n - a_{n-1} = 2(a_{n-1} - a_{n-2})$. Repeated use of this recurrence relation gives $a_{2017} - a_{2016} = 2^{2015}(a_2 - a_1) = -3 \\cdot 2^{2015}$."
] | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | -3 * 2^2015 | |
07sn | Circles $\Omega_1$, centre $Q$, and $\Omega_2$, centre $R$, touch externally at $B$. A third circle, $\Omega_3$, which contains $\Omega_1$ and $\Omega_2$, touches $\Omega_1$ and $\Omega_2$ at $A$ and $C$, respectively. Point $C$ is joined to $B$ and the line $BC$ is extended to meet $\Omega_3$ at $D$.
Prove that $QR$ a... | [
"Let $M$ be the centre of $\\Omega_3$. Then, $A$, $Q$, $M$ are collinear as are $C$, $R$, $M$. Let $H$ be the intersection point of the lines $QR$ and $AD$.\n\n\n\nBecause $|BR| = |CR|$ and $|DM| = |CM|$, we have $\\angle MDC = \\angle MCD = \\angle RCB = \\angle RBC$, hence $DM$ is paralle... | Ireland | IRL_ABooklet_2020 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0816 | Problem:
Un fiume è attraversato da due ponti $TS$ e $VM$; le due rive $TV$ e $SM$ sono due archi di circonferenza concentrici; i due ponti $TS$ e $VM$ sono allineati con il centro (si veda la figura). Una persona vuole arrivare in $V$ partendo da $T$ scegliendo il percorso più breve tra i due possibili:
(1) seguire ... | [
"Solution:\n\nLa risposta è (C). Misurando l'angolo $\\alpha$ in radianti, la lunghezza del primo percorso è $R \\alpha$, mentre la lunghezza del secondo è $2(R-r) + r \\alpha$. Perciò il primo percorso è quello più corto se e solo se\n$$\nR \\alpha < 2(R-r) + r \\alpha,\n$$\nciaè se e solo se\n$$\n(R-r)(\\alpha - ... | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | MCQ | C | |
0ann | Problem:
Find the equation of the line that contains the point $(1,0)$, that is of least positive slope, and that does not intersect the curve $4x^{2} - y^{2} - 8x = 12$. | [] | Philippines | AREA STAGE | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | y = 2x - 2 | |
0jxs | Problem:
Fisica and Ritmo discovered a piece of Notalium shaped like a rectangular box, and wanted to find its volume. To do so, Fisica measured its three dimensions using a ruler with infinite precision, multiplied the results and rounded the product to the nearest cubic centimeter, getting a result of $2017$ cubic c... | [
"Solution:\n\nIt is not difficult to see that the maximum possible value of $V$ can be achieved when the dimensions are $(0.5+\\epsilon) \\times (0.5+\\epsilon) \\times (8070-\\epsilon') = 2017.5-\\epsilon''$ for some very small reals $\\epsilon, \\epsilon', \\epsilon'' > 0$, which when measured by Ritmo, gives $V ... | United States | HMMT November | [
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | proof and answer | 7174 | |
0ksv | Problem:
Point $P$ is located inside a square $A B C D$ of side length $10$. Let $O_{1}, O_{2}, O_{3}, O_{4}$ be the circumcenters of $P A B$, $P B C$, $P C D$, and $P D A$, respectively. Given that $P A + P B + P C + P D = 23 \sqrt{2}$ and the area of $O_{1} O_{2} O_{3} O_{4}$ is $50$, the second largest of the lengt... | [
"Solution:\n\nNote that $O_{1} O_{3}$ and $O_{2} O_{4}$ are perpendicular and intersect at $O$, the center of square $A B C D$. Also note that $O_{1} O_{2}, O_{2} O_{3}, O_{3} O_{4}, O_{4} O_{1}$ are the perpendiculars of $P B, P C, P D, P A$, respectively. Let $d_{1} = O O_{1}$, $d_{2} = O O_{2}$, $d_{3} = O O_{3}... | United States | HMMT February 2022 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane G... | null | final answer only | 16902 | |
0f9p | Problem:
Given $2n$ genuine coins and $2n$ fake coins. The fake coins look the same as genuine coins but weigh less (but all fake coins have the same weight). Show how to identify each coin as genuine or fake using a balance at most $3n$ times. | [] | Soviet Union | 24th ASU | [
"Discrete Mathematics > Algorithms"
] | null | proof only | null | |
0kjl | A quadratic polynomial $p(x)$ with real coefficients and leading coefficient $1$ is called *disrespectful* if the equation $p(p(x)) = 0$ is satisfied by exactly three real numbers. Among all the disrespectful quadratic polynomials, there is a unique such polynomial $\tilde{p}(x)$ for which the sum of the roots is maxim... | [
"Suppose $p(x) = (x - r)(x - s)$. Observe that $p(x)$ must have (two) real roots in order for $p(p(x))$ to have any roots at all. More specifically, if $y$ is a root of $p(p(x))$, then $p(y) = r$ or $p(y) = s$. That is, the equations\n$$\n(x - r)(x - s) - r = 0 \\quad \\text{and} \\quad (x - r)(x - s) - s = 0\n$$\n... | United States | AMC 12 A | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | A | |
01l5 | Given an $n \times n$ table with one of two signs "+" or "-" in any of its cells. Per move one can replace the signs in all cells of some row (or of some column) by the opposite signs. At the beginning there are exactly two minuses in the table (all other signs are pluses). After some moves the table with exactly 9 min... | [
"Answer: $n = 5, n = 11$.\n\nNote that if the operation of the sign changes is applied even times to some row (column), then it is equivalent that the operation is not applied at all. If the operation of the sign changes is applied odd times to some row (column), then it is equivalent that the operation is applied ... | Belarus | Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | proof and answer | n = 5, n = 11 | |
0ius | Problem:
If $n$ is a positive integer such that $n^{3} + 2 n^{2} + 9 n + 8$ is the cube of an integer, find $n$. | [
"Solution:\n\nSince $n^{3} < n^{3} + 2 n^{2} + 9 n + 8 < (n+2)^{3}$, we must have $n^{3} + 2 n^{2} + 9 n + 8 = (n+1)^{3}$. Thus $n^{2} = 6 n + 7$, so $n = 7$."
] | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 7 | |
05t3 | Problem:
On répartit les entiers de $1,2, \ldots, 8$ en deux ensembles $A$ et $B$, puis on note $P_{A}$ le produit de tous les éléments de $A$ et $P_{B}$ le produit de tous les éléments de $B$.
Quelles sont les valeurs minimale et maximale que peut prendre la somme $P_{A}+P_{B}$ ?
Note : si un ensemble $E$ est vide, ... | [
"Solution:\n\nSoit $A$ et $B$ deux ensembles disjoints dont la réunion est égale à l'ensemble $E=\\{1, \\ldots, 8\\}$.\nTâchons tout d'abord de maximiser la somme $P_{A}+P_{B}$. Sans perte de généralité, on peut supposer que $P_{A} \\leqslant P_{B}$. Puis, si $A$ contient un entier $k \\geqslant 2$, on pose $A' = A... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | minimum 402, maximum 40321 | |
03j2 | Problem:
An acute-angled triangle has unit area. Show that there is a point inside the triangle whose distance from each of the vertices is at least $\frac{2}{\sqrt[4]{27}}$. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Ge... | null | proof only | null | |
0fou | Problem:
Sean $p$ y $n$ enteros positivos, tales que $p$ es primo, $n \geq p$, y $1+n p$ es un cuadrado perfecto. Probar que $n+1$ es suma de $p$ cuadrados perfectos no nulos. | [
"Solution:\n\nSea $1+n p=k^{2}$, con $k$ entero positivo. Entonces $n p=k^{2}-1=(k-1)(k+1)$. Ahora consideramos dos casos:\n\n1. Si el primo $p$ divide a $k-1$, entonces $k-1=p \\ell$ y $k=p \\ell+1$, con $\\ell$ entero positivo. Por tanto\n$$\n1+n p=k^{2}=(p \\ell+1)^{2}=p^{2} \\ell^{2}+2 p \\ell+1 \\Leftrightarro... | Spain | LI Olimpiada matemática Española (Concurso Final) | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0dxl | Problem:
Dokaži, da za nobeno realno število $x$ ne velja
$$
\frac{1}{9}<\frac{\tan 3 x}{\tan 2 x} \leq \frac{3}{2}
$$ | [
"Solution:\n\n1. način\n\nUvedimo novo spremenljivko $a=\\tan^{2} x$. Ker je $\\tan 2 x=\\frac{2 \\tan x}{1-\\tan^{2} x}$ in $\\tan 3 x=\\frac{3 \\tan x-\\tan^{3} x}{1-3 \\tan^{2} x}$, je\n$$\n\\frac{\\tan 3 x}{\\tan 2 x}=\\frac{(3-a)(1-a)}{2(1-3 a)}\n$$\nzato lahko neenakost prepišemo v\n$$\n\\frac{1}{9}<\\frac{(3... | Slovenia | 51. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
08ni | Problem:
Find all natural numbers $a$, $b$, $c$ for which $1997^{a} + 15^{b} = 2012^{c}$. | [
"Solution:\n$1997^{a} + 15^{b} = 2012^{c} \\Rightarrow 1 + (-1)^{b} \\equiv 0 \\pmod{4}$, so $b$ is an odd number.\n$1997^{a} + 15^{b} = 2012^{c} \\Rightarrow 1 + 0 \\equiv 2^{c} \\pmod{3}$, so $c$ is even, say $c = 2c_{1}$.\nWe intend to consider the given equation modulo $8$ and for this reason we discern two cas... | JBMO | JBMO Shortlist | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Grea... | null | proof and answer | a = b = c = 1 | |
0el6 | Problem:
Zlatar ima dve zlitini. Prva zlitina vsebuje $90\%$ zlata, druga pa $54\%$ zlata. Zlatar je zmešal $320~\mathrm{g}$ prve zlitine in $160~\mathrm{g}$ druge zlitine, da je dobil novo zlitino. Koliko odstotkov zlata vsebuje nova zlitina?
(A) 33
(B) 48
(C) 65
(D) 72
(E) 78 | [
"Solution:\n\nV $320~\\mathrm{g}$ prve zlitine je $320 \\cdot \\frac{90}{100} = 32 \\cdot 9 = 288~\\mathrm{g}$ zlata, v $160~\\mathrm{g}$ druge zlitine pa $160 \\cdot \\frac{54}{100} = 16 \\cdot \\frac{27}{5}~\\mathrm{g}$ zlata. Delež zlata v novi zlitini je torej enak\n$$\n\\frac{32 \\cdot 9 + 16 \\cdot \\frac{27}... | Slovenia | 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | MCQ | E | |
03w0 | Twenty-four volunteers will be allocated to three schools. The rule is that each school will accept at least one volunteer and all the schools will accept different numbers of volunteers. Then there are ______ different ways of allocating volunteers. | [
"We may use each space between every two consecutive bars ($|$) to represent a school and each asterisk ($*$) to represent a volunteer, as seen in the following example; the first, second and third schools receive $4$, $18$ and $2$ volunteers, respectively.\n$$\n| * * * * | * \\cdots * | * * |\n$$\nThen the allocat... | China | China Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | final answer only | 222 | |
0g3e | Problem:
Prove that there are infinitely many positive integers $n$ such that
$$
n^{2}+1 \mid n!
$$ | [
"Solution:\nSo we want all the factors of $n^{2}+1$ to be less than $n$. The best way to ensure that this is true would be by somehow factorising $n^{2}+1$. Aha, looks like a job for the factorising master Sophie-Germain.\n\nChoose $n=2k^{2}$. We then have\n$$\nn^{2}+1 = 4k^{4} + 4k^{2} + 1 - 4k^{2} = (2k^{2}+1)^{2... | Switzerland | IMO Selection | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof only | null | |
0fz5 | Problem:
Finde alle ganzen Zahlen $m, n \geq 2$, welche die folgenden zwei Bedingungen erfüllen:
(i) $m+1$ ist eine Primzahl von der Form $4k+3$ für eine ganze Zahl $k$.
(ii) Es existiert eine Primzahl $p$ und eine nichtnegative ganze Zahl $a$ mit
$$
\frac{m^{2^{n}-1}-1}{m-1}=m^{n}+p^{a}
$$ | [
"Solution:\n\nNach Voraussetzung ist $m \\equiv 2 \\pmod{4}$. Schreibe die Gleichung um zu\n$$\nm^{2^{n}-1}-1=(m-1)\\left(m^{n}+p^{a}\\right)\n$$\nDa $n \\geq 2$ erhält man $-1 \\equiv p^{a} \\pmod{4}$, also $p \\equiv 3 \\pmod{4}$ und $a \\geq 1$ ungerade. Nun ist $q=m+1$ prim und es gilt $-2 \\equiv -2\\left((-1)... | Switzerland | IMO Selektion | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All pairs (m, n) with n = 2 and m + 1 a prime congruent to 3 modulo 4. | |
0bzz | Consider integers of the form $\overline{abcd}$ such that:
a) $a \ge b \ge c \ge d$;
b) $a+b+c+d = 11$;
c) $\overline{abcd} - \overline{cba} = 8082$.
Determine integers of the form $\overline{bdca}$ that satisfy all these conditions. | [] | Romania | Shortlisted problems for the 69th NMO | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 2018 | |
0c91 | Let $a_1 = 2$ and, for every positive integer $n$, let $a_{n+1}$ be the smallest integer strictly greater than $a_n$ that has more positive divisors than $a_n$. Prove that $2a_{n+1} = 3a_n$ only for finitely many indices $n$.
North Macedonia | [
"Begin with a mere remark on the terms of the sequence under consideration.\n\n**Lemma 1.** Each $a_n$ is minimal amongst all positive integers having the same number of positive divisors as $a_n$.\n\n*Proof.* Suppose, if possible, that for some $n$, some positive integer $b < a_n$ has as many positive divisors as ... | Romania | Balkan Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
05p9 | Problem:
Soit $m, n$ des entiers positifs tels que $\operatorname{pgcd}(m, n)=1$, où $a \wedge b$ désigne le plus grand diviseur commun de $a$ et $b$. Quelle(s) valeur(s) peut prendre
$$
\left(2^{m}-2^{n} \wedge 2^{m^{2}+m n+n^{2}}-1\right) ?
$$ | [
"Solution:\nOn utilise la propriété suivante : si $m \\geq 1$ et $a, b \\in \\mathbb{N}$, alors $\\left(m^{a}-1\\right) \\wedge\\left(m^{b}-1\\right)= m^{a \\wedge b}-1$. On se ramène ainsi à étudier $\\left(m^{2}+m n+n^{2}\\right) \\wedge(m-n)$. Si $d \\mid m^{2}+m n+n^{2}$ et $d \\mid m-n$, alors $d \\mid m^{2}+m... | France | OCympiades Françaises de Mathématiques | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 1 and 7 | |
0gfn | 連續正整數的等幂次方和的定義是
$$
S_m(N) = \sum_{k=1}^{N-1} k^m
$$
特別是
$$
S_1(N) = 1 + 2 + \cdots + (N-1) = \frac{N(N-1)}{2}
$$
$$
S_2(N) = 1^2 + 2^2 + \cdots + (N-1)^2 = \frac{N(N-1)(2N-1)}{6}
$$
$$
S_3(N) = 1^3 + 2^3 + \cdots + (N-1)^3 = \frac{N^2(N-1)^2}{4}
$$
證明
$$
S_2(N)S_3(N) = \frac{7}{12}S_6(N) + \frac{5}{12}S_4(N)
$$ | [
"$$\nS_2(N)S_3(N) = \\sum_{j=1}^{N-1} j^2 \\sum_{k=1}^{N-1} k^3\n$$\n依 $k < j$, $k = j$, 以及 $k > j$ 而分成三部份\n即\n$$\n\\sum_{j=1}^{N-1} j^2 \\sum_{k=1}^{j-1} k^3 + S_5(N) + \\sum_{k=1}^{N-1} k^2 \\sum_{j=1}^{k-1} j^2\n$$\n或\n$$\n\\sum_{j=1}^{N-1} j^2 \\cdot \\frac{j^2(j-1)^2}{4} + S_5(N) + \\sum_{k=1}^{N-1} k^3 \\cdot... | Taiwan | 國際奧林匹亞競賽第三次訓練營 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | Chinese; English | proof only | null | |
0can | Problem:
Fixăm un număr întreg $n \geq 2$ şi considerăm $n^{2}$ numere reale strict pozitive $a_{ij}$, $i, j=1, \ldots, n$, care îndeplinesc simultan următoarele două condiţii:
(1) $a_{ii}=1$, $i=1, \ldots, n$; şi
(2) Pentru fiecare $j=2, \ldots, n$, numerele $a_{ij}$, $i=1, \ldots, j-1$, formează o permutare a numere... | [] | Romania | Olimpiada Nationala de Matematica 2022 baraj 2 de selectie seniori | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 1 | |
0gmi | In an acute triangle $ABC$ with $|BC| < |AC| < |AB|$, the points $D \in [AB]$ and $E \in [AC]$ satisfy the condition $|BD| = |BC| = |CE|$. Show that the circumradius of the triangle $ADE$ is equal to the distance between the incenter and the circumcenter of the triangle $ABC$. | [] | Turkey | X. NATIONAL MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",... | English | proof only | null | |
0638 | Problem:
Es sei $ABC$ ein spitzwinkliges Dreieck mit Umkreis $\omega$. Man beweise, dass es einen Punkt $J$ mit der folgenden Eigenschaft gibt: Ist $X$ ein innerer Punkt von $ABC$, treffen die Strahlen $AX$, $BX$ und $CX$ den Kreis $\omega$ erneut in den Punkten $A_{1}$, $B_{1}$ und $C_{1}$ und liegen die Punkte $A_{2... | [
"Solution:\n\nWir zeigen, dass der normalerweise $H$ genannte Höhenschnittpunkt $J$ des Dreiecks $ABC$ die beschriebene Eigenschaft aufweist. Hierzu sei $a$ die durch $A$ gezogene Parallele zu $BC$ und die Geraden $b$ und $c$ seien analog definiert. Keine zwei der drei Geraden $a$, $b$ und $c$ sind parallel und fol... | Germany | 1. Auswahlklausur | [
"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 > Homothety",
"Geometry > Plane Geometry > Transformation... | null | proof only | null | |
0bnn | Let $M$ be the midpoint of the side $[AC]$ of triangle $ABC$ and $N \in (AM)$. The parallel from $N$ at $AB$ intersects $BM$ in $P$, the parallel from $M$ at $BC$ intersects $BN$ in $Q$ and the parallel from $N$ at $AQ$ intersects $BC$ in $S$. Prove that the straight lines $PS$ and $AC$ are parallel.
Cosmin Manea, Dra... | [
"Denote $MQ \\cap AB = \\{E\\}$ and $\\{D\\} = NP \\cap ME$. Then $EA = EB$ and $ND = DP$. Since $ANPB$ is a trapezoid, points $A, Q, P$ are collinear.\n\nFrom $\\triangle ADP \\equiv \\triangle SDN$ (A.S.A.) follows $[AP] \\equiv [SN]$ and, since $AP \\parallel SN$, quadrilateral $ANSP$ is a parallelogram, hence $... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07z7 | Problem:
Qual è la cifra delle unità del numero $2^{3^{4}}$ ?
(A) 1
(B) 2
(C) 4
(D) 6
(E) 8 | [] | Italy | Progetto Olimpiadi di Matematica | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | MCQ | B | |
081s | Problem:
Sia dato un triangolo $ABC$. Si indichino con $M$ ed $N$ i punti medi rispettivamente dei lati $AC$ e $BC$. Siano inoltre $S$ e $T$ rispettivamente punti sui lati $AC$ e $BC$ tali che:
$$
AS = \frac{1}{3} AC \quad BT = \frac{1}{3} BC.
$$
Dimostrare che le bisettrici degli angoli $\angle AST$ e $\angle BTS$ si... | [
"Solution:\n\nSi supponga che le bisettrici degli angoli $\\angle AST$ e $\\angle BTS$ si incontrino in un punto $P$ del lato $AB$.\nDal momento che $\\frac{AS}{AC} = \\frac{BT}{BC}$, per il teorema di Talete, $ST$ e $AB$ sono segmenti paralleli, pertanto gli angoli $\\angle PST$ e $\\angle SPA$ sono uguali. Ora, p... | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
00s4 | Let $\triangle ABC$ be a triangle with $AB < AC$ inscribed into a circle $c$. The tangent of $c$ at the point $C$ meets the parallel from $B$ to $AC$ at the point $D$. The tangent of $c$ at the point $B$ meets the parallel from $C$ to $AB$ at the point $E$ and the tangent of $c$ at the point $C$ at the point $L$. Suppo... | [
"We will prove first that the circle $c_1$ is tangent to $AB$ at the point $B$. In order to prove this, we have to prove that $\\angle BDC = \\angle ABC$. Indeed, since $BD \\parallel AC$, we have that $\\angle DBC = \\angle ACB$. Additionally, $\\angle BCD = \\angle BAC$ (by chord and tangent), which means that th... | Balkan Mathematical Olympiad | BMO 2017 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0d2r | Ten students take a test consisting of 4 different papers in Algebra, Geometry, Number Theory and Combinatorics. First, the proctor distributes randomly the Algebra paper to each student. Then the remaining papers are distributed one at a time in the following order: Geometry, Number Theory, Combinatorics in such a way... | [
"First, since the proctor distributes randomly the Algebra paper to each student, he has $10!$ ways to do it depending on how he orders the students.\n\nFor the other three papers, we order the students from $1$ to $30$, each student receiving three positions corresponding to the three papers. Since the first posit... | Saudi Arabia | Preselection tests for the full-time training | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 10! · 30! / (3!)^10 | |
0bi2 | Denote $A = \{1000, 1001, 1002, \dots, 2014\}$. Find the maximum number of elements of a subset of $A$ which contains only perfect squares pairwise relatively prime. | [
"If $n \\in A$ and $n = p^2$, then $1000 \\le p^2 \\le 2014$, that is $32 \\le p \\le 44$. The largest subset of $A$ whose elements are perfect squares is\n$$\nB = \\{32^2, 33^2, 34^2, 35^2, 36^2, 37^2, 38^2, 39^2, 40^2, 41^2, 42^2, 43^2, 44^2\\}.\n$$\nWe must choose among them the maximum number of pairwise prime ... | Romania | 65th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 6 | |
06hp | Let $[x]$ denote the greatest integer not exceeding $x$. Find the last three digits of $[(\sqrt[3]{5}+2)^{2014} + (\sqrt[3]{5}-2)^{2014}]$.
設 $[x]$ 表示不超過 $x$ 的最大整數。求 $[(\sqrt[3]{5}+2)^{2014} + (\sqrt[3]{5}-2)^{2014}]$ 的最後三位數字。 | [] | Hong Kong | HONG KONG PRELIMINARY SELECTION CONTEST | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English; Chinese | final answer only | 375 | |
0191 | The side of a triangle is subdivided by the bisector of its opposite angle into two segments of lengths $1$ and $3$. Determine all possible values of the area of that triangle. | [
"Call the triangle $ABC$ and let $AP$ be the angle bisector, with $P$ on $BC$, $BP = 3$ and $CP = 1$. By the Angle Bisector Theorem, we get $\\frac{AB}{AC} = 3$. Fixing the points $B$ and $C$, the locus of all points $A$ satisfying this is an Apollonius circle, whose centre lies on the line $BC$. This circle passes... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | all positive real numbers less than or equal to 3 | |
0gsb | Let $a_1, a_2, a_3, a_4$ be positive integers such that in any circular arrangement of these numbers there are two adjacent non-coprime ones. What is the maximal possible number of ordered triples $(i, j, k)$; $i, j, k \in \{1, 2, 3, 4\}$ and $i \neq j, j \neq k, k \neq i$, such that $(\text{gcd}(a_i, a_j))^2 \mid a_k$... | [
"Answer: 16.\nNote that if $(a_1, a_2, a_3, a_4) = (1, 2, 3, 6)$, there are 16 triples satisfying conditions:\n(1, 2, 3), (1, 2, 6), (1, 3, 2), (1, 3, 6), (1, 6, 2), (1, 6, 3), (2, 3, 6), (2, 3, 1),\n(2, 1, 3), (2, 1, 6), (3, 1, 2), (3, 1, 6), (6, 1, 2), (6, 1, 3), (3, 2, 6), (3, 2, 1).\nNow we will show that the n... | Turkey | 26th Turkish Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 16 | |
06t5 | Let $ABC$ be a triangle with circumcircle $\Omega$ and incentre $I$. Let the line passing through $I$ and perpendicular to $CI$ intersect the segment $BC$ and the arc $BC$ (not containing $A$) of $\Omega$ at points $U$ and $V$, respectively. Let the line passing through $U$ and parallel to $AI$ intersect $AV$ at $X$, a... | [
"We start with some general observations. Set $\\alpha = \\angle A / 2$, $\\beta = \\angle B / 2$, $\\gamma = \\angle C / 2$. Then obviously $\\alpha + \\beta + \\gamma = 90^{\\circ}$. Since $\\angle UIC = 90^{\\circ}$, we obtain $\\angle IUC = \\alpha + \\beta$. Therefore $\\angle BIV = \\angle IUC - \\angle IBC =... | IMO | 55th International Mathematical Olympiad Shortlist | [
"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 > Inver... | null | proof only | null | |
0ha5 | Find the angles of at least one triangle, one bisector of which is twice bigger than another. | [
"Let's find an isosceles triangle. Let's pretend that the vertex angle is obtuse. Then let's denote the base angle as $2\\alpha$ (Fig. 49). Then we can easily find values of some angles (Fig. 49):\n$$\n\\angle ALC = \\pi - 3\\alpha, \\quad \\angle ALB = 3\\alpha, \\quad \\angle ABC = \\pi - 4\\alpha.\n$$\n\nLet's d... | Ukraine | The Problems of Ukrainian Authors | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 36°, 36°, 108° | |
00cc | Problem:
Un conjunto de números enteros positivos distintos se llama *singular* si, para cada uno de sus elementos, luego de tachar ese elemento, los restantes se pueden agrupar en dos conjuntos sin elementos comunes de modo que la suma de los elementos de los dos grupos sea la misma. Hallar el menor entero positivo $n... | [] | Argentina | Nacional OMA 2019 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | Spanish | proof and answer | 7 | |
09x7 | Let $m$ and $n$ be positive integers with $mn$ even. Jetze is going to cover an $m \times n$-board (with $m$ rows and $n$ columns) with domino tiles, in such a way that every domino tile covers exactly two squares, domino tiles do not protrude out of the board or overlap one another, and every square is covered by a do... | [
"First suppose that $n$ is odd. Then we must have $V \\ge 1$, as the difference must be odd. We show that $V = 1$ is always possible. Colour the vertical domino tiles in the odd numbered columns red and the vertical domino tiles in the even numbered columns blue. As in every row, every horizontal domino tile covers... | Netherlands | IMO Team Selection Test 3 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | V = 1 if n is odd; V = 2 if n ≡ 2 (mod 4); V = 0 if n ≡ 0 (mod 4). | |
01n2 | Pedestrian, Cyclist and Motorcyclist start at 12.00 from town $A$ to town $B$ simultaneously. When each of them arrives at $B$ he whip rounds and moves to $A$, when he arrives at $A$ he again whip rounds and moves to $B$, and so on. After the start of the movement the first meeting is the meeting of Cyclist and Motorcy... | [
"Answer: 13.30.\nLet the distance between $A$ and $B$ be equal to $S$ (km), the speeds of Pedestrian, Cyclist and Motorcyclist be equal to $a$, $b$ and $c$ (km/h), respectively. By condition (the first meeting is the meeting of Cyclist and Motorcyclist), it follows that $a < b < c$. Let Pedestrian arrive at point $... | Belarus | Belorusija 2012 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 13.30 | |
0gvh | Mykolka the numismatist possesses 241 coins of total amount 360 tugriks (value of each coin is a positive integer quantity of tugriks). May it be claimed for sure that all those coins can be divided into three heaps of equal amount? | [
"Розіб’ємо коло на 360 рівних частин, і 241 з 360 точок поділу відмітимо червоним кольором так, щоб 241 дуга з червоними кінцями відповідала — за довжиною — вартостям монет. Усі 360 точок поділу на колі позначимо по порядку таким чином: $A_1, A_2, ..., A_{120}, B_1, B_2, ..., B_{120}, C_1, C_2, ..., C_{120}$. За пр... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | Yes | |
0d3j | Let $a_{1}, \ldots, a_{n}$ be a non increasing sequence of positive real numbers. Prove that
$$
\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}} \leq a_{1}+\frac{a_{2}}{\sqrt{2}+1}+\cdots+\frac{a_{n}}{\sqrt{n}+\sqrt{n-1}}
$$
When does equality hold? | [
"We prove the inequality by induction on $n \\geq 2$.\n\nFor $n=2$, let $a_{1} \\geq a_{2}>0$. We have\n$$\n\\begin{aligned}\n\\sqrt{a_{1}^{2}+a_{2}^{2}} & =\\sqrt{2 a_{2}^{2}+\\left(a_{1}-a_{2}\\right)^{2}+2 a_{2}\\left(a_{1}-a_{2}\\right)} \\\\\n& \\leq \\sqrt{2 a_{2}^{2}+\\left(a_{1}-a_{2}\\right)^{2}+2 \\sqrt{2... | Saudi Arabia | SAMC | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English, Arabic | proof and answer | Equality holds if and only if all terms are equal. | |
0l21 | Problem:
Let $ABC$ be an equilateral triangle with side length $1$. Points $D$, $E$, $F$ lie inside triangle $ABC$ such that $A$, $E$, $F$ are collinear, $B$, $F$, $D$ are collinear, $C$, $D$, $E$ are collinear, and triangle $DEF$ is equilateral. Suppose that there exists a unique equilateral triangle $XYZ$ with $X$ o... | [
"Solution:\n\n\n\nFirst, note that point $X$ can be constructed from intersection of $\\odot(DOF)$ and side $\\overline{BC}$. Thus, if there is a unique equilateral triangle, then we must have that $\\odot(DOF)$ is tangent to $\\overline{BC}$. Furthermore, $\\odot(DOF)$ is tangent to $DE$, ... | United States | HMMT February 2024 Guts Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 1/(1+∛2) | |
0gy3 | Find all such positive integer $n$ that for every polynomial $g(x)$ with real coefficients there exist such polynomials $p_1(x), p_2(x),..., p_n(x)$ and $q_1(x), q_2(x),..., q_n(x)$ that $g(x) = \sum_{i=1}^{n} (p_i^2(x) + q_i^3(x))$. | [
"For $n \\ge 2$ let $q_1(x) = q_2(x) = ... = q_n(x) = 0$ and $p_2(x) = p_3(x) = ... = p_n(x) = 0$, $q_2(x) = -\\frac{x+1}{3}$, $q_1(x) = \\frac{x-2}{3}$, $p_1(x) = \\frac{1}{\\sqrt{3}}(x+1)$. Then\n\n$$\n\\sum_{i=1}^{n} (p_i^2(x) + q_i^3(x)) = x, \\quad (1)\n$$\nand changing $x$ by $g(x)$ for all polynomials $p_i(x... | Ukraine | The Problems of Ukrainian Authors | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | All positive integers n greater than or equal to 2 | |
0gwc | Assume that positive real numbers $a_1, a_2, \dots, a_{2006}$ satisfy the equality
$$
\frac{a_1}{(a_1+a_2)a_2} + \frac{a_2}{(a_2+a_3)a_3} + \dots + \frac{a_{2005}}{(a_{2005}+a_{2006})a_{2006}} + \frac{a_{2006}}{(a_{2006}+a_1)a_1} = 2006
$$
Find the value of the expression
$$
\frac{a_1}{a_{2006}(a_{2006} + a_1)} + \frac... | [
"Відповідь: 2006.\n\nПомітимо, що\n$$\n\\frac{a_i}{(a_i + a_{i+1})a_{i+1}} = \\frac{1}{a_{i+1}} - \\frac{1}{a_i + a_{i+1}}, \\quad \\frac{a_{i+1}}{a_i(a_i + a_{i+1})} = \\frac{1}{a_i} - \\frac{1}{a_i + a_{i+1}}\n$$\nде $i=1,2006$, $a_{2007} = a_1$. Додаванням цих рівностей одержимо, що значення сум, про які йдеться... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 2006 |
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