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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
054o | Find all positive integers $k$ for which the integers $1, 2, \dots, 2017$ can be divided into $k$ groups in such a way that the sums of numbers in these groups are $k$ consecutive terms of an arithmetic sequence. | [
"Let the arithmetic sequence have the first term $a$ and the common difference $d$. The sum of all terms equals the sum of numbers $1, 2, \\dots, 2017$, i.e.,\n$$\n\\frac{2a + (k-1)d}{2} \\cdot k = \\frac{2017 \\cdot 2018}{2}\n$$\nwhence\n$$\n(2a + (k-1)d) \\cdot k = 2017 \\cdot 2018 = 2 \\cdot 1009 \\cdot 2017.\n$... | Estonia | National Olympiad Final Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 1, 2, 1009, 2017 | |
04lr | An acute-angled triangle $ABC$ is given, such that $|AB| > |AC|$. Let $O$ be the centre of the circumcircle, and let $OQ$ be a diameter of the circumcircle of triangle $BOC$. The line parallel to $BC$ passing through $A$ intersects the line $CQ$ at $M$, while the line parallel to $CQ$ and passing through $A$ intersects... | [
"Let the line $AQ$ intersect the circumcircle of $BOC$ at point $Q$ and $T'$. Notice that $\\overline{OQ}$ is a diameter of this circle, so $\\angle OT'Q = 90^\\circ$.\nLet $P$ be the midpoint of the segment $\\overline{AC}$. Since $O$ is the centre of the circumcircle of $ABC$, we have $\\angle OPA = 90^\\circ$.\n... | Croatia | Mathematical competitions in Croatia | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
01a9 | In an acute triangle $ABC$ with $AC > AB$ let $D$ be the projection of $A$ on $BC$, let $E$ and $F$ be the projections of $D$ on $AB$ and $AC$ and let $G$ and $H$ be the second intersections of the line $AD$ with $EF$ and the circumcircle of triangle $ABC$. Prove
$$
AG \cdot AH = AD^2.
$$ | [
"\nFrom similar right triangles we get\n$$\n\\frac{BE}{AE} = \\frac{\\frac{BD}{AD}ED}{\\frac{AD}{BD}ED} = \\left(\\frac{BD}{AD}\\right)^2\n$$\nand analogously\n$$\n\\frac{CF}{AF} = \\left(\\frac{CD}{AD}\\right)^2.\n$$\nNow, because $AC > AB$, the lines $BC$ and $EF$ intersect in a point $X$... | Baltic Way | Baltic Way 2013 | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06np | Problem:
A number is called *Norwegian* if it has three distinct positive divisors whose sum is equal to $2022$. Determine the smallest Norwegian number.
(Note: The total number of positive divisors of a Norwegian number is allowed to be larger than $3$.) | [] | Hong Kong | IMO HK TST | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 1974 | |
0l1y | Problem:
The graphs of the lines
$$
y = x + 2, \quad y = 3x + 4, \quad y = 5x + 6, \quad y = 7x + 8, \quad y = 9x + 10, \quad y = 11x + 12
$$
are drawn. These six lines divide the plane into several regions. Compute the number of regions the plane is divided into. | [
"Solution:\n\n\nAll lines are of the form $y = xk + (k + 1)$. Note that all lines pass through the point $(-1, 1)$, since $1 = (-1)k + (k + 1)$ for all $k$.\nThus all lines pass through a single point, $(-1, 1)$. The first line divides the plane into two parts, and each subsequent line divi... | United States | HMMT November 2024 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 12 | |
02sa | Problem:
Na figura, $PQRS$ é um quadrado, $\overline{QE} = 17$, e $\overline{PH} = 12$.
Calcule $\overline{SE}$. | [
"Solution:\n\nPodemos completar o desenho de modo que, na figura seguinte, $HAQE$ seja um retângulo.\n\n\n\nSe chamamos $\\measuredangle HPS = a^\\circ$, então $\\measuredangle PQ = (90 - a)^\\circ$. Logo, $\\measuredangle PQA = a^\\circ$. Observe que, deste modo, os triângulos $PHS$ e $QAP... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 7 | |
0627 | Problem:
Wir betrachten ein $m \times n$-Rechteck aus $m n$ Einheitsquadraten. Zwei seiner Einheitsquadrate heißen benachbart, wenn sie eine gemeinsame Seitenkante haben, und ein Pfad ist eine Folge von Einheitsquadraten, in der je zwei aufeinander folgende Elemente benachbart sind.
Jedes Einheitsquadrat des Rechtecks... | [
"Solution:\n\nWir werden die Behauptung verallgemeinern. Dazu lassen wir zu, dass das $m \\times n$ Rechteck auf beiden Seiten gefärbt wird und dass einige der Einheitsquadrate transparent sind. Solche Felder brauchen nur auf einer Seite gefärbt zu werden und sehen dann auf beiden Seiten gleich aus. Ein nicht trans... | Germany | Auswahlwettbewerb zur Internationalen Mathematik-Olympiade | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0i11 | Problem:
Let $p$ be an odd prime and $a, b$ positive integers satisfying $(p+1)^{a} - p^{b} = 1$. Show that $a = b = 1$. | [
"Solution:\nConsider remainders modulo $p+1$. We know that $1 + (-1)^{b} \\equiv 1 + p^{b} = (p+1)^{a} \\equiv 0$, so (since $p+1 > 2$) $b$ is odd.\n\nNow suppose that $a > 1$. Then\n$$\n0 \\equiv (p+1)^{a-1} = \\frac{p^{b} + 1}{p+1} = p^{b-1} - p^{b-2} + p^{b-3} - \\cdots + 1 \\equiv 1 - (-1) + 1 - \\cdots + 1\n$$... | United States | Berkeley Math Circle | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0de2 | Let $N$ be a positive integer. Consider the sequence $a_1, a_2, ..., a_N$ of positive integers, none of which is a multiple of $2^{N+1}$. For $n \ge N+1$, the number $a_n$ is defined as follows: choose $k$ to be the number among $1, 2, ..., n-1$ for which the remainder obtained when $a_k$ is divided by $2^n$ is the sma... | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
013q | Problem:
A set $S$ of $n-1$ natural numbers is given ($n \geq 3$). There exists at least two elements in this set whose difference is not divisible by $n$. Prove that it is possible to choose a non-empty subset of $S$ so that the sum of its elements is divisible by $n$. | [
"Solution:\n\nSuppose to the contrary that there exists a set $X = \\{a_{1}, a_{2}, \\ldots, a_{n-1}\\}$ violating the statement of the problem, and let $a_{n-2} \\not\\equiv a_{n-1} \\pmod{n}$. Denote $S_{i} = a_{1} + a_{2} + \\cdots + a_{i}$, $i = 1, \\ldots, n-1$. The conditions of the problem imply that all the... | Baltic Way | Baltic Way | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
03aa | Let $M$ be the midpoint of the segment $AB$ and $C$ be an interior point of the segment $AB$, $C \neq M$. The isosceles triangles $ACK$ ($AK = CK$) and $BCL$ ($BL = CL$) lie in the same halfplane with respect to $AB$ and are such that the points $K, C, L$ and $M$ are concyclic. Prove that either $KL \parallel AB$ or $K... | [
"We may assume that $AC < BC$. Denote by $k$ the circumcircle of $\\triangle KCL$ and by $K_1$ and $L_1$ the midpoints of $AC$ and $BC$, respectively. Then $KK_1 \\parallel LL_1$, i.e. $KK_1L_1L$ is a trapezoid. If $T$ is the midpoint of $CM$ then\n$$\nCM = \\frac{BC - AC}{2}, \\quad CT = \\frac{BC - AC}{4},\n$$\n$... | Bulgaria | Winter Mathematical Competition | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0fsr | Problem:
Gegeben sind ein gleichseitiges Dreieck mit der Seitenlänge $1$ und fünf gleich grosse, gleichseitige Dreiecke der Seitenlänge $s < 1$. Zeige: Lässt sich das grosse Dreieck mit den fünf kleinen überdecken, so lässt es sich auch schon mit vier der kleinen überdecken. | [
"Solution:\n\nBetrachte folgende 6 Punkte im oder auf dem Rand des grossen Dreiecks: die drei Eckpunkte und die drei Seitenmittelpunkte. Wenn sich das grosse Dreieck mit den fünf kleinen überdecken lässt, muss nach dem Schubfachprinzip eines der kleinen Dreiecke mindestens 2 dieser 6 Punkte überdecken. Diese 6 Punk... | Switzerland | IMO - Selektion | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
06t7 | Let $c \geqslant 1$ be an integer. Define a sequence of positive integers by $a_{1}=c$ and
$$
a_{n+1}=a_{n}^{3}-4 c \cdot a_{n}^{2}+5 c^{2} \cdot a_{n}+c
$$
for all $n \geqslant 1$. Prove that for each integer $n \geqslant 2$ there exists a prime number $p$ dividing $a_{n}$ but none of the numbers $a_{1}, \ldots, a_{n-... | [
"Let us define $x_{0}=0$ and $x_{n}=a_{n} / c$ for all integers $n \\geqslant 1$. It is easy to see that the sequence ($x_{n}$) thus obtained obeys the recursive law\n$$\n\\begin{equation*}\nx_{n+1}=c^{2}\\left(x_{n}^{3}-4 x_{n}^{2}+5 x_{n}\\right)+1 \\tag{1}\n\\end{equation*}\n$$\nfor all integers $n \\geqslant 0$... | IMO | 55th International Mathematical Olympiad Shortlist | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0amg | Problem:
Find the value of $p$, where,
$$
p = \frac{16^{2}-4}{18 \times 13} \times \frac{16^{2}-9}{19 \times 12} \times \frac{16^{2}-16}{20 \times 11} \times \cdots \times \frac{16^{2}-64}{24 \times 7}
$$ | [] | Philippines | AREA STAGE | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 2 | |
0e23 | Problem:
Poišči vsa realna števila $x$ z intervala $[0,2\pi)$, za katera so vsi členi zaporedja s splošnim členom
$$
a_n=\frac{1}{\cos (n x)}
$$
cela števila. | [
"Solution:\n\nŠtevili $a_1$ in $a_2$ sta celi. Ker je $a_1=\\frac{1}{\\cos x}$, $a_2=\\frac{1}{\\cos (2 x)}$ in velja $\\cos (2 x)=2(\\cos x)^2-1$, sledi $a_2=\\frac{1}{2 \\cos ^2 x-1}=\\frac{a_1^2}{2-a_1^2}$. Ker je $a_2$ celo število, je $2-a_1^2$ delitelj $a_1^2$. Zato $2-a_1^2$ deli $2-a_1^2$ in $a_1^2$, torej ... | Slovenia | 54. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | x ∈ {0, π/3, 2π/3, π, 4π/3, 5π/3} | |
08me | Problem:
Find all pairs $(x, y)$ of real numbers such that $|x| + |y| = 1340$ and $x^{3} + y^{3} + 2010 x y = 670^{3}$. | [
"Solution:\nAnswer: $(-670, -670)$, $(1005, -335)$, $(-335, 1005)$.\n\nTo prove this, let $z = -670$. We have\n$$\n0 = x^{3} + y^{3} + z^{3} - 3 x y z = \\frac{1}{2}(x + y + z)\\left((x - y)^{2} + (y - z)^{2} + (z - x)^{2}\\right)\n$$\nThus either $x + y + z = 0$, or $x = y = z$. In the latter case we get $x = y = ... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (-670, -670), (1005, -335), (-335, 1005) | |
00ak | Point $D$ is chosen on side $BC$ of the acute triangle $ABC$ so that $AD = AC$. Let $P$ and $Q$ be respectively the feet of the perpendiculars from $C$ and $D$ to side $AB$. It is known that
$$AP^2 + 3BP^2 = AQ^2 + 3BQ^2.$$ Find $\hat{A}BC$. | [
"Write the condition $AP^2 + 3BP^2 = AQ^2 + 3BQ^2$ in the form $AQ^2 - AP^2 = 3(BP^2 - BQ^2)$. Express $AQ^2$ and $AP^2$ by Pythagoras theorem for the right-angled triangles $ADQ$ and $ACP$: $AQ^2 = AD^2 - DQ^2$, $AP^2 = AC^2 - CP^2$. Since $AC = AD$, it follows that $AQ^2 - AP^2 = CP^2 - DQ^2$. Likewise the right-... | Argentina | Argentine National Olympiad 2016 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 60° | |
0bja | An integer $n > 1$ will be called *p-periodic* if $\frac{1}{n}$ is a repeating decimal fraction, whose shortest period has length $p$ and begins immediately after the decimal point. For instance, $\frac{1}{9} = 0.111\dots$ is 1-periodic and $\frac{1}{11} = 0.090909\dots$ is 2-periodic.
a) Find all p-periodic numbers $... | [
"a) The first digit of the period is at least 1 if and only if $\\frac{1}{n} \\ge \\frac{1}{10}$, that is $n \\le 10$. Since the prime factors of $n$ must be different from 2 and 5, it follows that $n \\in \\{3, 7, 9\\}$; indeed, in this case $\\frac{1}{3} = 0,(3)$, $\\frac{1}{7} = 0,(142857)$ and $\\frac{1}{9} = 0... | Romania | 65th Romanian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a) 3, 7, 9; b) 101 | |
033f | Problem:
Let $A=\{1,2, \ldots, n\}, n \geq 4$. For any function $f: A \rightarrow A$ and any $a \in A$ define $f_{1}(a)=f(a)$, $f_{i+1}(a)=f\left(f_{i}(a)\right)$, $i \geq 1$. Find the number of the functions $f$ such that $f_{n-2}$ is a constant function but $f_{n-3}$ is not. | [
"Solution:\nDefine an oriented graph $G$ with vertices the elements of $A$ and oriented edge $xy$ if $f(x)=y$. We have to count the graphs $G$ such that:\n- there are no cycles with length greater than $1$;\n- there is a chain $a_{2} \\ldots a_{n}$ with length $n-2$ and there is no chain with length $n-1$;\n- the o... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | n!(2n-5)/2 | |
06q2 | In the coordinate plane consider the set $S$ of all points with integer coordinates. For a positive integer $k$, two distinct points $A, B \in S$ will be called $k$-friends if there is a point $C \in S$ such that the area of the triangle $A B C$ is equal to $k$. A set $T \subset S$ will be called a $k$-clique if every ... | [
"To begin, let us describe those points $B \\in S$ which are $k$-friends of the point $(0,0)$. By definition, $B=(u, v)$ satisfies this condition if and only if there is a point $C=(x, y) \\in S$ such that $\\frac{1}{2}|u y-v x|=k$. (This is a well-known formula expressing the area of triangle $A B C$ when $A$ is t... | IMO | 49th International Mathematical Olympiad Spain | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | 180180 | |
01p3 | Given positive real numbers $a$, $b$, $c$. Find the greatest real number $x$ such that there exist positive real numbers $p$, $q$, $r$ ($p + q + r = 1$) with $x \le \min\{ap/q, bq/r, cr/p\}$. | [
"Answer: $\\sqrt[3]{abc}$.\nWe can consider $x > 0$. Multiplying three inequalities $x \\le ap/q$, $x \\le bq/r$, $x \\le cr/p$, we obtain $x \\le \\sqrt[3]{abc}$.\nIt remains to show that the number $\\sqrt[3]{abc}$ satisfies the problem condition. It suffices to verify that the system of the equations (with unkno... | Belarus | BelarusMO 2013_s | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | sqrt[3]{abc} | |
05nj | Problem:
Soit $n \in \mathbb{N}^{*}$. On place des rectangles $1 \times n$ et $n \times 1$ sur un échiquier $2 n \times 2 n$, de telle manière que deux rectangles ne s'intersectent jamais et que chaque rectangle recouvre exactement $n$ cases. Un ensemble de tels rectangles est dit saturé s'il est impossible d'ajouter ... | [
"Solution:\n\nCommençons par le cas $n \\geqslant 3$ : la réponse est $k=2 n+1$. L'ensemble suivant (dans le cas $n=5$, la construction se généralisant facilement) convient :\n\n\n\nIl reste à montrer qu'avec $2 n$ rectangles il est toujours possible d'en ajouter un qui ne les intersecte pa... | France | OCympiades Françaises de Mathématiques - Envoi Numéro 4 - Combinatoire | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | k = 4 if n = 1; k = 6 if n = 2; k = 2n + 1 if n ≥ 3 | |
0fgf | Problem:
Calcular
$$
\prod_{k=1}^{14} \cos \left(\frac{k \pi}{15}\right)
$$ | [
"Solution:\nConsideremos las $15$-raíces complejas de la unidad, de las que sólo una de ellas es real, $z=1$, y las demás son $7$ parejas complejas conjugadas. Tenemos\n$$\nz^{15}-1=(z-1) \\prod_{1}^{7}\\left(z-\\epsilon_{k}\\right)\\left(z-\\bar{\\epsilon}_{k}\\right), \\quad \\text{con} \\ \\epsilon_{k}=e^{i \\fr... | Spain | OME 22 | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | -1/16384 | |
024w | Problem:
O revezamento em uma corrida - Numa competição de revezamento, em que cada equipe tem dois atletas, cada atleta corre $21~\mathrm{km}$ e o segundo atleta só inicia a corrida quando o primeiro atleta termina a sua parte e lhe passa o bastão. O recorde dessa competição é de 2 horas e 48 minutos. Na equipe de Jo... | [
"Solution:\n\nComo velocidade $=\\frac{\\text{distância}}{\\text{tempo}}$, ou seja, tempo $=\\frac{\\text{distância}}{\\text{velocidade}}$, o tempo gasto por João foi de\n$$\nt=\\frac{21}{12}~\\mathrm{h}=\\left(1+\\frac{9}{12}\\right)~\\mathrm{h}=1~\\mathrm{h}+\\frac{9}{12} \\times 60~\\mathrm{min}=1~\\mathrm{h}~45... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | greater than 20 km/h | |
06mh | Find all positive integers $n$ with the following property: the $k$ positive divisors of $n$ have a permutation $(d_1, d_2, \dots, d_k)$ such that for every $i = 1, 2, \dots, k$, the number $d_1 + d_2 + \dots + d_i$ is a perfect square. | [
"2. (IMO Shortlist 2021 N3) See the official solution."
] | Hong Kong | IMO HK TST | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)"
] | null | proof and answer | n = 1 or n = 3 | |
0kcv | Problem:
Compute the number of positive integers less than $10!$ which can be expressed as the sum of at most 4 (not necessarily distinct) factorials. | [
"Solution:\n\nSince $0! = 1! = 1$, we ignore any possible $0!$'s in our sums.\n\nCall a sum of factorials reduced if for all positive integers $k$, the term $k!$ appears at most $k$ times. It is straightforward to show that every positive integer can be written uniquely as a reduced sum of factorials. Moreover, by ... | United States | HMMO 2020 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 648 | |
029a | Problem:
A figura a seguir mostra um triângulo $ABC$ com lados $AB=13~\mathrm{cm}$, $BC=14~\mathrm{cm}$ e $CA=15~\mathrm{cm}$. A circunferência de centro $I$ tangencia os lados $AB$, $BC$ e $CA$ nos pontos $F$, $D$ e $E$, respectivamente. Os pontos $R$, $S$, $T$, $U$, $V$ e $X$ são marcados nos prolongamentos dos lados... | [
"Solution:\n(a) O Teorema do Bico diz que as distâncias de um ponto exterior a uma circunferência aos pontos onde suas tangentes tocam a circunferência são iguais. Aplicando este resultado aos pontos $A$, $B$ e $C$ na circunferência que tangencia os três lados do triângulo $ABC$, temos\n$$\n\\begin{aligned}\n& AE=A... | Brazil | NÍVEL 3 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 6 | |
0g4f | Problem:
For a positive integer $m$, we denote by $[m]$ the set $\{1,2, \ldots, m\}$. Let $n$ be a positive integer and let $\mathcal{S}$ be a non-empty collection of subsets of $[n]$. A function $f:[n] \rightarrow [n+1]$ is called kawaii if there exists $A \in \mathcal{S}$ such that for all $B \in \mathcal{S}$ with $... | [
"Solution:\n\nWe will prove there is an injection from functions $f:[n] \\rightarrow [n]$ and kawaii functions.\n\nLet $f$ be an arbitrary function in the latter set. Let $S$ be any set in $\\mathcal{S}$ whose sum of images is larger or equal to all other sums of images over sets in $\\mathcal{S}$. Now, add $1$ to ... | Switzerland | Switzerland Selection Solution | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
00ff | Let
$$
f(x) = a_{n} x^{n} + a_{n-1} x^{n-1} + \cdots + a_{0} \quad \text{and} \quad g(x) = c_{n+1} x^{n+1} + c_{n} x^{n} + \cdots + c_{0}
$$
be non-zero polynomials with real coefficients such that $g(x) = (x + r) f(x)$ for some real number $r$. If $a = \max \left(\left|a_{n}\right|, \ldots, \left|a_{0}\right|\right)$ ... | [
"Expanding $(x + r) f(x)$, we find that $c_{n+1} = a_{n}$, $c_{k} = a_{k-1} + r a_{k}$ for $k = 1, 2, \\ldots, n$, and $c_{0} = r a_{0}$. Consider three cases:\n- $r = 0$. Then $c_{0} = 0$ and $c_{k} = a_{k-1}$ for $k = 1, 2, \\ldots, n$, and $a = c \\Longrightarrow \\frac{a}{c} = 1 \\leq n+1$.\n\n- $|r| \\geq 1$. ... | Asia Pacific Mathematics Olympiad (APMO) | APMO 1993 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
06u6 | Let $q$ be a real number. Gugu has a napkin with ten distinct real numbers written on it, and he writes the following three lines of real numbers on the blackboard:
- In the first line, Gugu writes down every number of the form $a-b$, where $a$ and $b$ are two (not necessarily distinct) numbers on his napkin.
- In the ... | [
"Answer: $-2, 0, 2$.\n\nCall a number $q$ good if every number in the second line appears in the third line unconditionally. We first show that the numbers $0$ and $\\pm 2$ are good. The third line necessarily contains $0$, so $0$ is good. For any two numbers $a, b$ in the first line, write $a = x - y$ and $b = u -... | IMO | International Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Other"
] | English | proof and answer | -2, 0, 2 | |
01ue | Let $M$ be the midpoint of the hypotenuse $AB$ of the right triangle $ABC$. Point $P$ is chosen on the cathetus $CB$ so that $CP : PB = 1 : 2$. The straight line passing through $B$ meets the segments $AC, AP,$ and $PM$ at points $X, Y, \text{ and } Z$, respectively.
Prove that the bisector of the angle $PZY$ passes th... | [
"Let $D$ be symmetric to $A$ with respect to the vertex $C$ (see the Fig.). The segment $BC$ is the altitude and the median in the triangle $ABD$ so the triangle $ABD$ is isosceles. Likewise, the triangle $ADP$ is isosceles. Let $P'$ be the intersection point of the medians $DM$ and $BC$ of the triangle $ABD$. Sinc... | Belarus | Belarusian 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 > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0atb | Problem:
Let $p$ and $q$ be positive integers such that $p q = 2^{3} \cdot 5^{5} \cdot 7^{2} \cdot 11$ and $\frac{p}{q} = 2 \cdot 5 \cdot 7^{2} \cdot 11$. Find the number of positive integer divisors of $p$. | [] | Philippines | Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | final answer only | 72 | |
04nr | Let $ABC$ be an acute-angled triangle such that $|BC| > |AC|$. The perpendicular bisector of the segment $\overline{AB}$ intersects the side $BC$ at point $P$, and the line $AC$ at point $Q$. Let $R$ be the foot of the perpendicular from point $P$ to the side $\overline{AC}$, and let $S$ be the foot of the perpendicula... | [] | Croatia | Croatia_2018 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0hhl | Point $I$ is the incenter of triangle $ABC$, where $AB < AC$. On the external angle bisector of angle $ABC$, a point $X$ is chosen such that $IC = IX$. Let the tangent to the circumcircle of triangle $BXC$ at point $X$ intersect line $AB$ at point $Y$. Prove that $AC = AY$.
 | [
"From the fact that $XY$ is tangent to the circumcircle of $\\triangle BXC$, we have $\\angle BXY = \\angle BCX$. Combining this with the fact that $\\angle YBX = \\angle XBC$, we have the similarity $\\triangle BXY \\sim \\triangle BCX$. Therefore, we have $\\frac{BX}{BC} = \\frac{BY}{BX}$, so $BY = \\frac{BX^2}{B... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | English | proof only | null | |
0kb0 | Find all binary operations $\diamond$: $\mathbb{R}_{>0} \times \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ (meaning $\diamond$ takes pairs of positive real numbers to positive real numbers) such that for any real numbers $a, b, c > 0$,
* the equation $a \diamond (b \diamond c) = (a \diamond b) \cdot c$ holds; and
* if $a \ge ... | [
"**First solution using Cauchy FE** We prove:\n**Claim** — We have $a \\diamond b = af(b)$ where $f$ is some involutive and totally multiplicative function. (In fact, this classifies all functions satisfying the first condition completely.)\n*Proof*. Let $P(a, b, c)$ denote the assertion $a \\diamond (b \\diamond c... | United States | USA TSTST | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | a ◊ b = a b or a ◊ b = a / b | |
01gs | Let $ABC$ be a triangle with $AB > AC$. The bisector of $\angle BAC$ intersects the side $BC$ at $D$. The circles with diameters $BD$ and $CD$ intersect the circumcircle of $\triangle ABC$ a second time at $P \neq B$ and $Q \neq C$, respectively. The lines $PQ$ and $BC$ intersect at $X$. Prove that $AX$ is tangent to t... | [
"The key observation is that the circumcircle of $\\triangle DPQ$ is tangent to $BC$. This can be proved by angle chasing:\n$$\n\\begin{aligned}\n\\angle BDP &= 90^\\circ - \\angle PBD = 90^\\circ - \\angle PBC = 90^\\circ - (180^\\circ - \\angle CQP) \\\\\n&= \\angle CQP - 90^\\circ = \\angle DQP.\n\\end{aligned}\... | Baltic Way | Baltic Way 2020 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
073g | Prove that there are infinitely many pairs $(m, n)$ of positive integers such that $m < n$ and $\frac{(m+n)(m+n+1)}{mn}$ is an integer. | [
"If we take $m = 1$, $n = 2$, we see that $\\frac{(m+n)(m+n+1)}{mn} = 6$. (Or we can start with $m = 2$, $n = 3$ as well.)\n\nSuppose we have some pair $(m, n)$ of positive integers such that $m < n$ and\n$$\n\\frac{(m+n)(m+n+1)}{mn} = k\n$$\nis an integer. This may be written in the form\n$$\nnk = m + 2n + 1 + \\f... | India | Indija TS 2008 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
02a3 | Problem:
Qual o menor número inteiro positivo $n$ tal que
$$
\sqrt{n} - \sqrt{n-1} < 0,01 ?
$$ | [
"Solution:\n\nA inequação $\\sqrt{n} - \\sqrt{n-1} < 0,01$ é equivalente a $\\sqrt{n} < 0,01 + \\sqrt{n-1}$. Como os dois lados desta inequação são números positivos, podemos elevar esses dois membros ao quadrado para obter a inequação equivalente:\n$$\n(\\sqrt{n})^{2} < (0,01 + \\sqrt{n-1})^{2} \\Leftrightarrow n ... | Brazil | Nível 3 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 2501 | |
08bc | Problem:
Una pedina si trova inizialmente sulla casella centrale di una scacchiera $5 \times 5$. Un passo della pedina consiste nello spostarsi in una casella scelta a caso fra quelle che hanno esattamente un vertice in comune con la casella su cui si trova. Qual è la probabilità che dopo 12 passi la pedina si trovi i... | [
"Solution:\n\nLa risposta è (E). Si consideri la scacchiera colorata come in figura; un passo della pedina può solo condurla da una casella bianca a una casella grigia, e viceversa: ne consegue che, dopo un numero dispari di passi, la pedina non può che trovarsi in una delle quattro caselle bianche. Qualunque sia l... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | MCQ | E | |
0hsq | Problem:
Squares $A B D E$, $B C F G$ and $C A H I$ are drawn exterior to a triangle $A B C$. Parallelograms $D B G X$, $F C I Y$ and $H A E Z$ are completed. Prove that $\angle A Y B + \angle B Z C + \angle C X A = 90^{\circ}$. | [
"Solution:\n\nLet $\\rho$ be the $90^{\\circ}$ rotation about the center of square $A B D E$, counterclockwise (orienting $\\triangle A B C$ to have its vertices in counterclockwise order). Note that segments $C A$ and $Z E$ are congruent and perpendicular (thanks to square $C A H I$ and parallelogram $H A E Z$), s... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0cki | Consider a sequence of integers $a_1, a_2, a_3, \dots$ such that $a_1 > 1$ and $(2^{a_n} - 1)a_{n+1}$ is a square for all positive integers $n$. Is it possible that two terms of such a sequence be equal? | [
"The answer is in the negative. Notice first that, if $a_n > 1$, then $2^{a_n} - 1 \\equiv 3 \\pmod 4$; since $(2^{a_n} - 1)a_{n+1}$ is a perfect square, we should have $a_{n+1} \\equiv 0 \\pmod 4$ or $a_{n+1} \\equiv 3 \\pmod 4$, so in particular $a_{n+1} > 1$. As $a_1 > 1$, we conclude that all terms of the seque... | Romania | Seventeenth ROMANIAN MASTER OF MATHEMATICS | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof and answer | No | |
0ei2 | Problem:
Kateri interval je zaloga vrednosti funkcije $f(x)=\cos x+1-\pi$?
(A) $[-1+\pi, 2]$
(B) $[0, \pi+2]$
(C) $[-\pi, 2-\pi]$
(D) $[-\pi, 1+\pi]$
(E) $[1-\pi, 2-\pi]$ | [
"Solution:\n\nZaloga vrednosti funkcije kosinus je interval $[-1,1]$, zato sledi, da je zaloga vrednosti funkcije $f$ interval $[-\\pi, 2-\\pi]$."
] | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Precalculus > Trigonometric functions"
] | null | MCQ | C | |
0d30 | $\triangle ABC$ is a triangle and $I_{b}, I_{c}$ its excenters opposite to $B, C$. Prove that $\triangle ABC$ is right at $A$ if and only if its area is equal to $\frac{1}{2} AI_{b} \cdot AI_{c}$. | [
"First solution. Let $\\triangle ABC$ be any triangle. Considering triangle $A I_{c} C$, we have\n$$\n\\begin{aligned}\n\\measuredangle C I_{c} A & =180^{\\circ}-\\left(\\measuredangle A C I_{c}+\\measuredangle I_{c} A C\\right) \\\\\n& =180^{\\circ}-\\left(\\frac{1}{2} \\measuredangle A C B+\\measuredangle B A C+\... | Saudi Arabia | Preselection tests for the full-time training | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0la5 | In a triangle $ABC$, the angle $BEC$ is acute, where $E$ is the midpoint of $AB$. On the ray $EC$, choose a point $M$ such that $\widehat{BME} = \widehat{ECA}$. Denote by $\alpha$ the measure of the angle $BEC$. Find the ratio $\frac{MC}{AB}$ in terms of $\alpha$. | [] | Vietnam | Vijetnam 2008 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | cos(alpha) | |
03sr | Suppose positive integers $m$, $n$, $k$ satisfy $mn = k^2 + k + 3$. Prove that at least one of the following Diophantine equations
$$
x^2 + 11y^2 = 4m \text{ and } x^2 + 11y^2 = 4n
$$
has a solution $(x, y)$ with $x$, $y$ being odd numbers. | [
"First, we prove a lemma.\n\n**Lemma** The following Diophantine equation\n$$\nx^2 + 11y^2 = 4m\n$$\nhas a solution $(x_0, y_0)$, such that either $x_0$, $y_0$ are odd numbers or $x_0$, $y_0$ are even numbers with $x_0 \\equiv (2k+1)y_0 \\pmod m$.\n\nConsider the expression $x + (2k+1)y$, where $x$, $y$ are integer... | China | China Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | English | proof only | null | |
0abn | A square shaped orchard with side length $26\text{m}$ is fenced in with $3$ rows of wire. Is it possible with the same wire to fence in a rectangular shaped orchard with side lengths $95\text{m}$ and $60\text{m}$? | [
"For fencing in the square shaped orchard $3 \\cdot 4 \\cdot 26\\text{m} = 312\\text{m}$ of wire is used. To fence in the rectangular orchard $2 \\cdot (95\\text{m} + 60\\text{m}) = 310\\text{m}$ of wire is needed. So the answer is affirmative."
] | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | Yes | |
06kq | Given triangle $ABC$, let $D$ be an inner point of the segment $BC$. Let $P$ and $Q$ be distinct inner points of the segment $AD$. Let $K = BP \cap AC$, $L = CP \cap AB$, $E = BQ \cap AC$, $F = CQ \cap AB$. Given that $KL \parallel EF$, find all possible values of the ratio $BD : DC$. | [
"The only possible value is $1$.\nWe use projective geometry. Consider the following projection.\n$$\nAB(A, L, F, B) \\xrightarrow{\\{C\\}} AD(A, P, Q, D) \\xrightarrow{\\{B\\}} AC(A, K, E, C).\n$$\nThis shows $(A, L, F, B)$ and $(A, K, E, C)$ are perspective, and so $LK$, $FE$, $BC$ are concurrent. Since $KL \\par... | Hong Kong | null | [
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem"
] | null | proof and answer | 1 | |
0kso | A circle with radius $6$ is externally tangent to a circle with radius $24$. Find the area of the triangular region bounded by the three common tangent lines of these two circles. | [] | United States | AIME II | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 192 | |
0h6g | Let $A_1$ and $C_1$ be the points on the sides $BC$ and $AB$ of the triangle $ABC$ respectively, so that $AA_1 = CC_1$. Segments $AA_1$ and $CC_1$ meet at the point $F$. Prove that if $\angle CFA_1 = 2\angle ABC$ then $AA_1 = AC$. | [
"Consider translation by vector $\\vec{A_1A}$ (Fig. 40). Thus $AA_1CC_3$ is a parallelogram, $CC_1C_2C_3$ is a rhombus. Then $\\angle C_1CC_3 = \\angle C_1FA = \\angle CFA_1 = 2\\alpha$, therefore\n$$\n\\angle C_1AC_3 = 180^\\circ - \\angle ABC = 180^\\circ - \\frac{1}{2}\\angle CFA_1 = 180^\\circ - \\alpha.\n$$\nI... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0aqj | Problem:
Each of the integers $1,2,3, \ldots, 9$ is assigned to each vertex of a regular 9-sided polygon (that is, every vertex receives exactly one integer from $\{1,2, \ldots, 9\}$, and two vertices receive different integers) so that the sum of the integers assigned to any three consecutive vertices does not exceed... | [
"Solution:\n\nThere is an assignment of the integers $1,2,3, \\ldots, 9$ to the vertices of the regular nonagon that gives $n=16$. Let $S$ be the sum of all sums of the integers assigned to three consecutive vertices. If there are integers assigned to three consecutive vertices whose sum is at most $14$, then $S<13... | Philippines | 12th Philippine Mathematical Olympiad - Area Stage | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 16 | |
0kte | Problem:
Positive integers $a_{1}, a_{2}, \ldots, a_{7}, b_{1}, b_{2}, \ldots, b_{7}$ satisfy $2 \leq a_{i} \leq 166$ and $a_{i}^{b_{i}} \equiv a_{i+1}^{2} (\bmod 167)$ for each $1 \leq i \leq 7$ (where $a_{8}=a_{1}$). Compute the minimum possible value of $b_{1} b_{2} \cdots b_{7}\left(b_{1}+b_{2}+\cdots+b_{7}\right)... | [
"Solution:\n\nLet $B = b_{1} b_{2} \\cdots b_{7} - 128$. Since\n$$\na_{1}^{b_{1} b_{2} \\cdots b_{7}} \\equiv a_{2}^{2 b_{2} b_{3} \\cdots b_{7}} \\equiv a_{3}^{4 b_{3} b_{4} \\cdots b_{7}} \\equiv \\cdots \\equiv a_{1}^{128} \\quad(\\bmod 167)\n$$\nwe find that $a_{1}^{B} \\equiv 1 (\\bmod 167)$. Similarly, $a_{i}... | United States | HMMT February 2022 | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 675 | |
03al | Изпъкнал 2009-ъгълник е разбит на триъгълници чрез непресичащи се диагонали. Един от тези диагонали е оцветен в зелено. Разрешена е следната операция: за два триъгълника *ABC* и *BCD* от разбиването с обща страна $BC$ можем да заменим диагонала $BC$ с диагонала $AD$, като, ако замененият диагонал е бил зелен, той губи ... | [
"Първо ще докажем, че за даден връх на изпъкналия 2009-ъгълник и всяка триангулация, с прилагане на разрешената операция можем да получим триангулацията, получена от прекарването на всички диагонали през този връх. За произволен връх $A$, движейки се обратно на часовниковата стрелка, да означим с $B_1, B_2, \\dots,... | Bulgaria | 58. National mathematical olympiad Final round | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0j74 | Problem:
Each square in a $3 \times 10$ grid is colored black or white. Let $N$ be the number of ways this can be done in such a way that no five squares in an 'X' configuration (as shown by the black squares below) are all white or all black. Determine $\sqrt{N}$.
 | [
"Solution:\n\nNote that we may label half of the cells in our board the number $0$ and the other half $1$, in such a way that squares labeled $0$ are adjacent only to squares labeled $1$ and vice versa. In other words, we make this labeling in a 'checkerboard' pattern. Since cells in an 'X' formation are all labele... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 25636 | |
06vj | Let $ABC$ be an acute-angled triangle and let $D$, $E$, and $F$ be the feet of altitudes from $A$, $B$, and $C$ to sides $BC$, $CA$, and $AB$, respectively. Denote by $\omega_{B}$ and $\omega_{C}$ the incircles of triangles $BDF$ and $CDE$, and let these circles be tangent to segments $DF$ and $DE$ at $M$ and $N$, resp... | [
"Denote the centres of $\\omega_{B}$ and $\\omega_{C}$ by $O_{B}$ and $O_{C}$, let their radii be $r_{B}$ and $r_{C}$, and let $BC$ be tangent to the two circles at $T$ and $U$, respectively.\n\n\n\nFrom the cyclic quadrilaterals $AFDC$ and $ABDE$ we have\n$$\n\\angle MDO_{B} = \\frac{1}{2}... | IMO | IMO 2019 Shortlisted Problems | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate M... | English | proof only | null | |
04m5 | Prove that there are no positive integers $m$ and $n$ such that
$$
5m^3 = 27n^4 - 2n^2 + n.
$$ | [
"Let us write the equation in the form\n$$\n5m^3 = n(27n^3 - 2n + 1)\n$$\nand note that $n$ and $27n^3 - 2n + 1$ are relatively prime numbers. We consider two cases:\n\na) $n = 5a^3$ and $27n^3 - 2n + 1 = b^3$ for some positive integers $a$ and $b$.\nInequality $(3n-1)^3 = 27n^3 - 27n^2 + 9n - 1 < 27n^3 - 2n + 1$ i... | Croatia | Croatian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0js9 | Problem:
Let $ABC$ be a triangle with incenter $I$ whose incircle is tangent to $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ at $D, E, F$. Point $P$ lies on $\overline{EF}$ such that $\overline{DP} \perp \overline{EF}$. Ray $BP$ meets $\overline{AC}$ at $Y$ and ray $CP$ meets $\overline{AB}$ at $Z$. Point $Q$ is ... | [
"Solution:\n\nThe proof proceeds through a series of seven lemmas.\n\nLemma 1. Lines $DP$ and $EF$ are the internal and external angle bisectors of $\\angle BPC$.\n\nProof. Since $DEF$ is the cevian triangle of $ABC$ with respect to its Gergonne point, we have that\n$$\n-1=(\\overline{EF} \\cap \\overline{BC}, D ; ... | United States | HMMT February 2016 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformat... | null | proof only | null | |
010e | Problem:
Consider a ping-pong match between two teams, each consisting of $1000$ players. Each player played against each player of the other team exactly once (there are no draws in ping-pong). Prove that there exist ten players, all from the same team, such that every member of the other team has lost his game again... | [
"Solution:\n\nWe start with the following observation: In a match between two teams (not necessarily of equal sizes), there exists in one of the teams a player who won his games with at least half of the members of the other team.\n\nIndeed: suppose there is no such player. If the teams consist of $m$ and $n$ membe... | Baltic Way | Baltic Way 1998 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
06ww | For a positive integer $n$ we denote by $s(n)$ the sum of the digits of $n$. Let $P(x)= x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$ be a polynomial, where $n \geqslant 2$ and $a_{i}$ is a positive integer for all $0 \leqslant i \leqslant n-1$. Could it be the case that, for all positive integers $k$, $s(k)$ and $s(P(k)... | [
"With the notation above, we begin by choosing a positive integer $t$ such that\n$$\n10^{t} > \\max \\left\\{ \\frac{100^{n-1} a_{n-1}}{\\left(10^{\\frac{1}{n-1}} - 9^{\\frac{1}{n-1}}\\right)^{n-1}}, \\frac{a_{n-1}}{9} 10^{n-1}, \\frac{a_{n-1}}{9} (10 a_{n-1})^{n-1}, \\ldots, \\frac{a_{n-1}}{9} (10 a_{0})^{n-1} \\r... | IMO | International Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | No | |
06r3 | On some planet, there are $2^{N}$ countries $(N \geq 4)$. Each country has a flag $N$ units wide and one unit high composed of $N$ fields of size $1 \times 1$, each field being either yellow or blue. No two countries have the same flag.
We say that a set of $N$ flags is diverse if these flags can be arranged into an $... | [
"When speaking about the diagonal of a square, we will always mean the main diagonal.\n\nLet $M_{N}$ be the smallest positive integer satisfying the problem condition. First, we show that $M_{N}>2^{N-2}$. Consider the collection of all $2^{N-2}$ flags having yellow first squares and blue second ones. Obviously, bot... | IMO | 51st IMO Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof and answer | 2^{N-2}+1 | |
0256 | Problem:
Entre 10 e 99, quantos números existem tais que, invertendo a ordem de seus algarismos, obtemos um número maior do que o número original? | [
"Solution:\nDevemos contar os números $ab$ de dois algarismos que têm o algarismo $b$ da unidade maior do que o algarismo $a$ da dezena, ou seja, tais que $b > a$. Se $a = 1$, o algarismo $b$ da unidade pode ser $2, 3, 4, 5, 6, 7, 8$ ou $9$, portanto, temos oito possibilidades. Se $a = 2$, o algarismo $b$ da unidad... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 36 | |
02o1 | A positive integer $n$ is *clowny* if the number obtained by reversing its digits is greater than $n$. For example, $2009$ is clowny because $9002$ is greater than $2009$; however, $2010$ is not clowny because $0102 = 102$ is less than $2010$ and $3443$ is not clowny because it is equal to the number obtained by revers... | [
"Let $(abcd)$ be a four-digit number $a$, $b$, $c$, $d$ being its digits. So a number is clowny if and only if $(dcba) > (abcd)$. This means that either $d > a$ or $d = a$ and $c > b$. Notice that we cannot have both $d = a$ and $b = c$ because it would imply $(dcba) = (abcd)$.\n\nThere are $\\frac{9 \\cdot 8}{2} =... | Brazil | Brazilian Math Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 441 | |
05t6 | Problem:
Soit $ABC$ un triangle dont les angles sont aigus, et soit $D$ le pied de la hauteur issue du sommet $A$. Soit $X$ et $Y$ les deux points du cercle circonscrit à $ABC$ tels que $AX = AY = AD$, et tels que $B$ et $X$ se trouvent du même côté de la droite $(AD)$. Soit $E$ le point d'intersection des droites $(B... | [
"Solution:\n\nDans la suite, on note $\\mathcal{C}_{UVW}$ le cercle circonscrit à un triangle $UVW$. Soit $i$ l'inversion de centre $E$ qui laisse le cercle $\\mathcal{C}_{ABC}$ globalement invariant. Cette inversion échange les points $B$ et $X$, ainsi que les points $C$ et $Y$. Elle échange donc le cercle $\\math... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0k0t | Problem:
Consider a $2 \times 3$ grid where each entry is one of $0$, $1$, and $2$. For how many such grids is the sum of the numbers in every row and in every column a multiple of $3$? One valid grid is shown below.
$$
\left[\begin{array}{lll}
1 & 2 & 0 \\
2 & 1 & 0
\end{array}\right]
$$ | [
"Solution:\n\nAny two elements in the same row fix the rest of the grid, so $3^{2} = 9$."
] | United States | HMMT February 2018 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 9 | |
02vd | Problem:
Uma sequência de números reais $x_{n}$ é uma lista ordenada de reais em que o primeiro número da lista é o termo $x_{1}$, o segundo é o termo $x_{2}$ e assim por diante. Por exemplo, a sequência usual dos números inteiros positivos pode ser descrita como $x_{n}=n$ para todo inteiro positivo $n$. Algumas sequê... | [
"Solution:\n\na) Basta usar a equação dada para $n=2, n=3$ e $n=4$.\n$$\n\\begin{gathered}\nx_{2}=x_{1}+\\frac{1}{x_{1}^{2}}=1+\\frac{1}{1^{2}}=2 \\\\\nx_{3}=x_{2}+\\frac{1}{x_{2}^{2}}=2+\\frac{1}{2^{2}}=\\frac{9}{4} \\\\\nx_{4}=x_{3}+\\frac{1}{x_{3}^{2}}=\\frac{9}{4}+\\frac{1}{\\left(\\frac{9}{4}\\right)^{2}}=\\fr... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | x2 = 2, x3 = 9/4, x4 = 793/324. Moreover, one can take N = 2016^3 to ensure x_N > 2016. | |
01my | Several chess players took part in a chess tournament. Each participant played exactly one game with any other participant. A participant received $1$ point for a win, $0.5$ point for a draw, and $0$ point for a loss. Any two players received different numbers of points and the participant taking the last place receive... | [] | Belarus | Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 9 | |
0d0o | Sequence $(x_1, x_2, \dots)$ is defined as $x_1 = 20$, $x_2 = 12$,
$$
x_{n+2} = x_n + x_{n+1} + 2\sqrt{x_n x_{n+1} + 121},
$$
for $n \ge 1$.
1) Compute $x_{10}$.
2) Determine with justification if every term in the sequence is an integer? | [
"It is clear that $x_3 = 20 + 12 + 2 \\cdot 19 = 70$. We note that\n$$\n\\begin{aligned}\nx_{n+3} &= x_{n+1} + x_{n+2} + 2\\sqrt{x_{n+1}x_{n+2} + 121} \\\\\n&= x_{n+1} + x_{n+2} + 2\\sqrt{x_{n+1}(x_n + x_{n+1} + 2\\sqrt{x_n x_{n+1} + 121}) + 121} \\\\\n&= x_{n+1} + x_{n+2} + 2\\sqrt{x_{n+1}^2 + 2x_{n+1}\\sqrt{x_n x... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | x_10 = 49824; every term is an integer | |
07ih | Triangle *ABC* is an acute-angled triangle with orthocenter *H*. Point *P* is inside triangle $BHC$ such that $3\angle HBC = \angle HPC$ and $3\angle HCB = \angle HPB$. Denote by *X* and *Y*, the reflection of *P* with respect to $BH$ and $CH$, respectively. If *S* is the circumcenter of triangle $AXY$, prove that $\an... | [
"Let's assume that the feet of $A$, $B$, $C$ correspond to points $D$, $E$, $F$, respectively. We apply an inversion with radius $\\sqrt{AH \\cdot AD}$. In this inversion, points $D$, $H$, as well as $F$, $B$, and $E$, $C$, interchange with each other. We aim to prove that point $P$ goes to the nine point center of... | Iran | 40th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Circ... | null | proof only | null | |
0fus | Problem:
Sei $M$ eine Menge mit $n$ Elementen. Bestimme die Anzahl Möglichkeiten, drei Teilmengen $A, B, C$ von $M$ auszuwählen, sodass gilt
$$
\begin{gathered}
A \cap B \neq \emptyset, \quad B \cap C \neq \emptyset, \quad C \cap A \neq \emptyset \\
A \cap B \cap C=\emptyset
\end{gathered}
$$ | [
"Solution:\n\nBetrachte das Venn-Diagramm in der Abbildung. Ein Tripel $(A, B, C)$ wie in der Aufgabe zu wählen, ist dasselbe wie die $n$ Elemente von $M$ so auf die acht Felder in diesem Diagramm zu verteilen, dass das Feld $w$ leer ist, nicht aber die Felder $x, y$ und $z$.\n\nWir zählen zuerst die Anzahl Tripel ... | Switzerland | Vorrundenprüfung | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 7^n - 3*6^n + 3*5^n - 4^n | |
0d5g | Find all integer solutions of the equation $14^{x} - 3^{y} = 2015$. | [
"Notice that $14^{x} \\geq 2015$ which implies that $x \\geq 3$. On the other hand $3^{y} = 14^{x} - 2015 \\geq 14^{3} - 2015 = 729 = 3^{6}$. We deduce that $y \\geq 6$.\n\nWe have $3^{y} = 14^{x} - 2015 \\equiv 1 \\pmod{7}$. But $\\operatorname{Ord}_{7} 3 = 6$. We deduce that $y = 6k$ for some positive integer $k$... | Saudi Arabia | SAMC 2015 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English, Arabic | proof and answer | (x, y) = (3, 6) | |
08fe | Problem:
Tre circonferenze di raggio unitario sono tangenti tra loro e una quarta circonferenza è tangente a tutte e tre, e non le racchiude. Quanto vale il raggio della quarta circonferenza?
(A) $\frac{\sqrt{3}+1}{24}$
(B) $\frac{1}{8}$
(C) $\frac{\sqrt{3}}{12}$
(D) $\frac{2 \sqrt{3}}{3}-1$
(E) $\frac{\sqrt{3}-1}{4}$ | [
"Solution:\n\nLa risposta è (D). Chiamiamo $\\omega_{1}, \\omega_{2}, \\omega_{3}$ le prime tre circonferenze e $\\Omega$ la quarta, e siano $P_{1}, P_{2}, P_{3}, P$ i loro rispettivi centri. Il triangolo $P_{1} P_{2} P_{3}$ è evidentemente equilatero (ognuno dei suoi lati è pari al doppio del raggio delle circonfe... | Italy | Gara di Febbraio | [
"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 > Miscellaneous > Distance cha... | null | MCQ | D | |
09rt | Problem:
Zij $P(x)$ een polynoom met gehele coëfficiënten en graad $n \leq 10$ waarvoor geldt dat er voor elke $k \in \{1,2, \ldots, 10\}$ een gehele $m$ is met $P(m)=k$. Verder is gegeven dat $|P(10)-P(0)|<1000$. Bewijs dat er voor elke gehele $k$ een gehele $m$ is met $P(m)=k$. | [
"Solution:\n\nLaat voor $i=1,2, \\ldots, 10$ het gehele getal $c_{i}$ zo zijn dat $P\\left(c_{i}\\right)=i$. Er geldt voor $i \\in\\{1,2, \\ldots, 9\\}$ dat\n$$\nc_{i+1}-c_{i} \\mid P\\left(c_{i+1}\\right)-P\\left(c_{i}\\right)=(i+1)-i=1\n$$\ndus $c_{i+1}-c_{i}= \\pm 1$ voor alle $i \\in\\{1,2, \\ldots, 9\\}$. Verd... | Netherlands | IMO-selectietoets II | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
00gk | Let $ABC$ be an acute angled triangle with $\angle BAC = 60^\circ$ and $AB > AC$. Let $I$ be the incenter, and $H$ the orthocenter of the triangle $ABC$. Prove that
$$
2 \angle AHI = 3 \angle ABC.
$$ | [
"Let $D$ be the intersection point of the lines $AH$ and $BC$. Let $K$ be the intersection point of the circumcircle $O$ of the triangle $ABC$ and the line $AH$. Let the line through $I$ perpendicular to $BC$ meet $BC$ and the minor arc $BC$ of the circumcircle $O$ at $E$ and $N$, respectively. We have\n$$\n\\angle... | Asia Pacific Mathematics Olympiad (APMO) | XIX Asian Pacific Mathematics Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0hv7 | Problem:
Given five nonnegative real numbers with sum $1$, prove that it is possible to arrange them at the vertices of a regular pentagon such that no two numbers connected by a side of the pentagon have product exceeding $1/9$. | [
"Solution:\n\nLabel the numbers $a, b, c, d, e$ in increasing order. Place them around the pentagon in the order $e, a, d, c, b$. Then it is clear that the products of the numbers on the sides follow the inequalities\n$$\na d \\leq a e \\leq b e \\quad \\text{and} \\quad b c \\leq c d\n$$\nThus it suffices to prove... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof only | null | |
0b75 | a) $\{x \in \mathbb{R} \mid \log_2[x] = [\log_2 x]\} = \bigcup_{m \in \mathbb{N}} [2^m, 2^m + 1)$.
b) $\{x \in \mathbb{R} \mid 2^{\lfloor x \rfloor} = \lfloor 2^x \rfloor\} = \bigcup_{m \in \mathbb{N}} [m, \log_2 (2^m + 1))$.
(here, $[a]$ denotes the integer part (floor function) of the real number $a$). | [
"a) The existence of the logarithms requires $x > 1$.\nIf $[\\log_2 x] = m$, then $m \\in \\mathbb{N}$ and $2^m \\le x < 2^{m+1}$. From $\\log_2[x] = m$, follows $[x] = 2^m$, that is $2^m \\le x < 2^m + 1$.\nConversely, $x \\in [2^m, 2^m + 1)$, $m \\in \\mathbb{N}$ yields $[\\log_2 x] = \\log_2[x] = m$.\n\nb) If $[... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof and answer | a) { x ∈ ℝ | log₂([x]) = [log₂ x] } = ⋃_{m ∈ ℕ} [2^m, 2^m + 1).
b) { x ∈ ℝ | 2^{⌊x⌋} = ⌊2^x⌋ } = ⋃_{m ∈ ℕ} [m, log₂(2^m + 1)). | |
0bnq | Find all perfect squares $\overline{aabcd}$ with $a \neq 0$ and $d \neq 0$ such that the number $\overline{dcbaa}$ is also a perfect square. $(a, b, c, d, e$ are decimal digits) | [
"Answer: 44521 and 44944.\nSuppose $\\overline{aabcd}$ and $\\overline{dcbaa}$ are both squares. Then $a$ and $d$ must belong to $\\{1, 4, 5, 6, 9\\}$.\n\nThe numbers $\\overline{dcb55}$ are multiples of 5, but not of 25, hence they cannot be squares. The numbers $\\overline{dcb66}$ are even but not multiples of 4,... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 44521 and 44944 | |
0e30 | Problem:
Družina je naredila sneženega moža iz treh delov, ki so imeli obliko krogle. Polmeri teh krogel so tvorili geometrijsko zaporedje. Polmer najmajše krogle na vrhu snežaka je bil $8 \mathrm{dm}$, polmer največje krogle pa $18 \mathrm{dm}$. Koliko kubičnih metrov snega je bilo v tem sneženem možu? | [
"Solution:\n\nUpoštevamo lastnosti geometrijskega zaporedja $8 \\cdot q = x$ in $x \\cdot q = 18$. Zapišemo zvezo $8 q^{2} = 18$ in izračunamo količnik $q = \\frac{3}{2}$. Nato izračunamo polmer srednje krogle $r_{2} = 12 \\mathrm{dm}$. Izračunamo prostornino snežaka\n\n$$\nV = \\frac{4}{3} \\pi \\left(8^{3} + 12^{... | Slovenia | 10. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Geometry > Solid Geometry > Volume",
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | final answer only | 33.8119 m^3 | |
08la | Problem:
Kostas and Helene have the following dialogue:
Kostas: I have in my mind three positive real numbers with product $1$ and sum equal to the sum of all their pairwise products.
Helene: I think that I know the numbers you have in mind. They are all equal to $1$.
Kostas: In fact, the numbers you mentioned satisfy ... | [
"Solution:\nKostas is right according to the following analysis:\nIf $x, y, z$ are the three positive real numbers Kostas thought about, then they satisfy the following equations:\n$$\n\\begin{gathered}\nx y + y z + z x = x + y + z \\\\\nx y z = 1\n\\end{gathered}\n$$\nSubtracting (1) from (2) by parts we obtain\n$... | JBMO | 2008 Shortlist JBMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
00hv | Find all integers $n$ satisfying $n \geq 2$ and $\frac{\sigma(n)}{p(n)-1}=n$, in which $\sigma(n)$ denotes the sum of all positive divisors of $n$, and $p(n)$ denotes the largest prime divisor of $n$. | [
"Let $n=p_{1}^{\\alpha_{1}} \\cdot \\ldots \\cdot p_{k}^{\\alpha_{k}}$ be the prime factorization of $n$ with $p_{1}<\\ldots<p_{k}$, so that $p(n)=p_{k}$ and $\\sigma(n)=\\left(1+p_{1}+\\cdots+p_{1}^{\\alpha_{1}}\\right) \\ldots\\left(1+p_{k}+\\cdots+p_{k}^{\\alpha_{k}}\\right)$. Hence\n$p_{k}-1=\\frac{\\sigma(n)}{... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 6 | |
09b1 | Given an acute triangle $ABC$. Let $(\omega, I)$ be the inscribed circle of $ABC$, $(\Omega, O)$ be the circumscribed circle of $ABC$ and $A_0$ be the midpoint of $AH$ altitude. $\omega$ touches $BC$ at point $D$. $A_0D \cap \omega = P$ and the perpendicular from $I$ to $A_0D$ intersects $BC$ at the point $M$. $MR$ and... | [
"Let $\\omega$ touch $AB$ and $AC$ at $E$ and $F$ respectively. $(AD) \\cap \\omega = Q$, $(ID) \\cap \\omega = N$, $(EF) \\cap (BC) = T$. Since $E$, $F$ are points of tangency $(DEQF)$ is harmonic division. This implies $TQ$ is tangent to $\\omega$. Since $AA_0 = A_0H$, $(AA_0H\\infty)$ is harmonic division. From ... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | null | proof only | null | |
0ici | Problem:
Simplify $\prod_{k=1}^{2004} \sin (2 \pi k / 4009)$. | [
"Solution:\n$\\frac{\\sqrt{4009}}{2^{2004}}$\n\nLet $\\zeta = e^{2 \\pi i / 4009}$ so that $\\sin (2 \\pi k / 4009) = \\frac{\\zeta^{k} - \\zeta^{-k}}{2i}$ and $x^{4009} - 1 = \\prod_{k=0}^{4008} (x - \\zeta^{k})$. Hence $1 + x + \\cdots + x^{4008} = \\prod_{k=1}^{4008} (x - \\zeta^{k})$. Comparing constant coeffic... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | sqrt(4009)/2^2004 | |
02cl | Problem:
Cara ou Coroa - Jerônimo joga no tabuleiro ao lado da seguinte maneira: Ele coloca uma peça na casa "PARTIDA" e ele move a peça da seguinte maneira: ele lança uma moeda, se der CARA ele avança duas casas, e se der COROA ele recua uma casa. Jerônimo lançou a moeda 20 vezes e conseguiu chegar na casa CHEGADA. Q... | [
"Solution:\n\n12"
] | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 12 | |
083k | Problem:
Secondo una recente statistica, in Italia una persona ogni $76$ è allergica alle fragole e, tra quelli che lo sono, $2$ su $3$ sono donne. Sulla base di queste informazioni, e supponendo che in Italia il numero di donne sia uguale a quello degli uomini, si può concludere che è allergico alle fragole un uomo o... | [] | Italy | Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO BIENNIO | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | 114 | |
092f | Problem:
Let $N$ be a positive integer. In each of the $N^{2}$ unit squares of an $N \times N$ board, one of the two diagonals is drawn. The drawn diagonals divide the $N \times N$ board into $K$ regions. For each $N$, determine the smallest and the largest possible values of $K$.
Example with... | [
"Solution:\nA small triangle is a right-angled isosceles triangle whose area is $\\frac{1}{2}$ whose hypotenuse is a diagonal of a unit square. A board segment is a horizontal or a vertical segment on the boundary of the board. There are $4 N$ board segments and each of these segments belongs to the boundary of som... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | Minimum K = 2N; Maximum K = ceil(((N+1)^2)/2), i.e., ((N+1)^2 + 1)/2 for even N and ((N+1)^2)/2 for odd N. | |
04z6 | Given a convex quadrangle $ABCD$ with $|AD| = |BD| = |CD|$ and $\angle ADB = \angle DCA$, $\angle CBD = \angle BAC$, find the sizes of the angles of the quadrangle. (Juniors.) | [
"Denote $\\angle ADB = \\angle DCA = \\alpha$ and $\\angle CBD = \\angle BAC = \\beta$ (Fig. 1).\n\nIn triangle $DAC$ we have $|DA| = |DC|$ and therefore $\\angle DAC = \\angle DCA = \\alpha$; analogously in triangles $DAB$ and $DBC$, we have $\\angle DBA = \\angle DAB = \\alpha + \\beta$ and $\\angle DCB = \\angle... | Estonia | Estonija 2010 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | ∠DAB = 75°, ∠ABC = 120°, ∠BCD = 45°, ∠CDA = 120° | |
0apn | Problem:
If $A B C D E F$ is a regular hexagon with each side of length $6$ units, what is the area of $\triangle A C E$? | [
"Solution:\n$27 \\sqrt{3}$ square units\n\nNote that $\\triangle A C E$ is equilateral. Each interior angle of $A B C D E F$ measures $\\frac{1}{6}(6-2)\\left(180^{\\circ}\\right)=120^{\\circ}$. Using a property of a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle, we have\n$$\n\\frac{1}{2} A C=\\frac{\\sqrt{3}}{2} ... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 27√3 | |
08md | Problem:
Determine all four digit numbers $\overline{a b c d}$ such that
$$
a(a+b+c+d)\left(a^{2}+b^{2}+c^{2}+d^{2}\right)\left(a^{6}+2 b^{6}+3 c^{6}+4 d^{6}\right)=\overline{a b c d}
$$ | [
"Solution:\nFrom $\\overline{a b c d}<10000$ and\n$$\na^{10} \\leq a(a+b+c+d)\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)\\left(a^{6}+2 b^{6}+3 c^{6}+4 d^{6}\\right)=\\overline{a b c d}\n$$\nfollows that $a \\leq 2$. We thus have two cases:\n\nCase I: $a=1$.\n\nObviously $2000>\\overline{1 b c d}=(1+b+c+d)\\left(1+b^{2}+... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 2010 | |
0a9p | Problem:
Let $\left(a_{n}\right)_{n \geq 1}$ be a sequence with $a_{1}=1$ and
$$
a_{n+1}=\left\lfloor a_{n}+\sqrt{a_{n}}+\frac{1}{2}\right\rfloor
$$
for all $n \geq 1$, where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$. Find all $n \leq 2013$ such that $a_{n}$ is a perfect square. | [
"Solution:\nWe will show by induction that $a_{n}=1+\\left\\lfloor\\frac{n}{2}\\right\\rfloor\\left\\lfloor\\frac{n+1}{2}\\right\\rfloor$, which is equivalent to $a_{2m}=1+m^{2}$ and $a_{2m+1}=1+m(m+1)$. Clearly this is true for $a_{1}$.\n\nIf $a_{2m+1}=1+m(m+1)$ then\n$$\na_{2m+2}=\\left\\lfloor m^{2}+m+1+\\sqrt{m... | Nordic Mathematical Olympiad | Nordic Mathematical Contest | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | n = 1 | |
0d3r | Let $p \geq 2$ be a prime number and $\frac{a_{p}}{b_{p}} = 1 + \frac{1}{2} + \cdots + \frac{1}{p^{2}-1}$, where $a_{p}$ and $b_{p}$ are two relatively prime positive integers. Compute $\operatorname{gcd}\left(p, b_{p}\right)$. | [
"For $p = 2$ we have $\\frac{a_{2}}{b_{2}} = \\frac{11}{6}$, and therefore $\\operatorname{gcd}\\left(2, b_{2}\\right) = 2$.\n\nFor $p$ an odd prime we propose two solutions to this problem:\n\nFirst solution. We have\n$$\n\\frac{a_{p}}{b_{p}} = \\frac{\\sum_{k=1}^{p^{2}-1} \\frac{(p^{2}-1)!}{k}}{(p^{2}-1)!} .\n$$\... | Saudi Arabia | SAMC | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English, Arabic | proof and answer | gcd(p, b_p) = 2 if p = 2, and gcd(p, b_p) = 1 if p is odd | |
0az0 | Problem:
Find the remainder when $14^{100}$ is divided by $45$. | [
"Solution:\nLet $\\Phi$ denote the Euler's totient function. Since the prime factors of $45$ are $3$ and $5$ only, then\n$$\n\\Phi(45) = 45 \\left(1 - \\frac{1}{3}\\right) \\left(1 - \\frac{1}{5}\\right) = 24\n$$\nSince $\\gcd(14, 45) = 1$, by Euler's Theorem, $14^{24} \\equiv 1 \\pmod{45}$. So that\n$$\n14^{100} =... | Philippines | 20th Philippine Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | final answer only | 31 | |
0hy3 | Problem:
How many pairs of positive integers $(a, b)$ with $a \leq b$ satisfy $\frac{1}{a}+\frac{1}{b}=\frac{1}{6}$? | [
"Solution:\n$\\frac{1}{a}+\\frac{1}{b}=\\frac{1}{6} \\Rightarrow \\frac{a+b}{a b}=\\frac{1}{6} \\Rightarrow a b=6 a+6 b \\Rightarrow a b-6 a-6 b=0$.\nFactoring yields $(a-6)(b-6)=36$.\nBecause $a$ and $b$ are positive integers, only the factor pairs of $36$ are possible values of $a-6$ and $b-6$.\nThe possible pair... | United States | HMMT 1998 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | final answer only | 5 | |
0l97 | Let be given four positive integers $m, n, p, q$, with $p < m$ and $q < n$. Take the four points $A(0; 0)$, $B(p; 0)$, $C(m; q)$ and $D(m; n)$ in the coordinate plane.
Consider the paths $f$ from $A$ to $D$ and the paths $g$ from $B$ to $C$ such that when going along $f$ or $g$, one goes only in the positive direction... | [] | Vietnam | Vietnamese Team Selection Contest for the 44th IMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof only | null | |
05oq | Problem:
On considère 2017 droites du plan, qui se rencontrent deux à deux en des points distincts. On appelle $E$ l'ensemble de ces points d'intersection.
On veut attribuer une couleur à chacun des points de $E$ de sorte que deux quelconques de ces points qui appartiennent à une même droite et dont le segment qui les... | [
"Solution:\n\nLe minimum $m$ cherché est $m=3$ et, avec ce qui suit, il sera assez évident que le résultat reste vrai pour $n \\geq 3$ droites.\n\nTout d'abord, on note que, dans la configuration obtenue, il y a au moins une région non subdivisée qui est un triangle. En effet, trois droites non concourantes et deux... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Algorithms",
"Geometry > Plane Geometry > Analytic / Coordinate Methods >... | null | proof and answer | 3 | |
0gyz | Consider an arbitrary arrangement of the brackets. Numbers $2010$ and $2009$ always have signs '+' and '-', respectively, that's why the sum reaches the maximum value when the other terms will have '+' sign.
In the expression
$$
\frac{2010 - 2009 - 2010 - 2009 - 2010 - 2009 - \dots - 2010 - 2009}{2010 \text{ numbers}}... | [
"$$\n\\frac{2010 - (2009 - 2010 - 2009 - 2010 - 2009 - \\dots - 2010 - 2009)}{2010 \\text{ numbers}} = 4035077.\n$$\n\n**Solution.** Consider an arbitrary arrangement of the brackets. The first numbers $2010$ and $2009$ always have signs '+' and '-', respectively, that's why the sum reaches the maximum value when o... | Ukraine | 50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010) | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 4035077 | |
044z | Sequence $\{a_n\}$ satisfies $a_1 = a_2 = a_3$. Let
$$
b_n = a_n + a_{n+1} + a_{n+2} \quad (n \in \mathbb{N}_+).
$$
If $\{b_n\}$ is a geometric sequence with common ratio $3$, find the value of $a_{100}$. | [
"By the condition, we know that $b_n = b_1 \\cdot 3^{n-1} = 3^n$ $(n \\in \\mathbb{N}_+)$.\nThus,\n$$\na_{n+3} - a_n = b_{n+1} - b_n = 3^{n+1} - 3^n = 2 \\cdot 3^n \\quad (n \\in \\mathbb{N}_+).\n$$\nTherefore,\n$$\n\\begin{align*}\na_{100} &= a_1 + \\sum_{k=1}^{33} (a_{3k+1} - a_{3k-2}) \\\\\n&= 1 + \\sum_{k=1}^{3... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | (3^{100} + 10)/13 | |
0cdt | Given a positive even integer $a$ which is not a power of $2$, prove that at least one of the numbers $2^{2^n} + 1$ and $a^{2^n} + 1$ is composite for infinitely many non-negative integers $n$. | [
"Write $f_n = 2^{2^n} + 1$ and $a_n = a^{2^n} + 1$. Suppose now, if possible, that $f_n$ and $a_n$ are both prime for all but finitely many non-negative integers $n$, say, for all $n > N$. Clearly, we may and will assume that $f_n > a$ for all $n > N$.\n\nFix any $n > N$ and consider the (multiplicative) order of $... | Romania | THE Sixteenth STARS OF MATHEMATICS Competition | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0a4j | Problem:
How many positive integers less than $2019$ are divisible by either $18$ or $21$, but not both? | [
"Solution:\n\nFor any positive integer $n$, the number of multiples of $n$ less than or equal to $2019$ is given by\n$$\\left\\lfloor \\frac{2019}{n} \\right\\rfloor.$$\nSo there are $\\left\\lfloor \\frac{2019}{18} \\right\\rfloor = 112$ multiples of $18$, and $\\left\\lfloor \\frac{2019}{21} \\right\\rfloor = 96$... | New Zealand | NZMO Round One | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | final answer only | 176 | |
0lgf | Problem:
On a polynomial of degree three it is allowed to perform the following two operations arbitrarily many times:
(i) reverse the order of its coefficients including zeroes (for instance, from the polynomial $x^{3}-2 x^{2}-3$ we can obtain $-3 x^{3}-2 x+1$);
(ii) change polynomial $P(x)$ to the polynomial $P(x+1)... | [
"Solution:\n\nThe answer is no.\n\nSolution I. The original polynomial $x^{3}-2$ has a unique real root. The two transformations clearly preserve this property. If $\\alpha$ is the only real root of $P(x)$, then the first operation produces a polynomial with root $\\frac{1}{\\alpha}$, and the second operation gives... | Zhautykov Olympiad | XV International Zhautykov Olympiad in Mathematics | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
0iyu | Problem:
Five guys each have a positive integer (the integers are not necessarily distinct). The greatest common divisor of any two guys' numbers is always more than $1$, but the greatest common divisor of all the numbers is $1$. What is the minimum possible value of the product of the numbers? | [
"Solution:\n\nAnswer: $32400$\n\nLet $\\omega(n)$ be the number of distinct prime divisors of a number. Each of the guys' numbers must have $\\omega(n) \\geq 2$, since no prime divides all the numbers. Therefore, if the answer has prime factorization $p_{1}^{e_{1}} p_{2}^{e_{2}} \\ldots p_{k}^{e_{k}}$, then $e_{1}+... | United States | 2nd Annual Harvard-MIT November Tournament | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 32400 | |
05kz | Problem:
Soient $p$ et $q$ deux nombres premiers supérieurs ou égaux à $7$. Soit $x=\frac{p^{2012}+q^{2016}}{120}$. Calculer $x-[x]$, où $[x]$ désigne la partie entière de $x$, c'est-à-dire le plus grand entier inférieur ou égal à $x$. | [
"Solution:\n\nSi $a$ est un entier non divisible par $2$, $3$ et $5$, on vérifie facilement que $a^{4}$ est congru à $1$ modulo $3$, $5$ et $8$. Autrement dit, $a^{4}-1$ est divisible par $3$, $5$ et $8$, donc par $3 \\times 5 \\times 8 = 120$. On en déduit que $p^{4}$ et $q^{4}$ sont congrus à $1$ modulo $120$, do... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof and answer | 1/60 | |
0bw0 | Determine the 5-digit positive integers which are four times larger than their reversed numbers (the reversed of $abcde$ is $edcba$). | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | 87912 |
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