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values | competition stringlengths 3 108 ⌀ | topics_flat listlengths 0 12 | language stringclasses 18
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|---|---|---|---|---|---|---|---|---|---|
0c80 | A rectangular piece of paper of dimensions $20 \times 19$, divided into unit squares, is cut into several square pieces, the cuttings being made along the sides of the unit squares. Such a square piece is called an *odd square* if the length of its side is an odd number.
a) What is the minimum possible number of odd s... | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | a) 4; b) 80 | |
0bs0 | Let $n \ge 2$ be a positive integer and let $a_1, a_2, \dots, a_n$ be positive numbers such that $a_1 \le a_2$, $a_1 + a_2 \le a_3$, $a_1 + a_2 + a_3 \le a_4$, $\dots$, $a_1 + a_2 + \dots + a_{n-1} \le a_n$. Prove that
$$
\frac{a_1}{a_2} + \frac{a_2}{a_3} + \frac{a_3}{a_4} + \dots + \frac{a_{n-1}}{a_n} \le \frac{n}{2}.... | [
"Denote $x_1 = a_1$, $x_k = a_k - (a_{k-1} + \\dots + a_1)$, $k = \\overline{2, n}$, and observe that $x_{k+1} - x_k = a_{k+1} - 2a_k$, $k = \\overline{1, n-1}$. It results\n$$\n2 \\left( \\frac{a_1}{a_2} + \\frac{a_2}{a_3} + \\frac{a_3}{a_4} + \\dots + \\frac{a_{n-1}}{a_n} \\right) = \\sum_{i=1}^{n-1} \\left( 1 - ... | Romania | 67th Romanian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation"
] | English | proof only | null | |
05wd | Problem:
Soit $ABC$ un triangle tel que $\widehat{CAB} < \widehat{ABC} < \widehat{BCA} < 90^{\circ}$. Soit $\omega$ le cercle circonscrit à $ABC$, $\gamma_{a}$ le cercle de centre $A$ et de rayon $[AC]$, et $\gamma_{b}$ le cercle de centre $B$ et de rayon $[BC]$. Enfin, soit $D$ le point d'intersection, autre que $C$,... | [
"Solution:\n\n\n\nPar construction, $(CE)$ est l'axe radical des cercles $\\gamma_{a}$ et $\\gamma_{b}$, donc $C$ et $E$ sont symétriques l'un de l'autre par rapport à $(AB)$. En outre, on sait que $BC = BD = BE$. On dispose donc de multiples égalités d'angles et de longueurs, et la manière... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09r2 | Problem:
Bepaal alle paren positieve gehele getallen $(x, y)$ waarvoor
$$
x^{3}+y^{3}=4\left(x^{2} y+x y^{2}-5\right) .
$$ | [
"Solution:\nOplossing I. We kunnen de vergelijking als volgt herschrijven:\n$$\n(x+y)\\left(x^{2}-x y+y^{2}\\right)=4 x y(x+y)-20\n$$\nNu is $x+y$ een deler van de linkerkant en van de eerste term rechts, dus ook van de tweede term rechts: $x+y \\mid 20$. Omdat $x+y \\geq 2$, geeft dit voor $x+y$ de mogelijkheden $... | Netherlands | Dutch TST | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | [(1,3), (3,1)] | |
0i69 | Problem:
Prove for integers $n$ that
$$
\left\lfloor\frac{n}{2}\right\rfloor\left\lfloor\frac{n+1}{2}\right\rfloor=\left\lfloor\frac{n^{2}}{4}\right\rfloor .
$$ | [
"Solution:\nSuppose $n=2m$ is even; then $\\lfloor n / 2\\rfloor = \\lfloor m\\rfloor = m$ and $\\lfloor(n+1) / 2\\rfloor = \\lfloor m+1 / 2\\rfloor = m$, whose product is $m^{2} = \\left\\lfloor m^{2}\\right\\rfloor = \\left\\lfloor (2m)^{2} / 4 \\right\\rfloor$.\n\nOtherwise $n=2m+1$ is odd, so that $\\lfloor n /... | United States | HMMT 2002 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
07ya | Problem:
Diciamo che due polinomi a coefficienti interi $p$ e $q$ sono simili se hanno lo stesso grado e gli stessi coefficienti a meno dell'ordine.
a. Dimostrare che se $p$ e $q$ sono simili, allora $p(2007)-q(2007)$ è un multiplo di $2$.
b. Esistono degli interi $k>2$ tali che, comunque siano dati due polinomi sim... | [
"Solution:\n\na. Poiché $2007$ è un numero dispari, il valore di un polinomio in $2007$ è pari o dispari a seconda che il numero dei suoi coefficienti dispari sia pari o dispari. Ma se $p$ e $q$ sono simili, allora in particolare contengono lo stesso numero di coefficienti dispari, e quindi $p(2007)$ e $q(2007)$ so... | Italy | null | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | k = 2006 | |
0bxr | Consider the isosceles triangle $ABC$, with $m(\angle BAC) = 100^\circ$. Let $BD$ be the angle bisector of the angle $\widehat{ABC}$, with $D \in (AC)$, the point $E \in BD$ such that $D \in (BE)$ and $BE = BC$, and the point $F \in (BC)$ such that $AB = BF$. Prove that the lines $AC$ and $EF$ are orthogonal.
![](atta... | [
"Triangles $ABD$ and $FBD$ are congruent (S.A.S.), so that $m(\\angle FDB) = m(\\angle ADB) = 60^\\circ$, and $m(\\angle FDC) = 60^\\circ$.\n\nTriangle $EBC$ is isosceles ($BE = BC$), with $m(\\angle EBC) = 20^\\circ$, hence $m(\\angle BCE) = 80^\\circ$, and from the hypothesis we have $m(\\angle ACB) = 40^\\circ$,... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0al5 | At a round table are seated $n$ boys and $n$ girls, where $n > 3$. In every move, it is allowed to swap the sitting places of two adjacent children. The entropy of a sitting arrangement is the minimum number of moves resulting with each child having at least one neighbor of the same gender. Find the maximum possible en... | [] | North Macedonia | Macedonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof and answer | ceil(n/2) | |
02i6 | Problem:
Se um arco de $60^{\circ}$ num círculo I tem o mesmo comprimento que um arco de $45^{\circ}$ num círculo II, então a razão entre a área do círculo I com a do círculo II é:
(A) $16/9$
(B) $9/16$
(C) $4/3$
(D) $3/4$
(E) $6/9$ | [
"Solution:\n\nComo o arco de $60^{\\circ}$ do círculo I tem o mesmo comprimento que o arco de $45^{\\circ}$ no círculo II, concluímos que o raio do círculo I é menor que o do círculo II. Denotemos por $r$ e $R$ os raios dos círculos I e II respectivamente.\n\nNo círculo I o comprimento do arco de $60^{\\circ}$ é ig... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles"
] | null | MCQ | B | |
0i3r | Problem:
Point $D$ is drawn on side $B C$ of equilateral triangle $A B C$, and $A D$ is extended past $D$ to $E$ such that angles $E A C$ and $E B C$ are equal. If $B E = 5$ and $C E = 12$, determine the length of $A E$. | [
"Solution:\n\nBy construction, $A B E C$ is a cyclic quadrilateral. Ptolemy's theorem says that for cyclic quadrilaterals, the sum of the products of the lengths of the opposite sides equals the product of the lengths of the diagonals. This yields $$(B C)(A E) = (B A)(C E) + (B E)(A C).$$ Since $A B C$ is equilater... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 17 | |
0b9z | Let $(K, +, \cdot)$ be a finite field. Prove that:
a) if $K$ has $4k+1$ elements, then the polynomial $f = X^4 + 4$ has four roots in $K$;
b) the polynomial $g = X^8 - 16$ has at least a root in $K$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Algebra > Abstract Algebra > Field Theory",
"Algebra > Abstract Algebra > Group Theory",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0fvh | Problem:
Sei $n$ eine natürliche Zahl. Jede der Zahlen $\{1,2, \ldots, n\}$ ist weiss oder schwarz gefärbt. Man kann nun wiederholt eine Zahl auswählen und diese, sowie alle zu ihr nicht teilerfremden Zahlen umfärben. Anfangs sind alle Zahlen weiss. Für welche $n$ kann man erreichen, dass irgendwann alle Zahlen schwar... | [
"Solution:\n\nDies ist immer möglich. Wir verwenden Induktion nach $n$, der Fall $n=1$ ist klar. Nehme an, dies sei richtig für $n$ und betrachte die Zahlen $1, \\ldots, n+1$. Durch umfärben von gewissen Zahlen können wir annehmen, dass die Zahlen $1, \\ldots, n$ alle schwarz sind. Wir unterscheiden nun zwei Fälle ... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | all natural numbers | |
0djk | Given two circles $(O_1), (O_2)$ with different radii and intersecting at two points $A, B$. The tangent $CD$ of the two circles (closer to point $B$) with $C \in (O_1)$ and $D \in (O_2)$. Draw diameters $BP, BQ$ of $(O_1), (O_2)$. The line through $B$, and perpendicular to $CD$ cuts $PQ$ at $K$.
a) Prove that $B$ is ... | [
"a) Redefine the point $K$ as the orthocenter of the triangle $BCD$, then $B$ is also the orthocenter of the triangle $KCD$, we will show that $K \\in PQ$. Construct the parallelogram $BCDT$.\n\n\n\nSince $K$ is the orthocenter of triangle $BCD$, $BC \\perp KD$, and $BC \\parallel TD$ so $T... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"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 > Miscellaneous > Angle c... | English | proof only | null | |
09qp | Problem:
Bepaal alle gehele getallen $n$ waarvoor het polynoom $P(x)=3 x^{3}-n x-n-2$ te schrijven is als het product van twee niet-constante polynomen met gehele coëfficiënten. | [
"Solution:\n\nOplossing I. Stel dat $P(x)$ te schrijven is als $P(x)=A(x) B(x)$ met $A$ en $B$ niet-constante polynomen met gehele coëfficiënten. Omdat $A$ en $B$ niet constant zijn, hebben ze elk graad minstens 1. De som van de twee graden is gelijk aan de graad van $P$, dus gelijk aan 3. Dit betekent dat de twee ... | Netherlands | Dutch TST toets 8 juni | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Facto... | null | proof and answer | [-2, 26, 38, 130] | |
01lv | The lengths of the medians from the vertices of the acute angles of the right-angled triangle are equal to $19$ and $22$.
Find the length of the hypotenuse of the triangle. | [] | Belarus | 61st Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 26 | |
07oe | Over a period of $k$ consecutive days, a total of 2014 babies were born in a certain city, with at least one baby being born each day. Show that:
1. If $1014 < k \le 2014$, there must be a period of consecutive days during which exactly 100 babies were born.
2. By contrast, if $k = 1014$, such a period might not exist. | [
"Let $N_0 = 0$. For $1 \\le i \\le k$, let $N_i$ be the number of babies born on or before Day $i$, and let $n_i = N_i - N_{i-1}$ be the number of babies born on Day $i$. Let $S = \\{N_1, \\dots, N_k\\}$ and $T = \\{N_1 + 100, \\dots, N_k + 100\\}$. Because at least one baby is born each day, both sets contain $k$ ... | Ireland | Irska 2014 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0fkq | Problem:
Demuestra que $2222^{5555} + 5555^{2222}$ es múltiplo de $7$. | [
"Solution:\nTenemos las siguientes congruencias módulo $7$\n$$\n\\begin{aligned}\n2222^{0} &\\equiv 1, \\\\\n2222^{1} &\\equiv 3, \\\\\n2222^{2} &\\equiv 2, \\\\\n2222^{3} &\\equiv 6, \\\\\n2222^{4} &\\equiv 4, \\\\\n2222^{5} &\\equiv 5, \\\\\n2222^{6} &\\equiv 1, \\ldots \\\\\n5555^{0} &\\equiv 1, \\\\\n5555^{1} &... | Spain | FASE LOCAL DE LA XLIV OME | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof only | null | |
043f | Suppose $a, b \in \mathbb{R}$. If equation
$$(z^2 + az + b)(z^2 + az + 2b) = 0$$
about $z$ has four mutually different complex roots $z_1, z_2, z_3, z_4$ and their corresponding points in the complex plane are exactly four vertices of a square with side length $1$, then find the value of $|z_1| + |z_2| + |z_3| + |z_4... | [
"Denote quadratic equations $E_1: z^2 + az + b = 0$, $E_2: z^2 + az + 2b = 0$. Let $z_1, z_2$ be solutions of $E_1$ and $z_3, z_4$ be solutions of $E_2$.\nIf $z_1, z_2, z_3, z_4$ are all real numbers, then their corresponding points on the complex plane are all on the real axis, which is not consistent with the que... | China | China Mathematical Competition | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | sqrt(6) + 2 sqrt(2) | |
05me | Problem:
Soit $A$ un ensemble de 13 entiers entre 1 et 37. Montrer qu'il existe quatre nombres deux à deux distincts dans $A$ tels que la somme de deux d'entre eux est égale à la somme des deux autres. | [
"Solution:\n\n$A$ contient 13 entiers, donc il y a $\\frac{13 \\times 12}{2} = 78$ manières de choisir deux nombres de $A$ différents (13 manières de choisir le premier, 12 manières de choisir le deuxième et on divise par 2 car on a compté deux fois chaque ensemble de 2 nombres).\n\nOr, la somme de deux nombres de ... | France | OCympiades Françaises de Mathématiques - Envoi Numéro 4 - Combinatoire | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
01jv | Find all pairs of positive integers $a$ and $b$ such that
$$
ab = 160 + 90(a, b),
$$
where $(a, b)$ is the greatest common divisor of $a$ and $b$. | [
"By condition it follows that one of the numbers is divisible by $5$. Moreover, exactly one of the numbers is divisible by $5$, otherwise $(a, b)$ and $90(a, b)$ are divisible by $25$, and so $160$ is divisible by $25$, a contradiction. Let without loss of generality $a \\neq 5$, $b \\neq 5$. Then $a = 5c$, $(a, b)... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (125, 2), (2, 125), (250, 1), (1, 250), (10, 34), (34, 10), (170, 2), (2, 170) | |
0jyl | Problem:
Welcome to the USAYNO, where each question has a yes/no answer. Choose any subset of the following six problems to answer. If you answer $n$ problems and get them all correct, you will receive $\max (0,(n-1)(n-2))$ points. If any of them are wrong, you will receive 0 points.
Your answer should be a six-charac... | [] | United States | February 2017 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence rela... | null | MCQ | null | |
06d0 | On a table there are $1999$ tea cups with their mouths facing upward initially. In each move, $100$ of them are turned upside down. After a number of moves, can they be turned so that all their mouths face downward? Why?
Answer the above two questions for the case where the number of cups is $1998$. | [
"No. In a move, if we choose $100$ cups where $k$ of them are facing downward and $100-k$ of them are facing upward, then the number of cups facing downward is changed by $(100-k) - k = 100 - 2k$. Since there is an even number of cups facing downward initially, there is always an even number of cups facing downward... | Hong Kong | HKG TST | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 1999: No. 1998: Yes. | |
05ou | Problem:
Soit $P(x) = x^{4} - x^{3} - 3 x^{2} - x + 1$. Montrer qu'il existe une infinité d'entiers $n$ tels que $P\left(3^{n}\right)$ ne soit pas premier. | [
"Solution:\n\nOn observe que $3^{2} \\equiv -1 \\pmod{5}$ et $3^{4} \\equiv 1 \\pmod{5}$. Soit $n \\geqslant 1$. Soit $x = 3^{4n+1}$, alors $x = \\left(3^{4}\\right)^{n} \\times 3 \\equiv 3 \\pmod{5}$, donc $P(x) \\equiv 3^{4} - 3^{3} - 3^{2} - 3 + 1 \\equiv 1 + 3 + 3 - 3 + 1 \\equiv 0 \\pmod{5}$.\n\nD'autre part, ... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0c1n | On a blackboard one wrote the numbers $1, 2, 3, \dots, 27$. One step consists in erasing three numbers $a, b, c$ from the blackboard and writing instead the number $a + b + c + n$, where $n$ is a fixed positive integer. Determine $n$ knowing that, after 13 steps, the number $n^2$ is written on the blackboard. | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 27 | |
05o7 | Problem:
Trouver les couples d'entiers $(x, y) \in \mathbb{Z}$ solutions de l'équation $y^{2}=x^{5}-4$. | [
"Solution:\n\nPar le petit théorème de Fermat, pour tout entier $x$ premier avec $11$, $x^{10} \\equiv 1[11]$. Donc $11$ divise $x^{10}-1=\\left(x^{5}-1\\right)\\left(x^{5}+1\\right)$. Par le lemme de Gauss, $11$ divise $x^{5}-1$ ou $x^{5}+1$. Donc pour tout entier $x$, $x^{5}-4 \\equiv -5[11]$ ou $x^{5}-4 \\equiv ... | France | Envoi 1: Arithmétique | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | No integer solutions. | |
045x | Given a positive integer $n$, let $D$ denote the set of all positive divisors of $n$. Let $A$ and $B$ be subsets of $D$ satisfying: for any $a \in A$ and $b \in B$, we have $a \nmid b$ and $b \nmid a$. Prove that
$$
\sqrt{|A|} + \sqrt{|B|} \le \sqrt{|D|}.
$$ | [
"**Proof:** Decompose $D$ into the following disjoint unions $D = X \\sqcup Y \\sqcup Z \\sqcup W$, where\n$$\nX = \\{x \\in D : \\exists a | x, \\exists b | x\\}, \\quad Y = \\{x \\in D : \\exists a | x, \\nexists b | x\\},\n$$\n$$\nZ = \\{x \\in D : \\nexists a | x, \\exists b | x\\}, \\quad W = \\{x \\in D : \\n... | China | 2022 China Team Selection Test | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Discrete Mathematics > Combi... | null | proof only | null | |
046b | As shown below, a square grid of side length $8$ is constructed by $144$ sticks of length $1$. Find the least number of sticks to be removed so that the resulting figure contains no rectangle.
 | [
"The answer is $43$.\n\nFirst, we prove that at least $43$ sticks must be removed. Suppose the figure does not contain any rectangles, then each bounded connected area must consist of at least three unit squares, i.e., the area is at least $3$. Therefore, there can be at most $\\lfloor \\frac{64}{3} \\rfloor = 21$ ... | China | 22nd Chinese Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 43 | |
08jt | Problem:
Prove that, for all real numbers $x, y, z$ :
$$
\frac{x^{2}-y^{2}}{2 x^{2}+1}+\frac{y^{2}-z^{2}}{2 y^{2}+1}+\frac{z^{2}-x^{2}}{2 z^{2}+1} \leq (x+y+z)^{2}
$$
When does equality hold? | [
"Solution:\nFor $x = y = z = 0$ the equality is valid.\nSince $(x + y + z)^{2} \\geq 0$ it is enough to prove that\n$$\n\\frac{x^{2}-y^{2}}{2 x^{2}+1}+\\frac{y^{2}-z^{2}}{2 y^{2}+1}+\\frac{z^{2}-x^{2}}{2 z^{2}+1} \\leq 0\n$$\nwhich is equivalent to the inequality\n$$\n\\frac{x^{2}-y^{2}}{x^{2}+\\frac{1}{2}}+\\frac{... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Equality holds only when x = y = z = 0. | |
0iji | Problem:
Suppose $ABC$ is a triangle with incircle $\omega$, and $\omega$ is tangent to $\overline{BC}$ and $\overline{CA}$ at $D$ and $E$ respectively. The bisectors of $\angle A$ and $\angle B$ intersect line $DE$ at $F$ and $G$ respectively, such that $BF=1$ and $FG=GA=6$. Compute the radius of $\omega$. | [
"Solution:\nLet $\\alpha, \\beta, \\gamma$ denote the measures of $\\frac{1}{2} \\angle A, \\frac{1}{2} \\angle B, \\frac{1}{2} \\angle C$, respectively. We have $m \\angle CEF = 90^\\circ - \\gamma$, $m \\angle FEA = 90^\\circ + \\gamma$, $m \\angle AFG = m \\angle AFE = 180^\\circ - \\alpha - (90^\\circ + \\gamma... | United States | Harvard-MIT Mathematics Tournament | [
"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 > Miscellaneous > Angle c... | null | proof and answer | 2√5/5 | |
0cpm | Determine whether there exists a real $\alpha$ such that $\cos \alpha$ is irrational, while all the numbers $\cos 2\alpha$, $\cos 3\alpha$, $\cos 4\alpha$, $\cos 5\alpha$ are rational.
Существует ли такое вещественное $\alpha$, что число $\cos \alpha$ иррационально, а все числа $\cos 2\alpha$, $\cos 3\alpha$, $\cos 4\... | [
"Не существует.\n\nПредположим противное. Тогда число $A = \\cos \\alpha + \\cos 5\\alpha$ иррационально как сумма рационального и иррационального; с другой стороны, $A = 2 \\cos 2\\alpha \\cos 3\\alpha$ рационально как произведение трёх рациональных чисел. Противоречие."
] | Russia | Russian Mathematical Olympiad | [
"Precalculus > Trigonometric functions"
] | English, Russian | proof and answer | Does not exist | |
04q4 | Find all pairs $(a, b)$ of real (not necessarily positive) numbers such that $a^2 + b^2 = 25$, for which $ab + a + b$ attains the smallest possible value. | [
"Transforming the inequality $(a + b + 1)^2 \\ge 0$ yields $a^2 + b^2 + 1 + 2ab + 2a + 2b \\ge 0$, from which it follows that $2(ab + a + b) \\ge -(a^2 + b^2) - 1$, i.e. $ab + a + b \\ge -13$.\nThe equality is attained if and only if $a + b + 1 = 0$, i.e. $b = -a - 1$. Plugging this into $a^2 + b^2 = 25$ gives us\n... | Croatia | Croatian Junior Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | (-4, 3) and (3, -4) | |
031z | Problem:
Find all polynomials $P(x)$ with integer coefficients such that for any positive integer $n$ the equation $P(x) = 2^{n}$ has an integer solution. | [
"Solution:\nDenote by $m$ and $a$ the degree and the leading coefficient of $P(x)$, respectively. Let $x_{n}$ be an integer solution of the equation $P(x) = 2^{n}$. Since $\\lim_{n \\rightarrow \\infty} |x_{n}| = +\\infty$, then\n$$\n\\lim_{n \\rightarrow \\infty} \\frac{a |x_{n}|^{m}}{2^{n}} = 1\n$$\nand hence\n$$... | Bulgaria | 52. Bulgarian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All such polynomials are P(x) = a(x + b) with a ∈ {±1, ±2} and b any integer. | |
03h2 | Problem:
Let $a$ and $b$ be distinct real numbers. Prove that there exist integers $m$ and $n$ such that $a m + b n < 0$, $b m + a n > 0$. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
04pk | We define the weight of a pair of numbers $\{a, b\}$ as $|a-b|$. In how many ways can the set $\{1, 2, \dots, 12\}$ be divided into six pairs so that the total sum of weights of all pairs equals $30$?
(Japan 2018) | [] | Croatia | Croatian Mathematical Society Competitions | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 1104 | |
0fgy | Problem:
Calcular, para cualquier valor del parámetro entero $t$, soluciones enteras $x$, $y$ de la ecuación
$$
y^{2}=x^{4}-22 x^{3}+43 x^{2}+858 x+t^{2}+10452(t+39)
$$ | [
"Solution:\nEscribimos la expresión\n$$\ny^{2}=x^{4}-22 x^{3}+43 x^{2}+858 x+t^{2}+10452(t+39)\n$$\nen la siguiente forma:\n$$\ny^{2}=(x^{2}-11 x-5226)(x^{2}-11 x+5148)+(t+5226)^{2}\n$$\nLa expresión $x^{2}-11 x-5226$ se anula para $x=-67$ y para $x=78$, lo que da soluciones enteras $y= \\pm(t+5226)$."
] | Spain | OME 24 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | For every integer t, solutions include (x, y) = (-67, t + 5226), (-67, -(t + 5226)), (78, t + 5226), (78, -(t + 5226)). | |
08q4 | Problem:
Let $ABC$ be a right-angled triangle with $\hat{A}=90^{\circ}$ and $\hat{B}=30^{\circ}$. The perpendicular at the midpoint $M$ of $BC$ meets the bisector $BK$ of the angle $\hat{B}$ at the point $E$. The perpendicular bisector of $EK$ meets $AB$ at $D$. Prove that $KD$ is perpendicular to $DE$. | [
"Solution:\nLet $I$ be the incenter of $ABC$ and let $Z$ be the foot of the perpendicular from $K$ on $EC$. Since $KB$ is the bisector of $\\hat{B}$, then $\\angle EBC=15^{\\circ}$ and since $EM$ is the perpendicular bisector of $BC$, then $\\angle ECB=\\angle EBC=15^{\\circ}$. Therefore $\\angle KEC=30^{\\circ}$. ... | JBMO | Junior Balkan 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 > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane... | null | proof only | null | |
0g8o | 給定一個大於 1 的正整數 $k$。甲、乙兩人玩以下的數字遊戲：在遊戲開始時，有一個正整數 $n \ge k$ 被寫在黑板上。接著，從甲開始，兩人輪流進行以下動作：擦掉寫在黑板上的數 $m$，並在黑板上寫下一個與 $m$ 互質的正整數 $m'$，且 $k \le m' < m$。第一個無法寫下數字的人輸。
對於一開始在黑板上的數字 $n \ge k$, 如果乙有必勝法, 則稱 $n$ 是個好數字; 反之, $n$ 是個壞數字。
現在, 假設 $n, n' \ge k$, 且質數 $p \le k$ 整除 $n$ 若且唯若 $p$ 整除 $n'$。試證: $n$ 和 $n'$ 要不同時是好數字, 要不同時是壞數字。
Fix an i... | [
"為方便說明, 令 $n \\to x$ 表示擦掉 $n$, 寫上 $x$ 的動作; 依題意, 必有 $n > x \\ge k$ 且 $(n, x) = 1$.\n\n**Claim A.** 若 $m$ 為好數字, 而 $n > m$ 與 $m$ 互質, 則 $n$ 為壞數字。\n*Proof.* 因為甲只要選 $n \\to m$, 並複製乙從 $m$ 開始的必勝策略即可。\n\n**Claim B.** 任何兩個好數字不能互質。\n*Proof.* 由 Claim A 和 B 立得。\n\nClaim 1. 若 $n$ 為好數字且 $n|n'$. 則 $n'$ 為好數字。\n*Proof.* 若 $n'$ 為壞數字,... | Taiwan | 14-2J-M1 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0k0x | Problem:
Do there exist positive irrational numbers $x$ and $y$ such that $x+y$ and $x y$ are both rational? If so, give an example; if not, explain why not. | [
"Solution:\n\nSuch numbers do exist. One example is $x = 3 - \\sqrt{2}$ and $y = 3 + \\sqrt{2}$ (which are irrational due to the famous fact that $\\sqrt{2}$ is irrational). Then $x + y = 6$ and $x y = 7$."
] | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | x = 3 - sqrt(2), y = 3 + sqrt(2) | |
0bx7 | Given $x_1, x_2, \dots, x_n$ real numbers, prove that there exists a real number $y$ such that
$$
\{y - x_1\} + \{y - x_2\} + \dots + \{y - x_n\} \le \frac{n-1}{2}.
$$ | [
"As $\\{a\\} + \\{-a\\} \\le 1, \\forall a \\in \\mathbb{R}$ (with equality if $a$ is not an integer), we have\n$$\n\\sum_{1 \\le i \\ne j \\le n} \\{x_i - x_j\\} \\le \\frac{n(n-1)}{2},\n$$"
] | Romania | THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0lab | Determine the number of positive integers that are multiple of $9$ with at most $2008$ decimal digits, and among these digits at least two are $9$. | [] | Vietnam | Vijetnam 2008 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | (10^2008 + 8)/9 - 2017 * 9^2006 | |
0bt4 | Let $a \ge 2$ be a positive integer. Prove that the following statements are equivalent:
a) One can find positive integers $b, c$, such that $a^2 = b^2 + c^2$.
b) One can find a positive integer $d$, such that the equations $x^2 - a x + d = 0$ and $x^2 - a x - d = 0$ have integer roots. | [
"Let us suppose that $a^2 = b^2 + c^2$. The numbers $b$ and $c$ could not be both odd (the sum of two odd numbers is $4k + 2$ which is not a square). Then at least one is an even number and thus, the product $bc$ is even.\n\nOn the other hand, the discriminants of the two equations are $\\Delta_1 = a^2 - 4d$ and $\... | Romania | 67th Romanian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof only | null | |
05h6 | Problem:
Trouver tous les couples d'entiers strictement positifs $\left(m, n\right)$ pour lesquels :
$$
1+2^{n}+3^{n}+4^{n}=10^{m}
$$ | [
"Solution:\nCette équation est valable pour tous $n$, $m$, elle est donc valable en la passant modulo un entier $k$, c'est-à-dire en ne considérant que les restes de la division par rapport à $k$.\nOn commence par la regarder modulo $3$ :\n$$\n1+(-1)^{n}+0+1^{n} \\equiv 1^{m} \\pmod{3} \\text{ donc } (-1)^{n} \\equ... | France | ENVOi 3 : ARITHMÉTIQUE | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (m, n) = (1, 1) and (2, 3) | |
00vn | A social network has $2025$ users. Two different users are either friends or not friends. A user is considered *lonely* if they have no friends. Initially, there are no lonely users, and two users Alice and Bob are not friends. A user may swap all their friends, meaning if they were friends with someone before the swap... | [
"Let a longest path of the graph be $P = V_1V_2 \\dots V_k$.\n\n**Case 1:** $P$ consists of all vertices and there is no edge between $V_1$ and $V_k$.\nSwap vertices $V_i$ increasing $i$ from $1$ to $k$. $V_1$ and $V_k$ will never be lonely, as they will connect after the first swap and disconnect after the last sw... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlisted Problems | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Yes | |
0232 | Problem:
a. Sejam $\mathcal{C}$ uma circunferência com centro $O$ e raio $r$ e $X$ um ponto exterior a $\mathcal{C}$. Construímos uma circunferência de centro em $X$ passando por $O$, a qual intersecta $\mathcal{C}$ nos pontos $P$ e $Q$. Com centro em $P$ construímos uma circunferência passando por $O$ e com centro em... | [
"Solution:\n\na. Observe que os triângulos $XOP$ e $PYO$ são ambos isósceles, de bases $OP$ e $YO$, respectivamente. Estes triângulos possuem ângulos da base de mesma medida, pois o ângulo $P\\hat{O}X = Y\\hat{O}P$ é comum aos dois triângulos. Deste modo, os triângulos $XOP$ e $PYO$ são semelhantes e podemos escrev... | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
027r | Problem:
Seja $a_{n}$ o número de maneiras de preencher um tabuleiro $n \times n$ com os algarismos 0 e 1, de modo que a soma em cada linha e em cada coluna seja a mesma. Por exemplo, os tabuleiros $2 \times 2$ que satisfazem essa regra são:
| 0 | 0 |
| :--- | :--- |
| 0 | 0 |
| 1 | 0 |
| :--- | :--- |
| 0 | 1 |
| 0 |... | [
"Solution:\nPara os tabuleiros $3 \\times 3$, dividiremos a solução em casos de acordo com a quantidade de números 1 por linha:\n\n- Caso 1: não há números 1. Neste caso, só temos um tabuleiro:\n\n| 0 | 0 | 0 |\n| :--- | :--- | :--- |\n| 0 | 0 | 0 |\n| 0 | 0 | 0 |\n\n- Caso 2: temos um número 1 por linha. Neste cas... | Brazil | null | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | a3 = 14, a4 = 140 | |
0ga6 | 給定一正整數 $k$。設正整數數列 $a_0, a_1, \dots, a_n$ ($n > 0$) 滿足下列所有條件：
(i) $a_0 = a_n = 1$;
(ii) 對任何的 $i = 1, 2, \dots, n-1$, 都有 $2 \le a_i \le k$;
(iii) 對任何的 $j = 2, 3, \dots, k$, $j$ 在 $a_0, a_1, \dots, a_n$ 中皆出現 $\varphi(j)$ 次 ($\varphi(j)$ 代表不超過 $j$ 且與 $j$ 互質之正整數的個數);
(iv) 對任何的 $i = 1, 2, \dots, n-1$, $\text{gcd}(a_{i-1}, a_... | [
"$b_n - b_0$ 的最小值為 1。\n方便起見, 我們稱滿足題目條件的數列 $a_0, a_1, \\dots, a_n$ 為「$k$-好數列」。\n首先, 我們先證明 $k$-好數列是唯一的。為此, 我們將命題加強, 同時證明 $k$-好數列滿足下面這個條件:\n若 $(a, b) = 1, a + b \\ge k + 1$ 且 $1 \\le a, b \\le k$, 則存在唯一一個正整數 $i$ 滿足 $a_i = a, a_{i+1} = b$. ...... (*)\n對 $k$ 歸納。$k = 1$ 時, 若 $n \\ge 2$, 則 $1 \\le 1 \\le n-1$, 所以由 (1) 知 $... | Taiwan | 二〇一六數學奧林匹亞競賽第三階段選訓營 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Other"
] | null | proof and answer | 1 | |
0k15 | Problem:
One million bucks (i.e. one million male deer) are in different cells of a $1000 \times 1000$ grid. The left and right edges of the grid are then glued together, and the top and bottom edges of the grid are glued together, so that the grid forms a doughnut-shaped torus. Furthermore, some of the bucks are hone... | [
"Solution:\n\nNote that each honest buck has at most one honest neighbor, and each dishonest buck has at least two honest neighbors. The connected components of honest bucks are singles and pairs. Then if there are $K$ honest bucks and $B$ buckaroo pairs, we get $B \\geq 3K$. From the dishonest buck condition we ge... | United States | HMMT November 2018 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1200000 | |
0bb4 | Let $p$ be a positive integer and $A_p = \{x \in \mathbb{R} \mid p\{x\} = (p+1)[x]\}$. Find the cardinals of the sets $A_p$ and $A_1 \cup A_2 \cup \dots \cup A_p$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | card(A_p) = 1 and card(A_1 ∪ A_2 ∪ ⋯ ∪ A_p) = 1 | |
00uk | Find the greatest integer $k \le 2023$ for which, regardless of how Alice colors exactly $k$ numbers among $\{1, 2, \dots, 2023\}$ in red, Bob can color some of the remaining uncolored numbers in blue, such that the sum of the red numbers is the same as the sum of the blue ones. | [
"Answer: 592.\nFor $k \\ge 593$, Alice can color the greatest 593 numbers $1431, 1432, \\dots, 2023$ and any other $(k-593)$ numbers so that their sum $s$ would satisfy\n$$\ns \\ge \\frac{2023 \\cdot 2024}{2} - \\frac{1430 \\cdot 1431}{2} > \\frac{1}{2} \\cdot \\left( \\frac{2023 \\cdot 2024}{2} \\right),\n$$\nthus... | Balkan Mathematical Olympiad | BMO 2023 Short List | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 592 | |
0cvc | In a convex quadrilateral $P$, all the side lengths are integers, and the perimeter of $P$ equals $10^{100}$. Moreover, each side length divides the sum of the three other side lengths. Prove that $P$ is a rhombus. | [
"**Первое решение.** Пусть $d$ — наибольшая сторона. Согласно условию, $a+b+c$ делится на $d$, то есть $a+b+c = k d$ для некоторого натурального $k$. Ясно, что $a+b+c > d$ (длина отрезка меньше длины ломаной с теми же концами), поэтому $k > 1$. Кроме того, так как $a \\le d$, $b \\le d$ и $c \\le d$, имеем $a+b+c \... | Russia | Regional round | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Number Theory > Divisibility / Factorization"
] | English; Russian | proof only | null | |
0h86 | For each pair of integers $a, b$ a non-negative integer $a*b$ is defined such that it satisfies the following two conditions:
1) $(a+b)*b = a*b+1$;
2) $(a*b) \cdot (b*a) = 0$.
Find values of the expressions $2016*121$ and $2016*144$. | [
"**Answer:** $2016 * 121 = 16$, $2016 * 144 = 13$.\n\nSuppose there are two positive integers $a, b$. Let us rewrite them as follows: $a = bq + r$, where $q$ is a non-negative integer, $r$ is a positive integer and $r \\le b$. We prove that under such conditions $a*b = q$.\n\nIf in the second condition $\\forall a ... | Ukraine | UkraineMO | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 2016*121 = 16, 2016*144 = 13 | |
06c0 | One may perform the following two operations on a positive integer:
(a) multiply it by any positive integer; or
(b) delete zeros in its decimal representation.
Prove that for every positive integer $X$, one can perform a sequence of these operations that will transform $X$ to a one-digit number. | [
"By the pigeonhole principle, two of the numbers $1, 11, 111, \\dots$ leave the same remainder when divided by $X$. Their difference, which is of the form $11\\cdots100\\cdots0$, is divisible by $X$. Therefore, we can transform $X$ to $11\\cdots100\\cdots0$, and then to $11\\cdots1$ by deleting the zeros.\n\nIf the... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Other"
] | null | proof only | null | |
09ko | Let $ABCD$ be a quadrilateral. Let point $P$ be on side $BC$, and point $Q$ be on side $CD$ such that $2PB = AB$ and $2QD = AD$. Let $M$ be the midpoint of segment $BD$, and $N$ be the midpoint of segment $PQ$. If $4MN = BD$, prove that $ABCD$ is a cyclic quadrilateral. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08xs | Let $O$ be the point of intersection of two diagonals of a square $ABCD$. Points $P, Q, R, S$ lie on the line segments $OA, OB, OC, OD$, respectively, and satisfy $OP = 3$, $OQ = 5$, $OR = 4$. Here we denote for a line segment $XY$ its length also by $XY$. If the point of intersection of lines $AB$ and $PQ$, the point ... | [
"$\\boxed{\\frac{60}{23}}$\nLet $\\ell$ be the length of a side of square $ABCD$, and $OA = OB = OC = OD = r$, $OP = a$, $OQ = b$, $OR = c$, $OS = d$. Also let $X$, $Y$, $Z$ be the point of intersection of lines $AB$ and $PQ$, lines $BC$ and $QR$, lines $CD$ and $RS$, respectively. Then, by Menelaus' theorem, we ha... | Japan | Japan Mathematical Olympiad Initial Round | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 60/23 | |
03c2 | The hexagon $ABLCDK$ is inscribed and the line $LK$ intersects the segments $AD$, $BC$, $AC$ and $BD$ in points $M$, $N$, $P$ and $Q$, respectively. Prove that $NL \cdot KP \cdot MQ = KM \cdot PN \cdot LQ$. | [
"*First solution.* Denote $s = \\sin \\frac{\\hat{A}B}{2}$, $t = \\sin \\frac{\\hat{B}L}{2}$, $u = \\sin \\frac{\\hat{L}C + \\hat{A}K}{2}$, $v = \\sin \\frac{\\hat{C}K}{2}$, $w = \\sin \\frac{\\hat{D}K}{2}$ and $x = \\sin \\frac{\\hat{L}D + \\hat{A}K}{2}$. Then we consecutively have\n$$\n\\frac{NL \\cdot KP \\cdot ... | Bulgaria | Bulgarian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0dqy | Let $p$ be an odd prime. Prove that
$$
1^{p-2} + 2^{p-2} + 3^{p-2} + \dots + \left(\frac{p-1}{2}\right)^{p-2} \equiv \frac{2-2^p}{p} \pmod{p}.
$$ | [
"First, for each $i = 1, 2, \\dots, \\frac{p-1}{2}$,\n$$\n\\frac{2i}{p} \\binom{p}{2i} = \\frac{(p-1)(p-2)\\cdots(p-(2i-1))}{(2i-1)!} \\equiv \\frac{(-1)(-2)\\cdots(-(2i-1))}{(2i-1)!} \\equiv -1 \\pmod{p}.\n$$\nHence\n$$\n\\begin{aligned}\n\\sum_{i=1}^{(p-1)/2} i^{p-2} &\\equiv - \\sum_{i=1}^{(p-1)/2} i^{p-2} \\fra... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof only | null | |
0fap | Problem:
A graph has 17 points and each point has 4 edges. Show that there are two points which are not joined and which are not both joined to the same point. | [
"Solution:\n\nSuppose not. We will obtain a contradiction.\n\nTake any point $A$. Suppose the four edges at $A$ are $BA$, $CA$, $DA$, $EA$. If there is any other point $X$ not joined to any of $A$, $B$, $C$, $D$, $E$ then with $A$ it forms the required pair of points. Suppose the three other points joined to $B$ (a... | Soviet Union | 1st CIS | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
01d4 | Mobile operator has $n$ clients and holds the following advertising campaign. In the beginning it deposits $1$ € on account of each client. When two persons which have $a$ € and $b$ € in their accounts communicate by a phone the operator makes both accounts equal to $(a + b)$ €. It happens that after $h(m)$ phone calls... | [
"Let the product of the clients' values after the $k$-th call be $a_k$. Suppose the values of two persons before a call were $a$ and $b$. By the arithmetic-geometric mean inequality, $(a+b)(a+b) \\ge 4ab$. Therefore, regardless of the choice of a call, $a_k \\ge 4a_{k-1}$. Since the initial and final values of $a_k... | Baltic Way | Baltic Way 2016 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
07i8 | In a $n \times n$ table some of the cells are black and the rest of them are white. *Alice* and *Bob* each have a copy of this table and trying to make the whole table red in the following ways:
If Alice finds a cell that is the only black cell in its row, She changes the color of all the cells in its column to red. If... | [
"If a cell is the only black cell in its row and Alice uses it to make its column red we call that cell special. Because after using a special set every cell in its $n$ column becomes red, now two special cells are in a column. On the other hand, no two special cells are in a row. Therefore, if Alice can make the t... | Iran | 40th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
03on | Suppose $O$ is the circumcenter of an acute triangle $\triangle ABC$, $P$ is a point inside $\triangle AOB$, and $D$, $E$, $F$ are the projections of $P$ on three sides $BC$, $CA$, $AB$ of $\triangle ABC$ respectively. Prove that a parallelogram with $FE$ and $FD$ as adjacent sides lies inside $\triangle ABC$. (posed b... | [
"**Proof** As shown in the figure, we construct a parallelogram $DEFG$ with $FE$ and $FD$ as adjacent sides. To prove the proposition to be true, we need only to prove that $\\angle FEG < \\angle FEC$, and $\\angle FDG < \\angle FDC$. It is equivalent to proving: $\\angle BFD < \\angle BAC$, and $\\angle AFE < \\an... | China | China Western Mathematical 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 | |
016l | Let $ABC$ be a given triangle. Let $\Gamma_A$, $\Gamma_B$ and $\Gamma_C$ be circles with radius $\rho$, centers $A'$, $B'$ and $C'$ respectively, and both the legs of angle $\angle BAC$ are tangents to $\Gamma_A$, both legs of angle $\angle ABC$ are tangents to $\Gamma_B$, both legs of angle $\angle BCA$ are tangents t... | [
"A multiplication with $\\frac{r}{\\rho} (=k)$ from $A$ moves $A'$ to $I$ ($r$ is the radius of the incircle of triangle $ABC$). Multiplications with the same factor from $B$ and $C$ move $B'$ and $C'$, respectively, to $I$ (then $\\overrightarrow{AI} = k AA'$, $\\overrightarrow{BI} = k BB'$ and $\\overrightarrow{C... | Baltic Way | BALTIC WAY | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
030m | Problem:
Considere um quadrado $ABCD$ de centro $O$. Sejam $E, F, G$ e $H$ pontos no interior dos lados $AB, BC, CD$ e $DA$, respectivamente, tal que $AE = BF = CG = DH$. Sabe-se que $OA$ intersecta $HE$ no ponto $X$, $OB$ intersecta $EF$ no ponto $Y$, $OC$ intersecta $FG$ no ponto $Z$ e $OD$ intersecta $GH$ no ponto ... | [
"Solution:\n\n\n\na) Sejam $x$ e $y$ as medidas dos comprimentos de $AE$ e $AH$, respectivamente. Dado que $AH = EB$, $AE = BF$ e $\\angle HAE = \\angle EBF$, segue que os triângulos $AEH$ e $EBF$ são congruentes. Daí\n$$\n\\angle HEF = 180^\\circ - \\angle HEA - \\angle BEF = 180^\\circ - ... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | x^2 + y^2 = 1 and Area(ABCD) × Area(XYZW) = 1 | |
0igq | Problem:
Three fair six-sided dice, each numbered $1$ through $6$, are rolled. What is the probability that the three numbers that come up can form the sides of a triangle? | [
"Solution:\n$37 / 72$\n\nDenote this probability by $p$, and let the three numbers that come up be $x$, $y$, and $z$. We will calculate $1-p$ instead: $1-p$ is the probability that $x \\geq y+z$, $y \\geq z+x$, or $z \\geq x+y$. Since these three events are mutually exclusive, $1-p$ is just $3$ times the probabilit... | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 37/72 | |
0jva | Problem:
Let $S=\{(x, y) \mid x, y \in \mathbb{Z}, 0 \leq x, y \leq 2016\}$. Given points $A=(x_1, y_1)$, $B=(x_2, y_2)$ in $S$, define
$$
d_{2017}(A, B) = (x_1 - x_2)^2 + (y_1 - y_2)^2 \pmod{2017}
$$
The points $A=(5,5)$, $B=(2,6)$, $C=(7,11)$ all lie in $S$. There is also a point $O \in S$ that satisfies
$$
d_{2017... | [
"Solution:\n\nNote that the triangle is a right triangle with right angle at $A$. Therefore,\n$$\nR^2 = \\frac{(7-2)^2 + (11-6)^2}{4} = \\frac{25}{2} = (25)\\left(2^{-1}\\right) \\equiv 1021 \\pmod{2017}.\n$$\n(An equivalent approach works for general triangles; the fact that the triangle is right simply makes the ... | United States | HMMT February 2016 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | 1021 | |
082t | Problem:
Sia $x_{0}, x_{1}, x_{2}, \ldots$ la successione definita da $x_{0}=2$ e $x_{n+1}=5+\left(x_{n}\right)^{2}$ per ogni $n \geq 0$. Dimostrare che in tale successione non compaiono numeri primi diversi da 2. | [
"Solution:\n\nSi osservi che i numeri in questione sono alternativamente pari e dispari, perché si aggiunge $5$ al quadrato del precedente. Perciò solo i termini di posto dispari possono fornire altri primi. Ma questi sono tutti multipli di $3$ e maggiori o uguali a $x_{1}=9$. Infatti la successione è ovviamente cr... | Italy | Progetto Olimpiadi di Matematica 2003 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0bli | Find all integrable functions $f : [0, 1] \to \mathbb{R}$ with the property: for every $x \in [0, 1]$,
$$
\int_0^x f(t) dt = (f(x))^{2015} + f(x).
$$ | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Differential Calculus > Applications",
"Calculus > Differential Equations > ODEs"
] | null | proof and answer | f(x) ≡ 0 on [0, 1] | |
0725 | Problem:
Let $ABC$ be a triangle with sides $a, b, c$. Consider a triangle $A_1 B_1 C_1$ with sides equal to $a+\frac{b}{2}$, $b+\frac{c}{2}$, $c+\frac{a}{2}$. Show that
$$
\left[A_1 B_1 C_1\right] \geq \frac{9}{4}[ABC]
$$
where $[XYZ]$ denotes the area of the triangle $XYZ$. | [
"Solution:\n\nIt is easy to observe that there is a triangle with sides $a+\\frac{b}{2}$, $b+\\frac{c}{2}$, $c+\\frac{a}{2}$. Using Heron's formula, we get\n$$\n16[ABC]^2 = (a+b+c)(a+b-c)(b+c-a)(c+a-b)\n$$\nand\n$$\n16\\left[A_1 B_1 C_1\\right]^2 = \\frac{3}{16}(a+b+c)(-a+b+3c)(-b+c+3a)(-c+a+3b)\n$$\nSince $a, b, c... | India | INMO | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0i6s | Problem:
Bob Barker went back to school for a PhD in math, and decided to raise the intellectual level of The Price is Right by having contestants guess how many objects exist of a certain type, without going over. The number of points you will get is the percentage of the correct answer, divided by 10, with no points... | [
"Solution:\n\n$15! / \\left(3^{4} \\cdot 5^{3} \\cdot 7^{2} \\cdot 9\\right) = 292864$. These are Standard Young Tableaux."
] | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions"
] | null | final answer only | 292864 | |
0aum | Problem:
How many ways can 6 boys and 6 girls be seated in a circle so that no two boys sit next to each other? | [] | Philippines | 17th Philippine Mathematical Olympiad Area Stage | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 86400 | |
0de6 | There are $330$ seats in the first row of the auditorium. Some of these seats are occupied by $25$ viewers. Prove that among the pairwise distances between these viewers, there are two equal. | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0cq8 | On a lesson, Pete and Nick write in their exercise books two numbers each: Pete writes $1$ and $2$, while Nick writes $3$ and $4$. Then, at the beginning of each minute, each boy finds some quadratic polynomial such that the numbers in a boy's exercise book are the roots of this polynomial. Let $f(x)$ and $g(x)$ be the... | [
"**Первое решение.** Будем рядом с каждой парой писать какой-нибудь квадратный трёхчлен, корнями которого являются числа этой пары. Пусть в некоторый момент у мальчиков записаны трёхчлены $p(x)$ и $q(x)$. Тогда они решали уравнение вида $\\alpha p(x) = \\beta q(x)$, где $\\alpha, \\beta$ — какие-то ненулевые числа.... | Russia | Russian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English, Russian | proof and answer | 14/5 | |
0cwq | Initially $100$ numbers $1$ are arranged on a circle. Petya and Vasya play the following game, taking turns; each boy performs $10^{10}$ moves; Petya starts. By his move, Petya chooses $9$ consecutive numbers and decreases each of them by $2$. By his move, Vasya chooses $10$ consecutive numbers and increases each of th... | [
"Let the numbers written in a circle be denoted as $a_1, a_2, \\dots, a_{100}$. Vasya will track only ten numbers, which he will pair as follows: $(a_9, a_{18})$, $(a_{27}, a_{36})$, $\\dots$, $(a_{90}, a_{99})$. In one move, Petya can decrease at most one of these $10$ numbers. If Petya decreases one number in a p... | Russia | LI Всероссийская математическая олимпиада школьников | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Russian | proof only | null | |
058a | Do there exist real numbers $x$, $y$, $z$, $t$ that meet the following system of equations?
$$
\begin{cases}
1 + x^3 + y^2 = 0 \\
1 + y^3 + z^2 = 0 \\
1 + z^3 + t^2 = 0 \\
1 + t^3 + x^2 = 0 \\
x + y + z + t = 0
\end{cases}
$$ | [
"The first equation implies $x^3 = -y^2 - 1$. Thus $x^3 < 0$, implying also $x < 0$. Similarly from the second, third and fourth equations we obtain $y < 0$, $z < 0$ and $t < 0$, respectively. The sum of negative numbers $x$, $y$, $z$, $t$ is negative, contradicting the fifth equation.",
"Suppose that the system ... | Estonia | Estonian Math Competitions | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof only | null | |
0g2q | Problem:
Soit $\mathbb{P}$ l'ensemble de tous les nombres premiers et $M$ un sous-ensemble de $\mathbb{P}$ ayant au moins trois éléments. On suppose que pour tout entier $k \geq 1$ et pour tout sous-ensemble $A=\{p_{1}, p_{2}, \ldots, p_{k}\}$ de $M$ tel que $A \neq M$, tous les facteurs premiers du nombre $p_{1} \cdo... | [
"Solution:\n\nPremière solution: (Arnaud) Que peut-on dire des (au moins) trois éléments de $M$ ? Au moins deux d'entre eux, disons $p_{3}, p_{4} \\in M$, sont impairs. En appliquant la condition pour $A=\\{p_{3}\\}$, on obtient donc $2 \\in M$. A-t-on forcément $3 \\in M$ ? Supposons que $3 \\notin\\{p_{3}, p_{4}\... | Switzerland | SMO 2019 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
072e | Problem:
Let $p, q, r$ be positive real numbers, not all equal, such that some two of the equations
$$
p x^{2}+2 q x+r=0, \quad q x^{2}+2 r x+p=0, \quad r x^{2}+2 p x+q=0
$$
have a common root, say $\alpha$. Prove that
(a) $\alpha$ is real and negative; and
(b) the third equation has non-real roots. | [
"Solution:\nConsider the discriminants of the three equations\n$$\n\\begin{array}{r}\np x^{2}+2 q x+r=0 \\\\\nq x^{2}+2 r x+p=0 \\\\\nr x^{2}+2 p x+q=0\n\\end{array}\n$$\nLet us denote them by $D_{1}, D_{2}, D_{3}$ respectively. Then we have\n$$\nD_{1}=4\\left(q^{2}-r p\\right), \\quad D_{2}=4\\left(r^{2}-p q\\righ... | India | INMO | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
06k9 | Let $n \ge 2$ be an integer. A game is played on an $n \times n$ board by two players X and Y as follows.
* **Order:** The two players take turns to play. X plays in the 1st round, Y plays the 2nd round, then X plays the 3rd round and so on.
* **Rule:** In the $k$th round, one has to choose $k$ unmarked consecutiv... | [
"We claim that the smallest *unnatural* case is $n = 6$. In the following, we use $(i, j)$ to mean the cell in the $i$th row and the $j$th column, and use $(i, j) - (i, k)$ to mean the consecutive cells from $(i, j)$ to $(i, k)$ inclusive (similar for $(i, j) - (k, j)$).\n\nFor $n = 2$, X and Y can play in any way ... | Hong Kong | Pre-IMO 2017 Mock Exam | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 6 | |
01nn | $ABCD$ is a cyclic quadrilateral. Let the circle $\Gamma_1$ pass through points $A$ and $B$ and touch $CD$ at point $E$; let the circle $\Gamma_2$ pass through points $B$ and $C$ and touch $DA$ at point $F$; let the circle $\Gamma_3$ pass through points $C$ and $D$, and touch $AB$ at point $G$; let at last the circle $... | [
"If the quadrilateral $ABCD$ has the pair of parallel sides the result follows from the symmetry of the construction. So we suppose that $ABCD$ is different from a trapezoid and a rectangle.\nLet $X = BC \\cap AD$. Since $\\Gamma_2$ touches $AD$ at point $F$, we have $XF^2 = XB \\cdot XC$.\n![](attached_image_1.png... | Belarus | 62nd Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
05gc | Problem:
Soit $p$ un nombre premier. Trouver tous les entiers $n \geqslant 1$ tels que pour tout entier $a \geqslant 1$, si $a^{n}-1$ est divisible par $p$, alors $a^{n}-1$ est aussi divisible par $p^{2}$. | [
"Solution:\n\nMontrons que les entiers $n$ qui conviennent sont exactement les multiples de $p$. Pour cela, nous allons d'abord prouver que lorsque $n=p m$ pour un entier $m \\geqslant 1$, pour tout entier $a \\geqslant 1$, si $a^{n}-1$ est divisible par $p$, alors $a^{n}-1$ est aussi divisible par $p^{2}$. Puis, r... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES, ENVOI No. 3 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All positive integers n that are multiples of p | |
0id7 | Problem:
Find the area of the region of the $x y$-plane defined by the inequality $|x|+|y|+|x+y| \leq 1$. | [
"Solution:\nTo graph this region we divide the $x y$-plane into six sectors depending on which of $x, y, x+y$ are $\\geq 0$, or $\\leq 0$. The inequality simplifies in each case:\n\n| Sector | Inequality | Simplified inequality |\n| :---: | :---: | :---: |\n| $x \\geq 0, y \\geq 0, x+y \\geq 0$ | $x+y+(x+y) \\leq 1... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 3/4 | |
0cjt | Let $ABC$ be an acute triangle with $AB < AC$, let $O$ be its circumcentre and let $A'$ be the reflection of $A$ in $BC$. The parallel through $O$ to $BC$ crosses $AC$ at $F$, and the tangent at $F$ to circle $BFC$ crosses the parallel through $A'$ to $BC$ at $M$. Consider the point $K$ on the ray $AB$, emanating from ... | [
"Let $H$ be the orthocentre of triangle $ABC$, let $N$ be the midpoint of the segment $AH$ and let $D$ the foot of the perpendicular from $B$ to $AC$. We first prove that $\\angle BNF = 90^\\circ$. Let $F'$ be the intersection of line $AC$ and the perpendicular at $N$ to $BN$. The points $B, N, D$, and $F'$ lie on ... | Romania | 75th NMO Selection Tests | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations... | English | proof only | null | |
00wt | Problem:
Prove that there are no positive integers $n$ and $m>1$ such that $102^{1991}+103^{1991}=n^{m}$. | [
"Solution:\nFactorizing, we get\n$$\n102^{1991}+103^{1991}=(102+103)\\left(102^{1990}-102^{1989} \\cdot 103+102^{1988} \\cdot 103^{2}-\\cdots+103^{1990}\\right),\n$$\nwhere $102+103=205=5 \\cdot 41$. It suffices to show that the other factor is not divisible by $5$. Let $a_{k}=102^{k} \\cdot 103^{1990-k}$, then $a_... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0j6p | Problem:
Let $ABC$ be a triangle with $AB = 4$, $BC = 8$, and $CA = 5$. Let $M$ be the midpoint of $BC$, and let $D$ be the point on the circumcircle of $ABC$ so that segment $AD$ intersects the interior of $ABC$, and $\angle BAD = \angle CAM$. Let $AD$ intersect side $BC$ at $X$. Compute the ratio $AX / AD$. | [
"Solution:\n\nLet $E$ be the intersection of $AM$ with the circumcircle of $ABC$. We note that, by equal angles, $ADC \\sim ABM$, so that\n$$\nAD = AC \\left(\\frac{AB}{AM}\\right) = \\frac{20}{AM}\n$$\nUsing the law of cosines on $ABC$, we get that\n$$\n\\cos B = \\frac{4^2 + 8^2 - 5^2}{2(4)(8)} = \\frac{55}{64}\n... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles"
] | null | proof and answer | 9/41 | |
09d8 | $1, 2, 3, \ldots, 10$ тоонуудыг, энхний ангийн тоонуудын нийлбэр нь нөгөө ангийн тоонуудон үржвэртэй тэнцүү байхаар үл огтлөцөх 2 ангид хуваах бүх хуваалтыг ол. | [
"$1+2+\\ldots+10=55$ ба\n\n1. 2. 3. 4. 5 = 120 $\\Rightarrow$ $\\{1, 2, \\ldots, 10\\} = B \\cup C$ ба $B$-ийн элементүүдийн нийлбэр $C$-ийн элементүүдийн үржвэртэй тэншүү ($B \\cap C = \\varnothing$) гэлээ. $C$ олонлог 4-өөс олон элементтэй байж таарахгүй. Өөрөөр хэлбэр $|C| \\le 4$ болно.\n\n(1) $|C| = 1$ бол $C=... | Mongolia | ММО-48 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Mongolian | proof and answer | The three partitions are:
- B = {1, 2, 3, 4, 5, 8, 9, 10}, C = {6, 7}
- B = {2, 3, 5, 6, 7, 8, 9}, C = {1, 4, 10}
- B = {4, 5, 6, 8, 9, 10}, C = {1, 2, 3, 7} | |
0beu | Suppose $f: \mathbb{R} \to \mathbb{R}$ is a monotonic function.
a. Prove that $f$ has one-sided limits at any point $x_0 \in \mathbb{R}$.
b. Define the function $g: \mathbb{R} \to \mathbb{R}$, $g(x) = \lim_{t \to x} f(t)$, i.e. $g(x)$ is the left-sided limit at $x$ of the function $f$. Prove that $g$ is a continuous ... | [
"Suppose, without any loss, that $f$ is an increasing function.\n\na. Let $x_0 \\in \\mathbb{R}$. The set $\\{f(x) \\mid x < x_0\\}$ is upper bounded by $f(x_0)$, because $f$ is increasing. Set $L = \\sup\\{f(x) \\mid x < x_0\\}$. We claim that $L = f(x_0 - 0)$.\nTo this end, let $\\varepsilon > 0$ and notice that ... | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Precalculus > Limits",
"Precalculus > Functions"
] | null | proof only | null | |
022n | Problem:
Calcule a soma
$$
1+11+111+1111+\cdots+\underbrace{1111 \ldots 11}_{n \text{ uns }}
$$ | [
"Solution:\nUma solução pode ser feita usando soma de progressões geométricas. Mas daremos outra solução que não precisará disso! Observe. Chamemos de $S$ a soma que queremos calcular, ou seja,\n$$\nS=1+11+111+1111+\\cdots+\\underbrace{1111 \\ldots 11}_{n \\text{ uns }} .\n$$\nQuanto vale $9 \\times S$ ? Basta troc... | Brazil | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | S = (10/81)(10^n − 1) − n/9 | |
06iu | Let $k$ be a real number such that $-1 < k < 1$. The straight line $y = x + k$ meets the curve $y = 1 - x^2$ at $A$ and $B$. If $C$ denotes the point $(1, 0)$, find the greatest possible area of $\triangle ABC$. | [
"Since $A$ and $B$ both lie on the straight line $y = x + k$, we may let $A = (\\alpha, \\alpha + k)$ and $B = (\\beta, \\beta + k)$. Combining the equations of the straight line and the parabola, we get $x^2 + x + k - 1 = 0$. Its two roots are $\\alpha$ and $\\beta$ since the two graphs meet at $A$ and $B$. Hence ... | Hong Kong | Hong Kong Preliminary Selection Contest | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 3*sqrt(3)/4 | |
0b8d | For any integer $n \ge 2$ denote by $A_n$ the set of solutions of the equation
$$
x = \lfloor \frac{x}{2} \rfloor + \lfloor \frac{x}{3} \rfloor + \dots + \lfloor \frac{x}{n} \rfloor.
$$
a) Determine the set $A_2 \cup A_3$.
b) Prove that the set $A = \bigcup_{n \ge 2} A_n$ is finite and find $\max A$. | [
"Notice that $A_n \\subset \\mathbb{Z}$ for all $n \\in \\mathbb{N}$, $n \\ge 2$.\n\na) The elements of $A_2$ satisfy the inequalities $x - 2 < 2x \\leq x$. By inspection, we obtain $A_2 = \\{-1, 0\\}$. The elements of $A_3$ satisfy the inequalities $5x - 12 < 6x \\leq 5x$. By inspection, we have $A_3 = \\{-7, -5, ... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | A2 ∪ A3 = {-7, -5, -4, -3, -2, -1, 0}; the union over all n is finite and its maximum element is 23. | |
0b2j | Problem:
Alice tosses two biased coins, each of which has a probability $p$ of obtaining a head, simultaneously and repeatedly until she gets two heads. Suppose that this happens on the $r$th toss for some integer $r \geq 1$. Given that there is $36\%$ chance that $r$ is even, what is the value of $p$?
(a) $\frac{\sq... | [] | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | (a) | |
0fek | Problem:
Se sabe que el polinomio $p(x)=x^{3}-x+k$ tiene tres raíces que son números enteros.
Determínese el número $k$. | [
"Solution:\n\nPara $k=0$ tenemos $p(x)=x^{3}-x=x(x-1)(x+1)$, que tiene raíces $0,-1$ y $1$.\nSe demuestra que este es el único valor de $k$ para el cual $p(x)$ tiene tres raíces enteras. En efecto, si $a$, $b$, $c$ son enteros, y $p(x)=(x-a)(x-b)(x-c)$, resulta:\n$$\n\\left.\\begin{array}{c}\na+b+c=0 \\\\\nab+ac+bc... | Spain | null | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 0 | |
02w5 | Problem:
Na figura abaixo, estão desenhados 16 triângulos equiláteros de lado $1$. Dizemos que dois deles são vizinhos se possuem um lado em comum. Determine se é possível escrevermos os números de $1$ até $16$ dentro desses triângulos de modo que todas as diferença entre os números colocados em dois triângulos vizinh... | [
"Solution:\n\nObserve que entre quaisquer dois triângulos de lado $1$ da figura é possível construirmos um caminho formado por no máximo $5$ outros triângulos que são mutuamente vizinhos. Assim, se fosse possível fazer a distribuição dos $16$ números como indicado no enunciado, seria possível começar do triângulo c... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Not possible | |
0718 | Problem:
Given a set of 9 points in the plane, no three collinear, show that for each point $P$ in the set, the number of triangles containing $P$ formed from the other 8 points in the set must be even. | [
"Solution:\n\nJoin each pair of points, thus dividing the plane into polygonal regions. If a point $P$ moves around within one of the regions then the number of triangles it belongs to does not change. But if it crosses one of the lines then it leaves some triangles and enters others. Suppose the line is part of th... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
09km | Show that every positive integer $n$ can be expressed as a sum of positive integers, where the sum of their reciprocals is less than or equal to $4$. For instance, as $5 = 2 + 2 + 1$, we find that $1/2 + 1/2 + 1/1 = 2 \le 4$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
02mb | Problem:
O triângulo de moedas - Um menino tentou alinhar 480 moedas em forma de um triângulo, com uma moeda na primeira linha, duas moedas na segunda linha, e assim por diante. Ao final da tentativa, sobraram 15 moedas. Quantas linhas tem esse triângulo? | [] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 30 | |
0gav | 設 $C_1$ 及 $C_2$ 為兩同心圓, 其中 $C_2$ 在 $C_1$ 內部。從 $C_1$ 上一點 $A$ 向 $C_2$ 引切線 $AB$, 且點 $B$ 在 $C_2$ 上。令點 $C$ 為射線 $AB$ 與 $C_1$ 的另一個交點, 而點 $D$ 為 $\overline{AB}$ 的中點。作一條過 $A$ 的直線與 $C_2$ 交於 $E, F$ 兩點, 使得 $DE$ 的中垂線與 $CF$ 的中垂線交於 $AB$ 上的一點 $M$。試求 $AM/MC$ 的所有可能值。 | [
"$AM/MC = 5/3$。\n\n因為 $AC \\cdot AD = (2AB) \\cdot (\\frac{1}{2}AB) = AB^2 = AE \\cdot AF$, 故 $CDEF$ 四點共圓。又 $M$ 點位於 $CF$ 及 $DE$ 的中垂線上, 故 $M$ 點為 $CDEF$ 的外接圓圓心。因為 $CMD$ 在同一條直線上, 所以 $M$ 為 $CD$ 中點。故\n$$\n\\frac{AM}{MC} = \\frac{AD + DM}{MC} = \\frac{AD + \\frac{1}{2}CD}{\\frac{1}{2}CD} = \\frac{\\frac{1}{4}AC + \\frac{... | Taiwan | 二〇一七數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 5/3 | |
0eyv | Problem:
Given the equation $x^{3} + a x^{2} + b x + c = 0$, the first player gives one of $a$, $b$, $c$ an integral value. Then the second player gives one of the remaining coefficients an integral value, and finally the first player gives the remaining coefficient an integral value. The first player's objective is to... | [
"Solution:\nThe first player wins.\n\nThe first player starts by choosing $c = 0$. Now if the second player selects $a$, then the first player can take $b = a - 1$. Then the polynomial factorizes as: $x(x + 1)(x + a - 1)$ with integral roots $0$, $-1$, $1 - a$.\n\nIf the second player selects $b$, then the first pl... | Soviet Union | 3rd ASU | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | First player | |
0283 | Problem:
Em um torneio de xadrez, cada um dos participantes jogou exatamente uma vez com cada um dos demais e não houve empates. Mostre que existe um jogador $P$ tal que, para qualquer outro jogador $Q$, distinto de $P$, uma das situações a seguir ocorre:
i) $Q$ perdeu de $P$;
ii) $Q$ perdeu de alguém que perdeu de $... | [
"Solution:\n\nSeja $P$ o jogador que mais venceu partidas no torneio. Digamos que $P$ tenha vencido os jogadores do conjunto $S=\\{P_{1}, P_{2}, \\ldots, P_{k}\\}$. Considere um jogador qualquer $Q$ diferente de $P$. Se $Q$ perdeu para $P$, ele satisfaz o item $i$).\n\nSe além de vencer $P$, $Q$ também ganhou de to... | Brazil | null | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
003z | El círculo $\Gamma$ es tangente a los lados $AB$ y $AC$ del triángulo $ABC$ en $E$ y $F$, respectivamente. Sea $X$ la intersección de $BF$ y $EC$, y sea $H$ la intersección de $\Gamma$ con $AX$. Sean $Z$ y $T$ las intersecciones de $EH$ y $FH$ con $BC$, respectivamente. Las rectas $ET$ y $FZ$ se cortan en $Q$. Demostra... | [] | Argentina | XV Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem"
] | Español | proof only | null | |
0av4 | Problem:
In parallelogram $ABCD$, $\angle BAD = 76^{\circ}$. Side $AD$ has midpoint $P$, and $\angle PBA = 52^{\circ}$. Find $\angle PCD$.
 | [
"Solution:\n\nNote that $\\angle BPA = 180^{\\circ} - 76^{\\circ} - 52^{\\circ} = 52^{\\circ}$. Since $\\angle PBA = 52^{\\circ}$, then $\\triangle BPA$ is isosceles and $|AB| = |AP| = |PD|$. But $|AB| = |CD|$, so by transitivity $|PD| = |CD|$ and therefore $\\triangle PCD$ is also isosceles. Since $\\angle CDA = 1... | Philippines | 18th PMO National Stage Oral Phase | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 38° | |
0cwg | Let $P$ be a point lying inside triangle $ABC$. Let $Q$ be a point on the segment $AB$, and let $R$ be a point on the segment $AC$ such that both circles ($BPQ$) and ($CPR$) are tangent to line $AP$. Through $B$ and $C$ we draw the lines passing through the center of the circle ($BPC$), and through $Q$ and $R$ we draw ... | [
"Since $AB \\cdot AQ = AP^2 = AC \\cdot AR$, quadrilateral $BCRQ$ is cyclic. Let $O$ be the center of circle ($BCRQ$). Denote by $O_1$ and $O_2$ the centers of circles ($BPC$) and ($QPR$). We will show that lines $BO_1$, $CO_1$, $QO_2$, $RO_2$ are equidistant from $O$. Since $OB = OC = OQ = OR$, it suffices to esta... | Russia | LI Всероссийская математическая олимпиада школьников | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Russian | proof only | null | |
0ewn | Problem:
8 players compete in a tournament. Everyone plays everyone else just once. The winner of a game gets 1, the loser 0, or each gets $\frac{1}{2}$ if the game is drawn. The final result is that everyone gets a different score and the player placing second gets the same as the total of the four bottom players. Wh... | [
"Solution:\n\nThe bottom 4 played 6 games amongst themselves, so their scores must total at least 6. Hence the number 2 player scored at least 6. The maximum score possible is 7, so if the number 2 player scored more than 6, then he must have scored $6 \\frac{1}{2}$ and the top player 7. But then the top player mus... | Soviet Union | 3rd ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | The third player won against the seventh player. |
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