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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
09uj | We say that a number is a *child* of another number if we can get it by placing between any two digits of the other number either nothing, a $+$, or a $\times$. For example, $145$ and $5$ are children of $12121$ because $145 = 12 \times 12 + 1$ and $5 = 1 + 2 \times 1 + 2 \times 1$. The number $15$ is both a child of $... | [
"D) $42$"
] | Netherlands | Junior Mathematical Olympiad, September 2019 | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | D) 42 | |
0euf | There is an $n \times n$ chess board. Each of the $n^2$ small boxes can present a number from $0$ to $k$, for some positive integer $k$. In each row and column, there is a button and if we push the button in a row (or column), the number in each of the $n$ small boxes contained in that row (or column, respectively) inc... | [
"Let $a_{ij}$ be the number in the box at the intersection of the $i$-th row and $j$-th column in the present phase. For each $s$ and $t$ with $1 \\leq s, t \\leq n$, let $c_s$ be the number of times the button in the $s$-th row is pushed, and $d_t$ the number of times the button in the $t$-th column is pushed. The... | South Korea | Korean Mathematical Olympiad Final Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | kn | |
07wr | How many 4-digit numbers $ABCD$ are there with the property that
$$
|A - B| = |B - C| = |C - D|?
$$
Note that the first digit $A$ of a four-digit number $ABCD$ cannot be zero. | [
"**16.**\nLet $d = |A - B| = |B - C| = |C - D|$ then $d$ can take the values $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$. For $d = 0$ we find 9 numbers $1111, 2222, 3333, 4444, 5555, 6666, 7777, 8888, 9999$. For other values of $d$, we take each of the eight sequences\n$$\n+++, ++-, +−+, +−−, −+++, −+−, −−+, −−−\n$$\nand find p... | Ireland | IRL_ABooklet_2024 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 187 | |
0hvr | Problem:
Find all functions $f: \mathbb{Q} \rightarrow \mathbb{R}$ from rational to real numbers such that for all rational $p, q$,
$$
f(p+2q) - f(p-2q) = 2(f(p+q) - f(p-q)).
$$ | [
"Solution:\nLet $P(p, q)$ be the given condition. Expanding $P(p-q, q) + 2P(p, q) + P(p+q, q)$ gives\n$$\nf(p+3q) - f(p-3q) = 3(f(p+q) - f(p-q)).\n$$\nLet $F$ be the quadratic that equals $f$ at $-1, 0, 1$. Plugging in $p=0.5, q=0.5$, we get that it also intersects $f$ at $2$. Plugging in $p=1.5, q=0.5$ then gives ... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | All quadratic polynomials: f(x) = ax^2 + bx + c for real a, b, c. | |
0jhd | Problem:
Let triangle $ABC$ satisfy $2 BC = AB + AC$ and have incenter $I$ and circumcircle $\omega$. Let $D$ be the intersection of $AI$ and $\omega$ (with $A, D$ distinct). Prove that $I$ is the midpoint of $AD$. | [
"Solution:\n\nSince $AD$ is an angle bisector, $D$ is the midpoint of $\\operatorname{arc} BC$ opposite $A$ on $\\omega$. It is well-known that $B, I$, and $C$ lie on a circle centered at $D$. Thus $BD = DC = DI$. Applying Ptolemy's theorem to cyclic quadrilateral $ABDC$, we get\n$$\nAB \\cdot DC + AC \\cdot BD = A... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous ... | null | proof only | null | |
0bdk | Given an integer $n \ge 2$, determine all non-constant polynomials $f$ with complex coefficients satisfying the condition $1 + f(X^n + 1) = (f(X))^n$. | [
"It is readily checked that the polynomials in the above sequence all satisfy the condition in the statement.\n\nConversely, let $f$ be a polynomial with complex coefficients satisfying the condition\n$$\n1 + f(X^n + 1) = (f(X))^n. \\qquad (1)\n$$\nTo begin, we show that, if $f(0) = 0$, then $f = -X$ and $n$ must b... | Romania | 64th NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | For even n, there are no non-constant solutions. For odd n, all solutions are f(X) = −T_k(X), where T_0(X) = X and T_{k+1}(X) = (T_k(X))^n + 1 for k ≥ 0. | |
0ke2 | Problem:
For positive integers $n$ and $k$, let $\mho(n, k)$ be the number of distinct prime divisors of $n$ that are at least $k$. For example, $\mho(90,3)=2$, since the only prime factors of $90$ that are at least $3$ are $3$ and $5$. Find the closest integer to
$$
\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{\mho(n... | [
"Solution:\nA prime $p$ is counted in $\\mho(n, k)$ if $p \\mid n$ and $k \\leq p$. Thus, for a given prime $p$, the total contribution from $p$ in the sum is\n$$\n3^{7} \\sum_{m=1}^{\\infty} \\sum_{k=1}^{p} \\frac{1}{3^{p m+k}} = 3^{7} \\sum_{i \\geq p+1} \\frac{1}{3^{i}} = \\frac{3^{7-p}}{2}.\n$$\nTherefore, if w... | United States | HMMT February 2020 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 167 | |
0d9z | Let $ABC$ be an acute-angled triangle inscribed in circle $(O)$. Let $G$ be a point on the small arc $AC$ of $(O)$ and $(K)$ be a circle passing through $A$ and $G$. Bisector of $\angle BAC$ cuts $(K)$ again at $P$. The point $E$ is chosen on $(K)$ such that $AE$ is parallel to $BC$. The line $PK$ meets the perpendicul... | [
"Let $D$ be the second intersection of $AP$ and $(O)$ and $H$ be the intersection of $OD$ and $AE$. Note that $D$ is the midpoint of the minor arc $BC$ of $(O)$, then $OD$ is the perpendicular bisector of $BC$. Since $AE \\parallel BC$, we have $\\angle DHE = 90^{\\circ}$.\n\n\n\nOn the oth... | Saudi Arabia | Team selection tests for IMO 2018 | [
"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 chasing"
] | English | proof only | null | |
003e | El señor José tiene cuatro nietos, todos de edades diferentes. La diferencia de edades entre el mayor y el menor de los nietos es de $6$ años y la diferencia de edades entre los otros dos nietos es de $1$ año. Uno de los cuatro nietos tiene $12$ años. Se sabe que, haciendo dos cortes paralelos a los lados, como se mues... | [] | Argentina | XV Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Español | proof and answer | 6, 8, 9, 12 | |
05n4 | Problem:
Soit $ABC$ un triangle. On suppose que la médiane $(BM)$ et la bissectrice $(CD)$ se coupent en un point $J$ tel que $JB = JC$. Soit $H$ le pied de la hauteur issue de $A$. Montrer que $JM = JH$. | [
"Solution:\n\n\n\nNotons $\\gamma = \\widehat{ACB}$. On a $MA = MC = MH$, donc $\\widehat{CHM} = \\gamma$.\n\nOn a aussi $\\widehat{CHM} = 180^\\circ - \\widehat{MHB} = \\widehat{HBM} + \\widehat{BMH}$.\n\nDe plus, $\\widehat{HBM} = \\widehat{DCB} = \\gamma / 2$, donc $\\widehat{BMH} = \\wi... | France | Olympiades Françaises de Mathématiques | [
"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 > Distance chasing"
] | null | proof only | null | |
088p | Problem:
Un folletto vive nel mondo delle fate. Un certo giorno sceglie 12 coppie di numeri positivi: quelli della prima sono dispari, quelli della seconda danno resto 1 se divisi per 3, quelli della terza danno resto 1 se divisi per 4, e così via fino alla dodicesima. Poi calcola la differenza dei quadrati dei numeri... | [
"Solution:\n\nLa risposta è (B). La differenza dei quadrati di due numeri che danno lo stesso resto se divisi per $k$ è un multiplo di $k$: infatti se i due numeri sono $a = n k + r$ e $b = m k + r$,\n$$\na^2 - b^2 = (a - b)(a + b) = (n - m) k (n k + m k + 2 r)\n$$\n(nel nostro caso abbiamo $r = 1$ per tutti i $k$ ... | Italy | Olimpiadi di Matematica | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | B | |
0itq | Problem:
Let $S$ be the set of points $(a, b)$ with $0 \leq a, b \leq 1$ such that the equation
$$
x^{4}+a x^{3}-b x^{2}+a x+1=0
$$
has at least one real root. Determine the area of the graph of $S$. | [
"Solution:\nAnswer: $\\frac{1}{4}$\n\nAfter dividing the equation by $x^{2}$, we can rearrange it as\n$$\n\\left(x+\\frac{1}{x}\\right)^{2}+a\\left(x+\\frac{1}{x}\\right)-b-2=0\n$$\nLet $y = x + \\frac{1}{x}$. We can check that the range of $x + \\frac{1}{x}$ as $x$ varies over the nonzero reals is $(-\\infty, -2] ... | United States | 11th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 1/4 | |
07yb | Problem:
In una gara di matematica si propongono 3 problemi, ciascuno dei quali viene valutato con un punteggio intero compreso fra 0 e 7 (estremi inclusi). Si sa che, comunque si scelgano due concorrenti, c'è al più un problema su cui questi hanno ottenuto lo stesso punteggio (per esempio, non ci sono due concorrenti... | [
"Solution:\n\nLa risposta è 64. Indichiamo con $p_{1}, p_{2}, p_{3}$ i punteggi che un concorrente può ottenere nei problemi $1, 2, 3$, rispettivamente. In primo luogo osserviamo che sicuramente il numero di concorrenti non può essere superiore a 64. Infatti, considerando anche solo i primi due problemi, le coppie ... | Italy | null | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 64 | |
05pp | Problem:
Soit $ABC$ un triangle tel que $AB \neq AC$. Soit $E$ tel que $AE = BE$ et $(BE)$ perpendiculaire à $(BC)$ et soit $F$ tel que $AF = CF$ et $(CF)$ perpendiculaire à $(BC)$. Soit $D$ le point de $(BC)$ tel que $(AD)$ soit tangente au cercle circonscrit à $ABC$ en $A$.
Montrer que les points $D$, $E$, $F$ sont... | [
"Solution:\n\n\n\nSupposons (sans restreindre la généralité) que $AB < AC$.\n\nSoit $O$ le centre du cercle circonscrit à $ABC$ et $G$ le point de la droite $(AD)$ tel que $(AB)$ soit parallèle à $(GC)$.\n\nMontrons tout d'abord que les angles $\\widehat{BAD}$ et $\\widehat{BCA}$ sont de mê... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, in... | null | proof only | null | |
0flz | Find all integer positive numbers $r$ and $s$ such that the integer positive number $n = 2^r - 16^s$ gives remainder $5$ when divided by $7$. Find the smallest $n$ satisfying this condition. | [] | Spain | Spanija 2012 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof and answer | All positive integer pairs (r, s) with r ≡ 1 (mod 3), s ≡ 2 (mod 3), and r > 4s. The smallest n occurs at r = 10, s = 2 and equals 768. | |
0chj | Let $n \ge 2$ be a natural number, and $\mathcal{F}$ be the set of functions $f : \{1, 2, \dots, n\} \to \{1, 2, \dots, n\}$ such that $f(k) \le f(k+1) \le f(k)+1$, for every $k \in \{1, 2, \dots, n-1\}$.
a) Determine the cardinality of the set $\mathcal{F}$.
b) Determine the total number of fixed points of the funct... | [
"a) We count the functions in $\\mathcal{F}$ with $f(1) = k$, $k = 1, \\dots, n$. We associate to each $i = 2, \\dots, n$ the number $f(i)-f(i-1) \\in \\{0, 1\\}$, with the restriction that there can be at most $n-k$ occurrences of $1$. This association is bijective, and the number of possibilities of choosing the ... | Romania | 74th Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof and answer | a) (n+1) 2^{n-2}; b) n 2^{n-1} | |
0eh1 | Problem:
V nekem trenutku kmalu po 4. uri urni in minutni kazalec na uri oklepata kot $119^{\circ}$ (glej sliko). Koliko stopinj je velik kot, ki ga kazalca na uri oklepata natanko 1 uro in 20 minut po tem trenutku?
(A) 30
(B) 39
(C) 41
(D) 43
(E) 45
 | [
"Solution:\n\nKote bomo merili v smeri urinega kazalca. Veliki kazalec v eni uri opiše kot $360^{\\circ}$, v eni minuti pa kot $\\frac{360^{\\circ}}{60}=6^{\\circ}$. V 1 uri in 20 minutah se torej premakne za kot $20 \\cdot 6^{\\circ}=120^{\\circ}$. Mali kazalec v eni uri opiše kot $\\frac{360^{\\circ}}{12}=30^{\\c... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije Državno tekmovanje | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | B | |
054f | Let $n$ be a positive integer such that there exists a positive integer that is less than $\sqrt{n}$ and does not divide $n$. Let $(a_1, \dots, a_n)$ be an arbitrary permutation of $1, \dots, n$. Let $a_{i_1} < \dots < a_{i_k}$ be its maximal increasing subsequence and let $a_{j_1} > \dots > a_{j_l}$ be its maximal dec... | [
"The first phase of the solution consists in showing that $kl \\ge n$. For every $i = 1, \\dots, n$, let $f(i)$ denote the length of the longest increasing subsequence ending with $a_i$, and let $g(i)$ be the length of the longest decreasing subsequence ending with $a_i$. For distinct indices $i < j$, if $a_i < a_j... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof only | null | |
0fg7 | Problem:
Para cada número real $x_{1}$, se construye la sucesión $x_{1}, x_{2}, \ldots, x_{n}, \ldots$ haciendo
$$
x_{n+1} = x_{n}\left(x_{n} + \frac{1}{n}\right) \text{ para cada } n \geq 1
$$
Demostrar que existe exactamente un valor de $x_{1}$ para el cual $0 < x_{n} < x_{n+1} < 1$ para cada $n$. | [] | Spain | International Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0c7d | For any integer $n \ge 2$, denote by $s(n)$ the number of pairs of integers $(x, y)$, from $1, 2, \dots, n$, with $x > y$, such that $x$ and $y$ have exactly $x - y$ common divisors.
a)
Is there any $n$ such that $s(n) = 2019$?
b)
But for $s(n) = 2020$? | [] | Romania | 70th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | a) No.
b) Yes; n = 1348. | |
0jps | Problem:
Consider the function $z(x, y)$ describing the paraboloid
$$
z = (2x - y)^2 - 2y^2 - 3y.
$$
Archimedes and Brahmagupta are playing a game. Archimedes first chooses $x$. Afterwards, Brahmagupta chooses $y$. Archimedes wishes to minimize $z$ while Brahmagupta wishes to maximize $z$. Assuming that Brahmagupta wi... | [
"Solution:\n\nAnswer: $-\\frac{3}{8}$\n\nViewing $x$ as a constant and completing the square, we find that\n$$\n\\begin{aligned}\nz & = 4x^2 - 4x y + y^2 - 2y^2 - 3y \\\\\n & = -y^2 - (4x + 3) y + 4x^2 \\\\\n & = -\\left(y + \\frac{4x + 3}{2}\\right)^2 + \\left(\\frac{4x + 3}{2}\\right)^2 + 4x^2\n\\end{aligned}\n... | United States | HMMT February | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | final answer only | -3/8 | |
02qd | Problem:
Quatro times, entre os quais o Quixajuba, disputam um torneio de vôlei em que:
- cada time joga contra cada um dos outros uma única vez;
- qualquer partida termina com a vitória de um dos times;
- em qualquer partida, os times têm a mesma probabilidade de ganhar;
- ao final do torneio, os times são classifica... | [
"Solution:\n\na) O número total de partidas disputadas no torneio é $3+2+1=6$. Como $6$ não é divisível por $4$, o torneio não pode acabar com os quatro times tendo o mesmo número de vitórias.\n\nb) 1ª solução: Para que o Quixajuba termine isolado em primeiro lugar, ele deve ganhar todas as suas partidas. De fato, ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) No. b) 1/8. c) 1/8. | |
0jxq | Problem:
At a recent math contest, Evan was asked to find $2^{2016} \pmod{p}$ for a given prime number $p$ with $100 < p < 500$. Evan has forgotten what the prime $p$ was, but still remembers how he solved it:
- Evan first tried taking $2016$ modulo $p-1$, but got a value $e$ larger than $100$.
- However, Evan noted th... | [
"Solution:\nAnswer is $p=211$. Let $p=2d+1$, $50 < d < 250$. The information in the problem boils down to\n$$\n2016 = d + 21 \\pmod{2d}\n$$\nFrom this we can at least read off $d \\mid 1995$.\nNow factor $1995 = 3 \\cdot 5 \\cdot 7 \\cdot 19$. The values of $d$ in this interval are $57, 95, 105, 133$. The prime val... | United States | February 2017 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 211 | |
0au0 | Problem:
Let $r$ and $s$ be positive real numbers that satisfy the equation
$$
(r+s-r s)(r+s+r s)=r s
$$
Find the minimum values of $r+s-r s$ and $r+s+r s$. | [
"Solution:\nThe given equation can be rewritten into\n$$\n(r+s)^2 = r s(r s+1)\n$$\nSince $(r+s)^2 \\geq 4 r s$ for any $r, s \\in \\mathbb{R}$, it follows that $r s \\geq 3$ for any $r, s > 0$. Using this inequality, equation (1), and the assumption that $r$ and $s$ are positive, we have\n$$\n\\begin{aligned}\nr+s... | Philippines | 15th Philippine Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 2√3 − 3 and 3 + 2√3 | |
049e | Ivan, Stipe and Tonći take turns in throwing a die. Ivan throws first, then Stipe, then Tonći, then Ivan again and so on in the same order. When it's their turn, everyone throws the die once, until they get a “six”. After getting his first “six”, in every following turn Ivan throws the die four times. After his first “... | [] | Croatia | CroatianCompetitions2011 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | Ivan | |
02ey | Let $d(n)$ be the number of positive divisors of $n$. Show that
$$
n\left(\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right) \le d(1) + d(2) + \dots + d(n) \le n\left(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right)
$$ | [
"Let's count the number $N$ of pairs $(j, k)$ such that $j$ divides $k$ and $1 \\le j, k \\le n$. Fixing $j$, we must count the multiples of $j$, which is $\\lfloor \\frac{n}{j} \\rfloor$. So\n$$\nN = \\lfloor \\frac{n}{1} \\rfloor + \\lfloor \\frac{n}{2} \\rfloor + \\lfloor \\frac{n}{3} \\rfloor + \\dots + \\lfloo... | Brazil | XIV OBM | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
0l4p | Problem:
Let $a$, $b$, and $c$ be pairwise distinct positive integers such that $\frac{1}{a}$, $\frac{1}{b}$, $\frac{1}{c}$ is an increasing arithmetic sequence in that order. Prove that $\gcd(a, b) > 1$. | [
"Solution:\n\nObserve that $\\frac{1}{a} + \\frac{1}{c} = \\frac{2}{b}$, so $b(a + c) = 2ac$, and thus $a \\mid b(a + c)$. If we assume that $\\gcd(a, b) = 1$, then we must have $a \\mid a + c$, so $a \\mid c$. However, $\\frac{1}{a} < \\frac{1}{c}$, so $a > c$, contradiction. Thus, $\\gcd(a, b) > 1$, as desired.",... | United States | HMMT February 2025 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | proof only | null | |
0ct1 | In a country, there are $n$ cities, and every two of them are connected with a non-stop train operating in both directions. The ticket price for each train in both directions is the same, but for any two different trains these prices are different. Prove that a traveler may start from some city and take $n-1$ trains co... | [
"Первое решение. Уберём все экспресслы, а затем начнём запускать их обратно по одному в порядке возрастания цены (т. е. первым запустим самый дешёвый, вторым — самый дешёвый из остальных, и т. д.). В каждый момент в каждом городе будем писать максимальное количество экспрессов, на которых можно последовательно прое... | Russia | XL Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
03l1 | Problem:
Suppose $a_{1}, a_{2}, \ldots, a_{8}$ are eight distinct integers from $\{1,2, \ldots, 16,17\}$. Show that there is an integer $k>0$ such that the equation $a_{i}-a_{j}=k$ has at least three different solutions. Also, find a specific set of 7 distinct integers from $\{1,2, \ldots, 16,17\}$ such that the equat... | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | {1, 2, 5, 9, 14, 16, 17} | |
0b6p | Let $r$ be a positive integer and let $N_r$ be the smallest positive integer such that the numbers
$$
\frac{N_r}{n+r} \binom{2n}{n}, \quad n = 0, 1, 2, \dots,
$$
are all integers. Show that
$$
N_r = \frac{r}{2} \binom{2r}{r}.
$$ | [
"We first show that\n$$\nN_r \\le \\frac{r}{2} \\binom{2r}{r}\n$$\nby proving that\n$$\nK(n,r) = \\frac{r}{2(n+r)} \\binom{2n}{n} \\binom{2r}{r}\n$$\nis an integer for all $n \\ge 0$ and $r \\ge 1$. Notice that\n$$\nK(0, r) = \\binom{2r-1}{r} \\quad \\text{and} \\quad K(n, 1) = \\frac{1}{n+1} \\binom{2n}{n},\n$$\nt... | Romania | 2010 Eighth IMAR MATHEMATICAL COMPETITION | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | N_r = (r/2) * binomial(2r, r) | |
079y | Prove that for every positive integers $k$ and $n$ there exist $k$ monic polynomials $P_1(x), P_2(x),..., P_k(x)$ of degree $n$ with integer coefficient such that each two of them have no common factor and the sum of each arbitrary number of them has all its roots real. | [
"For each $1 \\le i \\le k$ we define\n$$\nP_i(x) = (x - i)(x - (k + i))\\cdots(x - ((n - 1)k + i)).\n$$\nWe claim that these polynomials satisfy the problem condition.\nFor each $1 \\le i \\le k$ and each $0 \\le j \\le n - 1$, $P_i(x)$ has exactly one simple root in the interval $(jk + \\frac{1}{2}, (j+1)k + \\fr... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
08be | Problem:
Sia $ABCDEF$ un esagono regolare di area $1$. Si considerino tutti i triangoli i cui vertici appartengono all'insieme $\{A, B, C, D, E, F\}$: quanto vale la somma delle loro aree?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7 | [
"Solution:\n\nLa risposta è (D). Consideriamo tutti i triangoli non degeneri i cui vertici sono anche vertici dell'esagono, distinguendoli in 3 tipi:\n\nTre vertici consecutivi. Ci sono $6$ triangoli di questo tipo (uno per ciascun vertice), e l'area di ognuno è pari a $1/6$ di quella dell'esagono (vedi figura).\n\... | Italy | Progetto Olimpiadi della Matematica | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | MCQ | D | |
0091 | For each positive integer $n$, let $s(n)$ be the sum of the digits of $n$. Find the smallest positive integer $k$ such that
$$
s(k) = s(2k) = s(3k) = \dots = s(2011k) = s(2012k).
$$ | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 9999 | |
0h7s | Is it possible to cut a regular triangle into:
a) three equal quadrilaterals;
b) three equal pentagons?
Convexity of quadrilaterals and pentagons is not required. | [
"a) Consider a regular triangle $\\triangle ABC$, denote its midpoints by $M, N, K$ respectively. (fig. 04). The segments $AN$, $BK$ and $CM$ intersect at $O$. It is easy to see that the quadrilaterals $BMON$, $CKON$ and $AMOK$ satisfy the condition.\n\nb) Now let us denote the midpoints of $OM$, $ON$ and $OK$ by $... | Ukraine | UkraineMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
09hb | Let $\omega$ be the circumcircle of a scalene triangle $ABC$. The tangents to $\omega$ at $A$ and $C$ meet in $P$, and the line $BP$ intersects $\omega$ in $D$. Let $BB'$ be a diameter of $\omega$. The exterior angle bisector of $\angle ABC$ and the lines $B'A$ and $B'C$ intersect in $A'$ and $C'$, respectively. Prove ... | [
"Since $PC$ is a tangent to $\\omega$ we have $\\triangle PCD \\cong \\triangle PBC$ and so $CD \\cdot PB = PC \\cdot CB$. Similarly, $AD \\cdot PB = PA \\cdot AB$. Thus\n$$\n\\frac{CD}{CB} = \\frac{AD}{AB}. \\qquad (*)\n$$\nsince $PA = PC$. We also have that $\\triangle AA'B \\sim \\triangle CC'B$ because of $\\an... | Mongolia | Mongolian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0748 | Problem:
Find all real numbers $x$ such that
$$
\left[x^{2}+2x\right] = [x]^{2} + 2[x]
$$
(Here $[x]$ denotes the largest integer not exceeding $x$.) | [
"Solution:\nAdding $1$ to both sides, the equation reduces to\n\n$$\n\\left[(x+1)^{2}\\right] = ([x+1])^{2}\n$$\n\nWe have used $[x] + m = [x + m]$ for every integer $m$.\n\nSuppose $x+1 \\leq 0$. Then $[x+1] \\leq x+1 \\leq 0$. Thus\n\n$$\n([x+1])^{2} \\geq (x+1)^{2} \\geq \\left[(x+1)^{2}\\right] = ([x+1])^{2}\n$... | India | Indian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | {-1, -2, -3, ...} ∪ ⋃_{n=-1}^{∞} [n, sqrt(1 + (n+1)^2) - 1) | |
00w8 | Problem:
Prove that none of the numbers
$$
F_{n} = 2^{2^{n}} + 1, \quad n = 0, 1, 2, \ldots,
$$
is a cube of an integer. | [
"Solution:\nAssume there exist such natural numbers $k$ and $n$ that $2^{2^{n}} + 1 = k^{3}$. Then $k$ must be an odd number and we have $2^{2^{n}} = k^{3} - 1 = (k - 1)(k^{2} + k + 1)$. Hence $k - 1 = 2^{s}$ and $k^{2} + k + 1 = 2^{t}$ where $s$ and $t$ are some positive integers. Now $2^{2s} = (k - 1)^{2} = k^{2}... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0cry | Натуральное число $n$ назовём хорошим, если каждый его натуральный делитель, увеличенный на 1, является делителем числа $n+1$. Найдите все хорошие натуральные числа. | [
"**Ответ.** Единица и все нечётные простые числа.\n\nЯсно, что $n = 1$ удовлетворяет условию. Также ему удовлетворяют все нечётные простые: если $n = p$, то его делители, увеличенные на 1, есть $2$ и $p+1$; оба они делят $p+1$.\n\nС другой стороны, у любого числа $n$, удовлетворяющего условию, есть делитель $1$; зн... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 1 and all odd primes | |
05fj | Problem:
Soit $ABC$ un triangle et $I$ le centre de son cercle inscrit. La perpendiculaire à la droite $(AI)$ passant par le point $I$ coupe la droite $(AB)$ en un point $D$ et la droite $(AC)$ en un point $E$. On suppose qu'il existe deux points $F$ et $G$ sur le segment $[BC]$ tels que $BA = BF$ et $CA = CG$. Soit $... | [
"Solution:\n\nSoient $M$ et $N$ les milieux respectifs des segments $[AD]$ et $[AE]$ et soit $X$ le point d'intersection de la droite $(MN)$ (qui est aussi la médiatrice du segment $[AI]$) avec la droite $(BC)$. Soient $O_1$ et $O_2$ les centres respectifs des cercles circonscrits aux triangles $ADF$ et $AEG$. On s... | France | ENVOI 1 : GÉOMÉTRIE Corrigé | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > ... | null | proof only | null | |
0ajk | A magical square of dimensions $3 \times 3$ is a square with side $3$, consisting of $9$ unit squares, so that the real numbers written in the unit squares (one number in each unit square) satisfy the property: the sum of the numbers in the unit squares in any row is equal to the sum of the numbers in the unit squares ... | [
"We consider the magical square:\n\n| $A_1$ | $A_2$ | $A_3$ |\n|-------|-------|-------|\n| $B_1$ | $B_2$ | $B_3$ |\n| $C_1$ | $C_2$ | $C_3$ |\n\n$$\n\\begin{aligned}\nA_1 + A_2 + A_3 &= B_1 + B_2 + B_3 = C_1 + C_2 + C_3 = A_1 + B_1 + C_1 \\\\\n&= A_2 + B_2 + C_2 = A_3 + B_3 + C_3 = A_1 + B_2 + C_3 = C_1 + B_2 + A_... | North Macedonia | Macedonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | If both dimensions are three, the maximum is nine; otherwise the maximum is one. | |
05s4 | Problem:
Soit $A, B, C$ et $P$ quatre points du plan tels que $ABC$ soit un triangle équilatéral et que $AP < BP < CP$. On suppose que la seule donnée des longueurs $AP, BP$ et $CP$ nous permet de déterminer, de manière unique, la longueur $AB$.
Démontrer que $P$ appartient au cercle circonscrit à $ABC$. | [
"Solution:\n\nSoit $\\Gamma_{a}$, $\\Gamma_{b}$ et $\\Gamma_{c}$ les cercles de centre $P$ et passant respectivement par $A$, $B$ et $C$. Soit également $C^{\\bullet}$ le symétrique de $C$ par rapport à $(AP)$. Notons que $C^{\\bullet}$ est le seul point de $\\Gamma_{c}$, autre que $C$ lui-même, tel que $AC = AC^{\... | France | Préparation Olympique Française de Mathématiques - Test du 15 Mai 2019 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0e65 | Let $D$ and $E$ be such points on the sides $BC$ and $AC$ of a triangle $ABC$, respectively, that the points $A$, $B$, $D$ and $E$ lie on the same circle. Let $L$ denote the center of the inscribed circle of the triangle $BCE$, and let $G$ denote the point of tangency of this inscribed circle with the side $EC$. Let $K... | [
"Let $K$ be a circle containing points $A$, $B$, $D$ and $E$. Because $EL$ is the bisector of the angle $\\angle BEC$, it bisects the arc $\\overarc{AB}$ that contains $E$. Similarly, the line $DK$ bisects the arc $\\overarc{AB}$ that contains $D$. Because $E$ and $D$ lie on the same side of the line $\\overarc{AB}... | Slovenia | National Math Olympiad 2012 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0doa | Problem:
Низови $\left(a_{n}\right)_{n=0}^{\infty}$ и $\left(b_{n}\right)_{n=0}^{\infty}$ дефинисани су рекурентним релацијама
$$
a_{0}=0, \quad a_{1}=1, \quad a_{n+1}=\frac{2018}{n} a_{n}+a_{n-1} \quad \text{ за } n \geqslant 1
$$
и
$$
b_{0}=0, \quad b_{1}=1, \quad b_{n+1}=\frac{2020}{n} b_{n}+b_{n-1} \quad \text{ за... | [
"Solution:\n\nДефинишимо низ $\\left(c_{m, n}\\right)\\left(m, n \\in \\mathbb{N}_{0}\\right)$ условима\n$$\nc_{m, 0}=0, \\quad c_{m, 1}=1, \\quad c_{m, n+1}=\\frac{2 m}{n} c_{m, n}+c_{m, n-1} \\quad \\text{ за } n \\geqslant 1\n$$\nТада је $a_{n}=c_{1010, n}$ и $b_{n}=c_{1009, n}$.\nВидимо да је нпр. $c_{1, n}=n, ... | Serbia | 13. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | proof only | null | |
0k8d | Problem:
Tessa has a figure created by adding a semicircle of radius $1$ on each side of an equilateral triangle with side length $2$, with semicircles oriented outwards. She then marks two points on the boundary of the figure. What is the greatest possible distance between the two points? | [
"Solution:\n\nNote that both points must be in different semicircles to reach the maximum distance. Let these points be $M$ and $N$, and $O_{1}$ and $O_{2}$ be the centers of the two semicircles where they lie respectively. Then\n$$\nMN \\leq MO_{1} + O_{1}O_{2} + O_{2}N\n$$\nNote that the right side will always be... | United States | HMMT February 2019 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 3 | |
0kvx | Problem:
Mr. Murgatroyd decides to throw his class a pizza party, but he's going to make them hunt for it first. He chooses eleven locations in the school, which we'll call $1,2, \ldots, 11$. His plan is to tell students to start at location $1$, and at each location $n$ from $1$ to $10$, they will find a message dire... | [
"Solution:\n\nIf the students never visit the same room twice, then their hunt lasts a finite number of steps. In that case, they must reach the pizza (since the hunt always continues if they have not yet reached the pizza).\n\nTherefore, the only way for the students to not reach the pizza is for them to visit the... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Other"
] | null | proof only | null | |
0ctb | We say that a non-empty set $A$ consisting of real numbers is complete if for any real $a$ and $b$ such that $a+b \in A$, the number $ab$ also lies in $A$ (the numbers $a$ and $b$ are not required to be distinct or to belong to $A$). Find all complete sets.
Назовём непустое (конечное или бесконечное) множество $A$, со... | [
"The only such set is the set of reals.\nTake any $a \\in A$. Show successively that (i) $0 = 0 \\cdot a \\in A$, (ii) $-x^2 = (-x) \\cdot x \\in A$ for all $x \\in \\mathbb{R}$, and (iii) $y^2 = (-y) \\cdot (-y) \\in A$ for all $y > 0$.",
"Как и в первом решении, выберем произвольный элемент $s \\in A$. Докажем,... | Russia | Russian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English; Russian | proof and answer | the set of all real numbers | |
0f13 | Problem:
Circles radius $r$ and $R$ touch externally. $AD$ is parallel to $BC$. $AB$ and $CD$ touch both circles. $AD$ touches the circle radius $r$, but not the circle radius $R$, and $BC$ touches the circle radius $R$, but not the circle radius $r$. What is the smallest possible length for $AB$? | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 4√(Rr) | |
0e2o | Let $H$ be the orthocentre of the acute triangle $ABC$ and let $D$ be a point inside the triangle $ABH$. A line that passes through the point $D$ and is parallel to the line $AH$ intersects the segments $BC$ and $AB$ at $K$ and $L$. A line that passes through the point $D$ and is parallel to the line $BH$ intersects th... | [
"Denote the angle $\\angle ABC$ by $\\beta$. The line $KL$ is parallel to the altitude to the side $BC$, so $KL$ is perpendicular to $BC$. Hence, $KLB$ is a right triangle and $\\angle KLB = \\frac{\\pi}{2} - \\beta$. We get $\\angle KLN = \\frac{\\pi}{2} - \\beta$. Since $KLMN$ is a cyclic quadrilateral, we have $... | Slovenia | National Math 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 | |
0fkj | Problem:
Se tienen en el plano $3n$ puntos: $n$ de color blanco, $n$ de color azul y $n$ de color negro. Cada uno de los puntos está unido con puntos de color distinto al suyo mediante $n+1$ segmentos exactamente. Probar que hay, al menos, un triángulo formado por vértices de distinto color. | [
"Solution:\n\nConsideramos el punto que está conectado con el número más alto de puntos de otro color. Supongamos que este punto $N$ es de color negro y que está conectado a $k$ puntos de color blanco. Como $k \\leq n$ y $N$ está conectado a $n+1$ puntos, existirá un punto $A$ de color azul al que está conectado $N... | Spain | XLV Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
01vt | 2018 people are registered in a social network, some of them are friends. It is known that Borya has the largest number of friends, and Zhenya has the smallest number of friends, and the total number of friends of Borya and Zhenya is not less than $k$. According to the rules established by the administrator, only frien... | [
"Answer : $k = 2016$.\nWe show that $k \\ge 2016$. Note that Borya and Masha together have more friends than Borya and Zhenya, since Zhenya has the least number of friends among all users of the social network. So Borya and Masha together have at least $2017$ friends, and except Borya and Masha there are in total $... | Belarus | Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2016 | |
02ot | Emerald wrote a list of positive integers. Renan noticed that each number in the list and any sum of any quantity of distinct numbers from the list were square-free (that is, not divisible by any perfect square except, of course, 1). What is the maximum quantity of numbers that Emerald's list can have? | [
"The smallest perfect square, apart from $1$, is $2^2 = 4$. So let $a_1, a_2, \\dots, a_k$ be the numbers on the list modulo $4$. We cannot have $a_i = 0$; also, there is at most one $a_i$ equal to $2$ and we cannot have $a_i = 1$ and $a_j = 3$ simultaneously.\n\nWe claim that among any four distinct numbers $a_1, ... | Brazil | Brazilian Math Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 3 | |
0gio | 試求最大的正整數 $L$ 使得存在正整數數列 $a_1, a_2, \dots, a_L$ 滿足：
(a) 數列中的每一項都小於或等於 $2^{2024}$；
(b) 不存在連續子數列 $a_i, a_{i+1}, \dots, a_j$ (其中 $1 \le i \le j \le L$) 使得我們可以適當選取 $s_i, s_{i+1}, \dots, s_j \in \{-1, 1\}$ 讓
$$
s_i a_i + s_{i+1} a_{i+1} + \dots + s_j a_j = 0.
$$
Determine the maximum positive integer $L$ such that there exis... | [
"答案為 $2^{2025}-1$；一般性地, 對於上限 $2^k$, 最大的 $L = 2^{k+1}-1$。\n\n構造：令 $v_2(x)$ 為 $x$ 對 $2$ 的幂，並取 $a_i = 2^{k-v_2(i)}$。顯然 $a_i \\le 2^k$。此外，我們有：\n\n引理：對任意 $1 \\le i \\le j \\le 2^{k+1}-1$，存在唯一的 $i \\le x \\le j$ 使得 $v_2(x) = \\max_{i \\le y \\le j} v_2(y)$。\n\n證明：若我們在 $x$ 和 $y$ 都取到最大值 $v$，表示 $x = p \\times 2^v$ 且 $y = q ... | Taiwan | IMO 3J, Mock Exam 2 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Chinese; English | proof and answer | 2^{2025}-1 | |
0dyy | Problem:
a. Pokaži, da vsota števk števila $10^{n} + 9 n$ ni deljiva z 2007 za nobeno naravno število $n$.
b. Poišči vsaj eno naravno število $n$, za katero je vsota števk števila $10^{n} + 9 n$ enaka 2008. | [
"Solution:\n\na. Število je deljivo z 9 natanko tedaj, ko je vsota njegovih števk deljiva z 9. Recimo, da je vsota števk števila $10^{n} + 9 n$ deljiva z 2007. Ker je 2007 večkratnik števila 9, je potem vsota števk števila $10^{n} + 9 n$ deljiva z 9. To pa pomeni, da je $10^{n} + 9 n$ deljivo z 9, kar pa ne velja.\... | Slovenia | 52. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | There is no natural number n for which the digit sum of 10^n + 9n is divisible by 2007. An example with digit sum 2008 is n equal to the integer consisting of 223 ones. | |
02gl | For any non-empty subset $A$ of $\{1, 2, \dots, n\}$ define $f(A)$ as the largest element of $A$ minus the smallest element of $A$. Find $\sum f(A)$ where the sum is taken over all non-empty subsets of $\{1, 2, \dots, n\}$. | [
"Let $m$ and $M$ be the sum of the minima and maxima of all subsets. Since the diameter of a set is the difference between its maximum and its minimum, the desired sum is $M - m$. We may include unitary subsets, since their minima and maxima coincide.\n\nThe number $k$, $1 \\le k \\le n$, is the minimum of all subs... | Brazil | XXIV OBM | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | (n - 3) * 2^n + n + 3 | |
067q | Let $AB\Gamma$ be an acute angled triangle with $AB < A\Gamma$ and circumcenter $O$. The altitudes $B\Delta$, $\Gamma E$ meet at point $H$. If $O_1$ is the circumcenter of the triangle $BH\Gamma$, prove that the quadrilateral $AHO_1O$ is parallelogram. | [
"Since $O_1$ belongs to perpendicular bisector $OM$ of the segment $B\\Gamma$ and $AH$, $OO_1$, it is enough to prove that $AH = OO_1$. Since $AH = 2OM$, it is enough to prove that $OM = MO_1$. The quadrilateral is cyclic, whereby $\\hat{BH}\\Gamma = 180^\\circ - \\hat{A}$. Moreover $\\hat{BO}_1\\Gamma = 2\\hat{A}$... | Greece | SELECTION EXAMINATION | [
"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 | |
0cai | Problem:
Fie $I$ centrul cercului înscris în triunghiul $A B C$. Cercul de centru $A$ şi rază $A I$ intersectează cercul circumscris triunghiului $A B C$ în punctele $M$ şi $N$.
Demonstraţi că dreapta $M N$ este tangentă la cercul înscris în triunghiul $A B C$. | [] | Romania | Primul test de selecţie pentru OBMJ | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06rn | Let $P(x)$ and $Q(x)$ be two polynomials with integer coefficients such that no nonconstant polynomial with rational coefficients divides both $P(x)$ and $Q(x)$. Suppose that for every positive integer $n$ the integers $P(n)$ and $Q(n)$ are positive, and $2^{Q(n)}-1$ divides $3^{P(n)}-1$. Prove that $Q(x)$ is a constan... | [
"First we show that there exists an integer $d$ such that for all positive integers $n$ we have $\\operatorname{gcd}(P(n), Q(n)) \\leq d$.\nSince $P(x)$ and $Q(x)$ are coprime (over the polynomials with rational coefficients), Euclid's algorithm provides some polynomials $R_{0}(x), S_{0}(x)$ with rational coefficie... | IMO | 52nd International Mathematical Olympiad 2011 Shortlist | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
05az | In the classroom there is a clock, whose minute and hour hands move with a constant angular velocity. The minute hand works correctly, but the hour hand moves at half the angular velocity of the minute hand. At 10:00, the clock shows the correct time. When will the clock show the correct time for the next time? | [
"Imagine a correctly working clock next to the broken clock. The hour hand of the working clock moves 12 times slower than its minute hand, which means that its hour hand moves 6 times slower than the hour hand of the broken clock. As the minute hand of the broken clock is always at the correct position, the broken... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 12:24 | |
0b7g | Let $a_1$, $a_2$, $a_3$, $a_4$, $a_5$ be five real numbers of zero sum, such that $|a_i - a_j| \le 1$, for all $i, j \in \{1, 2, 3, 4, 5\}$. Prove that $a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2 \le \frac{6}{5}$. | [
"Since $0 = (a_1 + a_2 + a_3 + a_4 + a_5)^2 = a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2 + 2 \\sum_{i<j} a_i a_j$, it follows that $\\sum_{i<j} (a_i - a_j)^2 = 4(a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2) - 2 \\sum_{i<j} a_i a_j = 5(a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2)$. On the other side, $\\sum_{i<j} (a_i - a_j)^2 \\le ... | Romania | NMO Selection Tests for the Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
01x2 | The sum of several (not necessarily different) real numbers from $[0, 1]$ does not exceed $S$.
Find the maximal value of $S$ such that these numbers can always be partitioned into two groups with sums $A \le 8$ and $B \le 4$. | [
"Answer: $\\max S = 11.2$.\n\nFirst we will show that if $S > 11.2$, the required partition can be impossible. Indeed, let $S = 11.2 + 14\\epsilon$, $\\epsilon > 0$. Suppose that we are given 14 numbers equal $0.8 + \\epsilon$. The sum of any ten of these numbers exceeds $8$ and the sum of any five of them is great... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 11.2 | |
08vv | Let us denote by $S(k)$ the sum of the digits of a positive integer $k$. Find the smallest positive integer $n$ for which $S(n^2) = S(n) - 7$. | [
"For integers $a, b, m$ we write $a \\equiv b \\pmod m$ if $a - b$ is divisible by $m$. First, let us note that $S(k) \\equiv k \\pmod 9$. From $S(n^2) = S(n) - 7$ it then follows that $n^2 \\equiv n - 7 \\pmod 9$. So, we must have $n \\equiv 2, 5, 8 \\pmod 9$.\nFrom $0 < S(n^2) = S(n) - 7$, we see that $S(n) \\ge ... | Japan | Japan Junior Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 149 | |
0bko | Let $n \ge 6$ be an integer. We have at our disposal $n$ colors. We color each of the unit squares of an $n \times n$ board with one of the $n$ colors.
a) Prove that, for any such coloring, there exists a path of a chess knight from the bottom-left to the upper-right corner, that does not use all the colors.
b) Prove... | [
"We associate coordinates to each unit square, the square in the bottom-left corner having coordinates $(1, 1)$, while the one in the upper-right corner has coordinates $(n, n)$. Notice that one can get from $(k, \\ell)$ to $(k+3, \\ell+3)$ in two moves: $(k, \\ell) \\to (k+2, \\ell+1) \\to (k+3, \\ell+3)$.\n\n• $n... | Romania | 65th NMO Selection Tests for JBMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
04vo | Suppose that $a, b$ are real numbers such that the expressions $a^2 + b$ and $a + b^2$ share the same value. What is the smallest possible shared value? | [] | Czech Republic | First Round (take-home) | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | -1/4 | |
06ea | Let $O$ and $H$ be respectively the circumcentre and orthocentre of $\triangle ABC$. Let $A'$, $B'$ and $C'$ be the midpoints of $BC$, $CA$ and $AB$ respectively and $D$, $E$ and $F$ be respectively the feet from $A$, $B$ and $C$ to the opposite sides. Show that $OA' \cdot HD = OB' \cdot HE = OC' \cdot HF$. | [
"It is well-known that $OA' = \\frac{1}{2} AH$, etc. Therefore, we need to prove\n$$\nAH \\times HD = BH \\times HE = CH \\times HF.\n$$\nIndeed, $AH \\times HD = BH \\times HE$ is true since $A$, $B$, $D$, $E$ are concyclic. By symmetry, the other equality also holds.\n\n"
] | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
04l0 | In a triangle $ABC$ the interior bisector at vertex $C$ meets the side $\overline{AB}$ at point $D$. Let $a$ and $b$ be the lengths of the sides $\overline{BC}$ and $\overline{AC}$ respectively. If $|CD| = \frac{ab}{a+b}$, determine $\angle ACB$. | [] | Croatia | Mathematical competitions in Croatia | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 120° | |
0fpg | Problem:
Sea $P = A_1A_2...A_k$ un polígono convexo en el plano. Los vértices $A_1, A_2, ..., A_k$ tienen coordenadas enteras y se encuentran sobre una circunferencia. Sea $S$ el área de $P$. Sea $n$ un entero positivo impar tal que los cuadrados de las longitudes de los lados de $P$ son todos números enteros divisibl... | [] | Spain | LVII Olimpiada Internacional de Matemáticas | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Number Theory > Modul... | Spanish | proof only | null | |
0155 | Problem:
Suppose that Romeo and Juliet each have a regular tetrahedron to the vertices of which some positive real numbers are assigned. They associate each edge of their tetrahedra with the product of the two numbers assigned to its end points. Then they write on each face of their tetrahedra the sum of the three numb... | [
"Solution:\nLet us prove that this conclusion can in fact be drawn. For this purpose we denote the numbers assigned to the vertices of Romeo's tetrahedron by $r_{1}, r_{2}, r_{3}, r_{4}$ and the numbers assigned to the vertices of Juliette's tetrahedron by $j_{1}, j_{2}, j_{3}, j_{4}$ in such a way that\n$$\n\\begi... | Baltic Way | Baltic Way 2008 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Yes | |
0dv7 | Problem:
Najmanj koliko zvezdic moramo narisati v tabelo velikosti $4 \times 4$, da bo po brisanju poljubnih 2 stolpcev in poljubnih 2 vrstic ostala v tabeli vsaj 1 zvezdica? | [
"Solution:\n\nSlika kaže, da zadošča 7 zvezdic. Dokažimo, da 6 zvezdic ne zadošča. V tem primeru je vsaj v 2 stolpcih največ 1 zvezdica. Če zbrišemo preostala 2 stolpca, ostaneta le še 2 zvezdici, ki ju lahko zbrišemo, če izpraznimo vrstici, v katerih ležita.\n\n"
] | Slovenia | 46. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 7 | |
004y | Dos equipos, $A$ y $B$, disputan el territorio limitado por una circunferencia.
$A$ tiene $n$ banderas azules y $B$ tiene $n$ banderas blancas ($n \ge 2$, fijo). Juegan alternadamente y $A$ comienza el juego. Cada equipo, en su turno, coloca una de sus banderas en un punto de la circunferencia que no se haya usado en u... | [] | Argentina | XXII Olimpiada Iberoamericana de Matemática | [
"Geometry > Plane Geometry > Circles",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | Spanish | proof only | null | |
08rr | Let $n$ be a positive integer. Two people $P$, $Q$ play a game in which they call an integer $m$ ($1 \le m \le n$) alternately. $P$ calls the first number. They cannot call the numbers which are already called by themselves or by their opponent. The game is over when neither can call numbers. If the sum of the numbers ... | [
"Let the number called by a player in the $m$th turn be $N_m$. Then sequence $(N_1, \\dots, N_l)$ is called \"history up to the $l$th turn\". We call $j$ which satisfies $j \\neq N_1, \\dots, N_l$ \"free in the $l+1$th turn\". We are going to prove a proposition that if $n \\equiv 0, 4, 5 \\pmod 6$, $P$ can absolut... | Japan | Japan 2007 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Other"
] | English | proof and answer | P can force a win if and only if n ≡ 0, 4, or 5 modulo 6. | |
0d3h | There are 14 students who have participated in a 3 hour test consisting of 15 short problems. Each student has solved a different number of problems and each problem has been solved by a different number of students. Prove that there exists a student who has solved exactly 5 problems. | [
"Because there are 14 students, the possible numbers of students solving a problem are $0, 1, \\ldots, 14$. Because there are 15 problems and each problem has been solved by a different number of students, there is a problem which has been solved by all the 14 students and another problem which has not been solved ... | Saudi Arabia | SAMC | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English, Arabic | proof only | null | |
0l53 | Let $A_1A_2A_3\ldots A_{11}$ be an 11-sided non-convex simple polygon with the following properties:
* For every integer $2 \leq i \leq 10$, the area of $\triangle A_iA_{i+1}$ is equal to $1$.
* For every integer $2 \leq i \leq 10$, $\cos(\angle A_iA_{i+1}) = \frac{12}{13}$.
* The perimeter of the 11-gon $A_1A_2A_3\ldo... | [] | United States | AIME II | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 19 | |
0cpf | 10 distinct real numbers are given. For each pair $a, b$ of given numbers, Basil writes down into his exercise book the number $(a-b)^2$, while Pete writes down into his exercise book the number $|a^2 - b^2|$. Determine whether it may appear that Basil and Pete obtain the same collections of 45 numbers in their exercis... | [
"Не могли.\n\nПредположим противное. Если среди исходных чисел есть ноль, то для любого другого числа $a$ имеем $a^2 - 0^2 = (a - 0)^2$. Значит, если вычеркнуть ноль, то останутся 9 чисел, также удовлетворяющих условию.\n\nИтак, можно считать, что исходных чисел 9 или 10, и все они ненулевые. Пусть среди них есть ч... | Russia | Russian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English, Russian | proof only | null | |
0av5 | Problem:
A circle is inscribed in a $2$ by $2$ square. Four squares are placed on the corners (the spaces between circle and square), in such a way that one side of the square is tangent to the circle, and two of the vertices lie on the sides of the larger square. Find the total area of the four smaller squares. | [
"Solution:\nLet the large square have side length $2$, and the inscribed circle has radius $1$ (since the diameter equals the side of the square).\n\nLet $s$ be the side length of one of the small squares. Place the large square with vertices at $(0,0)$, $(2,0)$, $(2,2)$, $(0,2)$, and the circle centered at $(1,1)$... | Philippines | 19th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 12 - 8√2 | |
041d | In a plane rectangular coordinate system $xOy$, points $A$, $B$ are on the parabola $y^2 = 4x$, satisfying $\vec{OA} \cdot \vec{OB} = -4$, and point $F$ is the focus of the parabola. Then $S_{\triangle OFA} \cdot S_{\triangle OFB} = \_\_\_\_\_\_$. | [
"Let $F(1, 0)$, $A(x_1, y_1)$, $B(x_2, y_2)$. Then $x_1 = \\frac{y_1^2}{4}$, $x_2 = \\frac{y_2^2}{4}$, and\n$$\n-4 = \\vec{OA} \\cdot \\vec{OB} = x_1x_2 + y_1y_2 = \\frac{1}{16}(y_1y_2)^2 + y_1y_2,\n$$\nfrom which we have $\\frac{1}{16}(y_1y_2 + 8)^2 = 0$, or $y_1y_2 = -8$.\nTherefore,\n$$\n\\begin{align*} \nS_{\\t... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof and answer | 2 | |
00hf | Let $\mathbb{Z}^{+}$ be the set of positive integers. Determine all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$ such that $a^{2} + f(a) f(b)$ is divisible by $f(a) + b$ for all positive integers $a$ and $b$. | [
"First we perform the following substitutions on the original relation:\n\n1. With $a = b = 1$, we find that $f(1) + 1 \\mid f(1)^{2} + 1$, which implies $f(1) = 1$.\n\n2. With $a = 1$, we find that $b + 1 \\mid f(b) + 1$. In particular, $b \\leq f(b)$ for all $b \\in \\mathbb{Z}^{+}$.\n\n3. With $b = 1$, we find t... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | f(n) = n for all positive integers n | |
07z5 | Problem:
100 delegati sono riuniti in congresso. Non tutti portano la cravatta, ma si sa che comunque se ne scelgano due, almeno uno dei due la porta. Quanti sono i congressisti con cravatta?
(A) Almeno 2, ma possono essere meno di 50
(B) Esattamente 50
(C) Più di 50, ma non si può dire esattamente quanti
(D) La situ... | [] | Italy | Progetto Olimpiadi di Matematica | [
"Discrete Mathematics > Logic"
] | null | MCQ | E | |
0gzd | $n \ge 5$ real numbers are written in a row. It turns out, that sum of any three consecutive numbers is positive and sum of any five consecutive numbers is negative. Find the largest $n$ for which it is possible?
**Answer:** $n = 6$. | [
"We first construct the example for $n = 6$: $3, -5, 3, 3, -5, 3$.\n\nSuppose that there exist $n \\ge 7$ real numbers, that satisfy the conditions of the problem and choose any 5 consecutive numbers $a, b, c, d, e$. Using the conditions we get: $a+b+c > 0$, $c+d+e > 0 \\Rightarrow (a+b+c+d+e) + c > 0$ and $a+b+c+d... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 23, 2010) | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 6 | |
0ise | Problem:
$5N$ teams participated in a national basketball championship in which every two teams played exactly one game. Of the $N$ teams, 251 are from California. It turned out that a Californian team Alcatraz is the unique Californian champion (Alcatraz has won more games against Californian teams than any other tea... | [
"Solution:\n\nWe will prove that $N=255$ is the smallest value.\n\n- Let us first construct a tournament with the described properties and 255 participating teams. First arrange 251 Californian teams in the circle and label them by $0,1, \\ldots, 250$ in the counter-clockwise direction (0 is Alcatraz). If each team... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 255 | |
06s0 | Let $ABC$ be a triangle with $\angle BCA = 90^{\circ}$, and let $C_0$ be the foot of the altitude from $C$. Choose a point $X$ in the interior of the segment $CC_0$, and let $K, L$ be the points on the segments $AX, BX$ for which $BK = BC$ and $AL = AC$ respectively. Denote by $M$ the intersection of $AL$ and $BK$. Sho... | [
"Let $C'$ be the reflection of $C$ in the line $AB$, and let $\\omega_1$ and $\\omega_2$ be the circles with centers $A$ and $B$, passing through $L$ and $K$ respectively. Since $AC' = AC = AL$ and $BC' = BC = BK$, both $\\omega_1$ and $\\omega_2$ pass through $C$ and $C'$. By $\\angle BCA = 90^{\\circ}$, $AC$ is t... | IMO | 53rd International Mathematical Olympiad Shortlisted Problems with Solutions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
01na | Suppose that $1000$ students are standing in a circle.
Prove that there exists an integer $k$ with $100 \le k \le 300$ such that in the circle there exists a contiguous group of $2k$ students, for which the first half contains the same number of girls as the second half. | [] | Belarus | Belorusija 2012 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0bdi | Fix a point $O$ in the plane and an integer $n \ge 3$. Consider a finite set $\mathcal{D}$ of closed unit discs in the plane such that:
(a) No disc in $\mathcal{D}$ contains the point $O$; and
(b) For each positive integer $k < n$, the closed disc of radius $k+1$ centred at $O$ contains the centres of at least $k$ di... | [
"For each disc $D$ in $\\mathcal{D}$, let $\\omega_D$ denote the centre of $D$, and let $\\alpha_D$ be the arc-length of the image of $D$ under radial projection from $O$ onto the unit circle centred at $O$. Clearly, $\\alpha_D/2 > \\sin(\\alpha_D/2) = 1/O\\omega_D$.\n\nNow, for each positive integer $k < n$, let $... | Romania | 64th NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
09zp | For how many integers $n$ with $1 \le n \le 800$ is the number $8n + 1$ a square? | [
"We want $8n + 1 = k^2$ for some integer $k$.\n\nThen $8n = k^2 - 1 = (k - 1)(k + 1)$.\n\nSo $n = \\dfrac{(k - 1)(k + 1)}{8}$.\n\nWe require $n$ to be an integer with $1 \\leq n \\leq 800$.\n\nLet us analyze when $n$ is an integer.\n\nNote that $k$ must be odd, since $k^2 \\equiv 1 \\pmod{8}$ only when $k$ is odd.\... | Netherlands | Second Round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | final answer only | 39 | |
033m | Problem:
Let $p$ be a prime number and let $0 \leq a_{1} < a_{2} < \cdots < a_{m} < p$ and $0 \leq b_{1} < b_{2} < \cdots < b_{n} < p$ be arbitrary integers. Denote by $k$ the number of different remainders of the numbers $a_{i} + b_{j}$, $1 \leq i \leq m$, $1 \leq j \leq n$, modulo $p$. Prove that:
a) if $m + n > p$... | [
"Solution:\n\na.\nLet $t \\in \\{0, 1, 2, \\ldots, p-1\\}$. Consider the remainders of $t - a_{i}$, $1 \\leq i \\leq m$ and $b_{j}$, $1 \\leq j \\leq n$, modulo $p$. Their number is $m + n > p$ and hence two of them are equal. Since the remainders of $t - a_{i}$ and $t - a_{j}$, $b_{i}$ and $b_{j}$, respectively, $... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0e5g | Find all 3-digit natural numbers $n$ with non-zero digits that satisfy the following condition: if we delete the left digit of $n$, the obtained number is a divisor of $n$. | [
"Let us write $n = \\overline{abc}$. The condition of the problem says that $\\overline{bc}$ divides $\\overline{abc} = 100a + \\overline{bc}$, hence it also divides $100a$. First suppose that $\\overline{bc}$ is not divisible by $5$. Then $\\overline{bc}$ must divide $4a$. Because $\\overline{bc}$ is a 2-digit nat... | Slovenia | National Math Olympiad 2012 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 125, 225, 312, 315, 325, 375, 416, 425, 525, 612, 615, 624, 625, 675, 714, 725, 728, 735, 816, 825, 832, 912, 915, 918, 925, 936, 945, 975 | |
04as | Let $n \ge 3$ be an integer. Determine the minimum number of points one has to mark inside a convex $n$-gon in order for the interior of any triangle with the vertices at vertices of the $n$-gon to contain at least one of the marked points. | [
"Since all diagonals from one vertex divide an $n$-gon into $n-2$ disjoint triangles, at least $n-2$ points are necessary.\n\nWe claim that it is possible to mark $n-2$ points so that the given condition is satisfied. Denote the vertices of the given $n$-gon with $A_1, A_2, \\dots, A_n$. Draw all the diagonals of t... | Croatia | Hrvatska 2011 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | n - 2 | |
0gn9 | Let $A$ and $B$ be distinct points on a circle $\Gamma$. For a variable point $P$ different from $A$ and $B$ on $\Gamma$, find the geometric locus of the point $M$ such that $PM$ is the opposite ray to the angle bisector of $\angle APB$ and $MP = AP + PB$. | [
"Draw the diameter $DL$ perpendicular to the secant $AB$. Let $\\alpha = \\angle LDA = \\angle LDB$. Assume that the point $P$ lies on the arc $ALB$ and between the points $A$ and $L$, and let $\\beta = \\angle LDP$. Also let $K$ be the point of intersection of the line $LD$ and the perpendicular to the line $DM$ t... | Turkey | Team Selection Test for IMO 2007 | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | Let D and L be the endpoints of the diameter of the circle perpendicular to AB. Define K on line DL so that KD = 2R(1 + 2 sin ∠LDA). When P ranges over the arc between A and L on the same side as B, the point M traces the arc of the circle with diameter DK lying inside ∠ADB. Similarly, define K′ on line DL on the same ... | |
0i1f | Problem:
Each vertex of a regular 17-gon is colored red, blue, or green in such a way that no two adjacent vertices have the same color. Call a triangle "multicolored" if its vertices are colored red, blue, and green, in some order. Prove that the 17-gon can be cut along nonintersecting diagonals to form at least two ... | [
"Solution:\n\nDenote the colors by $1, 2, 3$. Notice that all three colors must be used. This is true because if only two colors were used, the vertex coloring would be of the type $121212 \\ldots$ which is impossible, since $17$ is odd (this would make two adjacent vertices the same color). Hence there are three c... | United States | 3rd Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
05nb | Problem:
Soit $ABC$ un triangle, et $D, E, F$ les pieds des hauteurs de $A, B, C$ respectivement. On définit aussi $H$ l'orthocentre de $ABC$, $O$ le centre de son cercle circonscrit, et $X$ le point de la droite $(EF)$ qui vérifie $XA = XD$. Montrer que les droites $(AX)$ et $(OH)$ sont perpendiculaires. | [
"Solution:\n\n\n\nOn sait que l'axe radical de deux cercles est une droite perpendiculaire à la droite passant par les centres des deux cercles.\n\nLa condition $XA = XD$ signifie que $X$ est sur la médiatrice de $[AD]$, qui n'est autre que la droite des milieux $\\left(B', C'\\right)$ (ave... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle ... | null | proof only | null | |
02m9 | Problem:
O valor absoluto $|a|$ de um número $a$ qualquer é definido por
$$
|a|=\left\{\begin{array}{cl}
a & \text{ se } a>0 \\
0 & \text{ se } a=0 \\
-a & \text{ se } a<0
\end{array}\right.
$$
Por exemplo, $|6|=6$, $|-4|=4$ e $|0|=0$. Quanto vale $N=|5|+|3-8|-|-4|$ ?
(a) 4
(b) -4
(c) 14
(d) -14
(e) 6 | [
"Solution:\n\nTemos: $|5|=5$, $|3-8|=|-5|=5$ e $|-4|=4$. Logo, $N=5+5-4=6$."
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | e | |
08m9 | Problem:
A group of $n>1$ pirates of different ages owned a total of 2009 coins. Initially each pirate (except the youngest one) had one coin more than the next younger.
a) Find all possible values of $n$.
b) Every day a pirate was chosen. The chosen pirate gave a coin to each of the other pirates. If $n=7$, find the... | [
"Solution:\n\na) If $n$ is odd, then it is a divisor of $2009=7 \\times 7 \\times 41$. If $n>49$, then $n$ is at least $7 \\times 41$, while the average pirate has 7 coins, so the initial division is impossible. So, we can have $n=7$, $n=41$ or $n=49$. Each of these cases is possible (e.g. if $n=49$, the average pi... | JBMO | 2009 Shortlist JBMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a) n ∈ {2, 7, 14, 41, 49}. b) 1994 | |
0hpv | Problem:
There are $m$ friends with $n$ cupcakes each weighing 1 ounce. They wish to split the cupcakes equally by dividing each cupcake into some number of parts, and allocating some parts to each person.
a) Assume $m=3$ and $n=5$. Show that they may divide the cupcakes with all pieces being larger than $\frac{1}{3}... | [
"Solution:\n\na.\nEach person needs $\\frac{5}{3} = \\frac{20}{12}$ of a cupcake. If they cut four of the cupcakes into $\\frac{5}{12}$ and $\\frac{7}{12}$ and the last cupcake in half, then one person can take the four $\\frac{5}{12}$ pieces, giving them\n$$\n4 \\cdot \\frac{5}{12} = \\frac{20}{12}\n$$\nand the ot... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof only | null | |
0ccr | If $x$, $y$, $z$ are positive numbers with $x + y + z = 1$, show that:
$$
\begin{align*}
a)\ 1 - \frac{x^2-yz}{x^2+x} &= \frac{(1-y)(1-z)}{x^2+x}; \\
b)\ \frac{x^2-yz}{x^2+x} + \frac{y^2-zx}{y^2+y} + \frac{z^2-xy}{z^2+z} &\le 0.
\end{align*}
$$ | [
"a.\n$1 - \\frac{x^2 - yz}{x^2 + x} = \\frac{x + yz}{x^2 + x} = \\frac{1 - y - z + yz}{x^2 + x} = \\frac{(1 - y)(1 - z)}{x^2 + x}$.\n\nb.\nUsing a), the inequality is rewritten\n$$\n\\frac{(1-y)(1-z)}{x(x+1)} + \\frac{(1-z)(1-x)}{y(y+1)} + \\frac{(1-x)(1-y)}{z(z+1)} \\geq 3,\n$$\nthat is\n$$\n\\frac{(x+z)(x+y)}{x[(... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0bcr | Problem:
Un cuvânt este o secvenţă finită de litere dintr-un anume alfabet. Un cuvânt se zice repetitiv dacă este o concatenare de cel puţin două sub-cuvinte identice (de exemplu, $ababab$ şi $abcabc$ sunt repetitive, dar $ababa$ şi $aabb$ nu sunt). Demonstraţi că dacă un cuvânt are proprietatea că orice transpoziţie ... | [] | Romania | Olimpiada europeana de matematica a fetelor | [
"Discrete Mathematics > Other",
"Discrete Mathematics > Other"
] | null | proof only | null | |
07mr | Suppose $abc \neq 0$. Express in terms of $a$, $b$, and $c$, the solutions $x$, $y$, $z$, $u$, $v$, $w$ of the equations
$$
x + y = a, \quad z + u = b, \quad v + w = c, \quad ay = bz, \quad bu = cv, \quad cw = ax.
$$ | [
"Label the equations 1, 2, 3, 4, 5, 6 in the order of their appearance. Using equations 3, 6 eliminate $w$ from the system, thereby adding to equations 1, 2, 4, 5 the equation 7: $ax + cv = c^2$. Next, use this and 5 to eliminate $v$ to produce equation 8: $ax + bu = c^2$. Eliminate $z$ from 2, 4 to produce equatio... | Ireland | Ireland | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | Let α = (b^2 + c^2 − a^2) / (2bc), β = (c^2 + a^2 − b^2) / (2ca), γ = (a^2 + b^2 − c^2) / (2ab). Then x = cβ, y = bγ, z = aγ, u = cα, v = bα, w = aβ. | |
0dbq | How many integers $n$ satisfy the following conditions?
i) $219 \leq n \leq 2019$,
ii) there exist $x, y \in \mathbb{Z}$ such that $1 \leq x < n < y$ and $y$ is divisible by all integers from $1$ to $n$, except two numbers $x$ and $x+1$. | [
"The answer is $292$.\n\nWe can see that if $x = pq$ for some integers $p, q > 1$ and $\\gcd(p, q) = 1$ then $1 < p, q < x$ which implies that $p \\mid y$, $q \\mid y$, then $pq \\mid y$, contradiction.\n\nHence $x$ and $x+1$ must be the powers of primes. But one of these numbers is even so one of them must be the ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 292 | |
0869 | Problem:
Il So-poko è un nuovo gioco enigmistico che si gioca su una tabella quadrata di lato $203$ caselle. Le caselle sono colorate di bianco e di nero a cornici concentriche alternate; la cornice più esterna è nera, mentre la casella centrale è bianca (vedi a fianco un esempio $7 \times 7$). Qual è la differenza tr... | [
"Solution:\n\nLa risposta è $(\\mathbf{E})$. Consideriamo come aumenta la differenza tra caselle nere e bianche $d_{k}$ in un so-poko di lato $k$. Per $k=3$, la differenza è $7$. Ogni volta che vengono aggiunte all'esterno due cornici concentriche (una bianca più interna e una nera più esterna), $d_{k}$ aumenta di ... | Italy | Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | MCQ | E | |
03k7 | Problem:
In triangle $ABC$, the medians to the sides $AB$ and $AC$ are perpendicular. Prove that $\cot B + \cot C \geq \frac{2}{3}$. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Ine... | null | proof only | null | |
0klq | Problem:
Let $m, n > 2$ be integers. One of the angles of a regular $n$-gon is dissected into $m$ angles of equal size by $(m-1)$ rays. If each of these rays intersects the polygon again at one of its vertices, we say $n$ is $m$-cut. Compute the smallest positive integer $n$ that is both 3-cut and 4-cut. | [
"Solution:\n\nFor the sake of simplicity, inscribe the regular polygon in a circle. Note that each interior angle of the regular $n$-gon will subtend $n-2$ of the $n$ arcs on the circle. Thus, if we dissect an interior angle into $m$ equal angles, then each must be represented by a total of $\\frac{n-2}{m}$ arcs. H... | United States | HMMT Spring 2021 Guts Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 14 | |
02kr | Problem:
Números proporcionais - Se $\frac{x}{y}=\frac{3}{z}$, então $9 y^{2}$ é igual a:
(a) $\frac{x^{2}}{9}$
(b) $x^{3} z$
(c) $3 x^{2}$
(d) $x^{2} z^{2}$
(e) $\frac{1}{9} x^{2} z^{2}$ | [
"Solution:\n\nComo $\\frac{x}{y}=\\frac{3}{z}$, então $x z=3 y$. Elevando ao quadrado ambos os membros dessa igualdade obtemos $x^{2} z^{2}=9 y^{2}$. A opção correta é (d)."
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | d |
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