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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0gyh | On the sides $AB$, $BC$, $CA$ of triangle $ABC$ such points $C_1$, $A_1$, $B_1$ are chosen that the lines $AA_1$, $BB_1$ and $CC_1$ meet at point $K$. The perpendiculars are dropped from the point $K$ onto the sides of triangle. Lines $l_1$, $l_2$, $l_3$ are drawn through the feet of these perpendiculars, parallel to l... | [
"The points $A$, $M$, $K$, $N$ are cyclic, therefore\n$$\n\\begin{aligned}\n\\angle KAN &= \\angle NMK = \\\\\n&= 90^\\circ - \\angle AMN = \\angle MAP, \\text{ and so } AP \\perp MN. \\text{ Due to } l_1 \\parallel AP = u_1 \\text{ we have } l_1 \\perp MN, \\\\\n\\text{therefore the lines } l_1, l_2, l_3 \\text{ a... | Ukraine | The Problems of Ukrainian Authors | [
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point ci... | English | proof only | null | |
0ba8 | a) Find the largest possible value of the number
$$
x_1x_2 + x_2x_3 + \dots + x_{n-1}x_n,
$$
if $x_1, x_2, \dots, x_n$ ($n \ge 2$) are non-negative integers and their sum is $2011$.
b) Find the numbers $x_1, x_2, \dots, x_n$ for which the maximum value determined at a) is obtained. | [
"a.\nLet $x_1, x_2, \\dots, x_n$ be non-negative integers satisfying the conditions from the statement. We put $M = \\max_{1 \\le i \\le n} x_i$. If $x_j = M$ then\n$$\nx_1x_2 + x_2x_3 + \\dots + x_{n-1}x_n \\le x_1x_j + x_2x_j + \\dots + x_{j-1}x_j + x_jx_{j+1} + x_jx_{j+2} + \\dots + x_jx_n = x_j(2011-x_j) = M(20... | Romania | 62nd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | a) Maximum value: 1005 * 1006.
b) Maximizing sequences:
- For n = 2: (1005, 1006) or (1006, 1005).
- For n = 3: (k, 1005, 1006 − k) for k = 0, 1, ..., 1006; or (k, 1006, 1005 − k) for k = 0, 1, ..., 1005.
- For n ≥ 4: all entries are zero except for three consecutive positions, which equal (k, 1005, 1006 − k) with k =... | |
08y0 | Let $ABC$ be a triangle, which is not an isosceles triangle. Let $\Gamma$ be its circum-circle and $I$ be its in-center. Let $D$ and $E$ be the points of tangency of the in-circle of $ABC$ to the side $AB$ and $AC$, respectively. Let $P$ be the point of intersection, different from $B$, of $\Gamma$ and the circum-circl... | [
"Let $\\alpha = \\frac{1}{2} \\angle CAB$, $\\beta = \\frac{1}{2} \\angle ABC$, $\\gamma = \\frac{1}{2} \\angle BCA$. Since $P$ lies on the arc $AC$ opposite to $B$, we have $\\angle APB = \\angle ACB = 2\\gamma$. We will show that $\\angle APE = \\alpha + \\gamma$ holds, separating cases depending on the positions... | Japan | Japan 2015 | [
"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 > Miscellaneous > Angle c... | null | proof only | null | |
0e93 | Prove that every tangential quadrilateral whose diagonals intersect at right angles is a deltoid. | [
"Let us use the notation from the figure. By Pythagoras' theorem we have $a^2 = x^2 + y^2$, $b^2 = y^2 + z^2$, $c^2 = z^2 + w^2$ and $d^2 = w^2 + x^2$. This implies $a^2 + c^2 = b^2 + d^2$.\n\nBut the quadrilateral is tangential, so $a + c = b + d$. Squaring this equality and using the equalities obtained above we ... | Slovenia | National Math Olympiad 2013 - First Round | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
04ty | To a sequence of $n$ zeros and $n$ ones, we assign the number of maximal contiguous runs of equal digits in it. (For instance, sequence $00111001$ has four such runs: $00$, $111$, $00$, and $1$.) For a given $n$ we sum up all the numbers assigned to all such sequences. Prove that the resulting sum is equal to
$$
(n + 1... | [
"Consider one such sequence and let us count (from left to right) how many maximal contiguous runs (from now on, just runs) it contains. We count a new run when it ends, that is when we hit a different digit or the right end. The number of runs is thus one more than the number of digits that follow a different digi... | Czech Republic | 66th Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof and answer | (n + 1) * C(2n, n) | |
0gqa | In a convex quadrilateral $ABCD$, let $E$ be the intersection of the diagonals. It is given that $\angle EDC = \angle DEC = \angle BAD$. If $F$ is a point on the line segment $[BC]$ such that $\angle BAF + \angle EBF = \angle BFE$, then show that $A, B, F, D$ are concyclic. | [
"It is easy to observe that the point $F$ is unique when $A, B, C, D$ are fixed. Hence it is enough to show that the intersection of the circumcircle of $ABD$ and $[CB]$ satisfies the properties of the point $F$. Let the intersection be $F'$.\n$\\angle DBF' = \\angle DAF'$. Then $\\angle BAF' + \\angle EBX = \\angl... | Turkey | Team Selection Test for IMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
04sk | Real numbers $x, y, z$ satisfy
$$
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + x + y + z = 0
$$
and none of them lies in the open interval $(-1, 1)$. Find the maximum value of $x+y+z$. | [
"**Solution:** By changing $(x, y, z)$ to $(-x, -y, -z)$, the condition (1) remains valid and the value of $x+y+z$ changes sign. Since the ordering of the numbers $x, y, z$ is irrelevant and they can not be of the same sign because of (1), we can without the loss of generality assume $x > 0, y > 0$, and $z < 0$, an... | Czech Republic | 15th Czech-Polish-Slovak Mathematics Competition | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | sqrt(3) | |
0j7a | Problem:
Let $w = w_{1}, w_{2}, \ldots, w_{6}$ be a permutation of the integers $\{1, 2, \ldots, 6\}$. If there do not exist indices $i < j < k$ such that $w_{i} < w_{j} < w_{k}$ or indices $i < j < k < l$ such that $w_{i} > w_{j} > w_{k} > w_{l}$, then $w$ is said to be exquisite. Find the number of exquisite permutat... | [
"Solution:\nGiven a permutation $w = w_{1}, \\ldots, w_{n}$ for some $n$, call a sequence $w_{i_{1}}, w_{i_{2}}, \\ldots, w_{i_{m}}$ an increasing subsequence if $i_{1} < \\cdots < i_{m}$ and $w_{i_{1}} < \\cdots < w_{i_{m}}$. Define decreasing subsequences similarly. Let $is(w)$ denote the length of the longest in... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 25 | |
011g | Problem:
Given an isosceles triangle $A B C$ with $\angle A=90^\circ$. Let $M$ be the midpoint of $A B$. The line passing through $A$ and perpendicular to $C M$ intersects the side $B C$ at $P$. Prove that $\angle A M C=\angle B M P$. | [
"\nFigure 2\n\nChoose the point $K$ such that $A B K C$ is a square. Let $N$ be the point of intersection of $A P$ and $B K$ (see Figure 2). Since the lines $A N$ and $C M$ are perpendicular, $N$ is the midpoint of $B K$. Moreover, triangles $A M C$ and $B N A$ are congruent, which gives\n$... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0i59 | Problem:
We are given the values of the differentiable real functions $f$, $g$, $h$, as well as the derivatives of their pairwise products, at $x=0$:
$$
f(0)=1 ; \quad g(0)=2 ; \quad h(0)=3 ; \quad (g h)'(0)=4 ; \quad (h f)'(0)=5 ; \quad (f g)'(0)=6 .
$$
Find the value of $(f g h)'(0)$. | [
"Solution:\n16\n\nBy the product rule,\n$$(f g h)' = f' g h + f g' h + f g h' = \\left( (f g)' h + (g h)' f + (h f)' g \\right) / 2.$$\nEvaluated at $0$, this gives $16$."
] | United States | Harvard-MIT Math Tournament | [
"Calculus > Differential Calculus > Derivatives"
] | null | final answer only | 16 | |
0a0e | There is a power plant in the cell indicated by $C$. The cells indicated by $X$ are places to which the power plant should be connected. This requires electricity masts in some cells, which we can indicate by placing an $M$ in that cell. Two cells with a letter in them can be connected by a cable if the cells are adjac... | [
"B) 4"
] | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Other"
] | English | MCQ | B | |
0be8 | Determine all injective functions $f$ of the set of positive integers into itself satisfying the following condition: If $S$ is a finite set of positive integers such that $\sum_{s \in S} 1/s$ is an integer, then $\sum_{s \in S} 1/f(s)$ is also an integer. | [
"We shall prove that the identity is the unique function satisfying the conditions in the statement. Clearly, $f(1) = 1$, so $f(n) \\ge 2$ if $n \\ge 2$, by injectivity. We will use the following well-known result.\n\n**Egyptian fractions theorem.** For every positive rational $r$ and positive integer $n$, there ex... | Romania | 64th NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Other"
] | null | proof and answer | f(n) = n for all positive integers n | |
050o | There are 2012 points marked in a square with side length $11$. Prove that one can choose an equilateral triangle with side length $12$ which covers at least $671$ points. | [
"Place two equilateral triangles with side lengths $12$ on the square in such a way that both have one vertex lie on the side of the square and the opposite sides of these vertices partially coincide with the other side of the square and with each other (Fig. 16). The area common to both triangles forms an equilate... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0dbl | Let $a$, $b$, $c$ be positive real numbers such that $a^{2} + b^{2} + c^{2} = 1$. Find the minimum for
$$
S = \frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} - \frac{2\left(a^{3} + b^{3} + c^{3}\right)}{a b c}
$$ | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 3 | |
0k6y | Problem:
Let $N=2^{\left(2^{2}\right)}$ and $x$ be a real number such that $N^{\left(N^{N}\right)}=2^{\left(2^{x}\right)}$. Find $x$. | [
"Solution:\nWe compute\n$$\nN^{\\left(N^{N}\\right\\)}=16^{16^{16}}=2^{4 \\cdot 2^{4 \\cdot 2^{4}}}=2^{2^{2^{6}+2}}=2^{2^{66}}\n$$\nso $x=66$."
] | United States | HMMT February 2019 | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | 66 | |
07e2 | Let $n$ be an integer and $0 \le m \le n^2 - 1$. Prove that the total number of $m$ such that
$$
x^n + y^n \equiv m \pmod{n^2}
$$
has no integer solution, is at least $\frac{n(n-1)}{2}$. | [
"It is known that if $n$ divides $a - b$, then $a^n - b^n$ is divisible by $n^2$. Now, choose $z \\equiv x \\pmod{n}$ and $y \\equiv t \\pmod{n}$. Then\n$$\nx^n + y^n - (z^n + t^n) = (x^n - z^n) + (y^n - t^n) \\equiv 0 \\pmod{n^2}\n$$\nThus, for $z, t$ belonging to $\\{0, 1, \\dots, n-1\\}$, we must calculate maxim... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof only | null | |
0at5 | Problem:
How many integral solutions does the equation
$$
|x| + |y| + |z| = 9
$$ | [
"Solution:\n326"
] | Philippines | Philippines Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 326 | |
0f23 | Problem:
We call a number doubly square if (1) it is a square with an even number $2n$ of (decimal) digits, (2) its first $n$ digits form a square, (3) its last $n$ digits form a non-zero square. For example, $1681$ is doubly square, but $2500$ is not.
(1) Find all $2$-digit and $4$-digit doubly square numbers.
(2) ... | [] | Soviet Union | ASU | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | (1) 2-digit: 49; 4-digit: 1681. (2) Yes, for example 256036. (3) Yes, such a 20-digit number exists. (4) There are at least ten 100-digit such numbers. (5) Yes, such a 30-digit number exists. | |
0ijh | Problem:
Compute
$$
\sum_{n=1}^{\infty} \frac{2 n+5}{2^{n} \cdot\left(n^{3}+7 n^{2}+14 n+8\right)}
$$ | [
"Solution:\nFirst, we manipulate using partial fractions and telescoping:\n$$\n\\begin{aligned}\n\\sum_{n=1}^{\\infty} \\frac{2 n+5}{2^{n} \\cdot\\left(n^{3}+7 n^{2}+14 n+8\\right)} & =\\frac{1}{2} \\cdot \\sum_{n=1}^{\\infty} \\frac{1}{2^{n}}\\left(\\frac{2}{n+1}-\\frac{1}{n+2}-\\frac{1}{n+4}\\right) \\\\\n& =\\fr... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 137/24 - 8 ln 2 | |
0b5y | Show that a sequence $(\varepsilon_n)_{n \in \mathbb{N}}$ of plus and minus ones is periodic with period a power of $2$, if and only if $\varepsilon_n = (-1)^{P(n)}$, $n \in \mathbb{N}$, where $P$ is an integer-valued polynomial with rational coefficients. | [
"A polynomial $P$ of degree at most $k$ with complex coefficients is integer-valued if and only if\n$$\nP = \\sum_{j=0}^{k} a_j \\binom{X}{j} = \\sum_{j=0}^{k} \\frac{a_j}{j!} X(X-1) \\cdots (X-j+1),\n$$\nwhere the $a_j$ are all integer numbers, so its coefficients are rational.\n\nWe show that for such a $P$, the ... | Romania | 2010 Eighth IMAR MATHEMATICAL COMPETITION | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof only | null | |
0euo | For a trapezoid $ABCD$ with $AB // CD$, suppose that $A$, $B$, $C$, $D$ lie in the clockwise direction. Let $\Gamma_1$ be the circle centered at $A$, and passing through $B$. Let $\Gamma_2$ be the circle centered at $C$, and passing through $D$. Let $P$ be the intersection (distinct from $B$, $D$) of the line $BD$ and ... | [
"It suffices to show that $X$, $B$, $Y$, $Q$ are cyclic under the assumption that $Q$ is the intersection of the line $BD$ and $\\Gamma_2$. There are six cases depending on the ordering of $P$, $D$, $B$, $Q$. Although we should consider all those six cases, we here give the proof only for the case where the order i... | South Korea | 23rd Korean Mathematical Olympiad Final Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | English | proof only | null | |
022x | Problem:
As páginas do dicionário - Para numerar as páginas de um dicionário, imprimiu-se 1988 vezes o algarismo 1. Quantas páginas tem esse dicionário? | [
"Solution:\n\nObservemos que:\n\na. A cada dez números imprime-se uma vez o $1$ nas unidades;\n\nb. A cada cem números imprime-se dez vezes o $1$ nas dezenas;\n\nc. A cada mil números imprime-se cem vezes o $1$ nas centenas.\n\nAssim, de $1$ até $999$, imprime-se o algarismo $1$ um total de $300$ vezes, das quais $... | Brazil | Nível 2 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 3151 | |
0iip | Problem:
Compute $\lim_{x \rightarrow 0} \frac{e^{x \cos x}-1-x}{\sin \left(x^{2}\right)}$. | [
"Solution:\n\nLet's compute all the relevant Maclaurin series expansions, up to the quadratic terms:\n$$\nx \\cos x = x + \\ldots, \\quad e^{x \\cos x} = 1 + x + \\frac{1}{2} x^{2} + \\ldots, \\quad \\sin \\left(x^{2}\\right) = x^{2} + \\ldots\n$$\nso\n$$\n\\lim_{x \\rightarrow 0} \\frac{e^{x \\cos x}-1-x}{\\sin \\... | United States | Harvard-MIT Mathematics Tournament | [
"Precalculus > Limits",
"Precalculus > Trigonometric functions"
] | null | final answer only | 1/2 | |
0cqu | A positive integer is called *good* if it has exactly two different prime divisors. Determine whether there exist 18 consecutive positive integers all of which are good. (O. Podlipsky) | [
"Ответ. Не могут.\n\nПредположим, что нашлись 18 хороших чисел подряд. Среди них найдутся три числа, делящихся на $6$. Пусть это числа $6n$, $6(n+1)$ и $6(n+2)$. Поскольку эти числа — хорошие, и в разложение каждого из них на простые множители входят двойка и тройка, других простых делителей у них быть не может.\n\... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | No, such 18 consecutive integers do not exist. | |
06ls | In $\triangle ABC$, $AC = kAB$, with $k > 1$. The internal angle bisector of $\angle BAC$ meets $BC$ at $D$. The circle with $AC$ as diameter cuts the extension of $AD$ at $E$. Express $\frac{AD}{AE}$ in terms of $k$. | [
"The answer is $\\frac{2}{k+1}$.\n\nLet $CE$ meet $AB$ at $Q$. Let $P$ be the point on $AQ$ such that $PD // QC$. Note that $E$ is the midpoint of $QC$ since $\\triangle AEQ \\cong \\triangle AEC$. Also, we have $\\triangle APD \\sim \\triangle AQE$ and $\\triangle BPD \\sim \\triangle BQC$. Thus, we have\n$$\n\\fr... | Hong Kong | Year 2021 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 2/(k+1) | |
00ox | By a *strip* of breadth $b$ we mean a closed part of the plane consisting of all points that lie between two parallel lines at distance $b$ from each other. Let $S$ be a finite set of $n$ ($n \ge 4$) points in the plane, such that any three points from $S$ can be covered by a strip of breadth 1. Prove that $S$ can be c... | [
"Firstly we shall prove the following statement.\n\n**Lemma.** If a triangle can be covered by a strip of breadth $b$, then at least one altitude of the triangle is at most $b$ long.\n\n**Proof.** At least one of the perpendicular lines through the vertices of the triangle to the border lines of the strip meets the... | Balkan Mathematical Olympiad | BMO 2010 Shortlist | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0ev4 | For a triangle $ABC$ let $B_1$, $C_1$ be the excenters corresponding to $B$, $C$, respectively. The line $B_1C_1$ intersects the circumcircle of $ABC$ at $D$ ($\neq A$). Let $E$ be the intersection of the perpendiculars from $B_1$ to $CA$ and from $C_1$ to $AB$. Let $\omega$ be the circumcircle of the triangle $ADE$. T... | [
"We first show that $\\angle ADE = 90^\\circ$. $\\angle EB_1C_1 = \\frac{1}{2}\\angle A = \\angle EC_1B_1$, hence we have $EB_1 = EC_1$. Similarly, we have $EB_1 = EA_1$, that is, $E$ is the circumcenter of $\\triangle A_1B_1C_1$. On the other hand, $A$, $B$, $C$ are perpendicular feet of $\\triangle A_1B_1C_1$ bec... | South Korea | The 26th Korean Mathematical Olympiad Final Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07wj | An *Irish Set* is a set of positive integers such that each pair of integers in the set has a common factor greater than $1$, and there is no prime number that divides all integers in this set. Is there an Irish Set that contains exactly $2024$ integers, all smaller than $8000$? | [
"A key observation is that when we add to the Irish Set $S = \\{s_1, s_2, \\dots, s_n\\}$ any positive multiple of an element $s_i$ that is not yet contained in $S$ we obtain an Irish Set with $n + 1$ elements.\n\nOne possibility then is to start with an Irish Set of length $3$ and add multiples of these three numb... | Ireland | IRL_ABooklet_2024 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
07jf | Let $T$ be a given positive integer. Find all functions $f : \mathbb{Z} \to \mathbb{C}$ such that for every integer $m$, $f(m+T) = f(m)$, and for every three integers $a, b$, and $c$, the following equality holds:
$$
f(a)\overline{f(a+b)}f(a+c)\overline{f(a+b+c)} = 1.
$$
Note: $\bar{z}$ denotes the complex conjugate of... | [
"First, by substituting $b = c = 0$ into the given relation, we obtain:\n$$\nf(a) \\cdot \\bar{f}(a) \\cdot \\bar{f}(a) \\cdot f(a) = 1 \\Rightarrow |f(a)|^2 \\cdot |f(a)|^2 = 1 \\Rightarrow |f(a)|^4 = 1 \\Rightarrow |f(a)| = 1\n$$\nSince for every $x \\in \\mathbb{C}$, $x\\bar{x} = |x|^2$, and we know $|f(x)| = 1$... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | All solutions are f(n) = c ε^n where |c| = 1 and ε ∈ {1, −1} with ε^T = 1. Equivalently: if T is odd, f(n) = c (|c| = 1); if T is even, f(n) = c or f(n) = c(−1)^n with |c| = 1. | |
0032 | Sean $a > b > c > d$ números enteros positivos que satisfacen
$$
a + b + c + d = 502 \text{ y } a^2 - b^2 + c^2 - d^2 = 502.
$$
Calcular cuántos son los valores posibles de $a$. | [] | Argentina | Argentina 2006 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Español | proof and answer | 124 | |
025m | Problem:
A lei pirata estabelece que, para dividir as moedas de um tesouro, o capitão deve escolher um grupo de piratas (excluindo a si mesmo). Em seguida, o capitão deve distribuir a mesma quantidade de moedas a cada um dos piratas desse grupo, de tal modo que não seja possível dar a cada um deles nenhuma outra das mo... | [
"Solution:\na) Seja $N$ o número de moedas que há no tesouro. Quando Barbaroxa escolhe 99 piratas para dividir as $N$ moedas, sobram 51 moedas para ele, ou seja, ele consegue dividir $N-51$ moedas entre 99 piratas, ou equivalentemente, o número $N-51$ é divisível por 99. Em particular, o número $N-51+99=N+48$ é tam... | Brazil | null | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 645 coins; choose 93 pirates | |
00tu | Let $a$, $b$, $c$, $d$ be non-negative real numbers such that
$$
\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} + \frac{1}{d+1} = 3.
$$
Prove that
$$
3(ab + ac + ad + bc + bd + cd) + \frac{4}{a+b+c+d} \le 5.
$$ | [
"Let $S = a + b + c + d$. By AM-HM (or Cauchy-Schwarz) we have\n$$\nS + 4 = (a + 1) + (b + 1) + (c + 1) + (d + 1) \\ge \\frac{16}{\\frac{1}{a+1} + \\frac{1}{b+1} + \\frac{1}{c+1} + \\frac{1}{d+1}} = \\frac{16}{3}\n$$\ngiving $S \\ge \\frac{4}{3}$.\nMultiplying the given equality by $(a+1)(b+1)(c+1)(d+1)$ we get\n$$... | Balkan Mathematical Olympiad | BMO 2022 shortlist | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0jxh | Problem:
Does there exist a two-variable polynomial $P(x, y)$ with real number coefficients such that $P(x, y)$ is positive exactly when $x$ and $y$ are both positive? | [
"Solution:\n\nAnswer: No\nFor any $\\epsilon$ and positive $x$, $P(x, \\epsilon)>0$ and $P(x,-\\epsilon) \\leq 0$. Thus by continuity/IVT, $P(x, 0)=0$ for all positive $x$. Similarly $P(0, y)=0$ for all positive $y$. This implies $x y \\mid P(x, y)$, and so we can write $P(x, y)=x y Q(x, y)$. But then this same log... | United States | February 2017 | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | No | |
0l79 | Let $k$ be a real number such that the system
$$
\begin{aligned}
|25 + 20i - z| &= 5 \\
|z - 4 - k| &= |z - 3i - k|
\end{aligned}
$$
has exactly one complex solution $z$. The sum of all possible values of $k$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. Here $i ... | [
"Geometrically, $|a-b|$ represents the distance between complex numbers $a$ and $b$ in the complex plane. Thus $|(25+20i)-z| = 5$ means that the distance from $25+20i$ to $z$ is $5$. The set of solutions for $z$ is then a circle with radius $5$ and center $25+20i$. Similarly, $|z-(4+k)| = |z-(k+3i)|$ means that the... | United States | AIME I | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 77 | |
0fc7 | Problem:
Sean $D, E, F$ los puntos de tangencia del círculo inscrito al triángulo $A B C$ con los lados $B C, A C$ y $A B$ respectivamente. Demuestra que
$$
4 S_{D E F} \leq S_{A B C}
$$
donde $S_{X Y Z}$ denota el área del triángulo $X Y Z$. | [
"Solution:\nSea $I$ el incentro del triángulo $A B C$. Tenemos que $I D \\perp B C$, $I E \\perp A C$ e $I F \\perp A B$. Por otro lado, utilizando las notaciones usuales, $S_{A B C} = \\frac{1}{2} A B \\cdot A C \\operatorname{sen} A = \\frac{1}{2} b c \\operatorname{sen} A$ y $S_{E F I} = \\frac{1}{2} E I \\cdot ... | Spain | null | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > ... | null | proof only | null | |
05ce | Find all natural numbers whose last digit is not zero and deleting the first digit of which gives a number exactly 25 times smaller. | [
"Let $n$ be the desired number, and let $m$ be the number obtained by deleting the first digit. Since the last digit of the product $m \\cdot 25$ is not 0, it can only be 5. Since $5 \\cdot 25$ has three digits, it cannot be the number $n$ itself,\nso the number $m$ must have more digits. Consider a digit $d$ in bo... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 625, 3125, 9375 | |
0g9o | 令 $Z$ 表示所有整數所成的集合。試求所有函數 $f: Z \rightarrow Z$ 滿足：
$$
f(f(m) + n) + f(m) = f(n) + f(3m) + 2014 \text{對於所有整數 } m, n \text{ 皆成立。}
$$
Let $Z$ be the set of all integers. Determine all functions $f : Z \to Z$ satisfying
$$
f(f(m) + n) + f(m) = f(n) + f(3m) + 2014 \text{ for all integers } m \text{ and } n.
$$ | [
"答案：只有一個函數，$f(n) = 2n + 1007$\n\n解法：令 $f$ 為一個滿足原式的函數，令 $C = 1007$，然後定義函數 $g: Z \\to Z$，對於所有整數 $m, g(m) = f(3m) - f(m) + 2C$. 則有 $g(0) = 2C$, 原式可改寫為\n$$\nf(f(m) + n) = g(m) + f(n)\n$$\n對於所有 $m, n \\in Z$ 皆成立。使用歸納法可以得到\n$$\nf(tf(m) + n) = tg(m) + f(n) \\quad (1)\n$$\n對於所有 $m, n, t \\in Z$ 皆成立。對於任意的 $r \\in Z$，將 $(r, ... | Taiwan | 2015 Math Olympiad Second Stage Training Camp | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | f(n) = 2n + 1007 | |
02tr | Problem:
No quadrilátero $A B C D$, o lado $A D$ é tal que $A D = A B + C D$. Se $P$ é o ponto de encontro das bissetrizes de $\angle B A D$ e $\angle C D A$, mostre que $B P = P C$.
 | [
"Solution:\n\nConsidere um ponto $E$ sobre o lado $A D$ de modo que $A E = A B$. Consequentemente, $E D = A D - A E = A D - A B = C D$. Além disso, como $A P$ e $B P$ são bissetrizes, temos $\\triangle B A P \\equiv \\triangle A P E$ e $\\triangle P E D \\equiv \\triangle P D C$ pelo caso de congruência (L.A.L.). D... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0biw | $$
x + \log_2 \left( 1 + \sqrt{\frac{5^x}{3^x + 4^x}} \right) = 4 + \log_{1/2} \left( 1 + \sqrt{\frac{25^x}{7^x + 24^x}} \right).
$$ | [
"Rewrite the equation as\n$$\nx + \\log_2 \\left( 1 + \\sqrt{\\frac{1}{(3/5)^x + (4/5)^x}} \\right) = 4 + \\log_{1/2} \\left( 1 + \\sqrt{\\frac{1}{(7/25)^x + (24/25)^x}} \\right),\n$$\n\nand observe that the left hand side is an increasing function, while the right hand side is a decreasing one. We conclude that th... | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | 2 | |
0bac | Given integers $n \ge 3$ and $k \ge 1$, find all positive integers $a_1 < a_2 < \dots < a_n$ such that $\frac{1}{a_1^k} + \frac{1}{a_2^k} + \dots + \frac{1}{a_n^k} = 1$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | null | |
0h6b | Find all three prime numbers $p$, $q$, $r$ that satisfy
$$
\frac{q}{p-1} + \frac{r}{p+1} = \frac{q+r+1}{p}.
$$ | [
"Let's rewrite the given equality as follows:\n$$\n\\frac{q}{p-1} - \\frac{q}{p} = \\frac{r}{p} - \\frac{r}{p+1} + \\frac{1}{p} \\Leftrightarrow \\frac{q}{p(p-1)} = \\frac{r}{p(p+1)} + \\frac{1}{p} \\Leftrightarrow \\frac{q}{p-1} = \\frac{r}{p+1} + 1 \\Leftrightarrow q = \\frac{(p-1)r}{p+1} + p-1 \\Leftrightarrow q... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (p, q, r) = (2, 2, 3) and (3, 3, 2) | |
04od | Let $ABCD$ be an isosceles trapezium with bases $\overline{AB}$ and $\overline{CD}$. The diagonals of the trapezium meet at the point $S$, while $M$ denotes the midpoint of the leg $\overline{AD}$. The circle circumscribed to the triangle $BCM$ intersects $\overline{AD}$ again at the point $K$. Prove that the lines $SK... | [
"Instead of the original formulation, we will solve the following equivalent problem: Let $ABCD$ be an isosceles trapezium with bases $\\overline{AB}$ and $\\overline{CD}$, and let its diagonals meet at the point $S$, as in the original formulation.\nAdditionally, let $K$ be a point on the line $AD$ such that $SK \... | Croatia | Croatian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
07d8 | There is a sequence of $n$ distinct positive integers. Prove that these numbers can be fixed on a circle such that the sum of any two adjacent numbers is different from the other sums. | [
"Without loss of generality assume that $a_1 < a_2 < \\dots < a_n$. Consider these numbers on a circle, in the ascending order. All sums, other than $a_1 + a_n$ can be ordered as\n$$\na_1 + a_2 < a_2 + a_3 < \\dots < a_{n-2} + a_{n-1} < a_{n-1} + a_n\n$$\nWe're done if there's no two sums that are equal. But if the... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
0aq3 | Problem:
Three distinct diameters are drawn on a unit circle such that chords are drawn as shown in Figure 2. If the length of one chord is $\sqrt{2}$ units and the other two chords are of equal lengths, what is the common length of these chords?
Figure 2: Problem 60.1. | [
"Solution:\n\n$\\sqrt{2-\\sqrt{2}}$ units\n\nRefer to Figure 7. Let $\\theta$ be the central angle subtended by the chord of length $\\sqrt{2}$, and $\\alpha$ the central angle subtended by each of the chords of equal lengths (and let $x$ be this common length). By the Law of Cosines, we have\n$$\nx^{2}=1^{2}+1^{2}... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | sqrt(2 - sqrt(2)) | |
02l9 | Problem:
O raio da circunferência - Um arco de circunferência mede $300^{\circ}$ e o seu comprimento é $2~\mathrm{km}$. Qual o número inteiro mais próximo da medida do raio em metros?
(a) 157
(b) 284
(c) 382
(d) 628
(e) 764 | [
"Solution:\n\nSe o raio é $r$ então o comprimento de um arco de $\\theta$ graus é $2 \\pi \\frac{\\theta}{360} r$. Assim, no problema dado, temos que\n$$\n2 \\pi \\frac{300}{360} r = 2000~m \\Longrightarrow r = 2000 \\times \\frac{3}{5 \\pi} \\simeq 382,17~m\n$$\nLogo, a opção correta é (c).",
"Solution:\n\nComo ... | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Circles"
] | null | MCQ | c | |
0a6r | Problem:
Let $ABC$ be a triangle and let $D$ be a point inside the triangle $ABC$ such that $AD$ bisects $\angle BAC$. Let line $BD$ meet side $AC$ at $E$. Let line $CD$ meet side $AB$ at $F$. Let $T$ be the intersection of the (internal) angle bisectors of $\angle AED$ and $\angle AFD$. Prove that if $T$ lies on segme... | [
"Solution:\n\nAssume $T$ lies on $AD$. Then by the angle bisector theorem in triangles $AED$ and $AFD$, we have\n$$\n\\frac{AE}{ED} = \\frac{AT}{TD} = \\frac{AF}{FD}.\n$$\nApplying the angle bisector theorem in triangles $AEB$ and $AFC$, we also have:\n$$\n\\frac{AE}{ED} = \\frac{AB}{BD}\n$... | New Zealand | NZMO Round One | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
040y | An information station employs four different codes, $A$, $B$, $C$ and $D$, for communication, but each week uses only one of them. The code used in a definite week is randomly selected with equal chance among the three ones that have not been used in the last week. Suppose the code used in the first week is $A$. Then ... | [
"Let $P_k$ denote the probability that code $A$ is used in the $k$\\text{th}$ week. Then the probability that $A$ is not used in the $k$\\text{th}$ week is $1 - P_k$. Therefore, we have\n$$\nP_{k+1} = \\frac{1}{3}(1 - P_k).\n$$\nOr\n$$\nP_{k+1} - \\frac{1}{4} = -\\frac{1}{3}\\left(P_k - \\frac{1}{4}\\right).\n$$\nA... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | final answer only | 61/243 | |
07yy | Problem:
In un quadrilatero convesso $ABCD$ i lati $AB$, $BC$, $CD$ sono uguali. Inoltre $AC = BD = AD$. Quanto misura l'angolo in $D$? | [
"$\\overline{\\text { SOLUZIONE }}$"
] | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof and answer | 72° | |
00wv | Problem:
There are 20 cats priced from $\$ 12$ to $\$ 15$ and 20 sacks priced from 10 cents to $\$ 1$ for sale (all prices are different). Prove that each of two boys, John and Peter, can buy a cat in a sack paying the same amount of money. | [
"Solution:\n\nThe number of different possibilities for buying a cat and a sack is $20 \\cdot 20 = 400$ while the number of different possible prices is $1600 - 1210 + 1 = 391$. Thus by the pigeonhole principle there exist two combinations of a cat and a sack costing the same amount of money. Note that the two cats... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0cz8 | Let $n$ be a positive integer. Prove that all roots of the equation
$$
x(x+2)(x+4) \ldots(x+2n) + (x+1)(x+3) \ldots(x+2n-1) = 0
$$
are real and irrational. | [
"Consider the polynomial function\n$$\nf(x) = x(x+2)(x+4) \\ldots(x+2n) + (x+1)(x+3) \\ldots(x+2n-1)\n$$\nWe have $\\deg f = n+1$ and the leading coefficient is $1$.\n\nObserve that for $0 \\leq k < n$,\n$$\nf(-2k) = (-1)^k (2k-1)(2k-3) \\ldots 1 \\cdot 3 \\ldots (-2k+2n-1)\n$$\nand\n$$\n\\begin{gathered}\nf(-2k-2)... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0hsz | Problem:
A circle is tangent to the continuations of sides $CA$ and $CB$ of the triangle $ABC$, and is also tangent to the side $AB$ at point $P$. Prove that the radius of the circle tangent to $AP$, $CP$ and the circle circumscribed around $ABC$ is equal to the radius of the circle inscribed in $ABC$. | [
"Solution:\nLet $K$ and $M$ be the tangency points of the circle with $AP$ and $CP$ respectively, $L$ the point where it touches the circumscribed circle of $ABC$, $T$ the middle of the arc $AB$ of the circumscribed circle, $I$ the center of the inscribed circle of $ABC$.\n\nThe tangent to the circle $ABC$ at $T$ i... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid,... | null | proof only | null | |
07as | Find all functions $f : \mathbb{R}^+ \to \mathbb{R}^+$ such that for all positive real numbers $x$ and $y$,
$$
f\left(\frac{y}{f(x+1)}\right) + f\left(\frac{x+1}{x f(y)}\right) = f(y).
$$ | [
"Let $P(x, y)$ be the assertion $f\\left(\\frac{y}{f(x+1)}\\right) + f\\left(\\frac{x+1}{x f(y)}\\right) = f(y)$.\n\nRewriting the equation, we get $f\\left(\\frac{y}{f(x)}\\right) + f\\left(\\frac{x}{(x-1)f(y)}\\right) = f(y)$ which is valid for each $x > 1,\\ y \\in \\mathbb{R}^+$.\n\nIf there exists $a \\in \\ma... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 1/x for all positive x | |
01l9 | There are three chess boards in a chess club, one of them is peaceful, the other one is fair, and the third board is strange. If two chess players use the peaceful board, then the result of their game is always a draw, if they use the fair board, then a stronger player wins, but if players use the strange board, then a... | [
"We number chess players in accordance with their power: $P_1, P_2, P_3, P_4, P_5, P_6$ (if the $i$'th player is stronger than the $j$'th one, then $i < j$). Suppose, contrary to our claim, that the tournament can be organized. Let chess player $P_i$ get $a_i$ points ($i = 1, \\dots, 6$). By conditions, $a_i - a_{i... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | No, it is impossible. | |
0hv0 | Problem:
Find all ordered pairs $(m, n)$ of integers such that $231 m^{2} = 130 n^{2}$. | [
"Solution:\nThe unique solution is $(0, 0)$."
] | United States | null | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (0, 0) | |
03zu | Let $H$ be the orthocenter of an acute-angled $\triangle ABC$ with $\angle A > 60^\circ$. Let points $M$ and $N$ be on sides $AB$ and $AC$, respectively, such that $\angle HMB = 60^\circ = \angle HNC$. Let $O$ be the circumcenter of $\triangle HMN$. Let points $D$ and $A$ be on the same side of line $BC$, such that $\t... | [
"Let $T$ be the orthocenter of $\\triangle HMN$. Extended lines of $HM$ and $CA$ intersect at point $P$. Extended lines $HN$ and $BA$ intersect at point $Q$. It is easy to see that points $N$, $M$, $P$ and $Q$ are concyclic.\n\n\n\nBy $\\angle THM = \\angle OHN$, we see that $\\angle PQH - ... | China | China National Team Selection Test | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous... | English | proof only | null | |
0du9 | Problem:
Poišči vsa praštevila $p$, za katera je število $3^{p}-(p+2)^{2}$ praštevilo. | [
"Solution:\n\nZa $p=2$ je število $3^{2}-4^{2}$ negativno.\n\nČe je $p$ liho število, je $3^{p}-(p+2)^{2}$ sodo; torej enako 2. Enačba $3^{p}-(p+2)^{2}=2$ ima očitno rešitev $p=3$.\n\nDa ni drugih, dokažemo z indukcijo.\n\nNatančneje: Z indukcijo dokažemo, da za $n>3$ velja $3^{n}-(n+2)^{2}>2$ oziroma $3^{n}>n^{2}+... | Slovenia | 45. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Intermediate Algebra > Exponential functions",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 3 | |
0h3x | In Nottingham city, 100 riches reside, and each of them owns at least one million of gold coins. Robin Hood develops plans to raid them. Each plan consists in stealing same number of gold coins (not exceeding one million) from each of the riches. Prove that, no matter what wealths of the riches are, Robin Hood can deve... | [
"Без обмеження загальності можна вважати, що всі багатії мають різну кількість монет. Тоді задачу можна переформулювати так: нехай $A$ — довільна 100-елементна множина з натуральних чисел, кожне з яких не менше від мільйона. Тоді знайдуться такі різні числа $t_1, t_2, \\dots, t_{100}$, які не перевищують мільйона, ... | Ukraine | Ukrainian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
09d5 | Сүүлийн 2 цифр нь ижил бөгөөд бүхэл тооны квадрат болж байх 5 оронтой тоо хичнээн байх вэ? | [
"$n = \\overline{abcdd} = m^2$ гэж авъя.\n\n$\\Rightarrow d \\in \\{0, 1, 4, 5, 6, 9\\}$ байна. $n = 100 \\cdot \\overline{abc} + 11d$ ба $m^2 \\equiv 0, 1$ (mod 4) тул $n = m^2 = 3d \\equiv 0, 1$ (mod 4) $\\Rightarrow d = 0$ эсвэл $d = 4$ болно.\n\n(1) $d = 0$ бол $n = 100 \\cdot \\overline{abc} = 10^2 \\cdot \\ov... | Mongolia | ММО-48 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Mongolian | proof and answer | 31 | |
05k7 | Problem:
La ville de Gotham City est un grand rectangle, découpé en pâtés de maisons rectangulaires plus petits par des rues parallèles aux bords de la ville. On suppose qu'il y a au moins une rue, et qu'aucune rue ne traverse la ville de part en part.
Montrer qu'il existe un pâté de maison qui ne touche pas le bord ... | [
"Solution:\n\nBatman explore la ville selon le procédé suivant : il part du bord de la ville (du point rouge sur la figure) et suit jusqu'au bout une rue (l'Avenue du Joker par exemple) qui s'enfonce à l'intérieur. Comme l'Avenue de Joker ne traverse pas la ville de part en part, elle débouche sur une autre (appelo... | France | OFM 2013-2014 Envoi 2 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
0d05 | Three equal circles of radius $R$ are given such that each one passes through the centers of the other two. Find the area of the common region. | [
"\nLet $O_1$, $O_2$, $O_3$ be the centers of the three circles and $S$ the area of the common region. The three sectors with centers $O_1$, $O_2$, $O_3$ which subtend the arcs $O_2O_3$, $O_1O_3$, $O_2O_1$, respectively, cover the surface of area $S$ and twice more the surface of triangle $O... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof and answer | (1/2)(pi - sqrt(3)) R^2 | |
0cgj | Let $n \ge 2$ be an integer. We define the numbers $A = 33...3$, with $n$ digits $3$, and $B = 20 \cdot A + 6$. Find all the digits that form the number $A \cdot B$. | [
"Notice that $3 \\cdot A = 99...9 = 10^n - 1$. Then $A \\cdot B = (10^n - 1) \\cdot \\underbrace{22...2}_{n+1 \\text{ digits}} = \\underbrace{22...2}_{n+1 \\text{ digits}} \\underbrace{00...0}_{n \\text{ digits}} - \\underbrace{22...2}_{n+1 \\text{ digits}}$. Thus $A \\cdot B = \\underbrace{22...2}_{n-1 \\text{ dig... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 1, 2, 7, 8, 9 | |
0ewx | Problem:
The natural numbers $a$, $b$, $n$ are such that for every natural number $k$ not equal to $b$, $b - k$ divides $a - k^n$. Prove that $a = b^n$. | [
"Solution:\n\nWe have $k^n - a = b^n - a$ (mod $b - k$). Hence $b^n - a = 0$ (mod $b - k$) for every $k$ not equal to $b$. But if $b^n$ does not equal $a$, then by taking $k - b > b^n - a$ we could render the equation false."
] | Soviet Union | 4th ASU | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
01yr | The cells of the $2022 \times 2022$ table are filled with positive integers from $1$ to $2022^2$, each cell contains exactly one number, all numbers are used exactly once. For each line Vlad wrote out one number which is the second in descending order in this line. And Dima did the same for each column. It turned out t... | [
"**Answer:** $2020$.\n\nLet Dima wrote down the numbers $d_1 > d_2 > \\dots > d_{2022}$ and Vlad wrote down the numbers $v_1 > v_2 > \\dots > v_{2022}$. Suppose the answer in the problem is $2022$ or $2021$, then $d_{2022} > v_2$. Each of the $2021$ lines not containing $v_1$ contains at least $2021$ numbers not ex... | Belarus | Belarus2022 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2020 | |
081w | Problem:
Se $y=2x$ e $z=2y$, a cosa è uguale $x+y+z$?
(A) $5x$
(B) $4y$
(C) $3z$
(D) $\frac{7}{2}y$
(E) $\frac{7}{3}z$. | [
"Solution:\n\nLa risposta è (D). Si ha ovviamente $z=4x$ e $x=\\frac{y}{2}=\\frac{z}{4}$, da cui $x+y+z=7x=\\frac{7}{2}y=\\frac{7}{4}z$."
] | Italy | Progetto Olimpiadi di Matematica | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | (D) | |
054e | On the sides $AB$, $BC$ and $CA$ of triangle $ABC$, points $L$, $M$ and $N$ are chosen, respectively, such that the lines $CL$, $AM$ and $BN$ intersect at a common point $O$ inside the triangle and the quadrilaterals $ALON$, $BMOL$ and $CNOM$ have incircles. Prove that
$$
\frac{1}{AL \cdot BM} + \frac{1}{BM \cdot CN} +... | [
"$ALON$ is a circumscribed quadrilateral, hence $AL + ON = AN + OL$. Similarly $BM + OL = BL + OM$ and $CN + OM = CM + ON$. By adding the equations we obtain $AL + BM + CN = AN + BL + CM$. Lines $CL$, $AM$ and $BN$ intersect in one point, we get from Ceva's theorem $AL \\cdot BM \\cdot CN = AN \\cdot BL \\cdot CM$.... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals"
] | null | proof only | null | |
05lc | Problem:
On considère 6 points du plan, disposés de telle sorte que le triangle formé par trois quelconques d'entre eux ait trois côtés de longueurs distinctes. Montrer qu'il existe un triangle dont le plus court côté est également le côté le plus long d'un autre triangle. | [
"Solution:\n\nOn colorie en rouge le plus court côté de chaque triangle. Il suffit de montrer qu'on a un triangle rouge. Appelons $P_{1}, \\ldots, P_{6}$ les points. Parmi les cinq segments $P_{1} P_{2}, P_{1} P_{3}, P_{1} P_{4}, P_{1} P_{5}, P_{1} P_{6}$ ayant $P_{1}$ pour extrémité, il y en a au moins 3 rouges ou... | France | Envoi de combinatoire | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof only | null | |
07o7 | There are $100$ towns on the island of Hibernia. Every pair of towns is either connected by a single road, or is not connected. If we consider any subset of $98$ towns, the total number of roads connecting these towns to each other is a constant. There are $N$ roads in total in Hibernia.
Find all possible values of $N$... | [
"Consider the more general problem with $n$ towns (the specific case for this problem is $n = 100$). Let $K$ denote the (constant) number of roads connecting any subset of $n-2$ towns. Also, let $c_{i,j} \\in \\{0,1\\}$ denote the number of roads connecting town $i$ and town $j$. Finally, for $i = 1, 2, \\dots, n$,... | Ireland | Ireland | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Graph Theory",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 4950 | |
0fwt | Problem:
Sei $ABCD$ ein Sehnenviereck mit $CD < AD$ und $CD < BC$. Die Diagonalen $AC$ und $BD$ schneiden sich im Punkt $S$. Die Spiegelung der Gerade $AB$ an $AC$ sei $e$ und die Spiegelung der Geraden $AB$ an $BD$ sei $f$. Die Gerade $CD$ schneide $e$ und $f$ in den Punkten $E$ bzw. $F$. Beweise, dass das Dreieck $S... | [
"Solution:\n\nWir definieren $\\alpha = \\angle BAC$, $\\beta = \\angle DBA$ und $\\gamma = \\angle CAD$. Weil $ABCD$ ein Sehnenviereck ist, gilt $\\angle BDC = \\alpha$, $\\angle DCA = \\beta$ und $\\angle CBD = \\gamma$ (siehe Abbildung 1). Nach Konstruktion von $e$ und $f$ gilt $\\angle SAE = \\alpha$ bzw. $\\an... | Switzerland | Vorrundenprüfung | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0b9a | (a) Given a positive integer $k$, prove that there do not exist two distinct integers in the open interval $(k^2, (k+1)^2)$ whose product is a perfect square.
(b) Given an integer $n > 2$, prove that there exist $n$ distinct integers in the open interval $(k^n, (k+1)^n)$ whose product is the $n$-th power of an integer... | [
"(a) Suppose that two such integers, $a < b$, exist. Let $m$ be the squarefree part of $a$; $a = ms^2$, $s \\in \\mathbb{Z}$. Then $mb$ is a square, so since $m$ is squarefree, $b = mt^2$ for some integer $t \\ge s + 1$. Hence\n$$\n\\left(1 + \\frac{1}{s}\\right)^2 \\le \\frac{t^2}{s^2} = \\frac{b}{a} < \\frac{(k+1... | Romania | NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear a... | English | proof only | null | |
01wk | $$
Q(x_1, \dots, x_4) = 4(x_1^2 + x_2^2 + x_3^2 + x_4^2) - (x_1 + x_2 + x_3 + x_4)^2
$$
is represented as a sum of squares of four polynomials of four variables with integer coefficients.
a) Find at least one such representation.
b) Prove that for any such representation at least one of the four polynomials is identica... | [
"a) For example, take\n$$\nQ(x) = (x_1 + x_2 - x_3 - x_4)^2 + (x_1 - x_2 + x_3 - x_4)^2 + (x_1 - x_2 - x_3 + x_4)^2 + 0^2.\n$$\n\nb) Consider the representation from the problem conditions:\n$$\n4(x_1^2 + x_2^2 + x_3^2 + x_4^2) - (x_1 + x_2 + x_3 + x_4)^2 = P_1^2 + P_2^2 + P_3^2 + P_4^2. \\quad (1)\n$$\nThe equalit... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | One valid representation is: Q(x1,x2,x3,x4) = (x1 + x2 − x3 − x4)^2 + (x1 − x2 + x3 − x4)^2 + (x1 − x2 − x3 + x4)^2 + 0^2. Moreover, in any representation of Q as a sum of four squares of integer coefficient polynomials, at least one polynomial must be identically zero. | |
0eyc | Problem:
Given a triangle $ABC$, and $D$ on the segment $AB$, $E$ on the segment $AC$, such that $AD = DE = AC$, $BD = AE$, and $DE$ is parallel to $BC$. Prove that $BD$ equals the side of a regular 10-gon inscribed in a circle with radius $AC$. | [
"Solution:\n\n$DA = DE$, so $DAE$ is isosceles. $DE$ is parallel to $BC$, so $ABC$ is isosceles, so $BA = AC / (2 \\cos A)$. Hence $BD = AC / (2 \\cos A) - AC$. But $AE = 2 AC \\cos A$, so we have an equation for $c = \\cos A$: $4c^2 + 2c - 1 = 0$.\n\n$2\\pi/5, 4\\pi/5, 6\\pi/5, 8\\pi/5$ and $10\\pi/5$ are the root... | Soviet Union | 2nd ASU | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | null | proof only | null | |
0495 | Determine all $z \in \mathbb{C}$ such that $|z| = |z+1| = \left|\frac{1}{z}\right|$. | [] | Croatia | CroatianCompetitions2011 | [
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | null | proof and answer | z = -1/2 ± i√3/2 | |
0gd0 | 設 $p = 8k + 1$ 為質數且 $k$ 為正整數。令 $r$ 為 $\binom{4k}{k}$ 除以 $p$ 的餘數 (所以 $0 \le r < p$)。證明 $\sqrt{r}$ 不是整數。 | [
"Let $\\left( \\frac{\\cdot}{p} \\right)$ denote the Legendre symbol. We consider the two integers\n$$\nM = \\sum_{x=1}^{p-1} \\left( \\frac{1+x^4}{p} \\right) \\quad \\text{and} \\quad N = \\sum_{x=1}^{p-1} \\left( \\frac{1+x^8}{p} \\right).\n$$\nNotice that $M$ and $N$ are integers with $|M|, |N| < p$. Also, we c... | Taiwan | 二〇一九數學奧林匹亞競賽第三階段選訓營 | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Combinatorics... | null | proof only | null | |
0bd5 | Let $a$, $b$, $c$, $d > 0$ satisfying $abcd = 1$. Prove that
$$
\frac{1}{a+b+2} + \frac{1}{b+c+2} + \frac{1}{c+d+2} + \frac{1}{d+a+2} \le 1.
$$ | [
"We have $\\frac{1}{a+b+2} + \\frac{1}{c+d+2} \\le \\frac{1}{2\\sqrt{ab}+2} + \\frac{1}{2\\sqrt{cd}+2}$. Denoting $\\sqrt{ab} = x$, we get $\\sqrt{cd} = \\frac{1}{x}$, and the sum on the right hand side in the inequality above is $\\frac{1}{2} \\left( \\frac{1}{x+1} + \\frac{x}{1+x} \\right) = \\frac{1}{2}$.\n\nPro... | Romania | 64th NMO Selection Tests for the Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
083h | Problem:
All'interno dell'esagono regolare $ABCDEF$ si disegni il quadrato $ABGH$. Quanto vale l'ampiezza di $E\hat{H}F$ espressa in gradi? | [] | Italy | Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO BIENNIO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | 105 | |
0kcg | Problem:
Find the unique pair of positive integers $(a, b)$ with $a < b$ for which
$$
\frac{2020-a}{a} \cdot \frac{2020-b}{b} = 2
$$ | [
"Solution:\nIf either $a$ or $b$ is larger than $2020$, then both must be for the product to be positive. However, the resulting product would be less than $1$, so this case is impossible.\n\nNow, we see that $\\left(\\frac{2020-a}{a}, \\frac{2020-b}{b}\\right)$ must be in the form $\\left(\\frac{x}{y}, \\frac{2y}{... | United States | HMMT February 2020 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (505, 1212) | |
0505 | Prove that the set of integers $\{0, 1, 2, \dots, 2^n - 1\}$ can be partitioned into $n + 1$ disjoint subsets $A_0, A_1, \dots, A_n$ such that both of the following hold:
a) If $k + l = n$, then the subsets $A_k$ and $A_l$ have the same number of elements.
b) If $s$ and $t$ are non-negative integers and $s + t \le n$... | [
"Divide the set $A$ into subsets such that the subset $A_k$ consists of only those numbers which have exactly $k$ ones in their binary representation. Then $A_0 = \\{0\\}, A_1 = \\{1, 2, 4, \\dots, 2^{n-1}\\}, \\dots, A_n = \\{2^n - 1\\}$. Let us show that both conditions are met.\n\nThe first condition is met beca... | Estonia | Selected Problems from Open Contests | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
04hp | Let $n$ be an odd positive integer greater than $3$. Let $k$ denote the least positive integer such that $kn+1$ is a perfect square and $l$ denote the least positive integer such that $ln$ is a perfect square.
Prove that $n$ is a prime number if and only if $k > \frac{1}{4}n$ and $l > \frac{1}{4}n$.
(V.I. Bernik, I.K. ... | [
"If $n = p$ is a prime number, then $l = p$ so it is obvious that $l > \\frac{n}{4}$. For $k$ we have: $kp = (y - 1)(y + 1)$, so\n$$\n\\begin{cases} y - 1 = k_1, & y + 1 = k_2p, \\\\ y - 1 = k_1p, & y + 1 = k_2. \\end{cases} \n\\quad \\text{or}\n$$\nIn each case, adding these two equations gives us $2y > p$. If $4k... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
04s1 | Let us call by an "edge" any segment of length $1$ which is common to two adjacent fields of a given chessboard $8 \times 8$. Consider all possible cuttings of the chessboard into $32$ pieces $2 \times 1$ and denote by $n(e)$ the total number of such cuttings that involve the given edge $e$. Determine the last digit of... | [
"The number of edges, which are not involved in a given cutting, is equal to $32$, because each of these edges must coincide with the common segment of the two fields forming one of the $32$ resulting pieces $2 \\times 1$. Thus each cutting gives a contribution $112 - 32 = 80$ to the sum $S$ of all the numbers $n(e... | Czech Republic | 63rd Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 0 | |
00vi | For all $a$, $b$, $c$ positive real numbers with $a^2b + a^2c + b^2a + b^2c + c^2a + c^2b = 1$, show that
$$
\frac{ab + bc + ca}{1 + 2abc} \le \frac{(a + b + c)^2}{4}.
$$ | [
"Lemma. For positive $a$, $b$, $c$, $x$, $y$, $z$, it holds that $ax + by + cz + 2\\sqrt{ab + bc + ca}\\sqrt{xy + yz + zx} \\le (a + b + c)(x + y + z)$\n\n*Proof.* Use Cauchy-Schwarz inequality\n$$\n\\begin{aligned}\nax + by + cz + 2\\sqrt{ab + bc + ca}\\sqrt{xy + yz + zx} & \\le \\sqrt{a^2 + b^2 + c^2}\\sqrt{x^2 +... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlisted Problems | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
03fl | We call $n$ lines in the plane three-way if they can be separated into three nonempty sets, $X, Y, Z$. Every two lines from the same set are parallel to each other, no two lines from different sets are parallel to each other, and no three lines intersect at a point.
By $S_n$ we denote the maximum number of regions int... | [
"We will derive a general formula for the number of regions. Let the three sets have the number of elements $|X| = x$, $|Y| = y$ and $|Z| = z$, respectively. The first two divide the plane into a total of $(x+1)(y+1)$ regions. Each line of the third set intersects the others at $x+y$ points and is divided into $x+y... | Bulgaria | Bulgarian Spring Tournament | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 18 | |
00i0 | Consider a $100 \times 100$ table, and identify the cell in row $a$ and column $b$, $1 \leq a, b \leq 100$, with the ordered pair $(a, b)$. Let $k$ be an integer such that $51 \leq k \leq 99$. A $k$-knight is a piece that moves one cell vertically or horizontally and $k$ cells to the other direction; that is, it moves ... | [
"Cell $(x, y)$ is directly reachable from another cell if and only if $x-k \\geq 1$ or $x+k \\leq 100$ or $y-k \\geq 1$ or $y+k \\leq 100$, that is, $x \\geq k+1$ or $x \\leq 100-k$ or $y \\geq k+1$ or $y \\leq 100-k$ (*). Therefore the cells $(x, y)$ for which $101-k \\leq x \\leq k$ and $101-k \\leq y \\leq k$ ar... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | L(k) = 100^2 - (2k - 100)^2 if k is even; L(k) = (100^2 - (2k - 100)^2) / 2 if k is odd. | |
0l5p | Problem:
Let $a$, $b$, and $c$ be real numbers such that $a^2(b + c) = 1$, $b^2(c + a) = 2$, and $c^2(a + b) = 5$. Given that there are three possible values for $abc$, compute the minimum possible value of $abc$. | [
"Solution:\nLet $x = abc$. Multiplying all equations together and simplifying gives\n$$(abc)^2 (a + b)(b + c)(c + a) = 10,$$\n$$(abc)^2 \\left(a^2(b + c) + b^2(c + a) + c^2(a + b) + 2abc\\right) = 10,$$\n$$x^2 (1 + 2 + 5 + 2x) = 10,$$\n$$x^2 (x + 4) = 5.$$\nThe resulting cubic factors as $(x - 1)(x^2 + 5x + 5) = 0$... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | (-5 - sqrt(5))/2 | |
0azm | Problem:
A semiprime is a number that is a product of two prime numbers. How many semiprime numbers less than $2018$ can be expressed as $x^{3}-1$ for some natural number $x$? | [
"Solution:\n\n$x^{3}-1 = (x-1)(x^{2}+x+1)$, so this implies that both $x-1$ and $x^{2}+x+1$ have to be prime. Furthermore, this also means we only have numbers up to $12$ to work on, as $13^{3} > 2018$. Hence, we only have to check $x = 3, 4, 6, 8, 12$ and determine if $x^{2}+x+1$ is prime.\n\nThe values we get are... | Philippines | 20th Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 4 | |
00j6 | Let $k$ and $n$ be positive integers. Prove that, if $x_j$ are real numbers for $1 \le j \le n$, such that
$$
\sum_{j=1}^{n} \frac{1}{x_j^{2k} + k} = \frac{1}{k}
$$
holds, it follows that
$$
\sum_{j=1}^{n} \frac{1}{x_j^{2k+1} + k + 2} \le \frac{1}{k+1}
$$ | [
"**Solution:** We can, in fact, show that each of the expressions in the second sum is not greater than the corresponding expression in the first, multiplied by the factor $\\frac{k}{k+1}$.\nSubstituting $y := x_j^{2k}$, this means that we wish to show\n$$\n\\frac{1}{y^2 + k + 2} \\le \\frac{k}{k+1} \\cdot \\frac{1... | Austria | AustriaMO2011 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof only | null | |
0bif | Determine the smallest real constant $c$ such that
$$
\sum_{k=1}^{n} \left( \frac{1}{k} \sum_{j=1}^{k} x_j \right)^2 \le c \sum_{k=1}^{n} x_k^2,
$$
for all positive integers $n$ and all positive real numbers $x_1, \dots, x_n$. | [
"The best constant is $c = 4$. We first show that, if $n$ is a positive integer and $x_1, \\dots, x_n$ are positive real numbers, then\n$$\n\\sum_{k=1}^{n} \\left( \\frac{1}{k} \\sum_{j=1}^{k} x_j \\right)^2 + \\frac{2}{n} \\left( \\sum_{k=1}^{n} x_k \\right)^2 < 4 \\sum_{k=1}^{n} x_k^2,\n$$\nso $c \\le 4$. To prov... | Romania | 65th NMO Selection Tests for BMO and IMO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 4 | |
0jf3 | Problem:
Triangle $ABC$ has perimeter $1$. Its three altitudes form the side lengths of a triangle. Find the set of all possible values of $\min(AB, BC, CA)$. | [
"Solution:\n\nAnswer: $\\left(\\frac{3-\\sqrt{5}}{4}, \\frac{1}{3}\\right]$\n\nLet $a, b, c$ denote the side lengths $BC, CA$, and $AB$, respectively. Without loss of generality, assume $a \\leq b \\leq c$; we are looking for the possible range of $a$.\n\nFirst, note that the maximum possible value of $a$ is $\\fra... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | ((3 - sqrt(5)) / 4, 1/3] | |
0aaq | Let $a$ and $b$ be two side lengths of the triangle $\Delta ABC$. Their two corresponding medians are perpendicular. Evaluate the third side length using only $a$ and $b$. | [
"Let $T$ be the center of mass of the triangle $\\Delta ABC$ and $\\overline{AA_1} = t_a$, $\\overline{BB_1} = t_b$ are the two corresponding medians to the sides $BC$ and $AC$. From the right-angled triangle $\\Delta BA_1T$, using that $\\overline{A_1T} = \\frac{1}{3}t_a$ and $\\overline{TB} = \\frac{2}{3}t_b$, we... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | c = sqrt((a^2 + b^2)/5) | |
0a12 | A city wants to conduct a trial with alternative time measurements: a 24-hour day will be divided into 15 howers, an hower into 72 minotes, and a minote again into 80 seconds, which are as long as seconds in ordinary timekeeping. In other words, 15 howers of 72 minotes with 80 seconds last as long as 24 hours of 60 min... | [
"$16 : 13 : 30$"
] | Netherlands | Junior Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | final answer only | 16 : 13 : 30 | |
0fa4 | Problem:
(1) $r_1, r_2, \ldots, r_{100}, c_1, c_2, \ldots, c_{100}$ are distinct reals. The number $r_i + c_j$ is written in position $i, j$ of a $100 \times 100$ array. The product of the numbers in each column is $1$. Show that the product of the numbers in each row is $-1$.
(2) $r_1, r_2, \ldots, r_{2n}, c_1, c_2,... | [] | Soviet Union | 25th ASU | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | Part (1): The product in each row equals -1. Part (2): The products in all rows are equal and are the negative of the common column product. | |
0l7a | Problem:
Let $ABCD$ be an isosceles trapezoid such that $CD > AB = 4$. Let $E$ be a point on line $CD$ such that $DE = 2$ and $D$ lies between $E$ and $C$. Let $M$ be the midpoint of $\overline{AE}$. Given that points $A$, $B$, $C$, $D$, and $M$ lie on a circle with radius $5$, compute $MD$. | [
"Solution:\n\nLet $D'$ be the reflection of $D$ across $M$. Then, $ADED'$ is a parallelogram. Hence, $D'A = 2$, so $D'B = 6$. Thus, if $D'M = MD = x$, then Power of a Point at $D'$ gives $x \\cdot (2x) = 2 \\cdot 6$, so $x = \\sqrt{6}$."
] | United States | HMMT February 2025 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(6) | |
0dy5 | Find all real numbers $x$ for which the inequality
$$
|||2 - x| - x| - 8| \le 2008
$$
holds. | [
"We consider two cases depending on the sign of $2-x$.\n\nIf $x \\ge 2$, then we have $|2-x| = -(2-x)$ and the inequality can be rewritten as $||x-2-x|-8| \\le 2008$ or, equivalently, $6 \\le 2008$. So for all real numbers $x \\ge 2$ the inequality holds.\n\nNow, let $x < 2$. In this case we have\n$$\n||2 - 2x| - 8... | Slovenia | Slovenija 2008 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | x ≥ -1007 | |
04b0 | $$
S = \{0, 1, 1+2, 1+2+3, \dots, 1+2+3+\dots+(n-1)\}.
$$
a) If $n$ is a power of $2$, prove that all the elements of $S$ have different remainders modulo $n$.
b) If $n$ is not a power of $2$, prove that there exist two elements of $S$ with the same remainder modulo $n$. | [
"Let $k$ and $l$ be positive integers such that $0 \\le k < l \\le n-1$. Sums $1+2+\\dots+k$ and $1+2+\\dots+l$ give the same remainder modulo $n$ if and only if\n$$\n\\frac{1}{2} k (k + 1) \\equiv \\frac{1}{2} l (l + 1) \\pmod{n},\n$$\ni.e. if and only if $l(l+1) - k(k+1) = (l-k)(l+k+1)$ is divisible by $2n$.\nNot... | Croatia | CroatianCompetitions2011 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0l07 | What is $10! - 7! \cdot 6!$?
(A) -120 (B) 0 (C) 120 (D) 600 (E) 720 | [
"**Answer (B):** Note that\n$$\n10! - 7! \\cdot 6! = 7! \\cdot (10 \\cdot 9 \\cdot 8 - 6!) = 7! \\cdot (720 - 720) = 0.\n$$"
] | United States | AMC 10 B | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | B | |
09qd | Problem:
a. Laat $a$ en $b$ positieve gehele getallen zijn zodat $M(a, b) = a - \frac{1}{b} + b \left(b + \frac{3}{a}\right)$ een geheel getal is. Bewijs dat $M(a, b)$ een kwadraat is.
b. Vind gehele getallen $a$ en $b$, beide ongelijk aan nul, zodat $M(a, b)$ een positief geheel getal is, maar geen kwadraat. | [
"Solution:\n\na.\nOmdat $a + b^{2}$ een geheel getal is, is ook $-\\frac{1}{b} + \\frac{3b}{a}$ een geheel getal. Dit kunnen we schrijven als $\\frac{-a + 3b^{2}}{ab}$. We zien dat $ab$ een deler is van $3b^{2} - a$. In het bijzonder is $b$ een deler van $3b^{2} - a$ en dus geldt $b \\mid a$. Maar dat betekent dat ... | Netherlands | toets | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | a = 4, b = -2 | |
0fym | Problem:
In einem Land gibt es endlich viele Städte und endlich viele Strassen. Jede Strasse verbindet zwei verschiedene Städte und je zwei Städte sind durch höchstens eine Strasse verbunden. Alle Strassen können in beide Richtungen befahren werden und das Strassennetz ist so eingerichtet, dass man jede Stadt von jede... | [
"Solution:\n\nDie Aufgabe lässt sich in naheliegender Weise graphentheoretisch umformulieren: Die Städte und Strassen bilden die Ecken und Kanten eines endlichen Graphen $G$ (ohne Schlingen und Mehrfachkanten). Nach Voraussetzung ist $G$ zusammenhängend und jede Ecke besitzt gerade Ordnung.\n\nWir verwenden das fol... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
0aim | Let $n$ be a positive integer. A regular hexagon with side length $n$ is divided into equilateral triangles with side length $1$ by lines parallel to its sides.
Find the number of regular hexagons all of whose vertices are along the vertices of the equilateral triangles. | [
"By a lattice hexagon we will mean a regular hexagon whose sides run along edges of the lattice. Given any regular hexagon $H$, we construct a lattice hexagon whose edges pass through the vertices of $H$, as shown in the figure, which we will call enveloping lattice hexagon of $H$. Given a lattice hexagon $G$ of si... | North Macedonia | Balkan Mathematical Olympiads | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | (n(n+1)/2)^2 | |
06zy | Problem:
Show that there is a number $1 < b < 1993$ such that if $1994$ is written in base $b$ then all its digits are the same. Show that there is no number $1 < b < 1992$ such that if $1993$ is written in base $b$ then all its digits are the same. | [
"Solution:\n\nAny even number $2n$ can be written as $22$ in base $n - 1$. In particular, $1994 = 22_{996}$.\n\nWe have to show that we cannot write $1993 = aaa\\ldots a_b$. If the number has $n$ digits, then\n$$\n1993 = a(1 + b + \\ldots + b^{n-1}) = a\\frac{b^n - 1}{b - 1}.\n$$\nBut $1993$ is prime, so $a$ must b... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss... | null | proof only | null | |
0hx0 | Problem:
Let $a$ and $b$ be positive integers. Define a sequence $x_{0}, x_{1}, x_{2}, \ldots$ by $x_{0}=0$, $x_{1}=1$, and $x_{n+2}=a x_{n+1}+b x_{n}$ for $n \geq 0$. Prove that
$$
\frac{x_{m+1} x_{m+2} \cdots x_{m+n}}{x_{1} x_{2} \cdots x_{n}}
$$
is an integer for all positive integers $m$ and $n$. | [
"Solution:\n\nLet $c_{m, n}$ be the quotient in the problem. Inspired by the similarity to $\\left(\\begin{array}{c}m+n \\\\ m\\end{array}\\right)$, we seek a relation between $c_{m, n}$, $c_{m-1, n}$, and $c_{m, n-1}$. First we prove the addition identity:\nLemma 1. For $m \\geq 1$ and $n \\geq 0$,\n$$\nx_{m+n}=x_... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0aog | Problem:
The vertices $D$, $E$ and $F$ of the rectangle are midpoints of the sides of $\triangle ABC$. If the area of $\triangle ABC$ is $48$, find the area of the rectangle.
(a) 12
(b) 24
(c) 6
(d) $12 \sqrt{2}$ | [] | Philippines | QUALIFYING STAGE | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | MCQ | 12 |
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