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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0eu4 | Let $ABC$ be an acute triangle and $O$ be its circumcircle. Let $O'$ be the circle that is tangent to $O$ at $A$ and tangent to the side $BC$ at $D$. $O'$ intersects the lines $AB$ and $AC$ at $E$ and $F$, respectively. $O'$ intersects the lines $OO'$ and $EO'$ at $A' (\neq A)$ and $G (\neq E)$, respectively. The lines... | [
"Let $X$ be the intersection of the line $BO$ with the circle $O$. Since the line segments $EG$ and $BX$ are diameters of the circles $O$ and $O'$, respectively, $\\angle EAG = \\angle BAX = 90^\\circ$ and hence three points $A, G, X$ are collinear.\n\nSince $O'$ is tangent to $O$ at $A$, we have $\\angle AEG = \\a... | South Korea | 20th Korean Mathematical Olympiad Final Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ia3 | Problem:
Define the Fibonacci numbers by $F_{0}=0$, $F_{1}=1$, $F_{n}=F_{n-1}+F_{n-2}$ for $n \geq 2$. For how many $n$, $0 \leq n \leq 100$, is $F_{n}$ a multiple of $13$? | [
"Solution:\n\nThe sequence of remainders modulo $13$ begins $0, 1, 1, 2, 3, 5, 8, 0$, and then we have $F_{n+7} \\equiv 8 F_{n}$ modulo $13$ by a straightforward induction. In particular, $F_{n}$ is a multiple of $13$ if and only if $7 \\mid n$, so there are $15$ such $n$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 15 | |
0goh | Let $K$ be the set of the edges and the diagonals of a convex $2010$-gon in the plane. We say that a subset $A$ of $K$ is *intersecting* if any pair of line segments belonging to $A$ intersects. Determine the maximum possible number of elements in the union of two intersecting sets. | [
"The answer is $4019$. We will show that the answer is $2n-1$ for a $n$-gon if $n \\ge 5$.\nLet $A$ be an intersecting set with $|A| \\ge n$. Since the number of vertices in $A$ is not greater than the number of line segments in $A$, $A$ contains a cycle. Let $PQ$ and $QR$ be two line segments in this cycle. Every ... | Turkey | 18th Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 4019 | |
0bst | Let $\alpha$ and $\beta$ be real numbers. Find, with proof, the maximal value of the expression
$$
|\alpha x + \beta y| + |\alpha x - \beta y|,
$$
in each of the following cases:
a) $x, y \in \mathbb{R}$, such that $|x| \le 1$ and $|y| \le 1$;
b) $x, y \in \mathbb{C}$, such that $|x| \le 1$ and $|y| \le 1$. | [
"a) It is clear that for any real numbers $u, v$, we have $|u + v| + |u - v| \\in \\{\\pm 2u, \\pm 2v\\}$, and thus $|\\alpha x + \\beta y| + |\\alpha x - \\beta y| \\le \\max\\{2|\\alpha|, 2|\\beta|\\}$. As we get the equality for $x = y = 1$, the quantity on the right hand side is the maximal value.\n\nb) It is n... | Romania | 67th Romanian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | a) 2 max(|α|, |β|); b) 2√(α^2 + β^2) | |
09i7 | For nonnegative reals $x, y, z \ge 0$ satisfying $xy + yz + zx = 3$, prove
$$
(1 + 2x)(1 + y)^2 + (1 + 2y)(1 + z)^2 + (1 + 2z)(1 + x)^2 \ge 36.
$$ | [] | Mongolia | Mongolian Mathematical Olympiad Round 2 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0gkp | Determine the largest integer that divides $p^4 - 1$ for all primes $p$ greater than $3$. | [
"We first show that $3 \\mid (p^4 - 1)$. Since $p > 3$ is a prime, we have $3 \\nmid p$, and so $p \\equiv \\pm 1 \\pmod{3}$. Thus, $p^4 \\equiv 1 \\pmod{3}$, as required.\n\nWe next claim that $2^4 \\mid (p^4 - 1)$. Since $p$ is odd, we have $p \\equiv \\pm 1, \\pm 3, \\pm 5, \\pm 7 \\pmod{16}$. If $p \\equiv \\pm... | Thailand | The 10th Thailand Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 48 | |
0dn5 | Problem:
Нека композиција садржи $n>1$ вагона са златницима. Постоје две врсте наизглед истих златника: прави и лажни. У сваком вагону се налазе златници само једне врсте. Златници исте врсте су исте масе, док златници различитих врста немају исту масу. Маса правог златника је позната.
Одредити минималан број мерења н... | [
"Solution:\n\nДоказаћемо да је минималан број мерења једнак 2. Означимо тежине правог и лажног златника са $x$ и $y$ редом, и нека је $a_{i}=1$ ако су златници у $i$-том вагону лажни, а $a_{i}=0$ у супротном.\n\nУзмимо у првом мерењу по један златник из сваког вагона. Тада је $a_{1}+a_{2}+\\cdots+a_{n} = \\frac{n x... | Serbia | Serbian Mathematical Olympiad | [
"Discrete Mathematics > Algorithms",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 2 | |
059x | Little Juku writes all integers from $1$ to $n$ on a blackboard, but as he does not know the digit $4$ yet, he skips all numbers that contain $4$. Juku's sister Mari erases two numbers on the blackboard and writes the absolute value of the difference of these numbers on the blackboard. Then Mari again erases two number... | [
"*Answer:* (a) No; (b) No.\n\na. The difference and the sum of two integers have equal parity, likewise are an integer and its absolute value of equal parity. Thus Mari's every step keeps the parity of the sum of all numbers on the blackboard unchanged. Among $1, 2, \\ldots, 2021$, there are $1011$ odd numbers. As ... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | (a) No; (b) No | |
01bu | Does there exist positive real numbers $a$ and $b$ such that $\lfloor an + b \rfloor$ is a prime for each positive integer $n$? | [
"**Answer:** No, there are no such numbers.\n\nConsider the sequence $x_i = \\lfloor a i + b \\rfloor$. We can estimate the difference\n$$\nx_{i+1} - x_i = \\lfloor a(i+1) + b \\rfloor - \\lfloor a i + b \\rfloor < a(i+1) + b - (a i + b - 1) = a + 1\n$$\nand see that it is bounded. As there are arbitrary long seque... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | No | |
03cm | Let $f(x) = x^2 + bx + 1$, where $b$ is a real parameter. Find the number of the integer solutions of the inequality $f(f(x) + x) < 0$. | [
"If $|b| \\le 2$, then $f(x) \\ge 0$ for every $x$.\n\nLet $|b| > 2$ and $x_1 < x_2$ are the real zeros of $f$. Then\n$$\n(*) \\quad f(f(x) + x) = (f(x) + x - x_1)(f(x) + x - x_2) \\\\ = (x - x_1)(x - x_2 + 1)(x - x_2)(x - x_1 + 1).\n$$\nWe consider two cases.\n\nCase 1. If $x_2 - x_1 \\le 1$, then $2 < |b| \\le \\... | Bulgaria | 68. National Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | Number of integer solutions equals 0 for |b| ≤ 2; equals 1 when b = m + 1/m for an integer m ≥ 2; equals 2 in all other cases. | |
054w | Find the least positive integer $n$ for which there exists a positive integer $a$ such that both $a$ and $a+735$ have exactly $n$ positive divisors. | [
"If numbers $a$ and $a+735$ had exactly $2$ divisors, they would be primes. These numbers are of different parity, whence one of them is even. But if $a=2$ then $a+735 = 737 = 11 \\cdot 67$.\nIf numbers $a$ and $a+735$ had exactly $3$ divisors, each of them would be a square of a prime. Similarly to the previous ca... | Estonia | National Olympiad Final Round | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 4 | |
07r6 | There are some boys and some girls at a party. A set of boys is said to be *sociable* if every girl at the party knows at least one boy in that set, and similarly a set of girls is said to be *sociable* if every boy at the party knows at least one girl in that set.
Suppose that the number of sociable sets of boys is od... | [
"We say that a set $X$ of boys is separated from a set $Y$ of girls if no boy in $X$ knows any girl in $Y$. Similarly, a set $Y$ of girls is separated from a set $X$ of boys if no girl in $Y$ knows any boy in $X$. Since acquaintance is mutual, separation is symmetric: $X$ is separated from $Y$ if and only if $Y$ is... | Ireland | Ireland_2017 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
06jb | Let $a$ and $b$ be integers. If $a + b$ is a root of the equation $x^2 + a x + b = 0$, find the smallest possible value of $a b$. | [
"Since $a + b$ satisfies the given equation, we have $(a + b)^2 + a(a + b) + b = 0$. Rearranging gives $b^2 + (3a + 1)b + 2a^2 = 0$. This means $b$ is a root of the equation $x^2 + (3a + 1)x + 2a^2 = 0$. As $a$ and $b$ are integers, the discriminant $(3a + 1)^2 - 4 \\cdot 2a^2 = a^2 + 6a + 1$ must be a perfect squa... | Hong Kong | Hong Kong Preliminary Selection Contest | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | -54 | |
0dpo | Is there a positive integer sequence $a_1, a_2, \dots$ that contains every positive integer exactly once and such that
$$
\tau(na_{n+1}^n + (n+1)a_n^{n+1})
$$
is divisible by $n$ for every positive integer $n$? (Here $\tau(n)$ is the number of positive divisors of $n$.) | [
"From the statement we have\n$$\n\\tau(na_{n+1}^n + (n+1)a_n^{n+1}) \\vdash n. \\qquad (1)\n$$\nLet $a_1 = 1$ and $a_2 = 2$. Suppose that we have constructed the numbers $a_1, a_2, \\dots, a_m$ so that they are different natural numbers and (1) satisfied for each $n = 1, 2, \\dots, m-1$. Let $L$ be the smallest pos... | Silk Road Mathematics Competition | SILK ROAD MATHEMATICS COMPETITION XVII | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0hqg | Problem:
Let $F$ be a point, $d$ be a line, and let $\mathcal{P}$ be the parabola with focus $F$ and directrix $d$, i.e. the set of points which are equidistant from $F$ and $d$. Show that there is a line $\ell$ such that for any point $A$ on $\mathcal{P}$, the circle with diameter $A F$ is tangent to $\ell$. | [
"Solution:\n\nLet $\\ell$ be the line parallel to $d$ equidistant from $F$ and $d$,\n$V$ be the vertex of $\\mathcal{P}$, i.e., the foot of the altitude from $F$ to $\\ell$,\n$X$ be the foot of the altitude from $A$ to $d$,\n$O$ be the midpoint of $\\overline{F A}$,\n$B$ be the foot of the altitude from $A$ to $\\e... | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
09o2 | Let $I$ be the incenter of triangle $ABC$, which is inscribed in the circle $\omega$ centered at $O$. Suppose the line $BI$ intersects $\omega$ again at point $M$. Let $I_B$ be the reflection of $I$ over the line $AC$. Suppose that the line $MI_B$ intersects $\omega$ again at point $D \neq M$, and the line $DO$ interse... | [
"Note that since $OM \\parallel IIB$, it follows that $\\angle IMO = \\angle I_BIM$. Also, since $BO = OM$, we have\n$$\n\\angle IMO = \\angle IBO = \\angle I_BIM.\n$$\nLet us compute the power of point $I$ with respect to the circle $\\omega$. Since $BI \\cdot IM = R^2 - OI^2 = 2Rr$, we get\n$$\nBI \\cdot IM = 2Rr... | Mongolia | MMO2025 Round 4 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
06bh | Let $G$ be a simple graph with $n$ vertices and $m$ edges. Two vertices are called neighbours if there is an edge between them. It turns out that $G$ does not contain any cycle of length from $3$ to $2k$ (inclusive), where $k \ge 2$ is a given positive integer.
a. Prove that it is possible to pick a nonempty set $S$ o... | [
"a. Starting from the graph $G$, each time we remove a vertex with degree less than $\\frac{m}{n}$ together with its edges, until there does not exist such a vertex. Since each time we have removed less than $\\frac{m}{n}$ edges, it is impossible that we remove all $n$ vertices since otherwise there is no vertex le... | Hong Kong | 1997-2023 IMO HK TST | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
07qq | Determine the equations of the common tangents to the circles $x^2 + y^2 = 2r$ and $x^2 + y^2 = 2y$.
 | [
"The line $y = mx + c$ is a tangent to $x^2 + y^2 = 2x$ iff its distance from the centre of the circle is equal to the radius, namely, iff\n$$\n\\frac{|m + c|}{\\sqrt{1 + m^2}} = 1 \\iff |m + c| = \\sqrt{1 + m^2}.\n$$\nSimilarly, the same line is a tangent to $x^2 + y^2 = 2y$ iff $|1 - c| = \\sqrt{1 + m^2}$. So, we... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | y = -x + 1 + √2 and y = -x + 1 - √2 | |
07sv | Determine whether there is a finite set $\{a_1, a_2, \dots, a_n\}$ of distinct positive integers such that
$$
2020 = \sum_{j=1}^{n} \frac{1}{a_j}.
$$ | [
"Yes, there is such a set. We construct it in two stages.\nReaders who are familiar with Egyptian Fractions should be aware of the fact that any rational number $0 < x < 1$ can be expressed as a sum of reciprocals of distinct integers. Indeed, the greedy algorithm delivers such an expression: at each point choose t... | Ireland | IRL_ABooklet_2020 | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | Yes | |
01az | Let $a, b, c > 0$ and $abc = 1$. Prove that
$$
\frac{a^{2014}}{1+2bc} + \frac{b^{2014}}{1+2ca} + \frac{c^{2014}}{1+2ab} \ge \frac{3}{ab+bc+ca}
$$ | [
"Rewrite it as $(\\sum \\frac{a^{2014}}{1+2bc}) (\\sum \\frac{1}{a}) \\ge 3$. Observe that $\\frac{a^{2014}}{1+2bc} = \\frac{a^{2015}}{a+2}$ and $\\sum \\frac{1}{a} \\ge \\frac{1}{3} \\sum \\frac{a+2}{a}$. Thus\n$$\n\\left(\\sum \\frac{a^{2014}}{1+2bc}\\right) \\left(\\sum \\frac{1}{a}\\right) \\ge \\frac{1}{3} \\l... | Baltic Way | Baltic Way | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0iks | Problem:
Suppose $ABCD$ is an isosceles trapezoid in which $\overline{AB} \parallel \overline{CD}$. Two mutually externally tangent circles $\omega_1$ and $\omega_2$ are inscribed in $ABCD$ such that $\omega_1$ is tangent to $\overline{AB}$, $\overline{BC}$, and $\overline{CD}$ while $\omega_2$ is tangent to $\overlin... | [
"Solution:\n\nLet the radius of both circles be $r$, and let $\\omega_1$ be centered at $O_1$. Let $\\omega_1$ be tangent to $\\overline{AB}$, $\\overline{BC}$, and $\\overline{CD}$ at $P$, $Q$, and $R$ respectively. Then, by symmetry, $PB = \\frac{1}{2} - r$ and $RC = 3 - r$. By equal tangents from $B$ and $C$, $B... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 3/7 | |
0aoo | Problem:
Which is a set of factors of $(r-s)^{3}+(s-t)^{3}+(t-r)^{3}$ ?
(a) $\{r-s, s-t, t-r\}$
(b) $\{3 r-3 s, s+t, t+r\}$
(c) $\{r-s, s-t, t-2 r t+r\}$
(d) $\left\{r^{2}-s^{2}, s-t, t-r\right\}$ | [] | Philippines | QUALIFYING STAGE | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | (a) | |
0kq7 | Problem:
A number is chosen uniformly at random from the set of all positive integers with at least two digits, none of which are repeated. Find the probability that the number is even. | [
"Solution:\nSince the number has at least two digits, all possible combinations of first and last digits have the same number of possibilities, which is $\\sum_{i=0}^{8} \\frac{8!}{i!}$. Since the first digit cannot be zero, all of the last digits have 8 possible first digits, except for 0, which has 9 possible fir... | United States | HMMT November 2022 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 41/81 | |
0a1d | Dotty has drawn a paper with six dots on it, no three of which lie on a line. With a pen, she draws a straight line from a certain dot to another dot. Then, without lifting her pen, she draws a straight line to a dot she has not visited before. She continues like this until she has visited all the dots. Finally, she dr... | [
"60"
] | Netherlands | Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | final answer only | 60 | |
0e4h | Show that for all positive numbers $a$, $b$ and $c$ the following inequality holds:
$$
\frac{(a+b-c)^2}{c} + \frac{(b+c-a)^2}{a} + \frac{(c+a-b)^2}{b} \geq 6.
$$
When does the equality hold? | [
"Let us consider the given expression:\n$$\nS = \\frac{(a+b-c)^2}{c} + \\frac{(b+c-a)^2}{a} + \\frac{(c+a-b)^2}{b}.\n$$\n\nLet us use the AM-GM inequality and symmetry. First, note that all variables are positive.\n\nLet us try the case $a = b = c$:\n\nIf $a = b = c$, then\n$$(a+b-c) = a + b - c = a + a - a = a,$$\... | Slovenia | Selection Examinations for the IMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Equality holds when a = b = c = 2. | |
0j9c | Problem:
Let $ABC$ be a triangle with $AB < AC$. Let the angle bisector of $\angle A$ and the perpendicular bisector of $BC$ intersect at $D$. Then let $E$ and $F$ be points on $AB$ and $AC$ such that $DE$ and $DF$ are perpendicular to $AB$ and $AC$, respectively. Prove that $BE = CF$. | [
"Solution:\nNote that $DE$, $DF$ are the distances from $D$ to $AB$, $AC$, respectively, and because $AD$ is the angle bisector of $\\angle BAC$, we have $DE = DF$. Also, $DB = DC$ because $D$ is on the perpendicular bisector of $BC$. Finally, $\\angle DEB = \\angle DFC = 90^{\\circ}$, so it follows that $\\triangl... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0gvg | Two different points $P$ and $Q$ are chosen inside a parallelogram $ABCD$ in such a way that $\angle ABP = \angle ADP$, $\angle CBQ = \angle CDQ$, and those points do not lie on the diagonal $AC$. Prove that $\angle PAQ = \angle PCQ$. | [
"**10.4. Лема.** Якщо всередині паралелограма $ABCD$ відмічено таку точку $M$, що $\\angle ABM = \\angle ADM$, то $\\angle MAD = \\angle MCD$.\n\nДійсно, нехай $N$ — така точка, що відрізок $MN$ паралельний і рівний відрізку $BC$. Тоді чотирикутники $AMND$ і $MBCN$ є паралелограмами. Звідси встановлюємо, що\n$$\n\\... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06he | If a sequence $\{a_1, a_2, \dots, a_n\}$ of positive integers (where $n$ is a positive integer) has the property that the last digit of $a_k$ is the same as the first digit of $a_{k+1}$ (here $k = 1, 2, \dots, n$ and we define $a_{n+1} = a_1$), then the sequence is said to be a 'dragon sequence'. For example, $\{414\}$... | [
"If we take any 46 two-digit numbers, then only 44 two-digit numbers are not chosen. Hence we either have one of the 36 pairs $\\{\\overline{AB}, \\overline{BA}\\}$ where $1 \\leq A < B \\leq 9$, or one of the 9 numbers $11, 22, \\dots, 99$. In either case, we get a 'dragon sequence' by definition.\n\nNext, suppose... | Hong Kong | Hong Kong Preliminary Selection Contest | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 46 | |
09r8 | Problem:
Laat $A$ een verzameling van positieve gehele getallen zijn met de volgende eigenschap: voor elk positief geheel getal $n$ zit precies één van de drie getallen $n, 2 n$ en $3 n$ in $A$. Verder is gegeven dat $2 \in A$. Bewijs dat $13824 \notin A$. | [
"Solution:\nWe bewijzen de volgende twee beweringen:\n(i) Als $m \\in A$ met $2 \\mid m$, dan $6 m \\in A$.\n(ii) Als $m \\in A$ met $3 \\mid m$, dan $\\frac{4}{3} m \\in A$.\n\nEerst bewering (i). Neem aan dat $m \\in A$ met $2 \\mid m$. Door $n=\\frac{m}{2}$ te kiezen, zien we dat $\\frac{m}{2}$ en $\\frac{3}{2} ... | Netherlands | Dutch TST | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
07yj | Problem:
Nel trapezio $ABCD$ i lati $AB$ e $CD$ sono paralleli, gli angoli $\widehat{ABC}$ e $\widehat{BAD}$ sono acuti. Dimostrare che è possibile dividere il triangolo $ABC$ in 4 triangoli disgiunti $X_{1}, \ldots, X_{4}$ e il triangolo $ABD$ in 4 triangoli disgiunti $Y_{1}, \ldots, Y_{4}$ tali che i triangoli $X_{i... | [
"Solution:\n\nSia $D^{\\prime}$ il simmetrico di $D$ rispetto ad $AB$, e sia $E$ intersezione fra la retta $CD^{\\prime}$ e la retta $AB$. Dato che gli angoli $\\widehat{BAD}, \\widehat{CBA}$ sono acuti, $E$ giace all'interno del segmento $AB$. Inoltre dato che $C$ e $D^{\\prime}$ sono equidistanti da $AB$, si ha $... | Italy | null | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
027i | Problem:
Resolva a equação
$$
\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1
$$ | [
"Solution:\n\nTemos\n$$\n\\begin{aligned}\n\\sqrt{x+3-4 \\sqrt{x-1}}+\\sqrt{x+8-6 \\sqrt{x-1}} &= \\sqrt{(\\sqrt{x-1}-2)^{2}}+\\sqrt{(\\sqrt{x-1}-3)^{2}} \\\\\n&=|\\sqrt{x-1}-2|+|\\sqrt{x-1}-3|\n\\end{aligned}\n$$\nComo a distância entre 2 e 3 é 1, só podemos ter $|\\sqrt{x-1}-2|+|\\sqrt{x-1}-3|=1$ quando $\\sqrt{x... | Brazil | NÍVEL 3 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | [5, 10] | |
06o4 | Find all positive integer(s) $n$ such that $(2^n - 1)(5^n - 1)$ is a perfect square. | [
"$n = 1$.\nWhen $n = 1$, $(2^1 - 1)(5^1 - 1) = 4$ is a perfect square. Suppose $n > 1$.\n\n* If $n \\equiv 3 \\pmod{4}$, we have $(2^n - 1)(5^n - 1) \\equiv (2)(-1) \\equiv 3 \\pmod{5}$, which is not true for a square.\n\n* If $n \\equiv 1 \\pmod{4}$, we rewrite\n$$\n(2^n - 1)(5^n - 1) = (2^n - 1) \\cdot 4(5^{n-1} ... | Hong Kong | The 26th Hong Kong (China) Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Factorization techniques"
... | null | proof and answer | n = 1 | |
0d7s | Let $ABC$ be a triangle whose incircle ($I$) is tangent to $AB$, $AC$ at $D$, $E$ respectively. Denote by $\Delta_{b}$, $\Delta_{c}$ the lines symmetric to the lines $AB$, $AC$ with respect to $CD$, $BE$ correspondingly. Suppose that $\Delta_{b}$, $\Delta_{c}$ meet at $K$.
1. Prove that $IK \perp BC$.
2. If $I \in (KDE... | [
"1. Denote by $G$, $H$ the intersections of $KE$, $KF$ and $BC$ respectively. Consider triangle $CEG$: it is easy to see that\n- $CI$ is the internal bisector of $\\angle C$\n- $EI$ is the external bisector of $\\angle E$.\nHence, $I$ is the excenter of this triangle, which implies that $GI$ is the bisector of $\\a... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrila... | English | proof only | null | |
06zf | Problem:
$A$ and $B$ are opposite corners of an $n \times n$ board, divided into $n^{2}$ squares by lines parallel to the sides. In each square the diagonal parallel to $AB$ is drawn, so that the board is divided into $2 n^{2}$ small triangles. The board has $(n+1)^{2}$ nodes and a large number of line segments, each ... | [
"Solution:\n\nThe diagram above shows that $n=2$ is possible (the path is $AHEFGHCDIHB$). Now suppose $n > 2$.\n\nNote that if $X$ is any vertex except $A$ or $B$, then an even number of segments with endpoint $X$ must be in the path.\n\nLet $F$ be the bottom left-hand vertex. Two sides of the triangle $EFG$ are in... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | n = 2 | |
0dp1 | Find the maximal value of the real number $M$ such that for all positive real numbers $a$, $b$, $c$ the following inequality holds:
$$
a^3 + b^3 + c^3 - 3abc \ge M(|a-b|^3 + |b-c|^3 + |c-a|^3)
$$ | [
"Answer: $M = \\frac{1}{2}$.\nThe inequality $a^3 + b^3 + c^3 - 3abc \\ge M(|a-b|^3 + |b-c|^3 + |c-a|^3)$ for $M = 1/2$ implies from\n$$\n2(a^3 + b^3 + c^3 - 3abc) = (a + b + c)((a - b)^2 + (b - c)^2 + (c - a)^2)\n$$\nand from obvious inequalities\n$$\n(x + y + z)(x - y)^2 \\ge |x - y|^3\n$$\n\nThe value $M = 1/2$ ... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | M = 1/2 | |
04qb | Let $A$, $B$, $C$ and $D$ be points on a circle such that $|AB| = |BC| = |CD|$. The angle bisectors of $\angle ABD$ and $\angle ACD$ intersect at the point $E$.
If the lines $AE$ and $CD$ are parallel, find $\angle ABC$.
(Matko Ljulj) | [
"Due to symmetry, the isosceles triangles $ABC$ and $BCD$ are congruent, and the cyclic quadrilateral $ABCD$ is an isosceles trapezium.\n\n\nLet us denote by $x$ the measure of angles along the bases in $ABC$ and $BCD$. Then $\\angle ABC = \\angle BCD = 180^\\circ - 2x$ and $\\angle ABD = \... | Croatia | Croatian Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Triangles"
] | English | proof and answer | 180° × 5/7 | |
005h | Hallar todos los números primos $p, q$ tales que
$$
p^2 + q = 37q^2 + p.
$$
ACLARACIÓN: $p > 1$; $q > 1$. | [] | Argentina | Argentina 2008 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | Spanish | proof and answer | p = 43, q = 7 | |
0d66 | Let $a > b > c > d$ be positive integers such that $a^{2} + a c - c^{2} = b^{2} + b d - d^{2}$. Prove that $a b + c d$ is a composite number. | [
"By contrary, assume that $p = a b + c d$ is a prime. In particular $(b, c) = (b, d) = 1$, and $a b \\equiv -c d \\pmod{p}$. We get\n$$\n\\begin{aligned}\n0 & = b^{2} (b^{2} + b d - d^{2}) - b^{2} (a^{2} + a c - c^{2}) \\\\\n& = b^{2} (b^{2} + b d - d^{2}) - (a b)^{2} - (a b)(b c) + b^{2} c^{2} \\\\\n& \\equiv (b^{... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
04u2 | Let $k \neq 0$ be an integer. Prove that the number of ordered pairs $(x, y)$ of integers satisfying
$$
k = \frac{x^2 - xy + 2y^2}{x + y}
$$ | [
"Multiplying both sides by $x + y$ yields\n$$\nx^2 - xy + 2y^2 = k(x + y). \\quad (1)\n$$\nAny solution $(x, y)$ to the original equation is a solution to (1), but (1) can have extra solutions satisfying $x + y = 0$, tj. $y = -x$.\nA pair $(x, -x)$ is a solution to (1) if and only if $x^2 + x^2 + 2x^2 = k \\cdot 0$... | Czech Republic | 66th Czech and Slovak Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof only | null | |
08hz | Problem:
Find all five-digit numbers $\overline{abcde}$, written in decimal system, if it is known that $\overline{ab}cde - \overline{ebcda} = 69993$, $\overline{bcd} - \overline{dcb} = 792$, $\overline{bc} - \overline{cb} = 72$. | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | proof and answer | 78000, 88001, 98002, 79110, 89111, 99112 | |
085d | Problem:
Quanti simboli di radice quadrata, come minimo, devono comparire nell'espressione $\sqrt{\cdots \sqrt{\sqrt{123.456 .789}}}$ affinché il risultato sia minore di $2$?
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9. | [
"Solution:\n\nLa risposta è (A). Dato che, per numeri positivi o nulli, $\\sqrt{a}<b$ se e solo se $a<b^{2}$, la domanda chiede dopo quanti elevamenti al quadrato $\\left(\\cdots\\left(2^{2}\\right)^{\\cdots \\cdots}\\right)^{2}=2^{2^{n}}$, il risultato supera $123.456.789$. Dato che $2^{10}=1024>10^{3}$, si trova ... | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | MCQ | A | |
03p8 | Suppose points $I$ and $H$ are the incenter and orthocenter of an acute triangle $ABC$ respectively, and points $B_1$ and $C_1$ are the midpoints of sides $AC$ and $AB$ respectively. It is known that ray $B_1I$ intersects side $AB$ at $B_2$ ($B_2 \neq B$), and ray $C_1I$ intersects the extension of $AC$ at $C_2$: $B_2C... | [
"**Proof** First, we will prove that three points $A$, $I$ and $A_1$ are collinear $\\Leftrightarrow \\angle BAC = 60^\\circ$.\nAs shown in the figure, assume that $O$ is the circumcenter of $\\triangle ABC$. We join $BO$ and $CO$, then\n\n$$\n\\angle BHC = 180^\\circ - \\angle BAC,\n$$\n$$... | China | China Mathematical Olympiad | [
"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",
"Geometry > Plane Geometry > Miscellaneous > C... | English | proof only | null | |
0fx7 | Problem:
Sei $ABCDEF$ ein konvexes Sechseck mit einem Umkreis. Beweise, dass sich die Diagonalen $AD$, $BE$ und $CF$ genau dann in einem Punkt schneiden, wenn gilt
$$
\frac{AB}{BC} \cdot \frac{CD}{DE} \cdot \frac{EF}{FA}=1
$$ | [
"Solution:\n\nWir nehmen zuerst an, dass sich die Diagonalen in einem Punkt $S$ schneiden. Es gilt $\\Varangle ASB=\\Varangle DSE$ und $\\Varangle BAS=\\Varangle DES$, letzteres weil beides Peripheriewinkel über der Sehne $BD$ sind. Also sind die Dreiecke $ABS$ und $EDS$ ähnlich. Analog zeigt man die Ähnlichkeit de... | Switzerland | SMO Finalrunde | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
01a0 | The degree of each vertex in a graph $G$ does not exceed $100$. We remove edges of this graph. In one step we can remove an arbitrary set of edges without common endpoints which is maximal (in a sense that if we add to this set any of remaining edges, then there will appear two edges with common endpoint). Prove that a... | [
"It follows from the fact that for any edge $AB$, if it has not been removed yet, the operation decreases the sum of degrees $d(A) + d(B)$ by at least $1$."
] | Baltic Way | Baltic Way 2013 | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0acj | Find the digits $A$, $B$, $C$ such that the given multiplication procedure is correct:
$$
\begin{array}{c}
\overline{ABC} \cdot \overline{BAC} \\
\multicolumn{2}{c}{---} \\
--A & \\
\multicolumn{2}{c}{---B} \\
\end{array}
$$
Where $\overline{ABC} \cdot C = \overline{---A}$, $\overline{ABC} \cdot A = \overline{---A}$, $... | [
"The numbers $ABC$ and $BAC$ are three-digit, so $A \\neq 0$ and $B \\neq 0$. If $C=1$ the equality $\\overline{ABC} \\cdot C = \\overline{---A}$ is not possible because $\\overline{AB} \\cdot 1 = \\overline{AB} \\neq \\overline{AB} \\cdot 0$. If $A>3$, from the equality $\\overline{ABC} \\cdot A = \\overline{---A}... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | A = 2, B = 8, C = 6 | |
03kc | Problem:
Show that every positive integral power of $\sqrt{2}-1$ is of the form $\sqrt{m}-\sqrt{m-1}$ for some positive integer $m$. (e.g. $\left.(\sqrt{2}-1)^{2}=3-2 \sqrt{2}=\sqrt{9}-\sqrt{8}\right)$. | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Pell's equations",
"Algebra > Intermediate Algebra > Other"
] | null | proof only | null | |
0eua | Given a cyclic hexagon $ABCDEF$, let $BD$ and $CF$ meet at $G$, $AC$ and $BE$ meet at $H$, $AD$ and $CE$ meet at $I$. Suppose that $BD$ is perpendicular to $CF$ and $AI = CI$. Show that $CH = AH + DE$ if and only if $GH \cdot BD = BC \cdot DE$. | [
"Since $AI = CI$, $\\angle ACI = \\angle CAI$, and so $\\angle ACE = \\angle CAD$. Since $ACDE$ is cyclic, $\\angle CAD = \\angle CED$. Hence $\\angle ACE = \\angle CED$, and so $AC$ and $DE$ are parallel to each other.\n\nNow we will show that if $CH = AH + DE$, then $GH \\cdot BD = BC \\cdot DE$. Let $A'$ be the ... | South Korea | Korean Mathematical Olympiad Final Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0iwn | Problem:
Call an integer $n>1$ radical if $2^{n}-1$ is prime. What is the 20th smallest radical number? If $A$ is your answer, and $S$ is the correct answer, you will get $\max \left(25\left(1-\frac{|A-S|}{S}\right), 0\right)$ points, rounded to the nearest integer. | [
"Solution:\n\nAnswer: 4423"
] | United States | Harvard-MIT November Tournament | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | final answer only | 4423 | |
0hk5 | Problem:
The rectangle $M N P Q$ is inside the rectangle $A B C D$. The portion of the rectangle $A B C D$ outside of $M N P Q$ is colored in green. Using just a straightedge construct a line that divides the green figure in two parts of equal areas. | [
"Solution:\n\nThe line passing through the centers of $A B C D$ and $M N P Q$ divides each of the rectangles in two pieces of equal areas. Hence that line will divide the green part into equal areas. The line can be easily constructed using only a straightedge because we can construct the centers of the given recta... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | Draw the line through the centers of the two rectangles, where each center is found as the intersection point of its diagonals. | |
0dzn | Let $D$ be a point on the side $AB$ of an acute triangle $ABC$ such that the triangle $BCD$ is also acute. Denote the orthocentre of the triangle $BCD$ by $H$. Prove: if points $A$, $D$, $H$ and $C$ are concyclic, then the triangle $ABC$ is isosceles. | [
"Let $\\angle BAC = \\alpha$. Since $A$, $D$, $H$ and $C$ are concyclic, we have $\\angle DHC = \\pi - \\angle BAC = \\pi - \\alpha$.\n\nLet $E$ be the foot of the altitude from $C$ to the side $BD$. Since $\\angle DHE = \\pi - \\angle DHC = \\alpha$, we have $\\angle HDB = \\frac{\\pi}{2} - \\alpha$.\n\nSince the ... | Slovenia | Slovenija 2008 | [
"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 | |
01e9 | One of the cells of $20 \times 20$ torus contains a buried treasure. Today, in order to find the treasure we select several rectangles $1 \times 4$ or $4 \times 1$ on this torus and ask the sapper to investigate them by a mine detector. The results of all investigations will be known tomorrow, for each rectangle the sa... | [
"Answer: 160.\nIn our torus each cell is determined by coordinates $(i, j)$, $1 \\le i, j \\le 20$, the two cells being neighbours if one of their coordinates is the same, and the others differ by $\\pm 1 \\bmod 20$.\n\n**Example.** Select the following 160 rectangles\n$$\n(a-1,b); (a-2,b); (a-3,b); (a-4,b) \\pmod{... | Baltic Way | Baltic Way shortlist | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 160 | |
0c9r | Problem:
Considerăm mulţimea $M=\{1,2,3, \ldots, 2019,2020\}$. Găsiţi cel mai mic număr natural nenul $k$ ce verifică următoarea proprietate:
pentru orice submulţime $A$, cu $k$ elemente, a mulţimii $M$, există trei numere $a, b, c$ din $M$, distincte două câte două, astfel încât $a+b, b+c$ si $c+a$ aparţin mulţimii $A... | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 1012 | |
08j4 | Problem:
Natural numbers $1, 2, 3, \ldots, 2003$ are written in an arbitrary sequence $a_{1}, a_{2}, a_{3}, \ldots, a_{2003}$. Let $b_{1} = 1 a_{1}$, $b_{2} = 2 a_{2}$, $b_{3} = 3 a_{3}$, $\ldots$, $b_{2003} = 2003 a_{2003}$, and $B$ be the maximum of the numbers $b_{1}, b_{2}, b_{3}, \ldots, b_{2003}$.
a) If $a_{1} ... | [
"Solution:\n\na) Using the inequality between the arithmetical and geometrical mean, we obtain that $b_{n} = n (2004 - n) \\leq \\left( \\frac{n + (2004 - n)}{2} \\right)^{2} = 1002^{2}$ for $n = 1, 2, 3, \\ldots, 2003$. The equality holds if and only if $n = 2004 - n$, i.e. $n = 1002$. Therefore, $B = b_{1002} = 1... | JBMO | 7th JBMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | Part (a): B = 1002^2. Part (b): For any arrangement, B >= 1002^2. | |
052x | Inside a circle $c$ with the center $O$ there are two circles $c_1$ and $c_2$ which go through $O$ and are tangent to the circle $c$ at points $A$ and $B$ respectively. Prove that the circles $c_1$ and $c_2$ have a common point which lies in the segment $AB$. | [
"The radius $AO$ of the circle $c$ is perpendicular to the common tangent to circles $c$ and $c_1$ at the point $A$, hence $AO$ is a diameter of the circle $c_1$. Similarly $BO$ is a diameter of the circle $c_2$.\n\nIf the circles $c_1$ and $c_2$ are tangent at the point $O$ (Fig. 1), then the diameters $AO$ and $B... | Estonia | Open Contests | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0jim | Problem:
Solve for $x$ in the equation $20 \cdot 14 + x = 20 + 14 \cdot x$. | [
"Solution:\nAnswer: $20$\n\nBy inspection, $20 + 14 \\cdot 20 = 20 \\cdot 14 + 20$.\n\nAlternatively, one can simply compute $x = \\frac{20 \\cdot 14 - 20}{14 - 1} = 20$."
] | United States | HMMT November 2014 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 20 | |
01eh | Given a triangle $\Delta$ with circumradius $R$ and inradius $r$, prove that the area of the circle with radius $R + r$ is at least 5 times greater than the area of the triangle $\Delta$. | [
"Let the area of the triangle $\\triangle \\Delta$ be $S$. Among the triangles with fixed circumradius, the one with largest perimeter is equilateral (as can be easily seen from Jensen's inequality). Hence\n$$\nS = \\frac{a + b + c}{2} \\cdot r \\le \\frac{3\\sqrt{3}}{2} Rr.\n$$\nBy Euler's inequality, $R \\ge 2r$.... | Baltic Way | Baltic Way shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Jensen/smoothing",
"Algebra > Equations ... | English | proof only | null | |
0bjq | Let $n$ be a strictly positive integer and let $A = \{1, 2, \dots, n\}$. Find the number of the increasing functions $f : A \to A$ which satisfy the property
$$
|f(x) - f(y)| \le |x - y|,
$$
for any $x, y \in A$. | [
"We remark that $f(k+1) - f(k) = a_k \\in \\{0, 1\\}$, $k = 1, 2, \\dots, n-1$. Moreover, as $a_k \\in \\{0, 1\\}$, we have $|f(x) - f(y)| \\le |x - y|$ for any $x, y \\in A$. Therefore, any function $f$ which satisfies the conditions in the problem, is\n\ncompletely determined by the $n$-tuple $(f(0), a_1, a_2, \\... | Romania | 65th Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | (n+1) 2^{n-2} | |
0ecf | Problem:
Konstruiraj trikotnik $ABC$ s podatki: $a = 4~\mathrm{cm}$, $\beta = 75^\circ$, $t_c = 5~\mathrm{cm}$. Kot konstruiraj s šestilom in ravnilom. | [] | Slovenia | 15. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
05hd | Problem:
Soit $a$, $b$, $c$ des réels strictement positifs tels que : $ab + bc + ca = abc$. Montrer que :
$$
\frac{1}{a^{2}+b^{2}}+\frac{1}{b^{2}+c^{2}}+\frac{1}{c^{2}+a^{2}} \leqslant \frac{1}{6},
$$
et trouver les cas d'égalité. | [
"Solution:\n\nPosons $u = \\frac{1}{a}$, $v = \\frac{1}{b}$, $w = \\frac{1}{c}$. En divisant par $abc$, la condition se réécrit $u + v + w = 1$.\n\nL'inégalité voulue se réécrit\n$$\n\\frac{u^{2} v^{2}}{u^{2}+v^{2}} + \\frac{u^{2} w^{2}}{u^{2}+w^{2}} + \\frac{v^{2} w^{2}}{v^{2}+w^{2}} \\leqslant \\frac{1}{6}.\n$$\n... | France | ENVOI 2 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | The sum is at most 1/6, with equality when a = b = c = 3. | |
028z | Problem:
Seja $ABCD$ um trapézio retângulo de bases $AB$ e $CD$, com ângulos retos em $A$ e $D$. Dado que a diagonal menor $BD$ é perpendicular ao lado $BC$, determine o menor valor possível para a razão $\frac{CD}{AD}$. | [
"Solution:\n\nSeja $A\\widehat{B}D = B\\widehat{D}C = \\alpha$. Então temos que $DC = \\frac{BD}{\\cos \\alpha}$ e $AD = BD \\sen \\alpha$, donde\n\n$$\n\\frac{DC}{AD} = \\frac{\\frac{BD}{\\cos \\alpha}}{BD \\sen \\alpha} = \\frac{1}{\\sen \\alpha \\cos \\alpha} = \\frac{2}{\\sen 2\\alpha} \\geq 2.\n$$\n\nA igualda... | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 2 | |
0cmo | A positive integer $m$ is chosen so that the sum of all the digits of $2^m$ (in its decimal representation) equals 8. Determine if the last digit of $2^m$ can appear to be 6. (V. Senderov) | [
"Answer. It cannot.\n\nFirst solution. For $m = 1, 2, 3$ the last digit of $2^m$ is not 6. Suppose that the sum of the digits of $2^m$ for some $m > 3$ is 8, and it ends with 6. The number $2^m$ cannot end with 06 or 26, since in this case it is not divisible by 4. Therefore, it ends with 16 (otherwise the sum of d... | Russia | Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | English; Russian | proof and answer | It cannot. | |
0kqg | Problem:
Alice is bored in class, so she thinks of a positive integer. Every second after that, she subtracts from her current number its smallest prime divisor, possibly itself. After 2022 seconds, she realizes that her number is prime. Find the sum of all possible values of her initial number. | [
"Solution:\n\nLet $a_{k}$ denote Alice's number after $k$ seconds, and let $p_{k}$ be the smallest prime divisor of $a_{k}$. We are given that $a_{2022}$ is prime, and want to find $a_{0}$.\n\nIf $a_{0}$ is even, then $a_{n+1}=a_{n}-2$, since every $a_{n}$ is even. Then we need $a_{2022}=2$, so $a_{0}=4046$.\n\nIf ... | United States | HMMT November 2022 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 8093 | |
0jee | Let $ABCD$ be a quadrilateral with $AC = BD$. Diagonals $AC$ and $BD$ meet at $P$. Let $\omega_1$ and $O_1$ denote the circumcircle and circumcenter of triangle $ABP$. Let $\omega_2$ and $O_2$ denote the circumcircle and circumcenter of triangle $CDP$. Segment $BC$ meets $\omega_1$ and $\omega_2$ again at $S$ and $T$ (... | [
"**Note.** The result still holds without the assumption that both triangles $ABP$ and $CDP$ are acute. This assumption helps the contestants to focus on more specific configurations. Indeed, because triangles are acute, $O_1$ lies inside triangle $ABP$ and $O_2$ lies inside triangle $CDP$. Points $M$ and $N$ lie i... | United States | TSTST | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0eig | Problem:
Realni rešitvi enačbe $x^{4}-x^{3}-2 x-4=0$ sta prva dva člena padajočega aritmetičnega zaporedja s 40 členi.
a) Izračunaj prva dva člena in diferenco zaporedja.
b) Izračunaj zadnji člen in zapiši splošni člen $a_{n}$ zaporedja.
c) Izračunaj vsoto vseh členov zaporedja z lihimi indeksi. | [
"Solution:\n\na) Poiščemo realni rešitvi dane enačbe in zapišemo prva dva člena padajočega aritmetičnega zaporedja $a_{1}=2$ in $a_{2}=-1$. Diferenca zaporedja je $d=-3$.\n\nb) Izračunamo 40. člen zaporedja $a_{40}=a_{1}+39 d=2+39 \\cdot(-3)=-115$. Zapišemo splošni člen zaporedja $a_{n}=a_{1}+(n-1) d=2+(n-1) \\cdot... | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | a) a1 = 2, a2 = −1, d = −3; b) a40 = −115, an = 5 − 3n; c) sum of odd-indexed terms = −1100 | |
01yp | Palina marked $2021$ arbitrary points on the circle and drew $2021$ segments between them. By an intersection point we denote any point of intersection of two drawn segments if this point is the endpoint of none of them.
Find the maximal possible number of intersection points. (Mikhail Karpuk) | [
"$2021 \\cdot 1009 = 2039189$.\nLet's solve the problem for an arbitrary odd number $n$ of marked points and drawn segments. We enumerate the marked points on the circle by the numbers from $1$ to $n$ in the order of going around the circle. For each $i = 1, \\overline\\{n\\}$ we denote by $a_i$ the number of drawn... | Belarus | Belarus2022 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, ex... | English | proof and answer | 2039189 | |
05m0 | Problem:
Montrer que si $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n} \geqslant 0$ et $a_{1}+a_{2}+\cdots+a_{n}=1$ alors
$$
a_{1}^{2}+3 a_{2}^{2}+5 a_{3}^{2}+\cdots+(2 n-1) a_{n}^{2} \leqslant 1 .
$$ | [
"Solution:\nAfin de faire apparaître les termes $a_{i}^{2}$, il est tentant de commencer par élever l'expression $a_{1}+a_{2}+\\cdots+a_{n}=1$ au carré, puis de tout développer en espérant très fort qu'une simplification appraîtra. En outre, on voit bien qu'il faudra utiliser le caractère ordonné des termes $a_{i}$... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
02am | Problem:
Entre $1$ e $2$ - Complete os numeradores com inteiros positivos para satisfazer as condições: $\frac{a}{5}$ e $\frac{b}{7}$ são menores do que $1$, e $1 < \frac{a}{5} + \frac{b}{7} < 2$. | [
"Solution:\nComo as duas frações são positivas e menores do que $1$, seus numeradores devem ser respectivamente menores que seus denominadores, logo devemos ter:\n$$\n0 < a < 5 \\text{ e } 0 < b < 7\n$$\nTemos $\\frac{a}{5} + \\frac{b}{7} = \\frac{7a + 5b}{35}$, portanto:\n$$\n1 < \\frac{7a + 5b}{35} < 2 \\Rightarr... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | All solutions: (a,b) = (1,6), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6), (4,2), (4,3), (4,4), (4,5), (4,6). | |
0306 | Problem:
No planeta $X$, existem 100 países alienígenas com conflitos entre si. Para evitar uma guerra mundial, esses países se organizam em grupos de alianças militares para proteção mútua. Sabemos que as alianças seguem as seguintes regras:
1) Nenhuma aliança contém mais de 50 países.
2) Quaisquer dois países perten... | [
"Solution:\n\na) Não é possível. Suponha que o país $A$ pertence a no máximo duas alianças. Nesse caso, como cada aliança tem no máximo 50 países, o país $A$ é membro de uma mesma aliança com no máximo $49+49=98$ países. Como existem 99 países distintos de $A$, pelo menos um deles não estará em uma aliança com $A$ ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | a) No; b) 6 | |
0bou | Given a positive integer $n$, determine the largest real number $\mu$ satisfying the following condition: for every $4n$-point configuration $C$ in an open unit square $U$, there exists an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, and has an area greater t... | [
"The required maximum is $1/(2n+2)$. To show that the condition in the statement is not met if $\\mu > 1/(2n+2)$, let $U = (0,1) \\times (0,1)$, choose a small enough positive $\\varepsilon$, and consider the configuration $C$ consisting of the $n$ four-element clusters of points $(i/(n+1) \\pm \\varepsilon) \\time... | Romania | THE 2015 Seventh ROMANIAN MASTER OF MATHEMATICS | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 1/(2n+2) | |
09up | Mieke has a stack of $21$ cards. Mieke repeats the following operation:
She takes the top two cards from the stack, changes their order, and then puts them at the bottom of the stack (so the top card becomes the bottom card).
Mieke repeats this operation until the cards are back in their original order.
How many times... | [] | Netherlands | Junior Mathematical Olympiad, September 2019 | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Discrete Mathematics > Algorithms",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof and answer | 110 | |
08e6 | Problem:
Attorno a un tavolo rotondo sono sedute, a distanza costante l'una dalla successiva, 32 persone. In quanti modi è possibile scegliere 3 di loro in modo che a coppie non siano né adiacenti né diametralmente opposte?
(A) 2246
(B) 2480
(C) 3616
(D) 24128
(E) Nessuna delle precedenti | [
"Solution:\n\nLa risposta è (C). Selezioniamo le persone una alla volta. All'inizio non abbiamo vincoli, quindi possiamo scegliere la prima persona in 32 modi. Possiamo ora scegliere la seconda persona in 28 modi, escludendo dalle 32 la persona già selezionata, le due adiacenti ad essa e quella di fronte. Tuttavia ... | Italy | Olimpiadi della Matematica | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | MCQ | C | |
0iph | Problem:
Let $ABC$ be a triangle with $BC = 2007$, $CA = 2008$, $AB = 2009$. Let $\omega$ be an excircle of $ABC$ that touches the line segment $BC$ at $D$, and touches extensions of lines $AC$ and $AB$ at $E$ and $F$, respectively (so that $C$ lies on segment $AE$ and $B$ lies on segment $AF$). Let $O$ be the center ... | [
"Solution:\n\nAnswer: $2014024$\n\nLet line $AD$ meet $\\omega$ again at $H$. Since $AF$ and $AE$ are tangents to $\\omega$ and $ADH$ is a secant, we see that $DEHF$ is a harmonic quadrilateral. This implies that the pole of $AD$ with respect to $\\omega$ lies on $EF$. Since $\\ell \\perp AD$, the pole of $AD$ lies... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | null | proof and answer | 2014024 | |
07h5 | Find all functions $f : \mathbb{N} \to \mathbb{N}$ such that for all positive integers $a, b$ we have
$$
f^a(b) + f^b(a) \mid 2(f(ab) + b^2 - 1).
$$
By $f^a(x)$ we mean the $a$-times fold composition of $f$. | [
"Let $b = 1$ we have $f^a(1) + f(a)$ divides $2f(a)$. Hence, $2f(a) = C(f^a(1) + f(a))$, for some positive integer $C$. Thus, $C = 1$ and $f^a(1) = f(a)$. Whence, $f^a(b) = f^{a-1}(f(b)) = f^{a-1}(f^b(1)) = f^{a+b-1}(1) = f(a+b-1)$. By the same argument $f^b(a) = f(a+b-1)$. It then follows that $f(a+b-1)$ divides $... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | All solutions are exactly the following four functions:
1) f(n) = n + 1 for all positive integers n.
2) f(n) = 1 for all n.
3) f(n) = 1 if n is even, and f(n) = 2 if n is odd.
4) f(n) = 1 if n is even, and f(n) = 4 if n is odd. | |
0h9z | Country "U" has cities $A_1, A_2, \dots, A_n$. They are connected with flights so that one can get from any city to any other city with layovers. Let $d_{i,j}$ be the smallest amount of flights needed to get from city $A_i$ to city $A_j$, and let $d$ be the total number of flights in country "U" and let
$$
D = d_{1,1} ... | [
"Show that:\n$$\nd + \\frac{D}{n-1} \\le \\frac{1}{2}(n+1)(n+2).\n$$\n\nLet $i \\ne j$, and let the shortest path between $A_i$ and $A_j$ be the following: $A_i = A_{\\gamma_0}, A_{\\gamma_1}, \\dots, A_{\\gamma_k} = A_j$. Let $X$ be the set of such vertices ($k+1$ vertices), and let $Y$ be the set of all other ver... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
011c | Problem:
Let $n$ be a positive integer not divisible by $2$ or $3$. Prove that for all integers $k$, the number $(k+1)^{n} - k^{n} - 1$ is divisible by $k^{2} + k + 1$. | [
"Solution:\n\nNote that $n$ must be congruent to $1$ or $5$ modulo $6$, and proceed by induction on $\\lfloor n / 6 \\rfloor$. It can easily be checked that the assertion holds for $n \\in \\{1,5\\}$. Let $n > 6$, and put $t = k^{2} + k + 1$. The claim follows by:\n$$\n\\begin{aligned}\n(k+1)^{n} - k^{n} - 1 &= (t ... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
0drn | Find all integers $n$ such that $n^2 + 24n + 35$ is a square. | [
"Suppose $n^2 + 24n + 35$ is a square. Let $n^2 + 24n + 35 = m^2$. This equation can be written as $(2n + 2m + 24)(2n - 2m + 24) = 436$. Since $436 = 2^2 \\times 109$, we have $\\{2n + 2m + 24, 2n - 2m + 24\\} = \\{1, 436\\}, \\{-1, -436\\}, \\{2, 218\\}, \\{-2, -218\\}, \\{4, 109\\}, \\{-4, -109\\}$.\n\nFor instan... | Singapore | Singapur | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 43 and -67 | |
0a7p | Problem:
A piece of paper is the square $ABCD$. We fold it by placing the vertex $D$ on the point $D'$ of the side $BC$. We assume that $AD$ moves on the segment $A'D'$, and that $A'D'$ intersects $AB$ at $E$. Prove that the perimeter of the triangle $EBD'$ is one half of the perimeter of the square. | [
"Solution:\n(See Figure 6.) The fold gives rise to an isosceles trapezium $ADHG$. Because of symmetry, the distance of the vertex $D$ from the side $GH$ equals the distance of the vertex $H$ from side $AD$; the latter distance is the side length $a$ of the square. The line $GH$ thus is tangent to the circle with ce... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 8 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
053a | Find all prime numbers $p$ for which one can find a positive integer $m$ and non-negative integers $a_0, a_1, \dots, a_m$ less than $p$ such that
$$
\begin{cases} a_0 + a_1 p + \dots + a_{m-1} p^{m-1} + a_m p^m = 2013, \\ a_0 + a_1 + \dots + a_{m-1} + a_m = 11. \end{cases}
$$ | [
"*Answer:* 2003.\n\nSubtracting the second equation from the first one gives\n$$\na_1(p-1) + \\dots + a_m(p^m - 1) = 2002.\n$$\nAs the l.h.s. of the obtained equality is divisible by $p-1$, $2002 = 2 \\cdot 7 \\cdot 11 \\cdot 13$ must also be divisible by $p-1$. Thus $p-1$ equals one of $1, 2, 7, 11, 13, 14, 22, 26... | Estonia | IMO Team Selection Contest | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 2003 | |
05wp | Problem:
Soit $ABC$ un triangle acutangle (dont tous les angles sont aigus) avec $BA \neq BC$. Soit $O$ le centre de son cercle circonscrit. La droite $(AB)$ intersecte le cercle circonscrit à $BOC$ une deuxième fois en $P \neq B$. Montrer que $PA = PC$. | [
"Solution:\n\n\n\nTraçons la figure dans le cas où $BC < BA$, le cas $BC > BA$ étant totalement analogue. Il s'agit de montrer que $PA = PC$, c'est-à-dire que $\\widehat{ACP} = \\widehat{PAC} ( = \\widehat{BAC} )$. Or on a :\n$$\n\\begin{aligned}\n\\widehat{ACP} & = \\widehat{ACO} + \\wideh... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
0h1c | Find all pairs of natural numbers $a$ and $b$, such that the difference is $2011$ and the product is a complete square. | [
"Since $a - b = 2011$ and $2011$ is prime, then $(a, b)$ is $1$ or $2011$.\n\n1 Case. If $(a, b) = 1$ then $a = c^2$, $b = d^2$, and we have $a - b = c^2 - d^2 = (c - d)(c + d) = 2011 = 1 \\cdot 2011$ or\n$$\n\\begin{cases}\nc - d = 1, \\\\\nc + d = 2011.\n\\end{cases}\n$$\nSolving this system we get the answer.\n\... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (a, b) = (1006^2, 1005^2) | |
008n | Find the smallest positive integer $k$ for which the 100 fractions
$$
\frac{50}{k+49}, \frac{50}{k+51}, \frac{51}{k+50}, \frac{51}{k+52}, \dots, \frac{99}{k+98}, \frac{99}{k+100}
$$
are irreducible. | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 102 | |
0d09 | Let $G$ be the centroid of triangle $ABC$ with the side-lengths $a$, $b$, $c$. Prove that if $a + BG = b + AG$ and $b + CG = c + BG$, then triangle $ABC$ is equilateral. | [
"The relations are equivalent to\n$$\na + \\frac{2}{3}m_b = b + \\frac{2}{3}m_a \\quad \\text{and} \\quad b + \\frac{2}{3}m_c = c + \\frac{2}{3}m_b.\n$$\nWe will prove that if $a \\le b$, then $m_a \\ge m_b$. Indeed, we have\n$$\nm_a^2 - m_b^2 = \\frac{2(b^2 + c^2) - a^2}{4} - \\frac{2(a^2 + c^2) - b^2}{4} = \\frac... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"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 > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03fu | Given is a triangle $ABC$ and the points $M$, $P$ lie on the segments $AB$, $BC$, respectively, such that $AM = BC$ and $CP = BM$. If $AP$ and $CM$ meet at $O$ and $2\angle AOM = \angle ABC$, find the measure of $\angle ABC$. | [
"Let $D$ be the reflection of $P$ across $C$, so $AB = BD$ and $O'$ be the circumcenter of $\\triangle ABD$. As $\\triangle AO'B \\cong \\triangle BO'D$ and $MB = CD$, we have $\\angle O'CB = \\angle O'MA$, so $MBCO'$ is cyclic. Now $\\angle O'CM = \\angle O'BM = \\angle AOM$, hence $CO' \\parallel AP$. Therefore $... | Bulgaria | Bulgarian National Olympiad - Final Round | [
"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 and answer | 90° | |
0chg | Find all positive integers $a$ and $b$ such that $a^{4a} = b^b$. | [
"If $b \\ge 4a$, then $b > a$ and $b^b > a^{4a}$. Therefore, in this case the equality is impossible.\n\nIf $b < 4a$, we have $b^b = a^{4a} = a^{4a-b} a^b$, so $b^b$ is divisible by $a^b$. Therefore, $b$ is divisible by $a$. It follows that $b = n a$, with $n = 1$ (I), $n = 2$ (II), or $n = 3$ (III).\n\nWe get $(a^... | Romania | 74th Romanian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (1, 1), (2, 4), (27, 81) | |
0a3y | Let $\triangle ABC$ be an acute triangle such that $|AB| + |BC| = 4|AC|$ and $|AB| < |BC|$. Let $D$ be the intersection of the bisector of $\angle ABC$ with the side $AC$. Points $P$ and $Q$ lie on segment $BD$ such that $|BP| = 2|DQ|$. Let $\ell$ be the line through $P$ parallel to $AC$. The line through $Q$ perpendic... | [
"Let $R, S$ and $T$ be respectively the points where the ant is on $AC$ the first time, is on $\\ell$ and is on $AC$ the second time. Let $\\ell'$ and $S'$ be the reflections of $\\ell$ and $S$ in $AC$, and let $Y'$ be the reflection of $Y$ in $\\ell'$. Then, because of the triangle inequality for the total length ... | Netherlands | IMO Team Selection Test 1 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry ... | null | proof only | null | |
0b6l | Given an integer number $n \ge 2$, a positive real number $A$, and $n+1$ distinct points in the plane, $X_0, X_1, \dots, X_n$, show that the number of triangles $X_0X_iX_j$ of area $A$ does not exceed $4n\sqrt{n}$. | [
"Suppose that for some integer $n \\ge 2$, there exist $n+1$ distinct points in the plane, $X_0, X_1, \\dots, X_n$, such that the number of triangles $X_0 X_i X_j$ of area $A$ be greater than $4n\\sqrt{n}$. Choose the minimal such $n$, notice that $n \\ge 4$, and let $G$ be the graph whose vertices are $X_1, \\dots... | Romania | 2010 Eighth IMAR MATHEMATICAL COMPETITION | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
01np | Let $\triangle ABC$ be an acute triangle. Let $\omega$ be a circle whose center $L$ lies on the side $BC$. Suppose that $\omega$ is tangent to $AB$ at $B_1$ and to $AC$ at $C_1$. Suppose also that the circumcenter $O$ of the triangle $ABC$ lies on the shorter arc $B_1C_1$ of $\omega$.
Prove that the circumcircle of $AB... | [] | Belarus | Belorusija 2012 | [
"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"
] | English | proof only | null | |
0lbo | Let $a$, $b$ be two positive real numbers and $n$ be an integer greater than $1$. Prove that for all positive real number $x$ satisfying $x^n \leq a x + b$, we have
$$
x < n^{-1/2} \sqrt{a} + n^{1/2} \sqrt{b}.
$$ | [] | Vietnam | Vietnamese Mathematical Competitions | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0c1e | Let $p > 5$ be a prime number and $S = \{p - n^2 \mid n \in \mathbb{N}, n^2 < p\}$. Prove that $S$ contains two elements $a$ and $b$ such that $1 < a < b$ and $a$ divides $b$.
BMO, 1996 | [
"We show that the smallest element of $S$ that is greater than $1$ divides a larger element of $S$. If $p$ is of the form $m^2 + 1$, with $m \\in \\mathbb{N}$, we show that $p - (m - 1)^2 = 2m$ divides $p^2 - 1 = m^2$. Indeed, as $m$ is even, it follows that $2m \\mid m^2$.\n\nIf $p$ cannot be written as $m^2 + 1$,... | Romania | 69th NMO Selection Tests for JBMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0ajt | Let $n \ge 3$ and $a_1, a_2, \ldots, a_n$ be positive real numbers for which $\frac{1}{1+a_1^4} + \frac{1}{1+a_2^4} + \ldots + \frac{1}{1+a_n^4} = 1$ holds. Prove the inequality $a_1 a_2 \ldots a_n \ge (n-1)^{n/4}$. | [
"Let $a_i^2 = \\tan^2 x_i$, $x_i \\in [0, \\frac{\\pi}{2}]$, $i=1,2,\\ldots,n$. Then $\\sum_{i=1}^{n} \\cos^2 x_i = 1$.\nFrom the inequality between the arithmetical and the geometrical mean it follows that\n$$\n\\sin^2 x_i = 1 - \\cos^2 x_i \\ge (n-1) \\left( \\prod_{j=1,\\ j \\ne i}^{n} \\cos x_j \\right)^{2/(n-1... | North Macedonia | Macedonian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0b37 | Problem:
How many positive integers $n$ are there such that $\frac{n}{120-2n}$ is a positive integer?
(a) 2
(b) 3
(c) 4
(d) 5 | [] | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | b | |
0bqy | Let $f : [0, 1] \to \mathbb{R}$ be a continuous function with $\int_{0}^{1} f(x)dx = \int_{0}^{1} x f(x)dx$ and $f(0) = 0$. Prove that there exists $c \in (0, 1)$ so that
$$
\int_{0}^{c} x f(x)dx = \frac{c}{2} \int_{0}^{c} f(x)dx.
$$ | [] | Romania | 67th NMO Shortlisted Problems | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | English | proof only | null | |
0bra | Find all three-digit numbers which decrease 13 times when the tens' digit is suppressed. | [
"Let $\\overline{abc}$ be such a number. The condition is $\\overline{abc} = 13 \\cdot \\overline{ac}$, that is $100a + 10b + c = 130a + 13c$, same as $10b = 30a + 12c$. We notice that $5$ divides both $10b$ and $30a$, so $12c$ is divisible by $5$, hence $c \\in \\{0, 5\\}$.\n\nIf $c = 0$, then $b = 3a$. Since $a$ ... | Romania | 67th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 130, 195, 260, 390 | |
0ebl | Problem:
Za realni števili $x$ in $y$ velja
$$
x^{3}+x^{2}+x y+x+y+2=0 \quad \text{in} \quad y^{3}-y^{2}+3 y-x=0
$$
Določi vrednost izraza $x-y$. | [
"Solution:\n\n1. način. Enačbi odštejemo, da dobimo\n$$\nx^{3}-y^{3}+x^{2}+y^{2}+x y+2 x-2 y+2=0\n$$\nLevo stran preoblikujemo\n$$\n\\begin{aligned}\n& x^{3}-y^{3}+x^{2}+y^{2}+x y+2 x-2 y+2= \\\\\n& =(x-y)\\left(x^{2}+x y+y^{2}\\right)+\\left(x^{2}+x y+y^{2}\\right)+2(x-y+1)= \\\\\n& =(x-y+1)\\left(x^{2}+x y+y^{2}\... | Slovenia | 59. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | -1 | |
0f2c | Problem:
(1) The triangle $ABC$ is inscribed in a circle. $D$ is the midpoint of the arc $BC$ (not containing $A$), similarly $E$ and $F$. Show that the hexagon formed by the intersection of $ABC$ and $DEF$ has its main diagonals parallel to the sides of $ABC$ and intersecting in a single point.
(2) $EF$ meets $AB$ a... | [] | Soviet Union | ASU | [
"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 | |
0i0k | Problem:
Given $n+1$ distinct integers from the set $\{1,2, \ldots, 2n\}$ ($n \geq 1$), prove that some two of them are relatively prime. | [
"Solution:\n\nAs is well known, any two consecutive integers are relatively prime. (Proof: let $d$ be the greatest common divisor of $k$ and $k+1$ for some integer $k$; then, since $k+1$ and $k$ are both multiples of $d$, their difference, $1$, is also a multiple of $d$; so $d=1$.)\n\nNow partition the set $\\{1,2,... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0ivf | Problem:
Simplify
$$
2 \cos^{2}(\ln (2009) i) + i \sin (\ln (4036081) i)
$$ | [
"Solution:\n$$\n\\begin{aligned}\n2 \\cos^{2}(\\ln (2009) i) + i \\sin (\\ln (4036081) i)\n&= 1 + \\cos(2 \\ln (2009) i) + i \\sin(\\ln (4036081) i) \\\\\n&= 1 + \\cos(\\ln (4036081) i) + i \\sin(\\ln (4036081) i) \\\\\n&= 1 + e^{i^{2} \\ln (4036081)} \\\\\n&= 1 + \\frac{1}{4036081} \\\\\n&= \\frac{4036082}{4036081... | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | final answer only | 4036082/4036081 | |
0gc9 | 有 $n$ 對夫婦參加一場王國的盛宴, 丈夫們坐一圓桌, 妻子們坐另一圓桌。國王與皇后(不包含在這 $n$ 對夫婦中)要與他們握手。假設國王從某位男士開始握手, 而皇后從該男士的妻子開始握手。考慮兩種方式:
(i) 國王順時鐘依序和所有男士握手。當國王與某位男士握手時, 皇后順時鐘移動到該男士的妻子處, 與她握手。假設國王回到最初的那位男士處時, 皇后繞了圓桌 $a$ 圈。
(ii) 皇后順時鐘依序和所有女士握手。當皇后與某位女士握手時, 國王順時鐘移動到該女士的丈夫處, 與他握手。假設皇后回到最初的那位女士處時, 國王繞了圓桌 $b$ 圈。
試問: $|a - b|$ 的最大可能值。 | [
"答: 若 $n = m^2 - d, 0 \\le d < m$, 則最大值為 $(n - 2m)$; 若 $n = m^2 - m - d, 0 \\le d < m - 1$, 則最大值為 $(n - 2m + 1)$.\n\n- 估計: 不失一般性可以假設 $a > b$, 令男女主人的出發點為兩桌的基準點, 依序順時針編號 $0 \\sim n-1$, 並令 $i$ 號男士的妻子編號為 $x_i$, 考慮兩種方式中男女主人繞過 $0$ 號時的情況會得到:\n$$\na - 1 = |\\{i \\in [n-1] \\mid x_i > x_{i+1}\\}|\n$$\n$$\nb - 1 = |\\{i \\in... | Taiwan | 二〇一八數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | If n = m^2 − d with 0 ≤ d < m, then the maximum of |a − b| is n − 2m. If n = m^2 − m − d with 0 ≤ d < m − 1, then the maximum of |a − b| is n − 2m + 1. | |
0cmd | Problem:
Let $n$ be a positive integer. The kingdom of Zoomtopia is a convex polygon with integer sides, perimeter $6 n$, and $60^{\circ}$ rotational symmetry (that is, there is a point $O$ such that a $60^{\circ}$ rotation about $O$ maps the polygon to itself). In light of the pandemic, the government of Zoomtopia wou... | [
"Solution:\nLet $P$ denote the given polygon, i.e., the kingdom of Zoomtopia. Throughout the solution, we interpret polygons with integer sides and perimeter $6 k$ as $6 k$-gons with unit sides (some of their angles may equal $180^{\\circ}$). The argument hinges on the claim below:\n\nClaim. Let $P$ be a convex pol... | Romanian Master of Mathematics (RMM) | Romanian Master of Mathematics Competition | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0ajn | Find all the triples of integers $(a,b,c)$ such that the number
$$
N = \frac{(a-b)(b-c)(c-a)}{2} + 2
$$
is a power of $2016$.
(A power of $2016$ is an integer of the form $2016^n$, where $n$ is a non-negative integer). | [
"Let $a$, $b$, $c$ be integers and $n$ be a positive integer such that\n$$\n(a-b)(b-c)(c-a) + 4 = 2 \\cdot 2016^n.\n$$\nWe set $a-b = -x$, $b-c = -y$ and we rewrite the equation as\n$$\nxy(x+y) + 4 = 2 \\cdot 2016^n. \\tag{1}\n$$\nIf $n > 0$, then the right hand side is divisible by $7$, so we have that\n$$\nxy(x+y... | North Macedonia | Junior Balkan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All triples consisting of three consecutive integers in cyclic order: (k+2, k+1, k) for any integer k, together with its cyclic permutations. Equivalently, N = 1. |
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