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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
08op | Problem:
Let $x$, $y$, $z$ be positive real numbers that satisfy the equality $x^{2}+y^{2}+z^{2}=3$. Prove that
$$
\frac{x^{2}+y z}{x^{2}+y z+1}+\frac{y^{2}+z x}{y^{2}+z x+1}+\frac{z^{2}+x y}{z^{2}+x y+1} \leq 2
$$ | [
"Solution:\nWe have\n$$\n\\begin{aligned}\n& \\frac{x^{2}+y z}{x^{2}+y z+1}+\\frac{y^{2}+z x}{y^{2}+z x+1}+\\frac{z^{2}+x y}{z^{2}+x y+1} \\leq 2 \\Leftrightarrow \\\\\n& \\frac{x^{2}+y z+1}{x^{2}+y z+1}+\\frac{y^{2}+z x+1}{y^{2}+z x+1}+\\frac{z^{2}+x y+1}{z^{2}+x y+1} \\leq 2+\\frac{1}{x^{2}+y z+1}+\\frac{1}{y^{2}... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
08p1 | Problem:
Does there exist positive integers $a$, $b$ and a prime $p$ such that
$$
a^{3}-b^{3}=4 p^{2} ?
$$ | [
"Solution:\n\nThe given equality may be written as\n$$\n(a-b)\\left(a^{2}+a b+b^{2}\\right)=4 p^{2}\n$$\nSince $a-b<a^{2}+a b+b^{2}$, it follows from (1) that\n(2) $a-b<2 p$.\nNow consider two cases:\n\n1. $p=2$, and\n2. $p$ is an odd prime.\n\nCase 1: $p=2$. Then (1) becomes\n(3) $(a-b)\\left(a^{2}+a b+b^{2}\\righ... | JBMO | Junior Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequaliti... | null | proof only | null | |
06xs | Let $ABC$ be an equilateral triangle. Points $A_1$, $B_1$, $C_1$ lie inside triangle $ABC$ such that triangle $A_1B_1C_1$ is scalene, $BA_1 = A_1C$, $CB_1 = B_1A$, $AC_1 = C_1B$ and
$$
\angle BA_1C + \angle CB_1A + \angle AC_1B = 480^\circ.
$$
Lines $BC_1$ and $CB_1$ intersect at $A_2$; lines $CA_1$ and $AC_1$ intersec... | [
"Let $\\delta_A$, $\\delta_B$, $\\delta_C$ be the circumcircles of $\\triangle AA_1A_2$, $\\triangle BB_1B_2$, $\\triangle CC_1C_2$. The general strategy of the solution is to find two different points having equal power with respect to $\\delta_A$, $\\delta_B$, $\\delta_C$.\n\nClaim. $A_1$ is the circumcentre of $... | IMO | International Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneo... | null | proof only | null | |
0dko | Let $ABC$ be an acute triangle, $\Gamma$ is its circumcircle and $O$ is its circumcenter. Let $F$ be the point on $AC$ such that $\angle COF = \angle ACB$, such that $F$ and $B$ lie in opposite sides with respect to $CO$. The line $FO$ cuts $BC$ at $G$. The line parallel to $BC$ through $A$ intersects $\Gamma$ again at... | [] | Saudi Arabia | Saudi Booklet | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformat... | null | proof only | null | |
0a52 | Problem:
Determine all prime numbers $p$ such that $p^{2} - 6$ and $p^{2} + 6$ are both prime numbers. | [
"Solution:\nIf $p > 5$ then the units digit of $p$ must be 1, 3, 7 or 9. \n- If the units digit of $p$ is 1 or 9 then the units digit of $p^{2}$ is 1. Therefore the units digit of $p^{2} - 6$ is 5. Since $p^{2} - 6 > 5$ this means that $p^{2} - 6$ is not prime. \n- If the units digit of $p$ is 3 or 7 then the uni... | New Zealand | NZMO Round One | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | 5 | |
0kfs | Choose positive integers $b_1, b_2, \dots$ satisfying
$$
1 = \frac{b_1}{1^2} > \frac{b_2}{2^2} > \frac{b_3}{3^2} > \frac{b_4}{4^2} > \dots
$$
and let $r$ denote the largest real number satisfying $\frac{b_n}{n^2} \ge r$ for all positive integers $n$. What are the possible values of $r$ across all possible choices of th... | [
"The answer is $0 \\le r \\le 1/2$. Obviously $r \\ge 0$. In one direction, we show that\n**Claim (Greedy bound) —** For all integers $n$, we have\n$$\n\\frac{b_n}{n^2} \\le \\frac{1}{2} + \\frac{1}{2n}.\n$$\n*Proof.* This is by induction on $n$. For $n=1$ it is given. For the inductive step we have\n$$\n\\begin{al... | United States | USA IMO TST | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | [0, 1/2] | |
0ita | Problem:
The equation $x^{3}-9 x^{2}+8 x+2=0$ has three real roots $p, q, r$. Find $\frac{1}{p^{2}}+\frac{1}{q^{2}}+\frac{1}{r^{2}}$. | [
"Solution:\n\nFrom Vieta's relations, we have $p+q+r=9$, $pq+qr+rp=8$ and $pqr=-2$. So\n\n$$\n\\frac{1}{p^{2}}+\\frac{1}{q^{2}}+\\frac{1}{r^{2}}=\\frac{(pq+qr+rp)^{2}-2(p+q+r)(pqr)}{(pqr)^{2}}=\\frac{8^{2}-2 \\cdot 9 \\cdot(-2)}{(-2)^{2}}=25\n$$"
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 25 | |
066b | If $x$, $y$ are positive real numbers prove that: $(x + \frac{2}{y})(\frac{y}{x} + 2) \ge 8$.
When does equality hold? | [
"Since $x$, $y > 0$ the given inequality can be written as:\n$$\n\\begin{align*}\n(xy+2)(y+2x) \\ge 8xy &\\Leftrightarrow xy^2 + 2y + 2x^2y + 4x - 8xy \\ge 0 \\\\\n&\\Leftrightarrow (xy^2 - 4xy + 4x) + (2x^2y - 4xy + 2y) \\ge 0 \\\\\n&\\Leftrightarrow x(y^2 - 4y + 4) + 2y(x^2 - 2x + 1) \\ge 0 \\\\\n&\\Leftrightarro... | Greece | Selection Examination A | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | Equality holds at x = 1 and y = 2. | |
07vv | Let $ABCDE$ be a pentagon with $|AB| = |BC| = |CD| = |DE| = 1$, $\angle ABC = \angle CDE = 120^\circ$ and $\angle BCD = 100^\circ$. Determine whether $|EA|$ is smaller, equal or larger than 1. | [
"We connect $C$ to $A$ and $E$. Because triangles $ABC$ and $CDE$ are isosceles with an angle of $120^\\circ$, we have $\\angle BAC = \\angle BCA = \\angle DCE = \\angle DEC = 30^\\circ$.\n\n\n\nHence, $\\angle ACE = 40^\\circ$. Let $x = |AE|$ and $d = |CA| = |CE|$. The Cosine rule for $\\t... | Ireland | IRL_ABooklet_2023 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof and answer | larger than 1 | |
0anc | Problem:
Solve the inequality $\frac{1}{3^{x}}\left(\frac{1}{3^{x}}-2\right)<15$.
(a) $\left(-\frac{\log 5}{\log 3},+\infty\right)$
(c) $\left(\frac{\log 3}{\log 5}, 1\right)$
(b) $\left(-\infty, \frac{\log 5}{\log 3}\right)$
(d) $(\log 3, \log 5)$ | [] | Philippines | QUALIFYING STAGE | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | MCQ | (a) | |
0ge4 | 找出所有符合下列三個條件且定義在正整數上的非負整數值函數 $f$:
(1) 存在至少一個 $n$，有 $f(n) \neq 0$;
(2) 對於所有正整數 $x$ 和 $y$，有 $f(xy) = f(x) + f(y)$;
(3) 有無窮多個正整數 $n$ 對於所有 $k < n$ 有 $f(k) = f(n-k)$。
Determine all functions $f$ defined on the set of all positive integers and taking non-negative integer values, satisfying the three conditions:
(1) $f(n) \n... | [
"The sought functions are those of the form $f(n) = c \\cdot \\nu_p(n)$, where $p$ is some prime, $c$ is a positive integer, and $\\nu_p(n)$ denotes the exponent of $p$ in the prime decomposition of $n$.\n\n**Solution 1.** If a number $n$ is a product of primes, $n = p_1p_2\\cdots p_k$, then\n$$\nf(n) = f(p_1) + f(... | Taiwan | 2021 數學奧林匹亞競賽第三階段選訓營 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | f(n) = c · ν_p(n), where p is a prime, c is a positive integer, and ν_p(n) is the exponent of p in n. | |
0a7s | Problem:
Messages are coded using sequences consisting of zeroes and ones only. Only sequences with at most two consecutive ones or zeroes are allowed. (For instance the sequence $011001$ is allowed, but $011101$ is not.) Determine the number of sequences consisting of exactly $12$ numbers. | [
"Solution:\n\nLet $S_{n}$ be the set of acceptable sequences consisting of $2n$ digits. We partition $S_{n}$ in subsets $A_{n}, B_{n}, C_{n}$, and $D_{n}$, on the basis of the two last digits of the sequence. Sequences ending in $00$ are in $A_{n}$, those ending in $01$ are in $B_{n}$, those ending in $10$ are in $... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 9 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 466 | |
00lu | The nonnegative integers $2000$, $17$ and $n$ are written on a blackboard. Alice and Bob play the following game: Alice begins, then they play in turns. A move consists in replacing one of the three numbers by the absolute difference of the other two. No moves are allowed where all three numbers remain unchanged. A pla... | [
"If three numbers are written on the blackboard and one of them is replaced by the (positive) difference of the other two, then after this move one number on the blackboard will be the sum of the other two. Let $a$, $b$ and $a+b$ be the numbers on the blackboard; w.l.o.g. we assume that $b > a$. Because of $a+b-b=a... | Austria | 48th Austrian Mathematical Olympiad Regional Competition (Qualifying Round) | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | Alice | |
0etf | Jacob and Laban take turns playing a game. Each of them starts with a list of square numbers $1, 4, 9, \dots, 2021^2$, and there is a whiteboard in front of them with the number $0$ on it. Jacob chooses a number $x^2$ from his list, removes it from his list, and replaces the number $W$ on the whiteboard with $W + x^2$.... | [
"Since $n^2 \\equiv 0 \\pmod{4}$ if and only if $n$ is even, and $n^2 \\equiv 1 \\pmod{4}$ if and only if $n$ is odd, we can simplify notation by replacing the even squares by $0$ and the odd squares by $1$, in each of the two lists. Thus Jacob and Laban each has a pool of $1010$ zeros and $1011$ ones to choose fro... | South Africa | The South African Mathematical Olympiad, Third Round | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English, Afrikaans | proof and answer | 506 | |
03p4 | Assume that $S = (a_1, a_2, \dots, a_n)$ consists of $0$ and $1$ and is the longest sequence of numbers which satisfies the following condition: Every two sections of successive $5$ terms in the sequence of numbers $S$ are different, i.e., for arbitrary $1 \leq i < j \leq n-4$, $(a_i, a_{i+1}, a_{i+2}, a_{i+3}, a_{i+4}... | [
"**Proof** Noting that $S$ is the longest sequence of numbers satisfying the condition. Hence, if we add a term, $0$ or $1$, after the last term of $S$, there will occur two identical sections of successive $5$ terms in $S$, and that is, there exist $i \\neq j$ such that\n$$\n(a_i, a_{i+1}, \\dots, a_{i+4}) = (a_{n... | China | China Western Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
028x | Problem:
O triângulo de Reuleaux é um disco formado a partir de um triângulo equilátero, agregando arcos de circunferência com centros nos vértices do triângulo e raios iguais ao lado do triângulo.
Qual é a área de um triângulo de Reuleaux, se o triângulo equilátero tem lado de medida $1~\m... | [
"Solution:\n\nO triângulo de Reuleaux é formado por 4 regiões: um triângulo equilátero e três calotas. Cada calota é um sexto de um círculo de raio $1$ do qual foi retirado um triângulo equilátero de lado $1$.\n\nPelo Teorema de Pitágoras, a altura do triângulo equilátero é:\n$$\nh = \\sqrt{1 - \\left(\\frac{1}{2}\... | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | pi/2 - sqrt(3)/2 cm^2 | |
026e | Problem:
Potências de 3 - Se $3^{n}=2$ então quanto vale $27^{2 n}$ ? | [
"Solution:\n\nPotências de 3 - Temos: $27^{2 n} = \\left(3^{3}\\right)^{2 n} = 3^{6 n} = \\left(3^{n}\\right)^{6} = 2^{6} = 64$."
] | Brazil | null | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | final answer only | 64 | |
01ui | Points $K$ and $M$ are the midpoints of the sides $AB$ and $AC$ of triangle $ABC$, respectively. The equilateral triangles $AMN$ and $BKL$ are constructed on the sides $AM$ and $BK$ to the exterior of the triangle $ABC$. Point $F$ is the midpoint of the segment $LN$.
Find the value of the angle $KFM$. | [
"Answer: $90^\\circ$.\n\nLet points $A$, $B$, and $C$ lie in the same half-plane with respect to the line $LN$. Let $E$ and $D$ be the midpoints of the segments $AL$ and $AN$, respectively (see the Fig.). By condition, the triangle $BKL$ is equilateral and $K$ is the midpoint of the side $AB$, so $BK = KA = KL$. Th... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 90° | |
0f0t | Problem:
A collection of $n$ cards is numbered from $1$ to $n$. Each card has either $1$ or $-1$ on the back. You are allowed to ask for the product of the numbers on the back of any three cards. What is the smallest number of questions which will allow you to determine the numbers on the backs of all the cards if $n$... | [] | Soviet Union | ASU | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | (1) 30, (2) 31, (3) 32; circle part: 50 | |
06md | There are four positive integers. By computing the H.C.F. of two of them at a time, one gets six different values $1$, $2$, $3$, $4$, $5$, $k$. Find the smallest possible value of $k$. | [
"Answer: $14$\n\nFirst note that there are exactly three even numbers and one odd number. (If all four numbers are even then all H.C.F.'s would be even, whereas if there are at most two even numbers then at most one H.C.F. can be even.) Suppose $a$ is the odd number and $b, c, d$ are the even numbers. We use $(m, n... | Hong Kong | HongKong 2022-23 IMO Selection Tests | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 14 | |
05dc | Problem:
The numbers $p$ and $q$ are prime and satisfy
$$
\frac{p}{p+1}+\frac{q+1}{q}=\frac{2 n}{n+2}
$$
for some positive integer $n$. Find all possible values of $q-p$. | [
"Solution:\n\nRearranging the equation, $2 q n(p+1) = (n+2)(2 p q + p + q + 1)$. The left hand side is even, so either $n+2$ or $p+q+1$ is even, so either $p=2$ or $q=2$ since $p$ and $q$ are prime, or $n$ is even.\n\nIf $p=2$, $6 q n = (n+2)(5 q + 3)$, so $(q-3)(n-10) = 36$. Considering the divisors of $36$ for wh... | European Girls' Mathematical Olympiad (EGMO) | European Girls' Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | {2, 3, 5} | |
02rb | The *Somanacci* sequence is generated by three initial terms, each a positive integer smaller than $2012$. Each term of the sequence, beginning from the fourth term, is the sum of all preceding numbers. How many distinct Somanacci sequences have $2012$ as one of their terms? | [
"Let $S_n$ be the $n$-th Somanacci number. Then, for $n \\ge 5$, $S_n = S_{n-1} + S_{n-2} + \\dots + S_1 = S_{n-1} + S_{n-1} = 2S_{n-1}$, because $S_{n-1} = S_{n-2} + \\dots + S_1$. So, since $S_4 = S_3 + S_2 + S_1$, $S_n = (S_1 + S_2 + S_3) \\cdot 2^{n-4}$. Since $2012 = 2 \\cdot 1006 = 2^2 \\cdot 503$, the answer... | Brazil | Brazilian Math Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 630261 | |
02bn | Problem:
Fonte com 3 torneiras - Sílvia vai a uma fonte que tem três torneiras, encher os seus dez garrafões. Um dos garrafões demora um minuto para encher, outro dois minutos, outro três minutos e assim por diante. Como Sílvia deverá distribuir os garrafões pelas torneiras de modo a gastar o menor tempo possível? Qua... | [
"Solution:\n\nSolução 1: Para simplificar, numeramos os garrafões de acordo com os respectivos tempos que gastam para ficar cheios. A ideia é utilizar o \"tempo que sobra\" de um garrafão para encher outro garrafão, enchendo simultaneamente outros. As figuras ilustram a solução.\n\n\n\nNa f... | Brazil | null | [
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | Minimum time: 19 minutes. One optimal assignment is: Tap 1 fills the ten-minute and nine-minute containers; Tap 2 fills the eight-minute, seven-minute, and three-minute containers; Tap 3 fills the five-minute, four-minute, two-minute, and one-minute containers. | |
01o5 | Non-zero real numbers $a$, $b$, $c$, $d$ satisfy the equalities
$$
a + b + c + d = 0, \quad \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{1}{abcd} = 0.
$$
Find all possible values of the product $(ab - cd)(c + d)$. | [
"Answer: $-1$.\nBy condition,\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} = -\\frac{1}{abcd} \\Rightarrow bcd + cda + dab + abc = -1.\n$$\nSo\n$$\n\\begin{aligned}\n-1 = bcd + cda + dab + abc &= (bcd + cda) + (dab + abc) = cd(b + a) + ab(c + d) = \\\\\n&= [a + b + c + d = 0 \\Rightarrow a + b = -... | Belarus | Belorusija 2012 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | -1 | |
05aj | In Kükametsa School of Science, the number of secondary school participants in the "Kangaroo" competition was exactly one fourth greater than in the previous year. This year, the number of participants from the 11th and 12th grade was respectively 3 and 4 less than the number of participants from the 10th and 11th grad... | [
"Let $e$ be the number of secondary school participants in the \"Kangaroo\" competition in the previous year. This year, the number of participants was greater by $\\frac{1}{4}e$. Hence $e$ is divisible by 4. Since the total number of participants in the 11th and 12th grade this year was $3+4=7$ less than the numbe... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 106, 108, 110 | |
0a8i | Problem:
Let $ABCDEF$ be a convex hexagon, in which each of the diagonals $AD$, $BE$, and $CF$ divides the hexagon into two quadrilaterals of equal area. Show that $AD$, $BE$, and $CF$ are concurrent. | [
"Solution:\n\n(See Figure 12.) Denote the area of a figure by $|\\cdot|$. Let $AD$ and $BE$ intersect at $P$, $AD$ and $CF$ at $Q$, and $BE$ and $CF$ at $R$. Assume that $P$, $Q$, and $R$ are different. We may assume that $P$ lies between $B$ and $R$, and $Q$ lies between $C$ and $R$. Both $|ABP|$ and $|DEP|$ diffe... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 15 | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0jb2 | There are $2010$ students and $100$ classrooms in the Olympiad High School. At the beginning, each of the students is in one of the classrooms. Each minute, as long as not everyone is in the same classroom, somebody walks from one classroom into a different classroom with at least as many students in it (prior to his m... | [
"**Solution:** The answer is $63766$.\n\nWe label the classrooms $c_1, c_2, \\dots, c_{100}$. If $a_i$ students are in classroom $c_i$ for $1 \\le i \\le 100$ at a certain step in the process, denote by $(a_1, a_2, \\dots, a_{100})$ the configuration of the students.\n\n*Step 1:* We first show that $M$ is finite. C... | United States | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Algorithms"
] | null | proof and answer | 63766 | |
0izj | Problem:
Call an $2n$-digit number special if we can split its digits into two sets of size $n$ such that the sum of the numbers in the two sets is the same. Let $p_{n}$ be the probability that a randomly-chosen $2n$-digit number is special (we will allow leading zeros in the number).
a. The sequence $p_{n}$ converge... | [
"Solution:\n\nAnswer: $\\frac{1}{2}$\n\nWe first claim that if a $2n$-digit number $x$ has at least eight $0$'s and at least eight $1$'s and the sum of its digits is even, then $x$ is special.\n\nLet $A$ be a set of eight $0$'s and eight $1$'s and let $B$ be the set of all the other digits. We split $B$ arbitrarily... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 1/2 | |
046w | Find the largest real number $\lambda$ with the following property: for any 100th-order doubly stochastic matrix, it is always possible to select 150 elements from it and change the remaining 9850 elements to zero, such that the resulting matrix has row sums and column sums not less than $\lambda$.
Note: An "$n$th-ord... | [
"*Proof.* The answer is $\\frac{17}{1900}$.\n\nFirst, construct the following doubly stochastic matrix $M$: Divide $M$ into four submatrices, where the first 75 rows and the first 24 columns form submatrix $A$, the last 25 rows and the first 24 columns form submatrix $B$, the first 75 rows and the last 76 columns f... | China | China-TST-2023B | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Algebra > Linear Algebra > Matrices"
] | English | proof and answer | 17/1900 | |
0etv | Find all functions $f$ from integers to integers such that
$$
f(m+n) + f(m-n) - 2f(m) = 6mn^2
$$
for all integers $m$ and $n$. | [
"First note that if we replace $f(x)$ by $f(x) + ax + c$ for some constants $a$ and $c$, then we get another solution. Thus we may assume for now that $f(0) = 0$ and $f(1) = 1$.\nLet $P(m, n)$ be the statement that was given. Then $P(0, n)$ gives\n$$\nf(n) + f(-n) - 2f(0) = 0\n$$\nand since $f(0) = 0$, we get $f(-n... | South Africa | The South African Mathematical Olympiad Third Round | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All solutions are f(n) = n^3 + a n + c for integers a and c. | |
0d85 | How many ways are there to color the vertices of a square with $n$ colors $1, 2, \ldots, n$. (The colorings must be different so that we can't get one from the other by a rotation.) | [
"Consider the cases:\n1. Use 1 color: $\\binom{n}{1} = n$,\n\n2. Use 2 colors: $4 \\times \\binom{n}{2} = 2n(n-1)$,\n\n3. Use 3 colors: $9 \\times \\binom{n}{3} = 3 \\frac{n(n-1)(n-2)}{2}$,\n\n4. Use 4 colors: $6 \\times \\binom{n}{4} = \\frac{n(n-1)(n-2)(n-3)}{4}$\n\nThe sum up will give:\n$$\n\\frac{n^{4} + n^{2}... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | (n^4 + n^2 + 2n)/4 | |
04pj | Point $T$ is located inside the triangle $ABC$ so that $|AT| = 56$, $|BT| = 40$ and $|CT| = 35$. The feet of the altitudes from $T$ to the sides of $ABC$ are vertices of an equilateral triangle. Find $\triangle ABC$.
(Belarus 2018) | [] | Croatia | Croatian Mathematical Society Competitions | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Transformations > Inversion"
] | English | proof and answer | BC:CA:AB = 5:7:8 | |
0bhk | Denote $M$ the midpoint of the edge $[AD]$ of the cube $ABCD'A'B'C'D'$. Denote $\alpha$ the plane perpendicular in $B$ on the straight line $BM$. Find the tangent of the angle between the planes $\alpha$ and $(ACB')$.
Gabriel Popa | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | sqrt(6)/3 | |
0ctt | In a bazaar, there is a carpet-changer. If he gets from a client an $a \times b$ carpet, he can give instead either a $\frac{1}{a} \times \frac{1}{b}$ carpet, or two carpets of sizes $c \times b$ and $\frac{a}{c} \times b$ (at each such change, a number $c$ is chosen by the client). A traveler tells that initially he h... | [
"No.\n\nWe say that a carpet is *large* (resp., *small*) if both its side lengths are larger (resp., smaller) than $1$. Show that the total number of large and small carpets does not decrease.\n\nНазовём ковёр, все стороны которого больше $1$, большим, а ковёр, все стороны которого меньше $1$, — маленьким. Таким об... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English; Russian | proof and answer | No | |
0h98 | Given a foundation that is in a form of a rectangle $2m \times 2n$ that is divided into smaller $1 \times 1$ squares. There is a gap of length $1$ between any two adjacent squares. The foundation is covered by several layers of bricks of size $2 \times 1$. Every layer consists of $2mn$ bricks and each brick fully cover... | [
"Without loss of generality let $2 \\le m \\le n$. Consider a square $A$ of size $1 \\times 1$, that does not touch the sides of a big rectangle. All 4 sides of it have to be covered with different bricks, since half of each brick covers the square $A$ itself. Therefore, covering with less than 4 layers is not poss... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Minimum number of layers = 2 if m = n = 1; 3 if exactly one of m, n equals 1 and the other is greater than 1; 4 otherwise. | |
0cc3 | On the small arc $AB$ of the circumcircle of the equilateral triangle $ABC$ we consider a point $N$ such that the length of the arc $NB$ is $30^\circ$. Consider the perpendicular lines from $N$ to $AC$ and $AB$, respectively. These lines intersect again the circumcircle of the triangle $ABC$ in points $M$ and $I$, resp... | [
"a) Let $O$ be the circumcenter of the triangle $ABC$. Without loss of generality, we consider $O(0)$, while the vertices of the triangle are $A(1)$, $B(\\varepsilon)$, and $C(\\varepsilon^2)$, where $\\varepsilon = -\\frac{1}{2} + i\\frac{\\sqrt{3}}{2}$. Because the length of the arc $NB$ is $30^\\circ$, we have $... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0c8q | Denote $\mathbb{Z}_{>0} = \{1, 2, 3, \dots\}$ the set of all positive integers. Determine all functions $f : \mathbb{Z}_{>0} \to \mathbb{Z}_{>0}$ such that, for each positive integer $n$:
i) $\sum_{k=1}^{n} f(k)$ is a perfect square;
ii) $f(n)$ divides $n^3$. | [
"Induct on $n$ to show that $f(n) = n^3$ for all positive integers $n$. It is readily checked that this $f$ satisfies the conditions in the statement. The base case, $n=1$, is clear.\nLet $n \\ge 2$ and assume that $f(m) = m^3$ for all positive integers $m < n$. Then $\\sum_{k=1}^{n-1} f(k) = \\frac{n^2(n-1)^2}{4}$... | Romania | Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | f(n) = n^3 for all positive integers n | |
0dam | In square $ABCD$ with side $1$, point $E$ lies on $BC$ and $F$ lies on $CD$ such that $\angle EAB = 20^{\circ}$, $\angle EAF = 45^{\circ}$. Find the length of altitude $AH$ of $\triangle AEF$. | [
"Take point $G$ on the opposite ray of ray $DC$ such that $GD = BE$. Then two triangles $ABE$, $ADG$ are congruent, implies that $AE = AG$ and $\\angle EAB = \\angle DAG = 20^{\\circ}$. Thus\n$$\n\\angle FAG = \\angle FAD + \\angle DAG = \\angle EAB + \\angle DAG = 45^{\\circ}.\n$$\nThus two triangles $AEF$, $AGF$ ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 1 | |
0eng | The set $T$ consists of 66 points in the plane, and set $P$ consist of 16 lines in the plane. We say that a point $A \in T$ and a line $\ell \in P$ form an *incident pair* if $A \in \ell$. Show that the number of incident pairs cannot exceed 159, and that there is such a configuration with exactly 159 incident pairs. | [
"Denote by $A_1, A_2, \\dots, A_{66}$ the points from $T$ and by $a_i$ the number of lines from $P$ containing $A_i$. Then the number of pairs of lines intersecting at $A_i$ equals $\\binom{a_i}{2}$, and the number of incident pairs $I = \\sum_{i=1}^{66} a_i$. Since any two lines meet in at most one point, we have ... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Pappus theorem",
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 159 | |
0jcj | Find all pairs $(a, n)$ of positive integers with $n > 1$ such that for every prime $p$ dividing $a^n - 1$, there exists $0 < k < n$ such that $p$ divides $a^k - 1$. | [
"Define a *good pair* to be a pair $(a, n)$ of positive integers, with $n > 1$, such that any prime dividing $a^n - 1$ also divides $a^k - 1$ for some $0 < k < n$. We claim that the only good pairs are $(2, 6)$, $(1, n)$ for any $n$, and $(2^m - 1, 2)$ for any $m \\ge 2$.\n\nIt is straightforward to verify that all... | United States | TST | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (1, n) for all n > 1; (2^m − 1, 2) for all m ≥ 2; and (2, 6). | |
08kg | Problem:
Let $ABC$ be an isosceles triangle such that $AB = AC$ and $\angle \frac{A}{2} < \angle B$. On the extension of the altitude $AM$ we get the points $D$ and $Z$ such that $\angle CBD = \angle A$ and $\angle ZBA = 90^{\circ}$. $E$ is the foot of the perpendicular from $M$ to the altitude $BF$ and $K$ is the foot... | [
"Solution:\nThe points $A, B, K, Z$ and $C$ are co-cyclic.\nBecause $ME \\parallel AC$ so we have\n$$\n\\angle KEM = \\angle EAC = \\angle MBK\n$$\nTherefore the points $B, K, M$ and $E$ are co-cyclic. Now, we have\n$$\n\\begin{aligned}\n& \\angle ABF = \\angle ABC - \\angle FBC \\\\\n& = \\angle AKC - \\angle EKM ... | JBMO | OJBM | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03zj | Given a tetrahedron $ABCD$, it is known that $\angle ADB = \angle BDC = \angle CDA = 60^\circ$, $AD = BD = 3$ and $CD = 2$. Then the radius of the sphere circumscribing $ABCD$ is ______. | [
"Let the center of the sphere circumscribing $ABCD$ be $O$. Then $O$ is on the vertical line of plane $ABD$ through point $N$ the circumcenter of $\\triangle ABD$. It is known that $\\triangle ABD$ is regular, so $N$ is the center of it. Let $P$ and $M$ be the midpoints of $AB$ and $CD$, respectively. Then $N$ is o... | China | China Mathematical Competition | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof and answer | sqrt(3) | |
0h68 | Trapezoid $ABCD$ with $BC \parallel AD$ is given. On diagonals $AC$ and $BD$ denote points $P$ and $Q$ respectively, so that $AC$ bisects $\angle BPD$, $BD$ bisects $\angle AQC$. Prove that $\angle BPD = \angle AQC$. | [
"Let $M_1$ and $M_2$ be the middles of the diagonals (Fig. 26). Then $M_1M_2 \\parallel BC$. Consider circumcircles of $\\angle AQC$ and $\\angle BPD$. Let them intersect $BD$ and $AC$ in points $X$ and $Y$. Since $BD$ contains the bisector of $\\angle AQC$, then point $X$ is the middle of the bigger arc $AC$ of th... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0c0y | Prove that $2^{-x} + 2^{-1/x} \le 1$, for any real number $x > 0$. | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof only | null | |
03dp | Let $p = (a_1; a_2; \dots; a_{12})$ be a permutation of the numbers $1, 2, \dots, 12$. Denote by $S_p = |a_1 - a_2| + |a_2 - a_3| + \dots + |a_{11} - a_{12}|$. We will call $p$ "optimistic", if $a_i > \min(a_{i-1}; a_{i+1})$ for every $i = 2, \dots, 11$.
a) Determine the maximal possible value of $S_p$? For how many p... | [
"a) Place the elements of $p$ along the real axis. Then $S_p$ equals the length of the \"walk\", starting from $a_1$, then reaching $a_2$, then reaching $a_3$ and so on until we end at $a_{12}$. In such a walk, the segments:\n\n$1$-$2$ and $11$-$12$ appear at most twice (before and after $1$; before and after $12$)... | Bulgaria | Bulgaria 2022 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | a) Maximum S_p is 71; achieved by 28800 permutations. b) Number of optimistic permutations is 2048. c) For optimistic permutations, maximum S_p is 21; achieved by 1024 permutations. | |
0edu | Problem:
Koliko je $1+3,5+6+8,5+\ldots+2501$, če je razlika med zaporednima seštevancema stalna?
(A) 1249750
(B) 1251000
(C) 1251750,5
(D) 1252251
(E) 1253502,5 | [
"Solution:\nUgotovimo, da moramo sešteti $n$ členov aritmetičnega zaporedja z diferenco $d=2,5$. Upoštevamo $a_n=a_1+(n-1)d$ oziroma $2501=1+(n-1) \\cdot 2,5$. Rešimo enačbo in ugotovimo, da moramo sešteti $1001$ členov. Upoštevamo $S_n=\\frac{n}{2}\\left(a_1+a_n\\right)$ ali $S_n=\\frac{n}{2}\\left(2a_1+(n-1)d\\ri... | Slovenia | 16. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | D | |
0imq | Let $S$ be a set containing $n^2 + n - 1$ elements, for some positive integer $n$. Suppose that the $n$-element subsets of $S$ are partitioned into two classes. Prove that there are at least $n$ pairwise disjoint sets in the same class. | [
"In order to apply induction, we generalize the result to be proved so that it reads as follows:\n\n**Proposition.** If the $n$-element subsets of a set $S$ with $(n+1)m-1$ elements are partitioned into two classes, then there are at least $m$ pairwise disjoint sets in the same class.\n\n*Proof.* Fix $n$ and procee... | United States | USAMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0gx5 | Peter has several equal squares with dimensions: $4 \times 4$. Each square is divided into sectors $(1 \times 1)$. He paints each of these sectors red or blue so that there are no similar patterns in all columns and all rows of all the squares. Rotating the squares is forbidden. How many squares can Peter paint in that... | [
"Altogether there are $2^4 = 16$ different patterns of painted columns. Since each square has 4 columns, in all there can not be more than $\\frac{2^4}{4} = 4$ differently painted squares.\n\n| 0001 | 1110 | 0110 | 1001 |\n|------|------|------|------|\n| 0010 | 1101 | 0011 | 1100 |\n| 0100 | 1011 | 0101 | 1010 |\n... | Ukraine | Ukrajina 2008 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 4 | |
0kne | Problem:
$ABCDE$ is a cyclic convex pentagon, and $AC = BD = CE$. $AC$ and $BD$ intersect at $X$, and $BD$ and $CE$ intersect at $Y$. If $AX = 6$, $XY = 4$, and $YE = 7$, then the area of pentagon $ABCDE$ can be written as $\frac{a \sqrt{b}}{c}$, where $a, b, c$ are integers, $c$ is positive, $b$ is square-free, and $... | [
"Solution:\n\n\n\nSince $AC = BD$, $ABCD$ is an isosceles trapezoid. Similarly, $BCDE$ is also an isosceles trapezoid. Using this, we can now calculate that $CY = DY = DX - XY = AX - XY = 2$, and similarly $BX = CX = 3$. By applying Heron's formula we find that the area of triangle $CXY$ is... | United States | HMMT November 2021 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 2852 | |
09bu | Дурын $[a, b] \subset [0, 1]$ хэрчимийн хувьд $2^n x \in [a, b]$ байдаг $n$ гэсэн натурал тоо олддог бол $x \in [0, 1]$ цэгийг сайн цэг гэе. Тэгвэл төгсгөлгүй олон сайн цэг олдоно гэж батал. | [
"Өгүүлбэрт \\{$2^n x$\\} $\\in [a, b]$ буюу бутархай хэсгийн тэмдгийг орхигдуулсан байсанд хүлцэл өчье. Энэ үедээ сайн цэг байхгүй гэдэг нь илэрхий. Зөв томъёололтой үед бодьё. Хэрэв бид $x$ гэсэн нэг сайн цэг олбол $\\frac{x}{2^k}$, $\\forall k \\in \\mathbb{N}$ хувьд мөн сайн цэг гэдэг нь илэрхий. Иймд 1 сайн цэг... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Number Theory > Other"
] | Mongolian | proof only | null | |
0gy0 | a) Four positive integer numbers $a$, $b$, $c$, $d$ satisfy the condition: every number $ab$, $bc$, $cd$, $da$ is a perfect cube. Are all the numbers $a$, $b$, $c$, $d$ perfect cubes?
b) Five positive integer numbers $a$, $b$, $c$, $d$, $e$ satisfy the condition: every number $ab$, $bc$, $cd$, $de$, $ea$ is a perfect ... | [
"a) The example that it isn't necessary: $a = c = 2$, $b = d = 4$.\n\nb) It shows that if a positive integer number $n$ has its square $n^2$ as a perfect cube of an integer number, then $n$ itself is a perfect cube of some integer number. Really, consider a factorization of $n$ into prime numbers $n = p_1^{m_1} \\l... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | a) No; for example a=c=2 and b=d=4. b) Yes; all five numbers must be perfect cubes. | |
00na | We consider the sequences $(a_n)_{n \ge 0}$ and $(b_n)_{n \ge 0}$ which are defined by $a_0 = b_0 = 2$ and $a_1 = b_1 = 14$ and by
$$
a_n = 14a_{n-1} + a_{n-2},
$$
$$
b_n = 6b_{n-1} - b_{n-2}
$$
for $n \ge 2$.
Decide whether there are infinitely many integers which occur in both sequences. | [
"*Answer.* Yes.\n\nSequence $(a_n)$ starts with values $2$, $14$, $198$, $2786$, $39202$, $551614$. Sequence $(b_n)$ starts with values $2$, $14$, $82$, $478$, $2786$, $16238$, $94642$, $551614$. We therefore conjecture that $a_{2k+1} = b_{3k+1}$ holds for $k \\ge 0$.\nShifting the recurrence yields\n$$\na_{n+2} - ... | Austria | Austria2019 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | Yes | |
0bk6 | Let $ABCD$ be a quadrilateral with $\angle A + \angle C = 60^\circ$ and $AB \cdot CD = BC \cdot AD$. Prove that $AB \cdot CD = AC \cdot BD$.
 | [
"Let us construct the equilateral triangle $BCE$ in the half-plane determined by line $BC$ and point $D$.\nThen $\\frac{AB}{AD} = \\frac{BC}{CD} = \\frac{CE}{CD}$ and $\\angle DAB = \\angle DCE$, so $\\Delta DAB \\sim \\Delta DCE$.\nTherefore $\\frac{AD}{DC} = \\frac{DB}{DE}$ and $\\angle ADB = \\angle CDE$, hence ... | Romania | 65th NMO Selection Tests for JBMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
09hd | Let $I$ be the incenter of triangle $ABC$. Let $D$ be a point on side $BC$, and $E$ be a point on ray $BC$ such that $C$ lies between $E$ and $D$ and $\frac{BD}{DC} = \frac{BE}{EC}$. Let $H$ be the foot of perpendicular from $D$ to line $IE$. Prove that $\angle AHE = \angle IDE$.
(Proposed by B. Battsengel, G. Batzaya) | [
"Let $\\omega$ be the incircle of the triangle $ABC$, $A_1$, $B_1$, $C_1$ be the touching points of $\\omega$ with sides $BC$, $AC$ and $AB$ respectively. Let $\\omega'$ be the circle with diameter $EI$, and $K$ be the intersection point of $\\omega$ and $\\omega'$, different from $A_1$. Denote by $M$ be the inters... | Mongolia | Mongolian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry >... | English | proof only | null | |
0i20 | Problem:
The polynomial $3 x^{5}-250 x^{3}+735 x$ is interesting because it has the maximum possible number of relative extrema and points of inflection at integer lattice points for a quintic polynomial. What is the sum of the $x$-coordinates of these points? | [
"Solution:\n\nThe first derivative is $15 x^{4}-750 x^{2}+735$, whose roots (which give the relative extrema) sum to $750 / 15 = 50$. The second derivative is $60 x^{3}-1500 x$, whose roots (which give the points of inflection) sum to $1500 / 60 = 25$, for a grand total of $75$."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | final answer only | 0 | |
0jhp | Problem:
Express $-2013$ in base $-4$. | [
"Solution:\n\n$-2013 \\equiv 3 \\pmod{4}$, so the last digit is $3$; now $\\frac{-2013-3}{-4}=504 \\equiv 0$, so the next digit (to the left) is $0$; then $\\frac{504-0}{-4}=-126 \\equiv 2$; $\\frac{-126-2}{-4}=32 \\equiv 0$; $\\frac{32-0}{-4}=-8 \\equiv 0$; $\\frac{-8-0}{-4}=2$.\n\nThus $-2013_{10}=200203_{-4}$."
... | United States | HMMT November 2013 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Modular Arithmetic"
] | null | final answer only | 200203_{-4} | |
0c8m | Let $N$ be a positive integer, and let $\mathbf{a} = (a(1), \dots, a(N))$ and $\mathbf{b} = (b(1), \dots, b(N))$ be sequences of non-negative integers, each written on a circle (so we assume $a(i \pm N) = a(i)$ and $b(i \pm N) = b(i)$). We say $\mathbf{a}$ is **b-harmonic**, if each $a(i)$ is the arithmetic mean of the... | [
"Let $a = \\min_i a(i)$ and let $b = \\min_i b(i)$. Since **a** is not constant, there exists an $i$ such that $a = a(i) < a(i + 1)$.\n\n**Claim 1.** If $a = a(i) < a(i + 1)$, then $b(i) = 0$. Similarly, if $a = a(i) < a(i - 1)$, then $b(i) = 0$.\n\n*Proof.* Otherwise the sum in $(\\ast)$ contains a term $a(i+1) > ... | Romania | Romanian Master of Mathematics | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0k14 | Problem:
Let $ABC$ be a triangle such that $AB = 6$, $BC = 5$, $AC = 7$. Let the tangents to the circumcircle of $ABC$ at $B$ and $C$ meet at $X$. Let $Z$ be a point on the circumcircle of $ABC$. Let $Y$ be the foot of the perpendicular from $X$ to $CZ$. Let $K$ be the intersection of the circumcircle of $BCY$ with li... | [
"Solution:\n\nLet $\\omega_1$ denote the circumcircle of $ABC$ and $\\omega_2$ denote the circle centered at $X$ through $B$ and $C$. Let $\\omega_2$ intersect $AB$, $AC$ again at $B'$, $C'$. The (signed) power of $Y$ with respect to $\\omega_1$ is $-CY \\cdot YZ$. The power of $Y$ with respect to $\\omega_2$ is $X... | United States | HMMT February | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 147/10 | |
09pz | Problem:
Vind alle functies $f: \mathbb{Z} \rightarrow \mathbb{Z}$ die voldoen aan
$$
f(m+n)+f(m n-1)=f(m) f(n)+2
$$
voor alle $m, n \in \mathbb{Z}$. | [
"Solution:\nStel eerst dat er een $c \\in \\mathbb{Z}$ is met $f(n)=c$ voor alle $n$. Dan hebben we $2c = c^{2} + 2$, dus $c^{2} - 2c + 2 = 0$ en dat heeft geen oplossing voor $c$. Dus is $f$ niet constant.\n\nVul nu in $m=0$. Dat geeft $f(n) + f(-1) = f(n) f(0) + 2$, waaruit we concluderen dat $f(n)(1 - f(0))$ een... | Netherlands | Dutch TST | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | f(n) = n^2 + 1 for all integers n | |
0j40 | Problem:
A set of identical square tiles with side length $1$ is placed on a (very large) floor. Every tile after the first shares an entire edge with at least one tile that has already been placed.
- What is the largest possible perimeter for a figure made of $10$ tiles?
- What is the smallest possible perimeter for ... | [
"Solution:\nThe initial single tile has perimeter $4$. Each time a tile is added, at most three edges are added and at least one edge of the previous boundary is eliminated, so the perimeter increases by at most $2$. Thus for $n$ tiles the maximum possible perimeter is $2(n+1)$ and this can be achieved (among other... | United States | 13th Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | For 10 tiles: maximum perimeter 22, minimum perimeter 14. For 2011 tiles: maximum perimeter 4024, minimum perimeter 180. | |
0kp4 | Problem:
Let $N$ be the number of triples of positive integers $(a, b, c)$ satisfying
$$
a \leq b \leq c, \quad \operatorname{gcd}(a, b, c)=1, \quad a b c=6^{2020}
$$
Compute the remainder when $N$ is divided by 1000. | [
"Solution:\nLet $n=2020$. If we let $a=2^{p_{1}} \\cdot 3^{q_{1}},\\ b=2^{p_{2}} \\cdot 3^{q_{2}},\\ c=2^{p_{3}} \\cdot 3^{q_{3}}$, then the number of ordered triples $(a, b, c)$ that satisfy the second and third conditions is the number of nonnegative solutions to $p_{1}+p_{2}+p_{3}=n$ and $q_{1}+q_{2}+q_{3}=n$, w... | United States | HMMT February | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combin... | null | proof and answer | 602 | |
0azj | Problem:
Let $N$ be the smallest positive integer divisible by $20$, $18$, and $2018$. How many positive integers are both less than and relatively prime to $N$? | [
"Solution:\nFirst, find $N = \\operatorname{lcm}(20, 18, 2018)$.\n\nPrime factorizations:\n$20 = 2^2 \\times 5$\n$18 = 2 \\times 3^2$\n$2018 = 2 \\times 1009$\n\nSo,\n$N = \\operatorname{lcm}(2^2 \\times 5, 2 \\times 3^2, 2 \\times 1009)$\n\nTake the highest power of each prime:\n- $2^2$ (from $20$)\n- $3^2$ (from ... | Philippines | 21st PMO Area Stage | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | final answer only | 48408 | |
05aa | An acute-angled triangle $ABC$ with $AC > AB$ is given. The perpendicular bisector of side $BC$ intersects the lines $AC$ and $AB$ at points $D$ and $E$, respectively. The circle with diameter $DE$ intersects the lines $AC$ and $AB$ at points $K$ and $L$, respectively ($K \neq D, L \neq E$). Let $M$ be the midpoint of ... | [
"Since the points $E, K, L$, and $D$ are concyclic (Fig. 28), it follows that $\\angle CKL = \\angle DKL = \\angle DEL = \\angle MEB$. From the problem conditions, $\\angle CME = 90^\\circ$, and by Thales' theorem, $\\angle CKE = \\angle DKE = 90^\\circ$. Therefore, the points $C, M, K, E$ are also concyclic. Conse... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0dus | Problem:
Kje dosežeta funkciji $f(x)=\sin 4 x$ in $g(x)=-\cos 2 x$ enako vrednost? | [
"Solution:\n\nZapis $f(x)=g(x)$ ali $\\sin 4 x=-\\cos 2 x$\n\n$2 \\sin 2 x \\cos 2 x+\\cos 2 x=0$\n\nZapis v obliki produkta: $\\cos 2 x(2 \\sin 2 x+1)=0$\n\nZapis enačb:\n$\\cos 2 x=0$\n$\\sin 2 x=-\\frac{1}{2}$\n\nRešitev:\n$x_{1}=\\frac{\\pi}{4}+\\frac{k \\pi}{2}$\n$x_{2}=-\\frac{\\pi}{12}+k \\pi$\n$x_{3}=\\frac... | Slovenia | 2. matematično tekmovanje dijakov srednjih tehniških in strokovnih šol | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | x = π/4 + k·π/2, x = −π/12 + k·π, or x = 7π/12 + k·π, where k ∈ ℤ | |
092z | Problem:
Determine the smallest possible value of
$$
\left|2^{m}-181^{n}\right|,
$$
where $m$ and $n$ are positive integers. | [
"Solution:\nCalculating\n$$\n181^{2}=32,761\n$$\none should get the idea that this may be close to\n$$\n2^{15}=32,768\n$$\nso taking the difference of both we arrive at the minimum possible value $7$.\nAs we can clearly see that the difference must be positive and odd, we only need to eliminate the possibilities $1... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 7 | |
07cy | In triangle $ABC$, arbitrary points $P, Q$ lie on side $BC$ such that $BP = CQ$ and $P$ lies between $B, Q$. The circumcircle of triangle $APQ$ intersects sides $AB$ and $AC$ at $E$ and $F$, respectively. The point $T$ is the intersection point of $EP$ and $FQ$. Two lines passing through the midpoint of $BC$ and parall... | [
"Let $M$ be the midpoint of $BC$. Since $BP = CQ$, it is clear that $MP = MQ$ and $BQ = CP$. Let $Z$ be the second intersection point of circumcircles of triangles $MPX$ and $MQY$. We claim that the circumcircles of triangles $APQ$ and $XYT$ are tangent to each other at $Z$.\n\n\n\nWe have\... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09lr | The numbers $1, 2, \dots, 25$ are written in the cells of a $5 \times 5$ grid as shown below.
Is it possible to make all the numbers in the grid equal by performing a series of operations, where each operation consists of selecting either a row or a column and increasing all the numbers in tha... | [
"Let us denote the cell in row $i$ and column $j$ by $a_{i,j}$, where $1 \\leq i, j \\leq 5$. Initially, $a_{i,j} = 5(i-1) + j$.\n\nEach operation increases all numbers in a row or a column by $1$. Suppose we perform $r_i$ operations on row $i$ and $c_j$ operations on column $j$. Then the final value in cell $(i, j... | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | No | |
05co | Let $n \ge 3$ be an integer. In an $n \times n$ grid there are three invisible monsters: one in the upper right corner square and one in each of its neighboring squares. In a $2 \times 2$ area in the opposite corner of the grid some frogs are placed. A square may contain more than one frog.
The frogs and the monsters ... | [
"We color the square in 3 colors by descending diagonals (in Fig. 16 the colors are denoted by A, B, C and X, Y, Z, which are still the same colors, but their order depends on the value of $n$). Notice that both frogs and monsters can only jump on squares of the same color as the one they started from. As the monst... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | 1 | |
0c3m | Problem:
Demonstraţi că un număr natural $A$ este pătrat perfect dacă şi numai dacă, pentru orice număr natural nenul $n$, cel puţin una din diferenţele
$$
(A+1)^{2}-A,(A+2)^{2}-A,(A+3)^{2}-A, \ldots,(A+n)^{2}-A
$$
este divizibilă cu $n$. | [
"Solution:\nDacă $A$ este pătrat perfect, adică există $B \\in \\mathbb{N}$ astfel încât $A=B^{2}$, atunci $(A+k)^{2}-A=\\left(B^{2}+k\\right)^{2}-B^{2}=\\left(B^{2}+B+k\\right)\\left(B^{2}-B+k\\right)$ pentru orice $k=\\overline{1, n}$ și exact unul dintre numerele (consecutive) $B^{2}+B+1, B^{2}+B+2, \\ldots, B^{... | Romania | Al patrulea test de selecţie pentru OBMJ | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
05f0 | Problem:
Déterminer tous les triplets d'entiers $(a, b, n)$ strictement positifs vérifiant:
$$
a! + b! = 2^{n}
$$ | [
"Solution:\nLes factorielles ayant beaucoup de facteurs impairs en commun, on se dit directement qu'obtenir une puissance de 2 va être très contraignant.\nSupposons que $a, b \\geqslant 3$, $3$ divise donc la somme des factorielles et donc $2^{n}$ : c'est absurde. C'est à dire que l'un des deux est dans $\\{1,2\\}$... | France | ENVOi 3 : ARITHMÉTIQUE | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | [(1,1,1), (2,2,2), (2,3,3), (3,2,3)] | |
07lt | Suppose $a$, $b$, $c$ are the side lengths of a triangle. Prove that
$$
\frac{a}{1+a}, \frac{b}{1+b}, \frac{c}{1+c}
$$
are also the side lengths of a triangle. | [
"First of all, if $x$, $y > -1$, then\n$$\n\\frac{x}{1+x} < \\frac{y}{1+y} \\Leftrightarrow x+xy < y+yx,\n$$\ni.e., iff $-1 < x < y$. Hence, since $a < b + c$, and $b$, $c > 0$,\n$$\n\\frac{a}{1+a} < \\frac{b+c}{1+b+c} = \\frac{b}{1+b+c} + \\frac{c}{1+b+c} < \\frac{b}{1+b} + \\frac{c}{1+c},\n$$\nas required."
] | Ireland | Irska | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | proof only | null | |
0kq8 | Problem:
In three-dimensional space, let $S$ be the region of points $(x, y, z)$ satisfying $-1 \leq z \leq 1$. Let $S_{1}, S_{2}, \ldots, S_{2022}$ be 2022 independent random rotations of $S$ about the origin $(0,0,0)$. The expected volume of the region $S_{1} \cap S_{2} \cap \cdots \cap S_{2022}$ can be expressed as... | [
"Solution:\n\nConsider a point $P$ of distance $r$ from the origin. The distance from the origin of a random projection of $P$ onto a line is uniform from $0$ to $r$. Therefore, if $r < 1$ then the probability of $P$ being in all the sets is $1$, while for $r \\geq 1$ it is $r^{-2022}$. Therefore the volume is\n$$\... | United States | HMMT February | [
"Geometry > Solid Geometry > Volume"
] | null | final answer only | 271619 | |
0181 | The real numbers $x_1, \dots, x_{2011}$ satisfy
$$
x_1 + x_2 = 2x'_1, \quad x_2 + x_3 = 2x'_2, \quad \dots, \quad x_{2011} + x_1 = 2x'_{2011}
$$
where $x'_1, x'_2, \dots, x'_{2011}$ is a permutation of $x_1, x_2, \dots, x_{2011}$. Prove $x_1 = x_2 = \dots = x_{2011}$. | [
"For convenience we call $x_{2011}$ also $x_0$. Let $k$ be the largest of the numbers $x_1, \\dots, x_{2011}$, and consider an equation $x_{n-1} + x_n = 2k$, where $1 \\le n \\le 2011$. Hence we get $2 \\max(x_{n-1}, x_n) \\ge x_{n-1} + x_n = 2k$, so either $x_{n-1}$ or $x_n$ is $\\ge k$. Since $x_{n-1} \\le k$, we... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0hoj | Problem:
Six children are invited to a birthday party, and each pair of them are either mutual friends or mutual strangers. Prove that there are either three of them that are all friends or three of them that are all strangers to one another. | [
"Solution:\n\nBegin by letting $A$ be any person at the party, and note that $A$ must be either friends or strangers with at least three of the others, for otherwise there would only be at most $2+2=4$ other people at the party. Because of the symmetry between friends and strangers, we can assume that $A$ has three... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0kzz | Problem:
An ordered pair $(a, b)$ of positive integers is called spicy if $\operatorname{gcd}(a+b, a b+1)=1$. Compute the probability that both $(99, n)$ and $(101, n)$ are spicy when $n$ is chosen from $\{1,2, \ldots, 2024!\}$ uniformly at random. | [
"Solution:\n\nWe claim that $(a, b)$ is spicy if and only if both $\\operatorname{gcd}(a+1, b-1)=1$ and $\\operatorname{gcd}(a-1, b+1)=1$.\n\nTo prove the claim, we note that\n$$\n\\operatorname{gcd}(a+b, a b+1)=\\operatorname{gcd}(a+b, b(-b)+1)=\\operatorname{gcd}\\left(a+b, b^{2}-1\\right)\n$$\nHence, we have\n$$... | United States | HMMT February 2024 Guts Round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | 96/595 | |
01jp | Call a pair of positive integers $(a, b)$ *carroty* if $S(a^{b+1}) = a^b$, where $S(m)$ is the digit sum of $m$. Find all *carroty* pairs $(a, b)$. | [
"*Answer:* $(a, b) \\in \\{(1, b) \\mid b \\in \\mathbb{Z}^+\\} \\cup \\{(3, 2), (9, 1)\\}$.\n\nLet us consider the case where $k \\ge 2$. Then $k \\le 2(k-1)$ and note that $b+1 \\le 2b$ as $b \\ge 1$. Put $(k-1)b =: x$, then $k(b+1) \\le 4(k-1)b = 4x$. So $10^x \\le 36x$. It is obvious that the only solutions in ... | Baltic Way | Baltic Way 2023 Shortlist | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | (1, b) for all positive integers b; (3, 2); (9, 1) | |
0en4 | a. Prove that the unit square can be covered by three sets of diameter not exceeding $\frac{\sqrt{65}}{8}$.
b. Prove that the unit square can not be covered by three sets of diameter less than $\frac{\sqrt{65}}{8}$. | [
"a. Set the square on the coordinate plane with vertices $(\\pm\\frac{1}{2}, \\pm\\frac{1}{2})$. Then the three circles with diameter $\\frac{\\sqrt{65}}{8}$ and centres $(0, \\frac{7}{16})$, $(\\frac{1}{4}, -\\frac{1}{16})$ and $(-\\frac{1}{4}, -\\frac{1}{16})$ covers the unit square.\n\nb. Label the square $ABCD$... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0bpm | Problem:
Se consideră patrulaterul $ABCD$ cu diagonalele neperpendiculare şi laturile $AB$ şi $CD$ neparalele. Fie $O$ punctul de intersecţie a diagonalelor, $H_{1}$ ortocentrul triunghiului $AOB$ şi $H_{2}$ ortocentrul triunghiului $COD$. Se notează cu $M$ mijlocul laturii $[AB]$ şi cu $N$ mijlocul laturii $[CD]$. Ar... | [
"Solution:\n\nFie $A'$, $B'$ picioarele înălţimilor din $A$, respectiv $B$ în triunghiul $AOB$, şi $C'$, $D'$ picioarele înălţimilor din $C$, respectiv $D$ în triunghiul $COD$.\n\nEvident, $A'$ şi $D'$ aparţin cercului $\\mathcal{C}_1$ de diametru $AD$, iar $B'$ şi $C'$ aparţin cercului $\\mathcal{C}_2$ de diametru... | Romania | Primul test de selecţie pentru OBMJ | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0e3k | For what positive integers $n \ge 3$ does there exist a polygon with $n$ vertices (not necessarily convex) with the property that each of its sides is parallel to another one of its sides? | [
"If $n \\ge 3$ is even, $n = 2k$, then such a polygon exists since every regular $2k$-gon satisfies the condition.\n\nIf $n = 3$ or $n = 5$, then such a polygon does not exist. Indeed, no two sides in a triangle are parallel. If every side of a pentagon would be parallel to some other side, we could find three para... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | All integers n ≥ 3 except n = 3 and n = 5 | |
04o9 | In the picture below a chain consisted of $54$ unit squares is shown. Each unit square, except for the two at the front and the end, is connected to its neighbouring ones in opposite vertices.
It is allowed to put any unit square of the chain in any position in space as long as it is connect... | [
"No, it is not possible.\n\nLet us assume the opposite, i.e. that the chain fully covers the surface of a $3 \\times 3 \\times 3$ cube. Let us draw a diagonal joining two opposite vertices which are two meeting points of the chain in each unit square of the chain (including the first and the last unit square in the... | Croatia | Croatian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Solid Geometry > Other 3D problems"
] | English | proof and answer | No | |
0f0f | Problem:
You are given 14 coins. It is known that genuine coins all have the same weight and that fake coins all have the same weight, but weigh less than genuine coins. You suspect that 7 particular coins are genuine and the other 7 fake. Given a balance, how can you prove this in three weighings (assuming that you t... | [
"Solution:\n\nLet the coins you suspect to be genuine be $G_1$, $G_2$, ..., $G_7$, and the suspected fakes be $F_1$, $F_2$, ..., $F_7$.\n\nFirst, weigh $F_1$ against $G_1$. Assuming $F_1$ weighs less, you have proved that $F_1$ is fake and $G_1$ genuine.\n\nSecond, weigh $F_1$, $G_2$, $G_3$ against $G_1$, $F_2$, $F... | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Logic",
"Discrete Mathematics > Algorithms"
] | null | proof only | null | |
0cnx | In a parallelogram $ABCD$, angle $ABC$ is obtuse. Let $\omega$ be the circumcircle of triangle $ABC$. Line $AD$ intersects $\omega$ at points $A$ and $E$, and line $CD$ intersects $\omega$ at points $C$ and $F$. Prove that the circumcenter of triangle $DEF$ lies on $\omega$. (T. Emelyanova) | [
"Пусть $\\angle ABC = \\angle ADC = \\angle EDF = \\alpha$; по условию, $\\alpha > 90^\\circ$ (см. рис. 7).\n\nТак как $BC \\parallel AD$, то $ABCE$ — равнобокая трапеция, откуда $\\angle ECF = \\angle BCE - \\angle BCD = \\alpha - (180^\\circ - \\alpha) = 2\\alpha - 180^\\circ$.\n\nДалее, пусть $P$ — центр описанн... | Russia | Russian mathematical olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English; Russian | proof only | null | |
0fd4 | Problem:
Saber cuál es la última cifra de $2009^{2011}$ es muy fácil, pero ¿cuántos ceros preceden a esa última cifra? | [
"Solution:\n\nSi $n \\geq 1$,\n$$\n2009^{n} = (2000+9)^{n} = 9^{n} + 2000k\n$$\nPor tanto, las 3 últimas cifras de $2009^{n}$ coinciden con las de $9^{n}$. Por el desarrollo del binomio de Newton:\n$$\n\\begin{aligned}\n& 9^{2011} = (10-1)^{2011} = (-1)^{2011} + \\binom{2011}{1}(-1)^{2010} \\cdot 10 + \\\\\n& + \\b... | Spain | XLVII Olimpiada Matemática Española Primera Fase | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1 | |
0eiw | Problem:
Dan je izraz $A = |x + 3| + |x - 5|$.
a) Izračunaj vrednost izraza za $x = \sqrt{2}$.
b) Poišči vse možne vrednosti $x$, za katere velja $A = 4$. | [
"Solution:\n\na)\nZa $x$ vstavimo $\\sqrt{2}$ in izračunamo vrednost izraza:\n$$\nA = |\\sqrt{2} + 3| + |\\sqrt{2} - 5| = (\\sqrt{2} + 3) + (5 - \\sqrt{2}) = 8.\n$$\n\nb)\nZa $x < -3$ ima enačba:\n$$\nA = |x + 3| + |x - 5| = -(x + 3) - (x - 5) = -x - 3 - x + 5 = -2x + 2.\n$$\nPostavimo $A = 4$:\n$$\n-2x + 2 = 4 \\i... | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a) 8; b) no real solutions | |
01ul | The central area of a town has a form of the $(2n+1) \times m$ rectangle, which is formed by $1 \times 1$ tiles. To illuminate the area, one-lamp lampposts are used. The lampposts are placed at the corners of some tiles, including a corner at the lamppost position, and only those.
Find the smallest number of the lamppo... | [
"Answer: $2(n + 1) \\lfloor \\frac{m + 1}{2} \\rfloor$, where $\\lfloor x \\rfloor$ is the greatest integer not exceeding $x$.\n\nWe paint some tiles of the town square black (if $m = 2k + 1$ is odd, then see Fig. 1, if $m = 2k$ is even, then see Fig. 2).\n\nIt is easy to see that any lamp can illuminate at most on... | Belarus | Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 2(n + 1) ⌊(m + 1)/2⌋ | |
0282 | Problem:
Um grupo de rapazes e moças saiu para comer pizza em dois dias consecutivos. No restaurante em que foram, as pizzas são cortadas em doze pedaços iguais. Maria observou que no primeiro dia cada rapaz comeu 7 pedaços, e cada moça 3 pedaços. Já no segundo dia, cada rapaz comeu 6 pedaços e cada moça 2 pedaços. Cu... | [
"Solution:\n\nSejam $x$ e $y$ o número de rapazes e moças, respectivamente. Sabemos que o número total de pedaços consumidos foi no mínimo $49$ (4 pizzas e um pedaço da última pizza) e no máximo $59$ (4 pizzas mais 11 pedaços, lembre que sobrou pelo menos um pedaço da última pizza). Por outro lado,\n$$\n\\begin{ali... | Brazil | null | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 8 boys and 1 girl | |
0gya | Let $a, b \in [-1,1]$. Prove that $a\sqrt{1-b^2} + b\sqrt{1-a^2} \le 1$. | [
"By substitution: $a = \\sin \\alpha$, $b = \\sin \\beta$ the given inequality becomes: $\\sin \\alpha \\cos \\beta + \\sin \\beta \\cos \\alpha \\le 1$. This inequality holds for an arbitrary $\\alpha, \\beta$."
] | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Algebra > Equations and Inequalities"
] | English | proof only | null | |
0iwq | Let $\triangle ABC$ be a triangle with $AB = AC$. The angle bisectors of $\angle CAB$ and $\angle ABC$ meet the sides $BC$ and $CA$ at $D$ and $E$, respectively. Let $K$ be the incenter of triangle $ADC$. Suppose that $\angle BEK = 45^\circ$. Find all possible values of $\angle CAB$.
(This problem was suggested by Pet... | [
"Let $E_1$ be the reflection of $E$ across the line $CI$. Because $CI$ is the angle bisector of $\\angle BCA$, $E_1$ lies on ray $CB$. We consider two cases.\n\nIn the first case, we assume that $E_1 = D$. Then $CDIE$ is a kite with $CI$ as the symmetry axis. In particular, $\\angle IEC = \\angle IDC = 90^\\circ$; ... | United States | IMO 2009 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordin... | null | proof and answer | 60° or 90° | |
0ktv | Problem:
For any positive integer $a$, let $\tau(a)$ be the number of positive divisors of $a$. Find, with proof, the largest possible value of $4 \tau(n)-n$ over all positive integers $n$. | [
"Solution:\n\nLet $d$ be the number of divisors of $n$ less than or equal to $\\frac{n}{4}$. Then, $\\tau(n)-3 \\leq d \\leq \\frac{n}{4} \\Longrightarrow$ $4 \\tau(n)-n \\leq 12$. We claim the answer is $12$. This is achieved by $n=12$."
] | United States | HMMT February | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | 12 | |
0d35 | $A B C D E F$ is an equiangular hexagon of perimeter $21$. Given that $A B = 3$, $C D = 4$, and $E F = 5$, compute the area of hexagon $A B C D E F$. | [
"We extend sides $F A$ and $B C$ to intersect at $A'$, and sides $B C$ and $D E$ to intersect at $B'$, and sides $D E$ and $F A$ to intersect at $C'$.\n\n\n\nTriangles $A' B' C'$, $A' B A$, $B' D C$, and $C' F E$ are equilateral with side lengths $11$, $3$, $4$, and $5$ respectively. Theref... | Saudi Arabia | Selection tests for the Balkan Mathematical Olympiad 2013 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 71√3/4 | |
0d64 | Let $P_{i}(x) = x^{2} + b_{i} x + c_{i}$ for $i = 1, 2, \cdots, n$ be pairwise distinct polynomials of degree $2$ with real coefficients so that for any $0 \leq i < j \leq n$; $i, j \in \mathbb{N}$, the polynomial $Q_{i, j}(x) = P_{i}(x) + P_{j}(x)$ has only one real root. Find the greatest possible value of $n$. | [
"The answer $n = 3$. In fact, the polynomials $P_{1}(x) = x^{2} - 4$, $P_{2}(x) = x^{2} - 4x + 6$ and $P_{3}(x) = x^{2} - 8x + 12$ satisfy the conditions:\n\n$P_{1} + P_{2} = 2(x - 1)^{2}$,\n\n$P_{1} + P_{3} = 2(x - 2)^{2}$,\n\n$P_{2} + P_{3} = 2(x - 3)^{2}$.\n\nSuppose that there are four polynomials $P_{1}, P_{2}... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | 3 | |
08tw | Suppose the real numbers $x_1, x_2, x_3, x_4, x_5$ satisfy the following set of five equations:
$$
\begin{cases}
x_1x_2 + x_1x_3 + x_1x_4 + x_1x_5 = -1 \\
x_2x_1 + x_2x_3 + x_2x_4 + x_2x_5 = -1 \\
x_3x_1 + x_3x_2 + x_3x_4 + x_3x_5 = -1 \\
x_4x_1 + x_4x_2 + x_4x_3 + x_4x_5 = -1 \\
x_5x_1 + x_5x_2 + x_5x_3 + x_5x_4 = -1
... | [
"Suppose the real numbers $x_1, x_2, \\dots, x_5$ satisfy the given set of equations. Let $a = x_1 + x_2 + \\dots + x_5$. Then, each of the five numbers $x_j, j = 1, 2, \\dots, 5$ satisfies the quadratic equation $x^2 - a x - 1 = 0$. This equation has 2 distinct real roots $\\frac{a \\pm \\sqrt{a^2 + 4}}{2}$. Denot... | Japan | Japan Mathematical Olympiad First Round | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | ±√2, ±√2/2 | |
0482 | Given an odd prime $p$. Find the largest positive integer $n$ such that there exist $n$ integer-coordinate points $A_1, A_2, \dots, A_n$ in the plane, with no three collinear, and for any $1 \le i < j < k \le n$, twice the area of triangle $A_iA_jA_k$ is **not** divisible by $p$. | [
"It is well-known that the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is given by\n$$\n\\frac{1}{2} |(y_2 - y_1)(x_3 - x_1) - (y_3 - y_1)(x_2 - x_1)|\n$$\n(This equals zero if and only if the three points are collinear). Therefore, the condition that $(x_1, y_1)$, $(x_2, y_2)$, $(x_3,... | China | China-TST-2025A | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots >... | English | proof and answer | p+1 | |
0ee1 | The lines containing the chords $AC$ and $BD$ of a circle with radius $1$ intersect at an angle of $45^\circ$. The length of the chord $AB$ equals $1$ (see figure).
What is the size of the angle $\angle CBD$?
(A) $45^\circ$
(B) $60^\circ$
(C) $65^\circ$
(D) $75^\circ$
(E) $85^\circ$ | [
"Let $S$ denote the centre of the circle. Then the triangle $SBA$ is equilateral since all of its sides have length $1$. So, $\\angle BSA = 60^\\circ$. By the Central Angle Theorem $\\angle BCA = \\frac{1}{2} \\angle BSA = 30^\\circ$. The outer angle of the triangle is always equal to the sum of the non-adjacent in... | Slovenia | Slovenija 2016 | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | D | |
0fh1 | Problem:
Demostrar que
$$
\frac{1}{10 \sqrt{2}} < \frac{1 \cdot 3 \cdot 5 \cdots 99}{2 \cdot 4 \cdot 6 \cdots 100} < \frac{1}{10}
$$ | [
"Solution:\nPrimera solución\nVeamos la primera desigualdad\n$$\n\\begin{aligned}\n&\\left(\\frac{1}{2} \\frac{3}{4} \\frac{5}{6} \\ldots \\frac{99}{100}\\right)^{2} = \\frac{1}{2} \\frac{1}{2} \\frac{3}{4} \\frac{3}{4} \\ldots \\frac{99}{100} \\frac{99}{100} > \\\\\n&> \\frac{1}{2} \\frac{1}{2} \\frac{3}{4} \\frac... | Spain | OME 25 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
0iyl | Problem:
Two colorings are distinct if there is no way to relabel the colors to transform one into the other. Equivalently, they are distinct if and only if there is some pair of vertices which are the same color in one coloring but different colors in the other. For what pairs $(n, k)$ of positive integers does there... | [
"Solution:\n\nAnswer: $(1,1),\\left(2,2^{k}\\right)$ for integers $k \\geq 0$, and $(n, k)$ for $n>2, k>0$\n\nIf $n=1$, there is only one coloring. If $n=2$, then each connected component of the graph can be colored in two ways, because the color of any vertex in the graph determines the colors of all vertices conn... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | (1, 1); (2, 2^k) for integers k ≥ 0; and (n, k) for all n > 2 and k > 0 | |
0in4 | Problem:
Let $\omega$ denote the incircle of triangle $ABC$. The segments $BC$, $CA$, and $AB$ are tangent to $\omega$ at $D$, $E$, and $F$, respectively. Point $P$ lies on $EF$ such that segment $PD$ is perpendicular to $BC$. The line $AP$ intersects $BC$ at $Q$. The circles $\omega_1$ and $\omega_2$ pass through $B$... | [
"Solution:\n\nAnswer: 101. Construct $D'$ diametrically opposed to $D$, so that $\\angle DFD'$ and $\\angle DED'$ are right, and note that $P$ lies on $DD'$. By standard angle chasing, $m \\angle FDD' = \\beta$ (half angle $B$) and $m \\angle D'DE = \\gamma$. Thus, $m \\angle DD'F = 90^\\circ - \\beta$ and $m \\ang... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quad... | null | proof and answer | 101 | |
04oq | Base $\overline{BC}$ is the longest side of the isosceles triangle $ABC$. Let $M$ be a point on the side $\overline{BC}$ such that $|BM| = |AB|$. Point $N$ is the foot of the altitude from $M$ to $\overline{AB}$.
Prove that the triangle $BMN$ and the quadrilateral $ACMN$ have equal areas and equal perimeters. (United K... | [] | Croatia | Croatian Mathematical Society Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
00xm | Problem:
An equilateral triangle is divided into $n^{2}$ congruent equilateral triangles. A spider stands at one of the vertices, a fly at another. Alternately each of them moves to a neighbouring vertex. Prove that the spider can always catch the fly. | [
"Solution:\n\nAssume that the big triangle lies on one of its sides. Then a suitable strategy for the spider will be as follows:\n\n(1) First, move to the lower left vertex of the big triangle.\n\n(2) Then, as long as the fly is higher than the spider, move upwards along the left side of the big triangle.\n\n(3) Af... | Baltic Way | Baltic Way 1993 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Other"
] | null | proof only | null | |
0kzo | Problem:
Let $ABC$ be an acute isosceles triangle with orthocenter $H$. Let $M$ and $N$ be the midpoints of sides $\overline{AB}$ and $\overline{AC}$, respectively. The circumcircle of triangle $MHN$ intersects line $BC$ at two points $X$ and $Y$. Given $XY = AB = AC = 2$, compute $BC^2$. | [
"Solution:\n\n\n\nLet $D$ be the foot from $A$ to $BC$, also the midpoint of $BC$. Note that $DX = DY = MA = MB = MD = NA = NC = ND = 1$. Thus, $MNXY$ is cyclic with circumcenter $D$ and circumradius $1$. $H$ lies on this circle too, hence $DH = 1$.\n\nIf we let $DB = DC = x$, then since $\... | United States | HMMT February 2024 Guts Round | [
"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 | 2(√17 − 1) |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.