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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0avr | Problem:
There are two distinct real numbers which are larger than their reciprocals by $2$. Find the product of these numbers. | [
"Solution:\n\nLet $x$ be one of these real numbers. We have $x = \\frac{1}{x} + 2$, which is equivalent to $x^2 - 2x - 1 = 0$. The required real numbers are the two distinct roots of this quadratic equation, which have a product of $-1$."
] | Philippines | 18th PMO National Stage Oral Phase | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | -1 | |
061c | Problem:
Man beweise: Für die positiven reellen Zahlen $a$, $b$, $c$ gilt die Ungleichung
$$
\frac{a}{\sqrt{(a+b)(a+c)}} + \frac{b}{\sqrt{(b+a)(b+c)}} + \frac{c}{\sqrt{(c+a)(c+b)}} \leq \frac{3}{2}.
$$ | [] | Germany | Auswahlwettbewerb zur IMO 2001 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0be7 | Problem:
Egy $(G, \cdot)$ csoport $(P)$ tulajdonságú, ha a $G$ minden $f$ automorfizmusa esetén léteznek a $G$ olyan $g$ és $h$ automorfizmusai, amelyekre $f(x)=g(x) \cdot h(x)$, bármely $x \in G$ esetén. Igazold, hogy:
a) Minden $(P)$ tulajdonságú csoport kommutatív!
b) Minden kommutatív, páratlan rendű, véges csop... | [] | Romania | Matematika tantárgyverseny Megyei szakasz | [
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
04iy | Determine all values that the expression
$$
\frac{1 + \cos^2 x}{\sin^2 x} + \frac{1 + \sin^2 x}{\cos^2 x},
$$
can attain, where $x$ is a real number. | [
"**3.2.** If we write down the given equation in the form $2^m p^2 = n^5 - 1$ and factorise the right-hand side, we get\n$$\n2^m p^2 = (n-1)(n^4 + n^3 + n^2 + n + 1).\n$$\nFactor $n^4 + n^3 + n^2 + n + 1$ is odd, so $n-1$ is divisible by $2^m$.\nWe immediately see that $p$ is odd.\nOn the other hand, since $n$ is p... | Croatia | First round – City competition | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | [6, ∞) | |
08qz | Problem:
Prove that for all non-negative real numbers $x, y, z$, not all equal to $0$, the following inequality holds
$$
\frac{2x^{2}-x+y+z}{x+y^{2}+z^{2}}+\frac{2y^{2}+x-y+z}{x^{2}+y+z^{2}}+\frac{2z^{2}+x+y-z}{x^{2}+y^{2}+z} \geqslant 3
$$
Determine all the triples $(x, y, z)$ for which the equality holds. | [
"Solution:\nLet us first write the expression $L$ on the left hand side in the following way\n$$\n\\begin{aligned}\nL & = \\left(\\frac{2x^{2}-x+y+z}{x+y^{2}+z^{2}} + 2\\right) + \\left(\\frac{2y^{2}+x-y+z}{x^{2}+y+z^{2}} + 2\\right) + \\left(\\frac{2z^{2}+x+y-z}{x^{2}+y^{2}+z} + 2\\right) - 6 \\\\\n& = \\left(2x^{... | JBMO | Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Equality holds exactly for the triples: (t, t, t) with t > 0; (t, t, 1 − t) with t in [0, 1]; (t, 1 − t, t) with t in [0, 1]; and (1 − t, t, t) with t in [0, 1]. | |
0ajp | Solve the equation $xyz + yzt + xzt + xyt = xyzt + 3$ in the set of natural numbers. | [
"After dividing the equation by $xyzt$ we get $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}+\\frac{1}{t}=1+\\frac{3}{xyzt}$. Because of symmetry, without loss of generality, we can assume that\n$$\nx \\le y \\le z \\le t \\quad \\dots \\tag{1}\n$$\nfrom where it follows that $\\frac{1}{x} \\ge \\frac{1}{y} \\ge \\frac{1}... | North Macedonia | Macedonian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All permutations of (3,3,4,11), (3,3,5,7), (2,3,7,39), (2,3,9,17), and (1,1,1,1). | |
0jcm | Problem:
$ABC$ is a triangle with $AB = 15$, $BC = 14$, and $CA = 13$. The altitude from $A$ to $BC$ is extended to meet the circumcircle of $ABC$ at $D$. Find $AD$. | [
"Solution:\n\nAnswer: $\\boxed{\\dfrac{63}{4}}$\n\nLet the altitude from $A$ to $BC$ meet $BC$ at $E$. The altitude $AE$ has length $12$; one way to see this is that it splits the triangle $ABC$ into a $9$-$12$-$15$ right triangle and a $5$-$12$-$13$ right triangle; from this, we also know that $BE = 9$ and $CE = 5... | United States | HMMT November | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 63/4 | |
09uc | One hundred students wear shirts numbered from $1$ to $100$. The students are arranged in a square of ten rows by ten columns. It turns out that adding the ten shirt numbers of the students in any row or any column always yields the same outcome.
Determine that outcome. | [
"Let us consider the arrangement of the students in a $10 \\times 10$ square. The shirt numbers are $1, 2, \\ldots, 100$.\n\nThe sum of all shirt numbers is:\n$$\n1 + 2 + \\cdots + 100 = \\frac{100 \\times 101}{2} = 5050.\n$$\n\nThere are $10$ rows, and the sum of the numbers in each row is the same. Let $S$ be the... | Netherlands | Junior Mathematical Olympiad, September 2019 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 505 | |
0i9w | Problem:
Find the minimum distance from the point $(0, 5/2)$ to the graph of $y = x^{4}/8$. | [
"Solution:\nWe want to minimize $x^{2} + \\left(\\frac{x^{4}}{8} - \\frac{5}{2}\\right)^{2} = \\frac{x^{8}}{64} - \\frac{5 x^{4}}{8} + x^{2} + \\frac{25}{4}$, which is equivalent to minimizing $\\frac{z^{4}}{4} - 10 z^{2} + 16 z$, where we have set $z = x^{2}$.\n\nThe derivative of this expression is $z^{3} - 20 z ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | sqrt(17)/2 | |
0igm | Problem:
In an election, there are two candidates, $A$ and $B$, who each have $5$ supporters. Each supporter, independent of other supporters, has a $\frac{1}{2}$ probability of voting for his or her candidate and a $\frac{1}{2}$ probability of being lazy and not voting. What is the probability of a tie (which include... | [
"Solution:\n\nThe probability that exactly $k$ supporters of $A$ vote and exactly $k$ supporters of $B$ vote is $\\binom{5}{k}^2 \\cdot \\frac{1}{2^{10}}$. Summing over $k$ from $0$ to $5$ gives\n$$\n\\left(\\frac{1}{2^{10}}\\right)(1+25+100+100+25+1)=\\frac{252}{1024}=\\frac{63}{256}.\n$$"
] | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | 63/256 | |
04y9 | Each of the $4n^2$ unit squares of a $2n \times 2n$ board ($n \ge 1$) has been colored blue or red. A set of four different unit squares of the board is called *pretty* if these squares can be labeled $A, B, C, D$ in such a way that $A$ and $B$ lie in the same row, $C$ and $D$ lie in the same row, $A$ and $C$ lie in th... | [
"Let us index the unit squares of the board by pairs of integers $(a, b)$ with $1 \\le a, b \\le 2n$. We prove that the largest possible number of pretty sets is $n^4$.\n\nFor the upper bound, consider coloring all the unit squares $(a, b)$ with $a, b \\le n$ or $a, b \\ge n+1$ blue, and all the other unit squares ... | Czech-Polish-Slovak Mathematical Match | null | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | n^4 | |
0gx8 | $n$ natural numbers are written on the board. You can add only natural numbers in the form $\frac{a+b}{a-b}$ where $a$ and $b$ are the numbers already written on the board. It appears that by doing so you can make any natural number appear on the board. Calculate the least value of $n$ and find the numbers initially wr... | [
"As $(a + b) > (a - b)$, you can not obtain $1$ performing the operations allowed. Therefore it should be written on the board, but one number is not enough. Let's show that two numbers will be enough. Let another number of the two be $x$. $\\frac{x+1}{x-1}$ is the only number which can be obtained in the first ste... | Ukraine | Ukrajina 2008 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Minimum n is 2; the valid initial sets are {1,2} and {1,3}. | |
0ih7 | Problem:
A $k$th root of unity is any complex number $\omega$ such that $\omega^{k}=1$.
Let $x$ and $y$ be two $k$th roots of unity. Prove that $(x+y)^{k}$ is real. | [
"Solution:\n\nNote that\n$$\n\\begin{aligned}\n(x+y)^{k} & = \\sum_{i=0}^{k} \\binom{k}{i} x^{i} y^{k-i} \\\\\n& = \\frac{1}{2} \\sum_{i=0}^{k} \\binom{k}{i} \\left(x^{i} y^{k-i} + x^{k-i} y^{i}\\right)\n\\end{aligned}\n$$\nby pairing the $i$th and $(k-i)$th terms. But $x^{k-i} y^{i} = \\left(x^{i} y^{k-i}\\right)^... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | null | proof only | null | |
0epc | How many different numbers can be written as the product of two or more of the numbers $3$, $4$, $5$, $6$, $7$, $7$, $7$? | [
"The required numbers are of the form $3^a4^b5^c6^d7^e$, where\n$$\n0 \\le a \\le 1, \\quad 0 \\le b \\le 2, \\quad 0 \\le c \\le 2, \\quad 0 \\le d \\le 1, \\quad 0 \\le e \\le 3,\n$$\nand $a+b+c+d+e \\ge 2$. This gives two possible values of $a$, three of $b$, three of $c$, two of $d$, and four of $e$, making a t... | South Africa | South African Mathematics Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | proof and answer | 138 | |
00a1 | Rectangle $ABCD$ has sides $AB = 3$, $BC = 2$. Point $P$ on side $AB$ is such that the bisector of $C\hat{D}P$ passes through the midpoint of $BC$. Find $BP$. | [
"Let $M$ be the midpoint of $BC$, and let line $DM$ intersect\n\n\n\nline $AB$ at $Q$ (it is exterior to the segment $AB$). Then $BQM = CDM$ as $AB \\parallel CD$. On the other hand $CDM = PDM$ by hypothesis (DM is the bisector of $CDP$). So $PQD = PDC$ and hence $PQ = PD$. In addition $BQ ... | Argentina | Argentine National Olympiad 2015 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 1/3 | |
0agn | The number $2009^{2009^{2009}}$ is given, written in base 10. In one step we make the following operation: we delete the first and last digit, and we add their sum to the number which remained after deleting the first and last digit.
a) If after a finite number of steps there remains a two-digit number, is it possible... | [
"We will use the following well-known results:\n**Property. 1.** Every natural number, when divided by $3$ or $9$ gives the same remainder as the sum of its digits.\n**Property. 2.** A square of a natural number, when divided by $3$ gives the remainder $0$ or $1$.\nAnd we will show this property.\n**Property. 3.** ... | North Macedonia | Macedonian Junior Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a) No. b) 5 | |
0bjo | Let $f : [1, +\infty) \to (0, +\infty)$ be a continuous function having the following properties:
(i) The function $g : [1, +\infty) \to (0, +\infty)$ given by $g(x) = \frac{f(x)}{x}$ has limit at $+\infty$,
(ii) The function $h : [1, +\infty) \to (0, +\infty)$ given by $h(x) = \frac{1}{x} \int_{1}^{x} f(t) dt$ has fin... | [
"a.\nLet $\\ell = \\lim_{x \\to +\\infty} g(x)$. If $\\ell \\in (0, +\\infty)$, let $a > 0$ be such that $g(x) > \\ell/2$ for $x \\ge a$. Then\n$$\n\\begin{aligned}\nh(x) &= \\frac{1}{x} \\left( \\int_{1}^{a} f(t) \\, dt + \\int_{a}^{x} f(t) \\, dt \\right) \n\\ge \\frac{1}{x} \\int_{1}^{a} f(t) \\, dt + \\frac{\\e... | Romania | 65th Romanian Mathematical Olympiad | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Integral Calculus > Applications",
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Other"
] | null | proof only | null | |
0dhy | Prove that for every $n \ge 1$, then $v_3(u_n) = v_3(n)$, where $(u_n)$ is the sequence with $u_0 = 0$, $u_1 = 1$ and
$$
u_{n+2} = 2u_{n+1} + 2u_n, \forall n \ge 0.$$ | [
"We have the general formula of the given sequence as\n$$\nu_n = \\frac{(1 + \\sqrt{3})^n - (1 - \\sqrt{3})^n}{2\\sqrt{3}}.$$\nConsidering the periodicity of the remainder when divided by $3$ of the given sequence, we have $0, 1, 2, 0, 1, 2, \\ldots$ So obviously $u_{3k+1}$ and $u_{3k+2}$ are not divisible by $3$, ... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization"
] | English | proof only | null | |
0goz | Find the greatest constant $M$ such that
$$
a^2 + b^2 + c^2 + 3abc \geq M(ab + bc + ca)
$$
for all nonnegative real numbers $a, b, c$ satisfying $a + b + c = 4$. | [
"Letting $a = 0$ and $b = c = 2$ we obtain $2 \\ge M$. We will show that $M = 2$ works.\n\nWithout loss of generality we may assume that $\\max\\{a, b, c\\} = c$. Let $x = a + b$ and $y = ab$. We have $c \\ge \\frac{a+b+c}{3} = \\frac{4}{3}$ and hence $x = a+b \\le \\frac{8}{3}$.\n\nThen\n$$\na^2 + b^2 + c^2 + 3abc... | Turkey | Team Selection Test | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 2 | |
0iqc | Problem:
Let $a$ and $b$ be nonzero real numbers. Prove that at least one of the following inequalities is true:
$$
\begin{aligned}
& \left|\frac{a+\sqrt{a^{2}+2 b^{2}}}{2 b}\right|<1 \\
& \left|\frac{a-\sqrt{a^{2}+2 b^{2}}}{2 b}\right|<1
\end{aligned}
$$ | [
"Solution:\nAssume for the sake of contradiction that both inequalities are false, that is,\n$$\n1 \\leq\\left|\\frac{a+\\sqrt{a^{2}+2 b^{2}}}{2 b}\\right| \\text{ and } 1 \\leq\\left|\\frac{a-\\sqrt{a^{2}+2 b^{2}}}{2 b}\\right|.\n$$\nMultiplying these inequalities together yields\n$$\n1 \\leq\\left|\\frac{a+\\sqrt... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
04b9 | Let $ABCD$ be a parallelogram with an acute angle at vertex $A$. $G$ is a point on the line $AB$, different from $B$, such that $|BC| = |CG|$, and $H$ is a point on the line $BC$, different from $B$, such that $|AB| = |AH|$. Prove that the triangle $DGH$ is isosceles. | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof only | null | |
0dnp | Problem:
Дат је природан број $n$. Дефинишимо $f(0, j)=f(i, 0)=0$, $f(1,1)=n$ и
$$
f(i, j)=\left\lfloor\frac{f(i-1, j)}{2}\right\rfloor+\left\lfloor\frac{f(i, j-1)}{2}\right\rfloor
$$
за све природне бројеве $i$ и $j$, $(i, j) \neq (1,1)$. Колико има уређених парова природних бројева $(i, j)$ за које је $f(i, j)$ непа... | [
"Solution:\n\nЗа $m \\geqslant 2$ означимо $s_{m}=\\sum_{i+j=m} f(i, j)$. Како је остатак $f(i, j)$ при дељењу са 2 једнак $f(i, j)-2\\left[\\frac{f(i, j)}{2}\\right]$, број непарних међу бројевима $f(i, j)$ за $i, j \\geqslant 0$ и $i+j=m$ једнак је\n$$\n\\begin{aligned}\n\\sum_{i+j=m}\\left(f(i, j)-2\\left[\\frac... | Serbia | 10. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | n | |
05ul | Problem:
Trouver les fonctions $f: \mathbb{N}_{\geqslant 1} \mapsto \mathbb{N}_{\geqslant 0}$ vérifiant les deux conditions suivantes :
1. $f(x y)=f(x)+f(y)$ pour tous les entiers $x \geqslant 1$ et $y \geqslant 1$;
2. il existe une infinité d'entiers $n \geqslant 1$ tels que l'égalité $f(k)=f(n-k)$ est vraie pour tou... | [
"Solution:\n\nLes fonctions recherchées sont les fonctions de la forme $f: n \\mapsto c v_{p}(n)$, où $c$ est un entier naturel, $p$ est un nombre premier, et $v_{p}(n)$ est la valuation $p$-adique de $n$. Tout d'abord, il est clair que ces fonctions sont bien solutions du problème.\n\nRéciproquement, soit $f$ une ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All such functions are f(n) = c · v_p(n) for some prime p and some nonnegative integer c, where v_p is the p-adic valuation. | |
0kib | Problem:
Let $x, y, z$ be real numbers satisfying
$$
\frac{1}{x} + y + z = x + \frac{1}{y} + z = x + y + \frac{1}{z} = 3.
$$
The sum of all possible values of $x + y + z$ can be written as $\frac{m}{n}$, where $m, n$ are positive integers and $\operatorname{gcd}(m, n) = 1$. Find $100m + n$. | [
"Solution:\nThe equality $\\frac{1}{x} + y + z = x + \\frac{1}{y} + z$ implies $\\frac{1}{x} + y = x + \\frac{1}{y}$, so $x y = -1$ or $x = y$. Similarly, $y z = -1$ or $y = z$, and $z x = -1$ or $z = x$.\n\nIf no two elements multiply to $-1$, then $x = y = z$, which implies $2x + \\frac{1}{x} = 3$ and so $(x, y, ... | United States | HMMT November 2021 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 6106 | |
0ims | Problem:
Circles $\omega_{1}$, $\omega_{2}$, and $\omega_{3}$ are centered at $M$, $N$, and $O$, respectively. The points of tangency between $\omega_{2}$ and $\omega_{3}$, $\omega_{3}$ and $\omega_{1}$, and $\omega_{1}$ and $\omega_{2}$ are tangent at $A$, $B$, and $C$, respectively. Line $MO$ intersects $\omega_{3}$... | [
"Solution:\n\n$\\boxed{2 \\sqrt{3}}$. Note that $ONM$ is an equilateral triangle of side length $2$, so $m \\angle BPA = m \\angle BOA / 2 = \\pi / 6$. Now $BPA$ is a $30$-$60$-$90$ triangle with short side length $1$, so $AP = \\sqrt{3}$. Now $A$ and $B$ are the midpoints of segments $PR$ and $PQ$, so\n\n$$\n[PQR]... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 2√3 | |
018f | Call an *n*-tuple $(a_1, \dots, a_n)$ of real numbers *stable* if the sums $a_1 + a_2 + \dots + a_k$ where $0 < k \le n$, as well as the sums $a_n + a_{n-1} + \dots + a_{n-k}$ where $0 \le k < n$, are either all negative or all non-negative.
Let $k$ be any natural number. Consider all stable $(2k+1)$-tuples consisting... | [
"Answer: $k$.\n\nCall stable tuples, whose elements are alternately negative and non-negative, interesting. We first show that each interesting tuple contains at least one stable subtuple of 3 elements.\n\nFor that, consider elements whose absolute value is minimal in the tuple. If there exists a negative such elem... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | k | |
018w | Let $a$, $b$, and $c$ be the lengths of the sides of a triangle, $R$ the radius of its circumcircle, and $r$ the radius of its incircle. Prove that
$$
\frac{Rr}{(a+b+c)^2} \le \frac{1}{54}.
$$ | [
"We use the identities $2S = (a+b+c)r$ and $\\frac{abc}{4S} = R$, where $S$ denotes the area of the triangle. Multiplying them we obtain\n$$\n\\frac{abc}{2} = Rr(a+b+c) = \\frac{Rr}{(a+b+c)^2} \\cdot (a+b+c)^3.\n$$\nIt remains to show that $(a+b+c)^3 \\ge 27abc$. This follows directly from AM-GM inequality."
] | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Ineq... | null | proof only | null | |
0a37 | Eva looks at “words” consisting of $n$ characters, each equal to ‘L’ or ‘R’. In one turn, Eva may replace ‘RL’ anywhere in the word with ‘LR’. For example, in two turns, she takes the word ‘LRRLRRRLR’ to the word ‘LRLRRRLRR’. If there is no ‘L’ immediately to the right of an ‘R’, Eve cannot make a turn.
a. Eva has suc... | [
"a. Add up the positions of the characters 'L', where the leftmost character in the word has position 1 and the rightmost character has position $n$. We call this number the *L-sum* of a word. For each word, the L-sum is a non-negative integer. Furthermore, for every move Eva makes, the L-sum becomes one lower. Ind... | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | b) Maximum number of turns = ℓ(n − ℓ), achieved by placing all occurrences of L at the right end of the string. c) The maximizing number of L is n/2 if n is even; if n is odd, either (n − 1)/2 or (n + 1)/2. | |
0jtf | Problem:
A particular coin can land on heads (H), on tails (T), or in the middle (M), each with probability $\frac{1}{3}$. Find the expected number of flips necessary to observe the contiguous sequence HMMTHMMT...HMMT, where the sequence HMMT is repeated 2016 times. | [
"Solution:\n\nLet $E_{0}$ be the expected number of flips needed. Let $E_{1}$ be the expected number more of flips needed if the first flip landed on H. Let $E_{2}$ be the expected number more if the first two landed on HM. In general, let $E_{k}$ be the expected number more of flips needed if the first $k$ flips l... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | final answer only | (3^8068 - 81)/80 | |
0be6 | Problem:
Legyenek $a, b \in \mathbb{C}$. Igazold, hogy az $|a z + b \bar{z}| \leq 1$ egyenlőtlenség akkor és csak akkor áll fenn bármely egységnyi moduluszú $(|z|=1)$ $z$ komplex szám esetén, ha $|a| + |b| \leq 1$. | [] | Romania | Matematika tantárgyverseny Megyei szakasz | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
098h | Problem:
Determinați toate valorile parametrului real $a$, pentru care ecuația $2 \cdot |2 \cdot |x|-a^{2}| = x-a$ are exact trei soluții reale. | [
"Solution:\n\nDacă vom considera $a=0$, atunci obținem ecuația $4 \\cdot |x| = x \\Leftrightarrow x=0$.\nPentru $a=0$ ecuația are doar o soluție reală, deci $a \\neq 0$.\n\nConsiderăm funcția $f: \\mathbb{R} \\rightarrow \\mathbb{R}$, $f(x) = x - a$. Pentru orice valoare fixată a parametrului $a$, graficul funcției... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a ∈ { -2, -1/2 } | |
0f96 | Problem:
$ABC$ is a triangle. $A'$, $B'$, $C'$ are points on the segments $BC$, $CA$, $AB$ respectively. $\angle B'A'C' = \angle A$ and $\dfrac{AC'}{C'B} = \dfrac{BA'}{A'C} = \dfrac{CB'}{B'A}$. Show that $ABC$ and $A'B'C'$ are similar. | [] | Soviet Union | 23rd ASU | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
0ejm | Problem:
Naj bosta $a$ in $b$ taki realni števili, da ima polinom $p(x)=x^{2}+a x+b$ dve realni ničli, polinom $p\left(q(x)\right)$, kjer je $q(x)=x^{2}+2 x+7$, pa nima realnih ničel. Dokaži, da je $p(8)>4$. | [
"Solution:\n\n1. Naj bosta $x_{1}$ in $x_{2}$ realni ničli polinoma $p(x)$. Tedaj je $p(x)=\\left(x-x_{1}\\right)\\left(x-x_{2}\\right)$ in\n$$\np(q(x))=\\left(q(x)-x_{1}\\right)\\left(q(x)-x_{2}\\right)=\\left(x^{2}+2 x+\\left(7-x_{1}\\right)\\right)\\left(x^{2}+2 x+\\left(7-x_{2}\\right)\\right)\n$$\nPolinom $p(q... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
04yt | Prove the inequality
$$
2010 < \frac{2^2 + 1}{2^2 - 1} + \frac{3^2 + 1}{3^2 - 1} + \dots + \frac{2010^2 + 1}{2010^2 - 1} < 2010 \frac{1}{2} \quad \text{(Grade 9.)}
$$ | [
"Since $\\frac{n^2 + 1}{(n-1)(n+1)} = 1 + \\frac{1}{n-1} - \\frac{1}{n+1}$, the given sum can be rewritten in the form $1 + \\frac{1}{1} - \\frac{1}{3} + 1 + \\frac{1}{2} - \\frac{1}{4} + \\dots + 1 + \\frac{1}{2009} - \\frac{1}{2011}$.\n\n$= 2010 + \\frac{1}{2} - \\frac{1}{2010} - \\frac{1}{2011}$.\n\nBecause $0 <... | Estonia | Estonija 2010 | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0h1f | It is known, that $x_1, x_2, x_3$ are distinct real numbers.
a) $x_2, x_3$ are zeros of the function $f_1(x) = x^2 + p_1x + q_1$; $x_3, x_1$ are zeros of the function $f_2(x) = x^2 + p_2x + q_2$; $x_1, x_2$ are zeros of the function $f_3(x) = x^2 + p_3x + q_3$. Does the function $f(x) = f_1(x) + f_2(x) + f_3(x)$ alway... | [
"a) We write our function in the form\n$$\nf(x) = (x - x_2)(x - x_3) + (x - x_1)(x - x_3) + (x - x_1)(x - x_2),\n$$\nWLOG, $x_1 < x_2 < x_3$. Then $f(x_2) = (x_2 - x_1)(x_2 - x_3) < 0$, which is equivalent to the existence of the roots of the function.\n\n\nb) Consider the following three functions:\n$$\nf_1(x) = x... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | English | proof only | null | |
0fj1 | Problem:
Sea $ABCD$ un cuadrilátero inscrito en una circunferencia de radio $1$ de modo que $AB$ es un diámetro y el cuadrilátero admite circunferencia inscrita. Probar que $CD \leq 2\sqrt{5}-4$. | [
"Solution:\n\n\n\nSea $O$ el centro de la semicircunferencia. Pongamos $a = BC$; $b = AD$; $p = CD$; $2\\alpha = \\widehat{BOC}$; $2\\beta = \\widehat{AOD}$; $2\\gamma = \\widehat{COD}$.\n\nLa condición necesaria y suficiente para que $ABCD$ admita una circunferencia inscrita es\n$$\np + 2 ... | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geo... | null | proof only | null | |
02aq | Problem:
Circunferência e triângulo retângulo - Inscreve-se uma circunferência num triângulo retângulo. $O$ ponto de tangência divide a hipotenusa em dois segmentos de comprimentos $6~\mathrm{cm}$ e $7~\mathrm{cm}$. Calcule a área do triângulo. | [
"Solution:\n\nSeja $r$ o raio da circunferência inscrita. Usando o teorema de Pitágoras temos que $(6+7)^2 = (r+6)^2 + (r+7)^2 = r^2 + 12r + 36 + r^2 + 14r + 49 = 2(r^2 + 13r) + 85$, assim temos que $r^2 + 13r = \\frac{169 - 85}{2} = 42$.\n\nPor outro lado, a área do triângulo é\n$$\n\\frac{(r+6)(r+7)}{2} = \\frac{... | Brazil | null | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | 42 | |
0h2n | Let $P(x)$, $Q(x)$ be polynomials with real coefficients such that $P(0) > 0$ and all coefficients of the polynomial $S(x) = P(x)Q(x)$ are non-negative. Prove that for any positive $x$ the following inequality holds:
$$
S(x^2) - S^2(x) \le \frac{1}{4}(P^2(x^3) + Q(x^3)).
$$ | [
"If $S = 0$, then $Q = 0$, and the inequality is evident. Suppose now that $S$ is not identically zero. Then $\\forall x > 0$ $S(x) > 0$. If for some $y > 0$ $P(y) < 0$, then the polynomial $P$, and so the polynomial $S$, have roots on the interval $(0, y)$, which is impossible. So, $P$ and $Q$ are positive for $x ... | Ukraine | Problems of Ukrainian Authors | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0fot | Problem:
Sea $n \geq 2$ un entero positivo. Tenemos $2 n$ bolas, en cada una de las cuales hay escrito un entero. Se cumple que, siempre que formamos $n$ parejas con las bolas, dos de estas parejas tienen la misma suma.
(1) Demuestra que hay cuatro bolas con el mismo número.
(2) Demuestra que el número de valores di... | [
"Solution:\n\n(1) Sean los valores de las bolas, en orden no creciente, $a_{1} \\geq a_{2} \\geq \\cdots \\geq a_{2 n}$. Formemos la pareja $k$-ésima emparejando la bola $a_{2 k-1}$ con la bola $a_{2 k}$ para $k=1,2, \\ldots, n$, con lo que sus sumas son\n$$\ns_{1}=a_{1}+a_{2} \\geq s_{2}=a_{3}+a_{4} \\geq \\cdots ... | Spain | null | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
02az | Problem:
Cercando o Globo Terrestre - O raio do Globo Terrestre é aproximadamente $6670~\mathrm{km}$. Suponhamos que um fio esteja ajustado exatamente sobre o Equador, que é um círculo de raio aproximadamente igual a $6670~\mathrm{km}$.
Em seguida, suponhamos que o comprimento do fio seja a... | [
"Solution:\n\nCercando o Globo Terrestre - Como o raio da Terra é muito grande, e foi dado apenas um acréscimo de $1~\\mathrm{m}$ no comprimento do fio, parece que a folga entre o fio e o Equador é muito pequena. Mais ainda, se trocarmos o Globo Terrestre por Júpiter ou por uma bolinha de gude e realizarmos esta me... | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Circles"
] | null | proof and answer | Gap height = 1/(2π) meters ≈ 0.16 m (about 16 cm); only the ant can pass under. | |
0e24 | Problem:
Če med števki dvomestnega števila vrinemo ničlo, dobimo devetkrat večje število. Zapiši vsa takšna dvomestna števila. | [
"Solution:\n\nNaj bosta $a$ in $b$ števki iskanega dvomestnega števila. Nastavimo enačbo $100a + b = 9(10a + b)$. Odpravimo oklepaj in dobimo zvezo $4b = 5a$. Upoštevamo, da sta $a$ in $b$ števki, kar pomeni, da je edina možna rešitev $a = 4$ in $b = 5$. Iskano število je $45$."
] | Slovenia | 10. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 45 | |
0hyn | Problem:
King Arthur and his 100 knights are having a feast at a round table. Each person has a glass with white or red wine in front of him. Exactly at midnight, every person moves his glass in front his left neighbor if the glass has white wine, or in front of his right neighbor if the glass has red wine. It is know... | [
"Solution:\n\na. Suppose not. Label all seats as $1,2, \\ldots, 101$ by going around the table anticlockwise. If each person still has a glass after midnight, then no one has received glasses from both of his neighbors, i.e. any two people with exactly one person between must have the same color wine. Thus, the fol... | United States | BAMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | a) There will be at least one person without a glass after the move. b) No; with an even number of people an alternating assignment of colors lets everyone still have a glass. | |
0jbw | Problem:
Let $\pi$ be a permutation of the numbers from $1$ through $2012$. What is the maximum possible number of integers $n$ with $1 \leq n \leq 2011$ such that $\pi(n)$ divides $\pi(n+1)$? | [
"Solution:\nAnswer: $1006$\n\nSince any proper divisor of $n$ must be less than or equal to $n / 2$, none of the numbers greater than $1006$ can divide any other number less than or equal to $2012$. Since there are at most $1006$ values of $n$ for which $\\pi(n) \\leq 1006$, this means that there can be at most $10... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 1006 | |
0grf | Find all surjective functions $f: \mathbb{R} \to \mathbb{R}$ satisfying
$$
f(xf(y) + y^2) = f((x+y)^2) - x f(x)
$$
for all real numbers $x, y$. | [
"$f(x) = 2x$, $x \\in \\mathbb{R}$ is the only solution. Assume that there exists a real number $c \\neq 0$ such that $c = f(z) - 2z$ for some $z$. Let $P(x, y)$ be the assertion of the given equation.\n$$\nP(f(y) - 2y, y) \\rightarrow (f(y) - 2y) f(f(y) - 2y) = 0, \\forall y.\n$$\nThis shows that $f(c) = 0$. Since... | Turkey | Team Selection Test | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | f(x) = 2x for all real x | |
05y2 | Problem:
Soit $ABC$ un triangle et $\Gamma$ son cercle circonscrit. Soit $A_{1}$ le milieu de l'arc $\widehat{BC}$ ne contenant pas $A$ ; $B_{1}$ le milieu de l'arc $\overparen{CA}$ ne contenant pas $B$ ; $C_{1}$ le milieu de l'arc $\overparen{AB}$ ne contenant pas $C$. Enfin, soit $A_{2}$ le point pour lequel $AB_{1}... | [
"Solution:\n\n\n\nOn cherche tout d'abord à démontrer que $OA_{2} = OB_{2}$. Puisque $A_{1}B_{2} = BC_{1} = AC_{1} = B_{1}A_{2}$ et que $A_{1}O = B_{1}O$, l'égalité $OA_{2} = OB_{2}$ équivaut au fait que les triangles $OA_{1}B_{2}$ et $OB_{1}A_{2}$ soient isométriques l'un de l'autre, c'est... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocent... | null | proof only | null | |
0h5o | Is it possible to construct a triangle with sides $x$, $y$, $z$ satisfying the condition:
$$
3x^2y^2 + 3y^2z^2 + 3z^2x^2 = x^4 + y^4 + z^4?
$$ | [
"Rewrite the equation as:\n$$\n2x^2y^2 + 2y^2z^2 + 2z^2x^2 - x^4 - y^4 - z^4 = -(x^2y^2 + y^2z^2 + z^2x^2).\n$$\nThe left-hand side can be decomposed as:\n$$(x+y+z)(x+y-z)(y+z-x)(z+x-y) = -(x^2y^2 + y^2z^2 + z^2x^2).$$\nHence, the left-hand side is negative, therefore, at least one of the multipliers is negative to... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | No | |
0ffh | Problem:
Es bien sabido que si $\frac{p}{q} = \frac{r}{s}$, ambas razones son iguales a $\frac{p - r}{q - s}$. Escribimos ahora la igualdad
$$
\frac{3x - b}{3x - 5b} = \frac{3a - 4b}{3a - 8b}
$$
Por la propiedad anterior, ambas fracciones deben ser iguales a
$$
\frac{3x - 5b - 3a + 8b}{3x - b - 3a + 4b} = \frac{3x - 3... | [
"Solution:\n\nSucede que operando en la hipótesis tenemos\n$$\n\\frac{3x - b}{3x - 5b} = \\frac{3a - 4b}{3a - 8b} \\Longleftrightarrow b(x - a + b) = 0\n$$\ny por tanto para que se cumpla la hipótesis se debe dar necesariamente uno de estos casos:\n\na) $b = 0$. Entonces la hipótesis se reduce a $\\frac{3x}{3x} = \... | Spain | Olimpiadas Matemáticas Españolas | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof only | null | |
0czj | Find all integers $n \geq 2$ for which $\sqrt[n]{3^{n}+4^{n}+5^{n}+8^{n}+10^{n}}$ is an integer. | [
"We have $3^{3}+4^{3}+5^{3}+8^{3}+10^{3}=12^{3}$, that is $n=3$ satisfies the property. We shall prove that this is the unique solution. Consider the function $h:[2, \\infty) \\rightarrow \\mathbb{R}$,\n$$\nh(x)=\\left(\\frac{3}{12}\\right)^{x}+\\left(\\frac{4}{12}\\right)^{x}+\\left(\\frac{5}{12}\\right)^{x}+\\lef... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | 3 | |
0ent | Find all functions $f: \mathbb{Z} \to \mathbb{Z}$ such that, for all integers $a$, $b$, and $c$ satisfying $a + b + c = 0$, the following equality holds:
$$
f(a)^2 + f(b)^2 + f(c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).
$$ | [
"Substituting $a = b = c = 0$ yields $3f(0)^2 = 6f(0)^2$ which implies $f(0) = 0$. Now we can place $b = -a$, $c = 0$ to obtain $f(a)^2 + f(-a)^2 = 2f(a)f(-a)$, or, equivalently $(f(a) - f(-a))^2 = 0$ which implies $f(a) = f(-a)$.\n\nAssume now that $f(a) = 0$ for some $a \\in \\mathbb{Z}$. Then for any $b$ we have... | South Africa | International Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All such functions are exactly the following families, where c is any integer: (1) f(x) = c x^2 for all integers x; (2) f(x) = 0 for even x and f(x) = c for odd x; (3) f(x) = 0 for x divisible by four, f(x) = c for odd x, and f(x) = 4c for x congruent to two modulo four. | |
0j7s | Problem:
Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep a running tally of the sum of the results of all rolls made. A player wins if, after he rolls, the number on the running tally is a multiple of $7$. Play continues until either player wins, or else indefinitely. I... | [
"Solution:\n\nAnswer: $\\frac{5}{11}$\n\nFor $1 \\leq k \\leq 6$, let $x_{k}$ be the probability that the current player, say $A$, will win when the number on the tally at the beginning of his turn is $k$ modulo $7$. The probability that the total is $l$ modulo $7$ after his roll is $\\frac{1}{6}$ for each $l \\not... | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 5/11 | |
0exp | Problem:
There are an odd number of soldiers on an exercise. The distance between every pair of soldiers is different. Each soldier watches his nearest neighbour. Prove that at least one soldier is not being watched. | [
"Solution:\n\nThe key is to notice that no loops of size greater than two are possible. For suppose we have $A_1$, $A_2$, ..., $A_n$ with $A_i$ watching $A_{i + 1}$ for $0 < i < n$, and $A_n$ watching $A_1$. Then the distance $A_iA_i$ is greater than the distance $A_iA_{i + 1}$ for $1 < i < n$, and the distance $A_... | Soviet Union | 6th ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Other"
] | null | proof only | null | |
09hr | Let $n \ge 3$ be fixed. A sequence $a_1, a_2, \dots, a_n$ of real numbers is *nice* if
$$
0 \le a_1 + \dots + a_{k-1} + a_{k+1} + \dots + a_n \le 1
$$
for all $1 \le k \le n$. Let $m = \min\{a_1, a_2, \dots, a_n\}$ denote the minimum and let $M = \max\{a_1, a_2, \dots, a_n\}$ denote the maximum of the sequence $a_1, a_... | [
"Answer: (i) $\\max M = 1$, (ii) $\\min m = -\\frac{n-2}{n-1}$.\n\ni. For $(a_1, a_2, \\dots, a_n) = (0, \\dots, 0, 1)$, we have $M = 1$.\nNow we show $M \\le 1$ holds always. Suppose, on the contrary, that for some nice sequence $a_1, a_2, \\dots, a_n$, we have $M > 1$. Let $T = (a_1 + \\dots + a_n) - m - M$, then... | Mongolia | Round 3 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | proof and answer | (i) 1; (ii) -(n-2)/(n-1) | |
0duj | Problem:
Naj bo $k$ tako naravno število, da ima kvadratna enačba
$$
k x^{2}-(1-2 k) x+k-2=0
$$
racionalni rešitvi. Dokaži, da je $k$ zmnožek 2 zaporednih celih števil. | [
"Solution:\n\nDa bosta rešitvi kvadratne enačbe racionalni, mora biti diskriminanta $(1-2 k)^{2}-4 k(k-2)=1+4 k$ enaka $m^{2}$, $m \\in \\mathbb{N}$. Torej je\n$$\nk=\\frac{(m-1)(m+1)}{4}.\n$$\nKer je $k$ naravno število, mora biti $m$ liho število, večje od $1$, torej obstaja tako naravno število $n$, da je $m=2 n... | Slovenia | 46. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0df3 | We call a positive integer $n$ venerable if all its positive divisors less than $n$ (but including 1) add up to $n - 1$. Find all venerable numbers whose some power (with the exponent at least 2) is also venerable. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | All powers of two: n = 2^k for integer k ≥ 0. | |
06i7 | A real number is put inside each cell of an $n \times n$ table. Each time we may add a real number $x$ to each of the cells of a single row or a single column, where the real number $x$ may vary at each time. Find the maximum $k$ for which it is always possible to make $k$ of the cells zeros simultaneously after a fini... | [
"The maximum $k$ is $2n-1$.\n\nFirstly, we show that $k \\ge 2n-1$. Suppose the first entry in a row is $a$. By adding $-a$ to this row, we make the first entry 0. Similarly, we do this for each row until the first entry of each row is 0. Next, suppose the first entry in a column is $b$. By adding $-b$ to this colu... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2n-1 | |
0gwg | Eleven linguists were instructed to learn eleven foreign languages (initially, none of the linguists knew any of those languages). It became necessary to invite a Foreign Consultant who is able to teach (by means of hypnosis, of course!) any two linguists any two languages during one session (so that each one of those ... | [
"Кожен лінгвіст має взяти участь щонайменше в $6$ сеансах, інакше він не оволодіє всіма $11$ мовами. Оскільки в одному сеансі беруть участь два лінгвісти, то кількість сеансів не менша за $33 = \\frac{6 \\cdot 11}{2}$. Покажемо, що $33$ сеансів Консультантові насправді вистачить. Кожний сеанс будемо зображати у виг... | Ukraine | Ukrainian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 33 | |
0emw | Let $a$ and $b$ be two integers with $a > b$. If $ab - 1$ and $a+b$ are relatively prime, and $ab + 1$ and $a-b$ are relatively prime, prove that
$$
(ab + 1)^2 + (a - b)^2
$$
is not a perfect square. | [
"We know that\n$$\n(a-b)^2 + (ab+1)^2 = a^2b^2 + a^2 + b^2 + 1 = (a^2+1)(b^2+1),\n$$\nso it will be enough to show that $a^2+1$ and $b^2+1$ are relatively prime. Suppose that there is a prime number $p$ that divides both $a^2+1$ and $b^2+1$. Then $p$ divides $a^2-b^2$, and hence $p$ divides one of $a+b$ or $a-b$.\n... | South Africa | South-Afrika 2011-2013 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0hex | In the triangle $ABC$ $\angle A=75^\circ$ and $\angle C=45^\circ$. Points $P$ and $T$ are chosen on the segments $AB$ and $BC$ in such a way that quadrilateral $APTC$ is cyclic and $CT = 2AP$. Point $O$ is the circumcenter of $\triangle ABC$. The ray $TO$ crosses side $AC$ in a point $K$. Prove that $TO=OK$.
(Anton Tr... | [
"Let $CD$ be a diameter of $(ABC)$. Then $\\triangle ADC$ is a right triangle with an angle $60^\\circ$. Hence, $CD = 2AD$ (fig. 30) and $\\triangle DTC \\sim \\triangle DPA$ by two proportional sides and equal included angles. Therefore, $\\angle BPD = \\angle BTD$ and $PDBT$ is cyclic quadrilateral. We point out ... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
01ml | Given triangle $ABC$ with $\angle ACB = 120^\circ$. Point $L$ is marked on the side $AB$ so that $CL$ is the bisector of $\angle ACB$. Points $N$ and $K$ are marked on the sides $AC$ and $BC$, respectively, so that $CN + CK = CL$.
Prove that the triangle $KLN$ is equilateral. | [] | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ha4 | Capablanca and Alyokhin decided to play a match of 16 games according to the following rules. The winner of the first game received $1 = 3^0$ peso, the winner of the second one got $3 = 3^1$ peso, the winner of the third game received $9 = 3^2$ peso and so on.
If the game ended in a draw, then they split the prize poo... | [
"Without loss of generality, we can say that after each draw they both got 0 peso. Then Alyokhin in $k^{th}$ game could get $a_k \\cdot 3^{k-1}$, where $a_k \\in \\{-1; 0; 1\\}$. Let us show that his gain after the $k^{th}$ game could be from $A_k = 3^0 + 3^1 + \\dots + 3^{k-1}$ to $(-A_k)$. Moreover, each possible... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Number Theory > Other"
] | English | proof and answer | Each player won 3 games. | |
0bum | Problem:
Determinați numerele naturale $n$ știind că fracția $\frac{3 n+1}{2 n-7}$ este reductibilă. | [
"Solution:\nDacă fracția este reductibilă atunci există $d \\neq 1$ astfel încât $d \\mid 3 n+1$ și $d \\mid 2 n-7$. De aici avem $d \\mid 2(3 n+1)-3(2 n-7)$, adică $d \\mid 23$, prin urmare $d=23$. Acum $23 \\mid 3 n+1$ și $23 \\mid 2 n-7$ deducem că $23 \\mid n+8$, adică $n+8=23 k$, pentru $k \\in \\mathbb{N}^{*}... | Romania | Olimpiada de Matematică - Etapa Locală | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | All n with n ≡ 15 (mod 23), equivalently n = 23k − 8 for k a positive integer. | |
0668 | If $x$, $y$, $z$ are positive real numbers with sum $12$, prove that:
$$
\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + 3 \ge \sqrt{x} + \sqrt{y} + \sqrt{z}.
$$
**When is equality valid?** | [
"Since $x$, $y$, $z$ are positive integers with sum $12$, it is enough to prove that\n$$\n\\frac{x}{y} + \\frac{y}{z} + \\frac{z}{x} + \\frac{x+y+z}{4} \\geq \\sqrt{x} + \\sqrt{y} + \\sqrt{z}. \\quad (1)\n$$\nFrom the inequality of the arithmetic–geometric mean for the positive integers $x$, $y$, $z$ we get\n\n$$\n... | Greece | 28th Hellenic Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Equality holds when x = y = z = 4. | |
01ti | Let $a$ and $b$ be positive integers such that $a!b!$ is a multiple of $a! + b!$.
Prove that $3a \ge 2b + 2$. | [
"3. See IMO-2015 Shortlist, Problem N2."
] | Belarus | 66th Belarusian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof only | null | |
03rp | Find all ordered triples $(x, y, z)$ of real numbers such that
$$
\begin{cases} 5\left(x+\frac{1}{x}\right) = 12\left(y+\frac{1}{y}\right) = 13\left(z+\frac{1}{z}\right), \\ xy + yz + zx = 1. \end{cases} \quad \text{(posed by Zhu Huawei)}
$$ | [
"There are angles $A, B$, and $C$ in the interval $(0^\\circ, 180^\\circ)$ such that\n$$\nx = \\tan \\frac{A}{2}, \\quad y = \\tan \\frac{B}{2}, \\quad z = \\tan \\frac{C}{2}.\n$$\nBy the addition and subtraction formulas, the second equation in the given system becomes\n$$\n\\begin{aligned}\n1 &= \\tan \\frac{A}{2... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | (1/5, 2/3, 1) and (-1/5, -2/3, -1) | |
00ix | We consider permutations $f$ on the set $N$ of non-negative integers, i.e. bijective mappings $f$ from $N$ to $N$, with the following properties:
For all $n \in N$, we have $f(f(x)) = x$ and $|f(x) - x| \le 3$.
Furthermore, for all integers $n > 42$, we have
$$
M(n) = \frac{1}{n+1} \sum_{j=0}^{n} |f(j) - j| < 2.011
$$
... | [
"If an infinite number of such $K$ do not exist, there must exist some $K_o$, such that for all $K > K_o$ there exists an $n$ with $n \\le K < f(n)$. Since $|f(x) - x| \\le 3$ must always hold, such an $n$ can only be $K$, $K-1$ or $K-2$.\n\nThe same must hold for $K = f(n)$, and so on. This means that, from some $... | Austria | AustriaMO2011 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | English | proof only | null | |
05p3 | Problem:
On dit qu'un entier naturel $d$ est sympathique si, pour tout couple d'entiers $(x, y)$,
$$
d\left|(x+y)^{5}-x^{5}-y^{5} \Longleftrightarrow d\right|(x+y)^{7}-x^{7}-y^{7}
$$
Montrer qu'il existe une infinité de nombres sympathiques. Est-ce-que 2017, 2018 sont sympathiques ? | [
"Solution:\n\nOn remarque que :\n$$\n\\begin{aligned}\n& (x+y)^{5}-x^{5}-y^{5}=5 x y(x+y)\\left(x^{2}+x y+y^{2}\\right) \\\\\n& (x+y)^{7}-x^{7}-y^{7}=7 x y(x+y)\\left(x^{2}+x y+y^{2}\\right)^{2}\n\\end{aligned}\n$$\nAinsi, si on prend $p$ un nombre premier différent de $5$ et de $7$ :\n- si $p$ divise $(x+y)^{5}-x^... | France | OCympiades Françaises de Mathématiques | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | There are infinitely many: every prime distinct from five and seven is friendly (and any product of coprime friendly numbers is also friendly). Both 2017 and 2018 are friendly. | |
0ba4 | Let $n$ be an integer number greater than $2$, let $x_1, x_2, \dots, x_n$ be $n$ positive real numbers such that
$$
\sum_{i=1}^{n} \frac{1}{x_i + 1} = 1,
$$
and let $\alpha$ be a real number greater than $1$. Show that
$$
\sum_{i=1}^{n} \frac{1}{x_i^{\alpha} + 1} \geq \frac{n}{(n-1)^{\alpha} + 1}
$$ | [
"Let $y_i = 1/(x_i+1)$, $i = 1, 2, \\dots, n$, so the $y_i$ are positive real numbers that add up to $1$. Upon substitution, the left-hand member of the required inequality becomes\n$$\n\\sum_{i=1}^{n} \\frac{y_i^{\\alpha}}{(1 - y_i)^{\\alpha} + y_i^{\\alpha}} = \\sum_{i=1}^{n} \\frac{y_i^{\\alpha}}{\\left(\\sum_{j... | Romania | 62nd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
0cfz | Let $n \ge 3$ be a positive integer. Consider in the complex plane a regular polygon $P_1P_2\dots P_n$ inscribed in the unit circle, and let $A$ be the set of affixes of the vertices of the polygon. Let $a_1, a_2, \dots, a_{n-1}$ be complex numbers with the property:
$$
|z^{n-1} + a_1z^{n-2} + a_2z^{n-3} + \dots + a_{n... | [] | Romania | 74th NMO Shortlisted Problems | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | English | proof only | null | |
01c2 | Let $T(a)$ be the sum of digits of $a$. For which $R \in \mathbb{N}$ does there exist an $n \in \mathbb{N}$ such that $\frac{T(n^2)}{T(n)} = R$? | [
"Let $R \\in \\mathbb{N}$ and consider the number\n$$\nN = \\sum_{k=0}^{R-1} 10^{2^k}.\n$$\nWe see that $T(N) = R$. Now\n$$\nN^2 = \\left(\\sum_{k=0}^{R-1} 10^{2^k}\\right)^2 = \\sum_{0 \\le a, b < R} 10^{2^a + 2^b},\n$$\nand since $2^a + 2^b = 2^c + 2^d$ if and only if $(a, b) = (c, d)$ or $(a, b) = (d, c)$, there... | Baltic Way | Baltic Way | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All positive integers | |
0b7i | Given a positive integer $a$, prove that $\sigma(am) < \sigma(am + 1)$ for infinitely many positive integers $m$. (Here $\sigma(n)$ is the sum of all positive divisors of the positive integer number $n$.)
Vlad Matei | [
"Given an integer $N > a$, we claim that there exist an integer $d$ and a prime $p$, both greater than $N$, such that $d$ divides $ap+1$, $d$ and $(ap+1)/d$ are coprime, and $\\sigma(d)/d > \\sigma(a)$. In this case,\n$$\n\\sigma(ap + 1) = \\sigma\\left(\\frac{ap+1}{d} \\cdot d\\right) = \\sigma\\left(\\frac{ap+1}{... | Romania | NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Other"
] | English | proof only | null | |
0ehp | Problem:
Dani sta funkciji $f(x)=\log_{3}(x+3)+1$ in $g(x)=\log_{3}(3x)+1$.
a) Poišči absciso presečišča danih funkcij.
b) Izračunaj ničlo, začetno vrednost in zapiši enačbo asimptote za obe funkciji. Funkciji nariši v isti koordinatni sistem. | [
"Solution:\n\na) Poiščemo absciso presečišča. Izenačimo funkciji $f(x)=g(x)$, ju vstavimo v enačbo in dobimo $\\log_{3}(x+3)+1=\\log_{3}(3x)+1$. Enačbo uredimo in dobimo linearno enačbo $x+3=3x$, katere rešitev je $x=\\frac{3}{2}$.\n\nb) Za funkcijo $f(x)=\\log_{3}(x+3)+1$ izračunamo začetno vrednost $f(0)=\\log_{3... | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Odbirno tekmovanje | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | Intersection abscissa: x = 3/2. For f: zero at x = −8/3, value at zero is 2, vertical asymptote x = −3. For g: zero at x = 1/9, value at zero does not exist, vertical asymptote x = 0. | |
0j1u | Problem:
A zerg player can produce one zergling every minute and a protoss player can produce one zealot every $2.1$ minutes. Both players begin building their respective units immediately from the beginning of the game. In a fight, a zergling army overpowers a zealot army if the ratio of zerglings to zealots is more t... | [
"Solution:\nAt the end of the first minute, the zerg player produces a zergling and has a superior army for the $1.1$ minutes before the protoss player produces the first zealot. At this point, the zealot is at least a match for the zerglings until the fourth is produced $4$ minutes into the game. Then, the zerg ar... | United States | Harvard-MIT November Tournament | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | 1.3 | |
04sz | In the triangle $ABC$ denote $D$ point of contact of side $BC$ with the incircle. The incircle of the triangle $ABD$ is tangent to sides $AB$ and $BD$ at points $K$ and $L$. The incircle of the triangle $ADC$ is tangent to sides $DC$ and $AC$ at points $M$ and $N$. Prove that points $K, L, M, N$ lie on the same circle. | [] | Czech Republic | Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homot... | English | proof only | null | |
0dye | Problem:
V nekem podjetju je zaposlenih 150 ljudi. Direktor prejema mesečno plačo 12000 evrov, trije ožji sodelavci 5000 evrov, 12 najslabše plačanih delavcev dobi 500 evrov, preostali delavci zaslužijo bodisi 1500 bodisi polovico tega zneska. Koliko zaposlenih zasluži mesečno 1500 evrov in koliko polovico manj, če je... | [
"Solution:\n\nNaj bo $x$ število zaposlenih, ki zaslužijo $1500$ evrov, in $y$ število zaposlenih, ki zaslužijo $750$ evrov.\n\nZapišemo enačbi:\n\n$1 + 3 + 12 + x + y = 150$\n\n$\\frac{12000 + 3 \\cdot 5000 + 12 \\cdot 500 + x \\cdot 1500 + y \\cdot 750}{150} = 1010$\n\nEnačbi uredimo in dobimo sistem:\n\n$x + y =... | Slovenia | Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 24 employees earn 1500 euros and 110 employees earn 750 euros. | |
032a | Problem:
Is there a set $A \supset \{1,2, \ldots, 2004\}$ of positive integers such that the product of its elements is equal to the sum of their squares? | [
"Solution:\nThere exists. Let us take $a_{0}=1$, $a_{i}=2004!a_{0} a_{1} \\ldots a_{i-1}-1$, $i \\geq 1$ and $A_{i}=\\{2,3, \\ldots, 2004, a_{0}, a_{1}, \\ldots, a_{i}\\}$, $i \\geq 0$. Then\n$$\n\\begin{gathered}\n\\left(\\prod_{a \\in A_{i-1}} a-\\sum_{a \\in A_{i-1}} a^{2}\\right)-\\left(\\prod_{a \\in A_{i}} a-... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
019x | A capitalist returns from a business trip and brings $n$ gifts for his $n$ children. For $i \in \{1, 2, \dots, n\}$, his $i$-th oldest child considers $x_i$ of these items to be desirable. Assume that the numbers $x_1, \dots, x_n$ are positive and satisfy
$$
\frac{1}{x_1} + \dots + \frac{1}{x_n} \le 1.
$$
Prove that th... | [
"Evidently the age of the children is immaterial, so we may suppose\n$$\n1 \\le x_1 \\le x_2 \\le \\dots \\le x_n.\n$$\nLet us now consider the following procedure. First the oldest child chooses its favourite present and keeps it, then the second oldest child chooses its favourite remaining present, and so it goes... | Baltic Way | Baltic Way 2013 | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
0hrt | Problem:
A sequence $a_{1}, a_{2}, \ldots$ of positive integers satisfies
$$
a_{n+1}=a_{n}^{3}+103
$$
for every positive integer $n$. Prove that the sequence contains at most one perfect square. | [
"Solution:\nIt's easy to check that no two consecutive terms can be perfect squares, since the only squares which differ by $103$ are $51^{2}$ and $52^{2}$.\n\nNow, note that squares are $0$, $1$, or $4$ mod $8$. After a perfect square appears, the next term must be $-1$ or $0 \\bmod 8$, and thereafter all terms ar... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
06bx | Start with some positive integer. The following operation is performed on the number: its unit digit is split off and multiplied by $4$, then this product is added to the remaining number. (For example, $1997$ is changed to $7 \times 4 + 199 = 227$.) The operation is performed again and again. Prove that if the sequenc... | [
"If the current number is $10a + b$ where $0 \\le b \\le 9$, then the next number is $a + 4b$. Note that\n$$\n10(a + 4b) = (10a + b) + 39b.\n$$\nTherefore, $13 \\mid a + 4b$ if and only if $13 \\mid 10a + b$. Since $13 \\mid 1001$, all the numbers in the sequence are divisible by $13$. The only possible prime numbe... | Hong Kong | IMO HK TST | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
064y | In the triangle $ABC$, the angle $\alpha = \hat{A}$ and the side $a = |BC|$ are given. It is known that $a = \sqrt{rR}$, where $r$ is the inradius and $R$ is the circumradius. Determine all such triangles, that is, compute the sides $b$ and $c$ of all such triangles. | [
"According to the cosine rule we have\n$$\nb^2 + c^2 - 2bc \\cos A = a^2. \\quad (1)\n$$\nIf $\\tau = \\frac{a+b+c}{2}$ the given condition can be written as\n$$\na^2 = rR = \\frac{(ABC)}{\\tau} \\cdot \\frac{a}{\\sin A} = \\frac{bc \\sin A}{(a+b+c) \\sin A} \\Leftrightarrow a^2 = \\frac{abc}{a+b+c} \\\\\n\\Leftrig... | Greece | Mediterranean Mathematical Competition | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof and answer | b, c = (a/2) * [5 + 4 cos A ± sqrt(16 cos^2 A + 8 cos A − 23)], with the feasibility condition 16 cos^2 A + 8 cos A − 23 ≥ 0, i.e., cos A ≥ (√24 − 1)/4, so 0 < A ≤ arccos((√24 − 1)/4). | |
0jfz | Problem:
For how many unordered sets $\{a, b, c, d\}$ of positive integers, none of which exceed $168$, do there exist integers $w, x, y, z$ such that $(-1)^{w} a + (-1)^{x} b + (-1)^{y} c + (-1)^{z} d = 168$? If your answer is $A$ and the correct answer is $C$, then your score on this problem will be $\left\lfloor 25... | [
"Solution:\n\nAnswer: 761474\n\nAs an approximation, we assume $a, b, c, d$ are ordered to begin with (so we have to divide by $24$ later) and add to $168$ with a unique choice of signs; then, it suffices to count $e + f + g + h = 168$ with each $e, f, g, h$ in $[-168, 168]$ and then divide by $24$ (we drop the con... | United States | HMMT 2013 | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 761474 | |
01us | 1. The extension of the median *AM* of the triangle *ABC* intersects its circumcircle at *D*. The circumcircle of the triangle *CMD* intersects the line *AC* at *C* and *E*. The circumcircle of the triangle *AME* intersects the line *AB* at *A* and *F*.
Prove that *CF* is an altitude of the triangle *ABC*. | [
"**1.** It is enough to prove the equalities $FM = BM = MC$ from which it will follow\n\nthat $M$ is the midpoint of the hypotenuse of the right triangle $CFB$ and in particular $\\angle CFB = 90^\\circ$.\n\n\n\nSince the quadrilateral $ABDC$ is cyclic, it follows t... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
0dn3 | Problem:
Наћи све природне бројеве $n$ за које постоји пермутација $(p_{1}, p_{2}, \ldots, p_{n})$ бројева $(1,2, \ldots, n)$ таква да скупови $\{p_{i}+i \mid 1 \leqslant i \leqslant n\}$ и $\{p_{i}-i \mid 1 \leqslant i \leqslant n\}$ чине потпуне системе остатака по модулу $n$. (Марко Ђикић) | [
"Solution:\n\nПретпоставимо да таква пермутација постоји. Како је $\\{p_{i}+i \\mid 1 \\leq i \\leq n\\}$ потпун систем остатака по модулу $n$, важи $\\sum_{k=1}^{n} k \\equiv \\sum_{i=1}^{n}(p_{i}+i) \\equiv \\sum_{i=1}^{n} i+\\sum_{i=1}^{n} p_{i} \\equiv 2 \\sum_{k=1}^{n} k\\ (\\bmod\\ n)$, дакле $\\sum_{k=1}^{n}... | Serbia | Serbian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | All natural numbers n with gcd(n, 6) = 1 | |
00cs | En un tablero de $9 \times 9$ hay que colorear de rojo algunas casillas, por lo menos una. Para cada coloración, sea $P$ la cantidad de casillas, coloreadas o no, que tienen un número par de casillas vecinas rojas. (Dos casillas son vecinas si tienen un lado común.) Dar una coloración del tablero que tenga el menor val... | [] | Argentina | Nacional OMA 2019 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Linear Algebra > Matrices"
] | Spanish | proof and answer | 1 | |
0kps | Problem:
Compute the number of ways to color 3 cells in a $3 \times 3$ grid so that no two colored cells share an edge. | [
"Solution:\n\nIf the middle square is colored, then two of the four corner squares must be colored, and there are $\\binom{4}{2} = 6$ ways to do this.\n\nIf the middle square is not colored, then after coloring one of the 8 other squares, there are always 6 ways to place the other two squares. However, the number o... | United States | HMMT February 2022 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | final answer only | 22 | |
0dhu | In an acute-angled triangle $ABC$, point $O$ is the circumcenter and $H$ the orthocenter. Points $B'$ and $C'$ are the reflections of $B$ in line $AC$ and of $C$ in line $AB$, respectively. Point $K$ is the circumcenter of triangle $HB'C'$, point $D$ the midpoint of $KB$, and $S$ the intersection of line $OD$ with the ... | [
"Let $\\omega, \\omega_c, \\omega_b, \\Omega$ be the circumcircles of triangles $ABC, ABC', AB'C, AB'C'$ respectively. Denote $O_c, O_b$ as the circumcenters of $\\omega_c, \\omega_b$ and $R$ as the radius of $\\omega$. Let $f(X)$ be the reflection of the figure $X$ over the line $AB$.\n\nIt is easy to check that $... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety"
] | English | proof only | null | |
07m6 | In the triangle $ABC$ we have $|AB| = 1$ and $\angle ABC = 120^\circ$. The perpendicular line to $AB$ at $B$ meets $AC$ at $D$ such that $|DC| = 1$. Find the length of $AD$. | [
"Let $x = |AD|$ and $y = |BC|$. From $\\triangle ADB$ we obtain $\\sin(\\angle ADB) = \\frac{1}{x}$ and from $\\triangle BDC$ we get $\\frac{\\sin(\\angle BDC)}{y} = \\frac{\\sin(30^\\circ)}{1} = \\frac{1}{2}$. Since $\\sin(\\angle ADB) = \\sin(\\angle BDC)$, this gives $\\frac{1}{x} = \\frac{y}{2}$, hence $y = \\f... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | ∛2 | |
0910 | Problem:
Let $ABC$ be an acute-angled triangle with $AC > BC$ and circumcircle $\omega$. Suppose that $P$ is a point on $\omega$ such that $AP = AC$ and that $P$ is an interior point of the shorter arc $BC$ of $\omega$.
Let $Q$ be the point of intersection of the lines $AP$ and $BC$. Furthermore, suppose that $R$ is a... | [
"Solution:\n\nLet us denote $O$ the center of the circle $\\omega$ and $\\varphi = \\angle PAR$. Since the triangle $QAR$ is isosceles, we have $\\angle ARQ = \\varphi$ and $\\angle PQR = 2\\varphi$. The central angle theorem (applying to the chord $PR$) also yields $\\angle POR = 2\\varphi$. Thus the points $P, Q,... | Middle European Mathematical Olympiad (MEMO) | MEMO Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
04y7 | Let $k$ be a fixed positive integer. A finite sequence of integers $x_1, x_2, \dots, x_n$ is written on a blackboard. Pepa and Geoff are playing a game that proceeds in rounds as follows.
* In each round, Pepa first partitions the sequence that is currently on the blackboard into two or more contiguous subsequences (th... | [
"A finite sequence of integers is called a *word* and any its contiguous subsequence is called a *subword*. For a word $u$, by $\\sum u$ we denote the sum of numbers in $u$. A *prefix* of a word is a subword starting at the beginning of the word, and a prefix is *proper* if it is neither empty nor the whole word. A... | Czech-Polish-Slovak Mathematical Match | null | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
06sc | Let $m \neq 0$ be an integer. Find all polynomials $P(x)$ with real coefficients such that
$$
\left(x^{3}-m x^{2}+1\right) P(x+1)+\left(x^{3}+m x^{2}+1\right) P(x-1)=2\left(x^{3}-m x+1\right) P(x)
$$
for all real numbers $x$. | [
"Let $P(x)=a_{n} x^{n}+\\cdots+a_{0} x^{0}$ with $a_{n} \\neq 0$. Comparing the coefficients of $x^{n+1}$ on both sides gives $a_{n}(n-2 m)(n-1)=0$, so $n=1$ or $n=2 m$.\nIf $n=1$, one easily verifies that $P(x)=x$ is a solution, while $P(x)=1$ is not. Since the given condition is linear in $P$, this means that the... | IMO | International Mathematical Olympiad Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Intermediate V... | English | proof and answer | P(x) = t x for any real t | |
0k1j | Problem:
Given that $x$ is a positive real, find the maximum possible value of
$$
sin \left(\tan^{-1}\left(\frac{x}{9}\right)-\tan^{-1}\left(\frac{x}{16}\right)\right)
$$ | [
"Solution:\nConsider a right triangle $A O C$ with right angle at $O$, $A O = 16$ and $C O = x$. Moreover, let $B$ be on $A O$ such that $B O = 9$. Then $\\tan^{-1} \\frac{x}{9} = \\angle C B O$ and $\\tan^{-1} \\frac{x}{16} = \\angle C A O$, so their difference is equal to $\\angle A C B$.\n\nNote that the locus o... | United States | HMMT February 2018 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 7/25 | |
0bhd | The isosceles triangle $ABC$ has $AB = AC$ and points $M$ and $N$ are taken on $BC$ such that $M$ is between $B$ and $N$. Prove that the following properties are equivalent:
i) $m(\angle MAN) = \frac{1}{2} m(\angle BAC)$;
ii) the segments $[BM]$, $[MN]$ and $[MC]$ are the sides of a triangle in which the opposite ang... | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0isb | Problem:
Right triangle $X Y Z$, with hypotenuse $Y Z$, has an incircle of radius $\frac{3}{8}$ and one leg of length $3$. Find the area of the triangle. | [
"Solution:\n\nLet the other leg have length $x$. Then the tangents from $Y$ and $Z$ to the incircle have length $x - \\frac{3}{8}$ and $3 - \\frac{3}{8}$. So the hypotenuse has length $x + \\frac{9}{4}$, the semiperimeter of the triangle is $x + \\frac{21}{8}$, and the area of the triangle is $\\frac{3}{8}\\left(x ... | United States | 1st Annual Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | 21/16 | |
060f | Problem:
Montrer qu'il n'existe pas de réels $x, y, z$ strictement positifs tels que
$$
\left(2 x^{2}+y z\right)\left(2 y^{2}+x z\right)\left(2 z^{2}+x y\right)=26 x^{2} y^{2} z^{2}
$$ | [
"Solution:\nOn pourrait essayer de simplement appliquer l'inégalité arithmético-géométrique sur chacun des facteurs du côté gauche, mais ceci donne que le terme de gauche est supérieur ou égal à $16 \\sqrt{2} x^{2} y^{2} z^{2}$ ce qui ne suffit pas pour conclure (car $16 \\sqrt{2}<26$ ). En fait, ceci vient du fait... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 2 : AlgèBre | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0epg | One of the factors of $48$ is chosen at random. What is the probability that the chosen factor is NOT a multiple of $4$?
(A) $40\%$
(B) $30\%$
(C) $25\%$
(D) $20\%$
(E) $10\%$ | [
"There are ten factors: $1$, $2$, $3$, $4$, $6$, $8$, $12$, $16$, $24$, $48$. Four of them are not multiples of $4$, viz. $1$, $2$, $3$, $6$. The probability is therefore $\\frac{4}{10}$ or $40\\%$."
] | South Africa | South African Mathematics Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | MCQ | A | |
041c | Find all the positive real number pairs $(a, b)$, such that $f(x) = ax^2 + b$ satisfies $f(xy) + f(x + y) \ge f(x)f(y)$ (for any real numbers $x, y$). | [
"The given condition is equivalent to\n$$\n(ax^2 y^2 + b) + (a(x + y)^2 + b) \\ge (ax^2 + b)(ay^2 + b). \\quad ①\n$$\nIn ①, let $y = 0$. We have $b + (ax^2 + b) \\ge (ax^2 + b) \\cdot b$, or\n$$ (1 - b)ax^2 + b(2 - b) \\ge 0. $$\nAs $a > 0$ and $ax^2$ can be sufficiently large, then $1 - b \\ge 0$, i.e., $0 < b \\l... | China | China Mathematical Competition | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | {(a, b) | 0 < a < 1, 0 < b ≤ 1, and 2a + b ≤ 2} | |
00gt | Consider the following operation on positive real numbers written on a blackboard:
Choose a number $r$ written on the blackboard, erase that number, and then write a pair of positive real numbers $a$ and $b$ satisfying the condition $2 r^{2}=a b$ on the board.
Assume that you start out with just one positive real numbe... | [
"Using AM-GM inequality, we obtain\n$$\n\\frac{1}{r^{2}}=\\frac{2}{a b}=\\frac{2 a b}{a^{2} b^{2}} \\leq \\frac{a^{2}+b^{2}}{a^{2} b^{2}} \\leq \\frac{1}{a^{2}}+\\frac{1}{b^{2}} . \\tag{*}\n$$\nConsequently, if we let $S_{\\ell}$ be the sum of the squares of the reciprocals of the numbers written on the board after... | Asia Pacific Mathematics Olympiad (APMO) | XXI Asian Pacific Mathematics Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0ep1 | How many pairs of non-negative integers $x$ and $y$ are solutions of $\frac{x}{20} + \frac{y}{15} = 1$? | [
"**6**\n\nWe can re-write $\\frac{x}{20} + \\frac{y}{15} = 1$ in the form $3x + 4y = 60$. Then we can see that $3x$ must be divisible by $4$, so $x$ must be, and trying successive possible values we see that only the following combinations of $x$- and $y$-values are acceptable: $(0; 15)$, $(4; 12)$, $(8; 9)$, $(12;... | South Africa | South African Mathematics Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | final answer only | 6 | |
0d33 | Determine if there exists an infinite sequence of positive integers
$$
a_{1}, a_{2}, a_{3}, \ldots
$$
such that
(i) each positive integer occurs exactly once in the sequence, and
(ii) each positive integer occurs exactly once in the sequence $\left|a_{1}-a_{2}\right|, \left|a_{2}-a_{3}\right|, \ldots, \left|a_{k}-a_{k+... | [
"We will construct such a sequence by induction:\nDefine $a_{1}=1$ and $a_{2}=2$. In this case we have $b_{1}=|a_{1}-a_{2}|=1$.\nAssume that $a_{1}, a_{2}, \\ldots, a_{2n}$ are defined such that there is no positive integer which occurs at least twice neither in the finite sequence $a_{1}, a_{2}, \\ldots, a_{2n}$ n... | Saudi Arabia | Selection tests for the International Mathematical Olympiad 2013 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Yes, such an infinite sequence exists. | |
0hih | Positive integers $a_1, a_2, ..., a_{101}$ are such that $a_i + 1$ are divisible by $a_{i+1}$ for $1 \le i \le 101$ (we assume that $a_{102} = a_1$). What is the largest value that the maximum of these numbers can attain? | [
"Without loss of generality, let $a_{101}$ be the largest of these numbers (or one of the largest). It is clear that $a_i + 1 \\ge a_{i+1}$ for any $i = 1, 100$. If we add all these 100 inequalities, we get $a_1 + a_2 + ... + a_{100} + 100 \\ge a_2 + a_3 + ... + a_{101}$, that is $a_1 \\ge a_{101} - 100$. From the ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 201 | |
0k4l | Problem:
A $5 \times 5$ grid of squares is filled with integers. Call a rectangle corner-odd if its sides are grid lines and the sum of the integers in its four corners is an odd number. What is the maximum possible number of corner-odd rectangles within the grid?
Note: A rectangle must have four distinct corners to b... | [
"Solution:\nAnswer: 60\n\nConsider any two rows and the five numbers obtained by adding the two numbers which share a given column. Suppose $a$ of these are odd and $b$ of these are even. The number of corner-odd rectangles with their sides contained in these two rows is $a b$. Since $a+b=5$, we have $a b \\leq 6$.... | United States | HMMT November 2018 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 60 |
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