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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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0e9a | Prove that arbitrary real numbers $a$ and $b$ satisfy the inequality
$$
(a + ab - b^2)^2 + ab^2(a + 2) \geq 0.
$$
When does the equality hold? | [
"Expanding the left-hand side of the inequality we get $a^2 + a^2b^2 + b^4 + 2a^2b - 2ab^3 + a^2b^2$. This can be rearranged into $a^2(1+b)^2 + b^2(b-a)^2$, and the desired inequality now follows. At the same time we see that the equality holds if and only if $a = b = 0$ or $a = b = -1$."
] | Slovenia | National Math Olympiad 2013 - First Round | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | Equality holds if and only if a = b = 0 or a = b = -1. | |
0jea | For a positive integer $n \ge 3$ plot $n$ equally spaced points around a circle. Label one of them $A$, and place a marker at $A$. One may move the marker forward in a clockwise direction to either the next point or the point after that. Hence there are a total of $2n$ distinct moves available; two from each point. Let... | [
"**Solution 1** (By Sam Vandervelde). We will show that $a_n = \\frac{1}{3}(2^{n+1} + (-1)^n)$. This would be sufficient, since then we would have\n$$\na_{n-1} + a_n = \\frac{1}{3}(2^n + (-1)^{n-1}) + \\frac{1}{3}(2^{n+1} + (-1)^n) = \\frac{1}{3}(2^n + 2 \\cdot 2^n) = 2^n.\n$$\n\n**Lemma 1.** For all positive integ... | United States | USAMO | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
01ow | For two positive integers $a$ and $b$ the number $\overline{a.b}$ is equal to the decimal fraction which we have if after the number $a$ we put the decimal point and then write the number $b$. For example, for $a = 20$, $b = 13$ we get $\overline{a.b} = 20.13$, and $\overline{b.a} = 13.2$.
Prove that there are infinite... | [
"Let the decimal representation of $a$ consist of $k$ digits. Set $b = 5 \\cdot 10^m$, $m \\in \\mathbb{N}$. Then\n$$\n\\begin{align*}\n\\overline{a,b} \\cdot \\overline{b,a} &= n \\iff \\left(a + \\frac{b}{10^k}\\right)\\left(b + \\frac{a}{10^m}\\right) = n \\iff \\\\\n&\\iff \\frac{(2a + 1) \\cdot 5 \\cdot 10^m}{... | Belarus | BelarusMO 2013_s | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0kre | Problem:
Let $x < 0.1$ be a positive real number. Let the foury series be $4 + 4x + 4x^{2} + 4x^{3} + \ldots$, and let the fourier series be $4 + 44x + 444x^{2} + 4444x^{3} + \ldots$. Suppose that the sum of the fourier series is four times the sum of the foury series. Compute $x$. | [
"Solution:\n\nThe sum of the foury series can be expressed as $\\frac{4}{1-x}$ by geometric series. The fourier series can be expressed as\n$$\n\\begin{aligned}\n& \\frac{4}{9}\\left((10-1)+(100-1)x+(1000-1)x^{2}+\\ldots\\right) \\\\\n& = \\frac{4}{9}\\left(\\left(10+100x+1000x^{2}+\\ldots\\right)-\\left(1+x+x^{2}+... | United States | HMMT November | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | proof and answer | 3/40 | |
05kq | Problem:
Calculer
$$
\sqrt{1+\frac{1}{1^{2}}+\frac{1}{2^{2}}}+\sqrt{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}}+\cdots+\sqrt{1+\frac{1}{2014^{2}}+\frac{1}{2015^{2}}}
$$ | [
"Solution:\n\nOn réduit au même dénominateur\n$$\n\\begin{aligned}\n1+\\frac{1}{n^{2}}+\\frac{1}{(n+1)^{2}} & =\\frac{n^{2}(n+1)^{2}+(n+1)^{2}+n^{2}}{n^{2}(n+1)^{2}} \\\\\n& =\\frac{n^{4}+n^{2}(2 n+1)+n^{2}+(n+1)^{2}}{n^{2}(n+1)^{2}} \\\\\n& =\\frac{n^{4}+2 n^{2}(n+1)+(n+1)^{2}}{n^{2}(n+1)^{2}} \\\\\n& =\\frac{\\le... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 2015 - 1/2015 | |
0cia | Let $n \in \mathbb{N}^*$. Show that for any choice of $x_1, x_2, \dots, x_n \in \mathbb{N}^*$ such that $x_k \le k$ for all $k \in \{1, 2, \dots, n\}$ and the sum $x_1 + x_2 + \dots + x_n$ is odd, there exist $\varepsilon_1, \varepsilon_2, \dots, \varepsilon_n \in \{-1, 1\}$ such that $\sum_{k=1}^n \varepsilon_k \cdot ... | [] | Romania | 75th NMO | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0itx | Problem:
Let $n \geq 3$ be a positive integer. A triangulation of a convex $n$-gon is a set of $n-3$ of its diagonals which do not intersect in the interior of the polygon. Along with the $n$ sides, these diagonals separate the polygon into $n-2$ disjoint triangles. Any triangulation can be viewed as a graph: the verti... | [
"Solution:\nWe will show that all triangulations have chromatic number $3$, by induction on $n$.\n\nAs a base case, if $n=3$, a triangle has chromatic number $3$.\n\nNow, given a triangulation of an $n$-gon for $n>3$, every edge is either a side or a diagonal of the polygon. There are $n$ sides and only $n-3$ diago... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0b2l | Problem:
Find the remainder when $\sum_{n=2}^{2021} n^{n}$ is divided by $5$.
(a) $1$
(b) $2$
(c) $3$
(d) $4$ | [] | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | MCQ | d | |
0i2n | Problem:
What is the size of the largest rectangle that can be drawn inside of a $3$-$4$-$5$ right triangle with one of the rectangle's sides along one of the legs of the triangle? | [
"Solution:\n\nClearly one vertex of the rectangle will be at the right angle. Position the triangle with the leg of length $4$ along the $x$-axis and the leg of length $3$ along the $y$-axis. Then the hypotenuse is along the line $y = 3 - (3/4)x$.\n\nSuppose the rectangle has a side of length $y$ along the leg of l... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 3 | |
0bbt | Let $f: \mathbb{R} \to \mathbb{R}$ be a function with the property: $|f(x)-f(y)| \le |\sin x - \sin y|$, for each $x, y \in \mathbb{R}$.
a) Prove that there exists an unique $c \in \mathbb{R}$ such that $f(c) = c$.
b) Consider the sequence $(x_n)_{n \in \mathbb{N}}$ with $x_0 = 0$ and $x_{n+1} = f(x_n)$ for every $n ... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
004o | Se tienen $100$ enteros positivos tales que su suma es igual a su producto.
Determinar la mínima cantidad de números $1$ que hay entre los $100$ enteros. | [] | Argentina | XVIII Olimpiada Matemática del Cono Sur | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Español | proof and answer | 95 | |
0id4 | Problem:
A mouse is sitting in a toy car on a negligibly small turntable. The car cannot turn on its own, but the mouse can control when the car is launched and when the car stops (the car has brakes). When the mouse chooses to launch, the car will immediately leave the turntable on a straight trajectory at $1$ meter p... | [
"Solution:\nThe mouse can wait while the table rotates through some angle $\\theta$ and then spend the remainder of the time moving along that ray at $1~\\mathrm{m}/\\mathrm{s}$. He can reach any point between the starting point and the furthest reachable point along the ray, $(1-\\theta/\\pi)$ meters out. So the a... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Integral Calculus > Applications",
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | final answer only | π/6 | |
0f6i | Problem:
$ABCD$ is a parallelogram. A circle through $A$ and $B$ has radius $R$. A circle through $B$ and $D$ has radius $R$ and meets the first circle again at $M$. Show that the circumradius of $AMD$ is $R$. | [] | Soviet Union | 19th ASU | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fdy | Problem:
Se considera un triángulo $A B C$ con $\angle B A C=45^{\circ}$ y $\angle A C B=30^{\circ}$. Si $M$ es el punto medio del lado $B C$, se pide demostrar que $\angle A M B=45^{\circ}$ y que $B C \cdot A C=2 \cdot A M \cdot A B$. | [
"Solution:\n\nSea $D$ el punto de $A C$ tal que $B D \\perp A C$.\n\nPuesto que $\\angle D B A=90^{\\circ}-\\angle B A D=90^{\\circ}-45^{\\circ}=45^{\\circ}$, el triángulo $A D B$ es isósceles con $A D=D B$.\nPues $\\triangle C D B$ es rectángulo en $D, C M=M D$ y, por tanto, $\\angle C D M... | Spain | null | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0kkf | Problem:
Let gcd mean the greatest common divisor of two numbers and lcm their least common multiple. Suppose the three numbers $A$, $B$, $C$ satisfy
$$
\begin{array}{ll}
\operatorname{gcd}(A, B)=2, & \operatorname{lcm}(A, B)=60 \\
\operatorname{gcd}(A, C)=3, & \operatorname{lcm}(A, C)=42
\end{array}
$$
Determine the... | [
"Solution:\nFrom the given information, $A$ must be a multiple of $2$ and $3$, and thus a multiple of $\\operatorname{LCM}(2,3)=6$. It also must be a factor of $60$ and $42$, and thus a factor of $\\operatorname{GCD}(60,42)=6$. The only possibility is $A=6$.\n\nSince $\\operatorname{LCM}(A, B)$ is divisible by $5$ ... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | A=6, B=20, C=21 | |
08wi | Triangles $PAB$ and $PCD$ are placed on a plane. Suppose that $PA = PB$, $PC = PD$ are satisfied and that the 3 points $P$, $A$, $C$ lie on a straight line in this order and the same is true for the 3 points $B$, $P$, $D$. Suppose further that a circle $S_1$ going through $A$, $C$ and the circle $S_2$ going through $B$... | [
"Given a circle $S$ and a point $Z$ in the plane, we define the power $p = p_S(Z)$ of $Z$ with reference to $S$ in the following way: Let $p = 0$ if the point $Z$ lies on the circumference of $S$. Otherwise draw a line through $Z$ and intersecting at the points $I_1$ and $I_2$ with the circle $S$. Then, $p = p_S(Z)... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
04li | There are seven people in a room. Four of them each know exactly one other person, while the other three each know exactly two people. All acquaintances are mutual. What is the probability that two randomly chosen people do not know each other? (Graduate Management Admission Test) | [] | Croatia | Mathematical competitions in Croatia | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 16/21 | |
0aqo | Problem:
Let $A$, $B$, $C$, $D$ (written in the order from left to right) be four equally-spaced collinear points. Let $\omega$ and $\omega'$ be the circles with diameters $AD$ and $BD$, respectively. A line through $A$ that is tangent to $\omega'$ intersects $\omega$ again at point $E$. If $AB = 2\sqrt{3}\ \mathrm{cm}... | [] | Philippines | 12th Philippine Mathematical Olympiad - Area Stage | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 9 | |
0kxj | Problem:
A lattice point in the plane is a point with integer coordinates. Let $T$ be a triangle in the plane whose vertices are lattice points, but with no other lattice points on its sides. Furthermore, suppose $T$ contains exactly four lattice points in its interior. Prove that these four points lie on a straight l... | [
"Solution:\n\nLet us begin with some preliminaries. In the solution to follow, we treat points freely as vectors, e.g. writing $n A$ to mean the point whose coordinates are $n$ times the coordinates of $A$, or $A+B$ to mean the point which is the coordinate-wise sum of $A$ and $B$.\nA basic result from vector geome... | United States | Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Linear Algebra > Determinants",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-po... | null | proof only | null | |
0kip | Problem:
Joey wrote a system of equations on a blackboard, where each of the equations was of the form $a+b=c$ or $a \cdot b=c$ for some variables or integers $a, b, c$. Then Sean came to the board and erased all of the plus signs and multiplication signs, so that the board reads:
$$
\begin{array}{ll}
x & z=15 \\
x & ... | [
"Solution:\n\nThe bottom line gives $x=-6$, $x=6$ or $x=18$. If $x=-6$, $y$ can be $-2$ or $18$ and $z$ must be $21$, so the possible values for $100 x+10 y+z$ are $-599$ and $-399$. If $x=6$, $y$ can be $2$ or $6$ and $z$ must be $9$, so the possible values are $629$ and $669$. If $x=18$, $y$ must be $-6$ and $z$ ... | United States | HMMT November 2021 Team Round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 2037 | |
0323 | Problem:
Let $ABC$ be an isosceles triangle with $AC = BC$ and let $k$ be a circle with center $C$ and radius less than the altitude $CH$, $H \in AB$. Lines through $A$ and $B$ are tangent to $k$ at points $P$ and $Q$ lying on the same side of the line $CH$. Prove that the points $P$, $Q$ and $H$ are collinear. | [
"Solution:\n\nFirst solution. Since $CP = CQ$, $CA = CB$ and $\\angle APC = \\angle BQC = 90^\\circ$, then $\\triangle APC \\cong BQC$. Hence $\\angle CAP = \\angle CBQ$. Setting $AP \\cap BQ = T$, it follows that the quadrilateral $ABTC$ is cyclic. Then $\\angle BAC = \\angle QTC$ and now $\\angle TQC = \\angle AH... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0cn0 | In a tetrahedron $ABCD$, none of the face angles is $90^\circ$, and the orthocenters of the faces $ABC$, $ABD$, $ACD$ are collinear. Prove that the circumcenter of this tetrahedron is coplanar with the midpoints of edges $AB$, $AC$ and $AD$. (I. Bogdanov) | [
"Пусть $AB_1$, $AC_1$, $AD_1$ — высоты граней $ACD$, $ABD$, $ABC$. Точки пересечения высот этих граней лежат на прямых $AB_1$, $AC_1$, $AD_1$ и отличны от точки $A$. Поскольку они лежат на одной прямой $\\ell$, то прямые $AB_1$, $AC_1$, $AD_1$ лежат в плоскости $\\alpha$, содержащей $\\ell$ и $A$ (ясно, что $A$ не ... | Russia | Russian mathematical olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof only | null | |
0koj | Problem:
On a board the following six vectors are written:
$$
(1,0,0), \quad(-1,0,0), \quad(0,1,0), \quad(0,-1,0), \quad(0,0,1), \quad(0,0,-1)
$$
Given two vectors $v$ and $w$ on the board, a move consists of erasing $v$ and $w$ and replacing them with $\frac{1}{\sqrt{2}}(v+w)$ and $\frac{1}{\sqrt{2}}(v-w)$. After some... | [
"Solution:\nFor a construction, note that one can change\n$$\n(1,0,0),(-1,0,0) \\rightarrow (\\sqrt{2}, 0,0),(0,0,0) \\rightarrow (1,0,0),(1,0,0)\n$$\nand similarly for $(0,1,0),(0,-1,0)$ and $(0,0,1),(0,0,-1)$. Then $u=(2,2,2)$.\n\nFor the bound, argue as follows: let the vectors be $v_{1}, \\ldots, v_{6}$, $n=(x,... | United States | HMMT February 2022 | [
"Algebra > Linear Algebra > Vectors",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 2√3 | |
0knv | Problem:
A straight ladder starts upright against a vertical wall, and slides down until it is horizontal such that the top is always along the wall and the bottom on the floor. What shape does its midpoint trace out? | [
"Solution:\nThe ladder always forms a right triangle with the wall whose hypotenuse has a fixed length. Since the length of the median to the hypotenuse in a right triangle is always half the hypotenuse, the distance from the midpoint of the ladder to the corner where the wall meets the floor remains fixed at half ... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | A quarter circle centered at the corner where the wall meets the floor, with radius equal to half the length of the ladder. | |
0lfl | Let $\alpha \in (1, +\infty)$ be a real number, and let $P(x) \in \mathbb{R}[x]$ be a monic polynomial with degree $24$, such that
(i) $P(0) = 1$.
(ii) $P(x)$ has exactly $24$ positive real roots that are all less than or equal to $\alpha$.
Show that $|P(1)| \le \left(\frac{19}{5}\right)^5 (\alpha - 1)^{24}$. | [
"Let $x_1, x_2, \\dots, x_{24}$ be the roots of the polynomial $P(x)$. Then $0 < x_1, x_2, \\dots, x_{24} \\le \\alpha$, $x_1x_2 \\cdots x_{24} = 1$ and we have to prove\n$$\n|(x_1 - 1)(x_2 - 1) \\cdots (x_{24} - 1)| \\le \\left(\\frac{19}{5}\\right)^5 (\\alpha - 1)^{24}.\n$$\nIf among the numbers $x_1, x_2, \\dots... | Vietnam | Team selection tests | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
0hst | Problem:
Let $\ominus$ be an operation on the set of real numbers such that
$$
(x \ominus y)+(y \ominus z)+(z \ominus x)=0
$$
for all real $x$, $y$, and $z$. Prove that there is a function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
x \ominus y=f(x)-f(y)
$$
for all real $x$ and $y$. | [
"Solution:\nFirst we plug in $x=y=z=0$ to get\n$$\n(0 \\ominus 0)+(0 \\ominus 0)+(0 \\ominus 0)=0\n$$\nthat is, $0 \\ominus 0=0$. Then we plug $y=z=0$, keeping $x$ undetermined, into the original equation to get\n$$\n(x \\ominus 0)+0+(0 \\ominus x)+0=0 .\n$$\nSo $0 \\ominus x=-(x \\ominus 0)$. Define $g(x)=x \\omin... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
0i8a | Problem:
A particle moves along the $x$-axis in such a way that its velocity at position $x$ is given by the formula $v(x) = 2 + \sin x$. What is its acceleration at $x = \frac{\pi}{6}$? | [
"Solution:\nAcceleration is given by\n$$\na = \\frac{dv}{dt} = \\frac{dv}{dx} \\cdot \\frac{dx}{dt} = \\frac{dv}{dx} \\cdot v = \\cos x \\cdot (2 + \\sin x)\n$$\nAt $x = \\frac{\\pi}{6}$:\n$$\na = \\cos\\left(\\frac{\\pi}{6}\\right) \\cdot \\left(2 + \\sin\\left(\\frac{\\pi}{6}\\right)\\right) = \\frac{\\sqrt{3}}{2... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Related Rates"
] | null | final answer only | 5√3/4 | |
04jp | Let $M$ and $N$ be, respectively, the feet of the altitudes from vertices $A$ and $B$ of the acute triangle $ABC$. Let $Q$ be the midpoint of the segment $\overline{MN}$ and let $P$ be the midpoint of the segment $\overline{AB}$. If $|MN| = 10$ and $|AB| = 26$, determine the length of the segment $\overline{PQ}$. | [] | Croatia | Croatia Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 12 | |
0gxi | Distinct real numbers $a, b, c$ satisfy the condition $a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a}$. Find all possible values of the product $abc$:
1) for all real $a, b, c$;
2) for positive real $a, b, c$? | [
"a) See problem 9.5\n\nb) We will show that there are no such positive numbers that satisfy the condition of the problem. Without loss of generality we suppose that $a > b$, then from the equality $(a - b) = \\frac{b - a}{b - c}$ we conclude that $b > c$. From the condition $b - c = \\frac{c - a}{a - c}$, we can ge... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Other"
] | English | proof and answer | For real a, b, c: the product can be −1 or 1. For positive a, b, c: no such triples exist, so there are no possible values. | |
0knp | Let $n \ge 2$ be an integer. An $n \times n$ board is initially empty. Each minute, you may perform one of three moves:
* If there is an L-shaped tromino region of three cells without stones on the board (see figure; rotations not allowed), you may place a stone in each of those cells.
* If all cells in a column have a... | [] | United States | USAMO 2021 Day I | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | none | |
0c35 | Determine digits $x$ and $y$ such that $\frac{5xy23}{4xy24} = \frac{523}{424}$. | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | x=2, y=8 | |
0f93 | Problem:
The incircle of $ABC$ touches $AB$ at $M$. $N$ is any point on the segment $BC$. Show that the incircles of $AMN$, $BMN$, $ACN$ have a common tangent. | [] | Soviet Union | 23rd ASU | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
038x | One cuts a paper strip of length $2007$ into two parts of integer lengths and writes down the two integers on the board. Then cuts one of the two parts into two parts of integer lengths and writes down the two integers on the board. The cutting stops when all parts are of length $1$. A cut is called *bad* if the two pa... | [
"a) Let the length of the strip be $n$. Denote by $g(n)$ and $f(n)$ respectively the number of $1$'s in the binary representation of $n$ and the minimum possible number of bad cuts. Let $n = 2^{k_1} + 2^{k_2} + \\cdots + 2^{k_l}$. Consider the following sequence of cuts: first cut a strip of length $2^{k_1}$, next ... | Bulgaria | Winter Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) 8. b) For any optimal cutting achieving the minimum number of bad cuts, the number of distinct integers recorded equals 2 g(n) + k − 1, where g(n) is the number of ones in the binary expansion of n and k is the largest exponent of two appearing in that expansion. | |
05ql | Problem:
Trouver tous les entiers $n \geqslant 3$ tels que, si $a_{1}, \ldots, a_{n}$ sont des réels strictement positifs tels que $\max \left(a_{1}, \ldots, a_{n}\right) \leqslant n \cdot \min \left(a_{1}, \ldots, a_{n}\right)$, alors il existe nécessairement trois de ces réels qui sont les longueurs des côtés d'un t... | [
"Solution:\n\nSans perte de généralité, on suppose que $1=a_{1} \\leqslant a_{2} \\leqslant \\ldots \\leqslant a_{n}$. On note alors que trois réels $a_{i}, a_{j}$ et $a_{k}$, avec $i<j<k$, sont les longueurs des côtés d'un triangle acutangle si et seulement si $a_{i}^{2}+a_{j}^{2}>a_{k}^{2}$ (on peut le voir par e... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | all integers n ≥ 13 | |
0egx | Problem:
Štiri pozitivna števila so v razmerju $1:2:3:4$. Vsota kvadratov najmanjših treh števil je za $1$ manjša od vsote največjih treh števil. Največ koliko nizov takih števil lahko najdemo?
(A) $0$
(B) $1$
(C) $2$
(D) $3$
(E) $4$ | [
"Solution:\n\nIz razmerja $a : b : c : d = 1 : 2 : 3 : 4$ sledi, da je $a = \\frac{1}{10} t$, $b = \\frac{2}{10} t$, $c = \\frac{3}{10} t$, $d = \\frac{4}{10} t$.\n\nIz enakosti $a^{2} + b^{2} + c^{2} + 1 = b + c + d$ dobimo $14 t^{2} - 90 t + 100 = 0$.\n\nRešitvi kvadratne enačbe sta $t_{1} = \\frac{10}{7}$, $t_{2... | Slovenia | Državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | MCQ | C | |
0jwd | Problem:
Kevin and Yang are playing a game. Yang has $2017 + \binom{2017}{2}$ cards with their front sides face down on the table. The cards are constructed as follows:
- For each $1 \leq n \leq 2017$, there is a blue card with $n$ written on the back, and a fraction $\frac{a_{n}}{b_{n}}$ written on the front, where $... | [
"Solution:\n\nWe will provide an algorithm to determine which red card has the highest value in $2016 + 2015$ turns.\n\nWe start with a lemma: If the blue card with the largest fraction is labeled $k$ and the red card with the largest fraction is labeled $(i, j)$, then $i = k$ or $j = k$.\n\nTo prove this lemma, we... | United States | HMIC | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0cy3 | Find all triples $(x, y, z)$ of integers such that
$$
x^{2} y + y^{2} z + z^{2} x = 2010^{2} \text{ and } x y^{2} + y z^{2} + z x^{2} = -2010.
$$ | [
"We replace $2010$ by $p-1$, for some odd prime. Subtracting the first equation from the second, we obtain\n$$\n(x-y)(y-z)(z-x) = -p(p-1)\n$$\nWe have $(x-y)+(y-z)+(z-x)=0$ and $(x-y)(y-z)(z-x)<0$, so precisely two of them are positive. Assume that $x-y>0$ and $y-z>0$. Without loss of generality, suppose $x-y \\leq... | Saudi Arabia | SAMC | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Symmetric funct... | English | proof and answer | (1, 0, -2010), (-2010, 1, 0), (0, -2010, 1) | |
09cz | Дугуй ширээ тойрон суусан $8$ хүүхдийн аль ч хүүхэд ба түүний хөрш хоёр хүүхдэд байгаа нийт чихрийн тоо сондгой бол хүүхэд бүрт заавал сондгой тооны чихэр байх албатай ю? | [
"Дугуй ширээ тойрон суусан хүүхдүүдийг $1, 2, \\ldots, 7, 8$ гэж дугаарлая. Хэрэв $i$-р хүүхдэд тэгш тооны чихэр байвал $i \\to 1$-ийг, сондгой тооны чихэр байвал $i \\to 0$-ийг тус харгалзуулъя. Хэн нэгэн, жишээлбэл $1$-р хүүхэд тэгш тооны чихэртэй гэж саная. Тэгвэл $1245678$ эсвэл $12345678$ байхаас өөршгүй ба $8... | Mongolia | ММО-48 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Logic"
] | Mongolian | proof and answer | Yes, every child must have an odd number of candies. | |
009y | The lower row of a $2 \times 13$ rectangle is filled up with 13 markers labeled $1, 2, \ldots, 13$ in this order. An operation is moving a marker from its cell to an adjacent (by side) empty cell. The task is to rearrange the markers in the reverse order, in the lower row again. Do this with a minimal number of operati... | [
"Marker $1$ needs at least $12$ horizontal operations to reach its final position $13$; by symmetry the same holds for marker $13$. Similarly markers $2$ and $12$ need at least $10$ horizontal operations each. The analogous observation about the pairs of markers $3, 11$; $4, 10$; $5, 9$; $6, 8$ implies that at leas... | Argentina | Argentine National Olympiad 2015 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | 108 | |
01cm | Let $a_1, \dots, a_n$ be real numbers, fulfilling $0 \le a_i \le 1$ for $i = 1, \dots, n$. Prove the inequality
$$
(1 - a_1^n)(1 - a_2^n) \cdots (1 - a_{n-1}^n) \le (1 - a_1 a_2 \cdots a_n)^n.
$$ | [
"$$\n\\begin{aligned}\n(1 - a_1^n)(1 - a_2^n) \\cdots (1 - a_n^n) &\\le \\left( \\frac{(1 - a_1^n) + (1 - a_2^n) + \\cdots + (1 - a_n^n)}{n} \\right)^n \\\\\n&= \\left( 1 - \\frac{a_1^n + \\cdots + a_n^n}{n} \\right)^n.\n\\end{aligned}\n$$\nBy applying AM-GM again we obtain\n$$\na_1 a_2 \\cdots a_n \\le \\frac{a_1^... | Baltic Way | Baltic Way 2015 Shortlisted Problems | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
031f | Problem:
Consider the polynomial $f(x) = 4x^{4} + 6x^{3} + 2x^{2} + 2003x - 2003^{2}$. Prove that:
a) the local extrema of $f'(x)$ are positive;
b) the equation $f(x) = 0$ has exactly two real roots and find them. | [
"Solution:\n\na) Since $\\lim_{x \\rightarrow +\\infty} f'(x) = +\\infty$ and $\\lim_{x \\rightarrow -\\infty} f'(x) = -\\infty$, it is enough to show that the local minimum $m$ of $f'(x)$ is positive. Since the equation $f''(x) = 0$ has two real roots $x_{1} > x_{2}$, it follows that $m = f'(x_{1}) > 0$. Now it is... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (-1 + sqrt(4007))/2 and (-1 - sqrt(4007))/2 | |
0eym | Problem:
The sides and diagonals of $ABCD$ have rational lengths. The diagonals meet at $O$. Prove that the length $AO$ is also rational. | [
"Solution:\n\n$AB = AO \\cos OAB + BO \\cos OBA$. We can derive a rational expression for $\\cos OAB$ using the cosine rule for triangle $ABC$. Similarly for $\\cos OBA$ using the cosine rule for triangle $DAB$. So $OA = r_1 + r_2 OB$, where $r_1$ denotes a rational number. Similarly, $OB = r_3 + r_4 OC$, so $OA = ... | Soviet Union | 2nd ASU | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
01hd | Let $a$ be a real number. Prove that there exist real numbers $b$ and $c$ such that the inequalities
$$ \min \{\sin(x), \sin(a+x)\} \le b \sin(x+c) \le \max \{\sin(x), \sin(a+x)\} $$
hold for all real numbers $x$ and the equalities hold only if $\sin(x) = \sin(a+x)$. | [
"We recall that two real numbers $r$ and $s$ always satisfy the trigonometric identity $\\frac{1}{2}[\\sin(r+s) + \\sin(r-s)] = \\cos(s) \\sin(r)$.\nSubstituting $r = x + \\frac{a}{2}$ and $s = \\frac{a}{2}$, we see that $\\cos(\\frac{a}{2}) \\sin(x + \\frac{a}{2})$ is for all real numbers $x$ the arithmetic mean o... | Baltic Way | Baltic Way 2021 Shortlist | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
04t1 | $1 \cdot 2 \cdot 3 \cdot \dots \cdot n$
*is written on a blackboard. For which positive integers* $n \ge 2$ *can we append the exclamation mark to some factors and change it to factorials in such a way that the final product will be a square?* | [
"Let us denote $v_p(n)$ the highest power of a prime $p$ which divides positive integer $n$. This function has obviously the following properties:\n- For all primes $p$ and positive integers $n$ is $v_p(n)$ non-negative integer.\n- For all positive integers $m, n$ and all primes $p$ is $v_p(mn) = v_p(m) + v_p(n)$.\... | Czech Republic | 65th Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | All composite integers n ≥ 2 | |
0abk | Let $S$ be a subset of $\{1,2,\ldots,9\}$, such that sums of every two elements of $S$ are distinct. For example: the set $\{1,2,3,5\}$ has that property but the set $\{1,2,3,4,5\}$ doesn't because $\{2,3\}$ and $\{1,4\}$ both have sum $5$. How many elements at most can $S$ contain? Explain your answer. | [
"It is easy to check that $\\{1,2,3,5,8\\}$ satisfies the desired condition. We will prove that $S$ can't contain more than five elements. Let $S$ contain at least six elements. Then the smallest possible sum of pairs is $3$ and the largest is $8+9=17$, i.e. the only possible sums of pairs are $3,4,5,\\ldots,17$ wh... | North Macedonia | Macedonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 5 | |
09x8 | Let $\Gamma$ be the circumcircle of a triangle $ABC$ and let $D$ be a point on segment $BC$. The circle that passes through $B$ and $D$ and is tangent to $\Gamma$ and the circle that passes through $C$ and $D$ and is tangent to $\Gamma$, intersect at a point $E \neq D$. The line $DE$ intersects $\Gamma$ at two points, ... | [
"\n\nWe consider the configuration as in the figure, where $E$ is at least as close to $B$ as it is to $C$. The proof in the case of the configuration in which this is the other way around, is analogous.\n\nLet $O$ be the centre of $\\Gamma$. The angle between the line $BC$ and the common t... | Netherlands | IMO Team Selection Test 2 | [
"Geometry > Plane Geometry > Circles > Tangents",
"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 c... | English | proof only | null | |
0c7l | Determine all nonnegative integers $x, y, z$ which satisfy the equation
$$
2^x + 3 \cdot 11^y = 7^z.
$$ | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof and answer | [(2, 0, 1), (4, 1, 2)] | |
0cwo | A pair of polynomials $F(x, y)$, $G(x, y)$ with integer coefficients is called *important*, if the following condition holds: if for some integers $a, b, c, d$ both $F(a, b) - F(c, d)$ and $G(a, b) - G(c, d)$ are divisible by 100, then both $a - c$ and $b - d$ are divisible by 100. Determine if there exist an important... | [
"**Ответ.** Does not exist.\n\n**Решение.** Let $F$ and $G$ be an important pair of polynomials. Consider pairs of residues modulo 100 of numbers $F(a,b)$ and $G(a,b)$, where $a,b$ range over all integer pairs from 0 to 99. According to the problem's condition, all such residue pairs are distinct. Since there are $... | Russia | LI Всероссийская математическая олимпиада школьников | [
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | Russian | proof and answer | Does not exist. | |
0hlq | Problem:
The sequence
$$
5, 9, 49, 2209, \ldots
$$
is defined by $a_{1} = 5$ and $a_{n} = a_{1} a_{2} \cdots a_{n-1} + 4$ for $n > 1$. Prove that $a_{n}$ is a perfect square for $n \geq 2$. | [
"Solution:\n\nThis is clear for $n = 2$. We use the relation\n$$\n\\begin{gathered}\na_{n-1} = a_{1} a_{2} \\cdots a_{n-2} + 4 \\\\\na_{1} a_{2} \\cdots a_{n-2} = a_{n-1} - 4\n\\end{gathered}\n$$\nfor $n \\geq 3$ to transform $a_{n}$:\n$$\n\\begin{aligned}\na_{n} & = a_{1} a_{2} \\cdots a_{n-2} a_{n-1} + 4 \\\\\n& ... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
01at | We let $a_0 = a > 0$ be an integer and $a_n = 5a_{n-1} + 4$. Can we choose $a$ so that $a_{54}$ is a multiple of 2013? | [
"Let $x_n = \\frac{a_n}{5^n}$. Then $x_0 = a$ and $5^n x_n = a_n = 5a_{n-1} + 4 = 5^n x_{n-1} + 4$. So $x_n = x_{n-1} + \\frac{4}{5^n}$. By induction,\n$$\nx_n = x_0 + \\left( \\frac{4}{5} + \\frac{4}{5^2} + \\dots + \\frac{4}{5^n} \\right) = a + \\frac{4}{5} \\left( 1 + \\frac{1}{5} + \\dots + \\frac{1}{5^{n-1}} \... | Baltic Way | Baltic Way 2013 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | Yes | |
0ehg | Problem:
Naj bo $AB$ premer krožnice $\mathcal{K}$, očrtane tetivnemu štirikotniku $ABCD$. Premici $AD$ in $BC$ se sekata v točki $E$, tangenti na krožnico $\mathcal{K}$ v točkah $C$ in $D$ pa se sekata v točki $F$. Dokaži, da sta premici $EF$ in $AB$ pravokotni. | [
"Solution:\n\n\n\nOznačimo $\\alpha=\\angle BAE$ in $\\beta=\\angle EBA$, in naj bo $O$ središče krožnice $\\mathcal{K}$. Potem je $\\angle AEB=\\pi-\\alpha-\\beta$. Zaradi tetivnosti štirikotnika $ABCD$ je $\\angle DCB=\\pi-\\alpha$ in $\\angle ADC=\\pi-\\beta$. Ker je $\\angle OCB=\\angle... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije Državno tekmovanje | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0ec0 | Problem:
Naj bo $P$ razpolovišče stranice $AB$ trikotnika $ABC$. Zrcalna slika poltraka $PC$ pri zrcaljenju čez premico $AB$ seka trikotniku $ABC$ očrtano krožnico v točki $D$. Naj bo $E$ drugo presečišče premice $CP$ s trikotniku $ABC$ očrtano krožnico. Dokaži, da je $|AE| = |BD|$.
 | [
"Solution:\n\n1. način. Naj bo $\\mathcal{K}$ trikotniku $ABC$ očrtana krožnica. Ker je poltrak $PD$ zrcalna slika poltraka $PC$ pri zrcaljenju čez premico $AB$, velja $\\Varangle DPB = \\Varangle BPC = \\Varangle APE$. Oglejmo si zrcaljenje preko simetrale stranice $AB$. Točka $A$ se prezrcali v točko $B$. Ker je ... | Slovenia | 59. matematično tekmovanje srednješolcev Slovenije | [
"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 | |
08xb | Two circles $O_1$ and $O_2$ intersect at two distinct points $P$ and $Q$. The tangent line to the circle $O_1$ at the point $P$ intersects the circle $O_2$ at $R$, different from $P$, and the tangent line to the circle $O_2$ at the point $Q$ intersects the circle $O_1$ at $S$, different from $Q$. Let $X$ be the point o... | [
"\n\nIn view of a well-known theorem on angles subtended by arcs on a circle, we have $\\angle PSQ = \\angle QPR$, and $\\angle SQP = \\angle PRQ$. This implies that the triangles $PSQ$ and $QPR$ are similar triangles. Since the circles $O_1$ and $O_2$ are circum-circles of the triangles $P... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | (2/9)^(1/3) | |
0ajz | Determine all positive integers $n$ such that $(n^3+39n-2)n+17-21^n+5$ is a full square. | [
"Let us denote $a_n = (n^3+39n-2)n+17-21^n+5$.\n\nIf $n \\ge 4$, then $8 \\mid n!$. Moreover,\n$$\na_n \\equiv 5^n + 5 \\pmod{8}.\n$$\nIf $n$ is an even number, then $5^n \\equiv 1 \\pmod{8}$, so $a_n \\equiv 6 \\pmod{8}$. But, all full squares have remainder $0,1$ or $4$ when divided by $8$. So, if $n \\ge 4$ and ... | North Macedonia | Macedonian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | 1 | |
065z | A Mediterranean polynomial has only real roots and it is of the form
$$P(x) = x^{10} - 20x^9 + 135x^8 + a_7x^7 + a_6x^6 + a_5x^5 + a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0,$$ with real coefficients $a_0, a_1, \ldots, a_7$. Determine the largest real number that occurs as a root of some Mediterranean polynomial. | [] | Greece | Mediterranean Mathematical Competition | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | 11 | |
0agf | A trapezoid $ABCD$ is given, such that $\overline{AB} = \overline{AC} = \overline{BD}$. Let $M$ be the midpoint of $CD$. Find the angles of the trapezoid if $\angle MBC = \angle CAB$. | [
"By the conditions of the task it follows that the trapezoid is isosceles. Let $K$ be the midpoint of $AD$, and let $\\angle CAB = \\angle MBC = \\varphi$. Then\n$$\n\\angle MKA = 180^\\circ - \\angle KAC = 180^\\circ - \\angle MBA.\n$$\nTherefore the quadrilateral $ABMK$ is inscribed. Then, by the conditions we ha... | North Macedonia | Macedonian Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | ∠A = 75°, ∠B = 75°, ∠C = 105°, ∠D = 105° | |
0btg | A convex hexagon $A_1B_1A_2B_2A_3B_3$ is inscribed in a circle $\Omega$ of radius $R$. The diagonals $A_1B_2$, $A_2B_3$, and $A_3B_1$ concur at $X$. For $i = 1, 2, 3$, let $\omega_i$ be the circle tangent to the segments $XA_i$ and $XB_i$, and to the arc $A_iB_i$ of $\Omega$ not containing other vertices of the hexagon... | [
"a) Let $\\ell_1$ be the tangent to $\\Omega$ parallel to $A_2B_3$, lying on the same side of $A_2B_3$ as $\\omega_1$. The tangents $\\ell_2$ and $\\ell_3$ are defined similarly. The lines $\\ell_1$ and $\\ell_2$, $\\ell_2$ and $\\ell_3$, $\\ell_3$ and $\\ell_1$ meet at $C_3$, $C_1$, $C_2$, respectively (see Fig. ?... | Romania | 2016 Eighth Romanian Master of Mathematics | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
01di | Let $ABC$ be a triangle and let $P$ be a point such that $AP$ is the angle bisector of $\angle BAC$ and segment $BC$ bisects segment $AP$. Prove that perimeter of triangle $ABC$ is greater than or equal to perimeter of triangle $PBC$. | [
"If $AB = AC$ then the perimeters are equal. Further we assume that $AB < AC$.\n\nLet $D$ be the reflection of $A$ about the midpoint of $BC$. Let $E$ be the reflection of $A$ about $BC$. Then $BCDE$ is an isosceles trapezoid. Moreover,\n\n$P$ lies on the segment $DE$ because the angle bisector lies between the alt... | Baltic Way | Baltic Way 2016 | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09cf | $ABC$ гурвалжны $AB$ ба $AC$ талууд дээр харгалзан $D$ ба $E$ цэгүүдийг $DE \parallel BC$ байхаар авав. $P$ нь $ADE$ гурвалжин дотор орших дурын цэг ба $PB$ ба $PC$ хэрчмүүд нь $DE$ хэрчимтэй харгалзан $F$ ба $G$ цэгүүдэд огтлолцдог. $\triangle DPG$ ба $\triangle FPE$-г багтаасан тойргийн төвүүд нь харгалзан $O_1$ ба $... | [
"$\\triangle DPG$-г багтаасан тойрог $AB$ талтай $M$ цэгт, $AP$-тэй $Q_1$ цэгт огтлолцдог байг.\n\nХарин $\\triangle FPE$-г багтаасан тойрог $AC$ талтай $N$ цэгт, $AP$-тэй $Q_2$ цэгт огтлолцдог байг. $ENPF$ тойрогт багтах ба $ED \\parallel CB$\n\n---\n\nучраас $\\angle ANP = \\angle PFE = \... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Mongolian | proof only | null | |
0hij | Find the smallest real number $C$ for which the following condition is true: for any different positive integers $x, y$ the inequality holds
$$
\{\{\sqrt{x^2 + 2y}\}, \{\sqrt{y^2 + 2x}\}\} < C.
$$
Here, $\{a\} \in [0; 1)$ denotes the fractional part of the number $a$, that is, there exists an integer $n$ for which the... | [
"We will show that the desired $C$ is the positive root of the equation $x^2 + x = 1$, $C = \\frac{\\sqrt{5}-1}{2}$.\n\nFirst, suppose that for some positive integers $x < y$ the inequalities $\\sqrt{x^2 + 2y} > C$, $\\{\\sqrt{y^2 + 2x}\\} > C$ hold. Note that $y^2 < y^2 + 2x < (y + 1)^2$, so we have $y^2 + 2x > (y... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | (\sqrt{5}-1)/2 | |
06f3 | Let $ABCDEF$ be a convex hexagon such that $AB \parallel DE$, $BC \parallel EF$ and $CD \parallel FA$. Let $H$, $I$, $J$, $K$, $L$ and $M$ be the midpoints of sides $AB$, $BC$, $CD$, $DE$, $EF$ and $FA$, respectively. Prove that lines $HK$, $IL$ and $JM$ are concurrent. | [
"Let $AD$ meet $BE$ at $X$, $BE$ meet $CF$ at $Y$, and $CF$ meet $AD$ at $Z$. Since $AB \\parallel ED$, we know that $\\triangle XAB \\sim \\triangle XDE$. Also, the midpoints $H$ and $K$ of $AB$ and $DE$ are corresponding points under this similarity. Thus, $X$ lies on $HK$. Similarly, $Y$ lies on $IL$, and $Z$ li... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
03np | Problem:
A function $f$ from the positive integers to the positive integers is called Canadian if it satisfies
$$
\operatorname{gcd}(f(f(x)), f(x+y)) = \operatorname{gcd}(x, y)
$$
for all pairs of positive integers $x$ and $y$.
Find all positive integers $m$ such that $f(m) = m$ for all Canadian functions $f$. | [
"Solution:\nDefine an $m \\in \\mathbb{N}$ to be good if $f(m) = m$ for all such $f$. It will be shown that $m$ is good if and only if $m$ has two or more distinct prime divisors. Let $P(x, y)$ denote the assertion\n$$\n\\operatorname{gcd}(f(f(x)), f(x+y)) = \\operatorname{gcd}(x, y)\n$$\nfor a pair $x, y \\in \\ma... | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | All positive integers with at least two distinct prime divisors. | |
070p | Problem:
$k$ is the positive root of the equation $x^{2} - 1998x - 1 = 0$. Define the sequence $x_{0}, x_{1}, x_{2}, \ldots$ by $x_{0} = 1$, $x_{n+1} = \left[ k x_{n} \right]$. Find the remainder when $x_{1998}$ is divided by $1998$. | [
"Solution:\n\nPut $p(x) = x^{2} - 1998x - 1$. Then $p(1998) = -1$, $p(1999) = 1998$, so $1998 < k < 1999$. Also $k$ is irrational (using the formula for the root of a quadratic).\n\nWe have $x_{n} = \\left[ k x_{n-1} \\right]$, so $x_{n} < k x_{n-1}$ and $> k x_{n-1} - 1$. Hence $x_{n} / k < x_{n-1} < x_{n} / k + 1... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 1000 | |
0io3 | Problem:
Let the incircle of $ABCD$ be tangent to sides $AB$, $BC$, $CD$, and $AD$ at points $P$, $Q$, $R$, and $S$, respectively. Show that $ABCD$ is cyclic if and only if $PR \perp QS$. | [
"Solution:\n\nLet the diagonals of $PQRS$ intersect at $T$. Because $\\overline{AP}$ and $\\overline{AS}$ are tangent to $\\omega$ at $P$ and $S$, we may write $\\alpha = \\angle ASP = \\angle SPA = \\angle SQP$ and $\\beta = \\angle CQR = \\angle QRC = \\angle QPR$. Then $\\angle PTQ = \\pi - \\alpha - \\beta$. On... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0gwj | There are $n \ge 3$ soldiers in the captain Petrenko's squad, and none two of them have the same height. The captain has drawn them all up into a single rank (not necessarily sorted by height). We call a "wave" any subsequence of soldiers in this rank (they are not supposed to stand next to each other) such that the fi... | [
"Спочатку покажемо, що одна з найдовших \"хвиль\" шеренги містить найвищого солдата $A$ цієї шеренги.\nДійсно, нехай солдат $A$ не входить до жодної найдовшої \"хвилі\". Якщо вся найдовша \"хвиля\" знаходиться справа від $A$, то першого солдата \"хвилі\" замінюємо на $A$. Якщо вся \"хвиля\" знаходиться зліва від $A... | Ukraine | Ukrainian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | They are equal; the numbers are the same for every n. | |
08lw | Problem:
Determine the minimal prime number $p > 3$ for which no natural number $n$ satisfies
$$
2^{n} + 3^{n} \equiv 0 \pmod{p}
$$ | [
"Solution:\nWe put $A(n) = 2^{n} + 3^{n}$. From Fermat's little theorem, we have $2^{p-1} \\equiv 1 \\pmod{p}$ and $3^{p-1} \\equiv 1 \\pmod{p}$, from which we conclude $A(n) \\equiv 2 \\pmod{p}$. Therefore, after $p-1$ steps at most, we will have repetition of the power. It means that in order to determine the min... | JBMO | 2008 Shortlist JBMO | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | 19 | |
0cl9 | Let $ABC$ be an acute triangle, with $AB \neq AC$. Let $D$ be the midpoint of $BC$ and $I, J, K$ be the feet of the altitudes from $A, B$ and $C$, respectively, in the triangle $ABC$. The perpendicular from $A$ to the line $AD$ meets the lines $BJ$ and $CK$ at points $N$ and $Q$, respectively, and the parallel to $BC$ ... | [
"Let $H$ be the orthocenter of the triangle $ABC$. Quadrilaterals $BIHK$ and $CIHJ$ are cyclic, so $\\angle HIK = \\angle HBK = \\angle ABJ = \\angle ACK = \\angle JCH = \\angle HIJ$.\n\nTherefore, $IA$ is the angle bisector of $JIK$. Since $IA \\perp BC$ and $MP \\parallel BC$, it follows that $IA \\perp MP$. Beca... | Romania | 75th Romanian 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 > Miscellaneous > Angle chasing"
] | English | proof only | null | |
09zb | In a sequence $a_1, a_2, \dots, a_{1000}$ consisting of $1000$ distinct numbers, a pair $(a_i, a_j)$ with $i < j$ is called *ascending* if $a_i < a_j$ and *descending* if $a_i > a_j$. Determine the largest positive integer $k$ with the property that every sequence of $1000$ distinct numbers has at least $k$ non-overlap... | [
"We will prove that the greatest $k$ is $333$. First consider the sequence $1000, 999, 998, \\ldots, 669, 668, 1, 2, 3, \\ldots, 666, 667$. The first $333$ numbers in the sequence are not usable in an ascending pair, because for each of these numbers the numbers left of it are all greater and the numbers right of i... | Netherlands | IMO Team Selection Test 1 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 333 | |
0e4q | Let $m$ and $n$ be positive integers, such that $19 \le m \le 49$, $51 \le n \le 101$. What is the greatest possible value of the expression $\frac{n+m}{n-m}$?
(A) 20
(B) 30
(C) 40
(D) 50
(E) 60 | [
"We have $\\frac{n+m}{n-m} = \\frac{n-m+2m}{n-m} = 1 + 2\\frac{m}{n-m}$, so the value is maximal when $n-m = 2$ and $m = 49$. Then, the value is $50$."
] | Slovenia | National Math Olympiad | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | D | |
04r7 | A number $n$ is a product of three (not necessarily distinct) prime numbers. Adding 1 to each of them, after multiplication we get a larger product $n + 963$. Determine the original product $n$. (Pavel Novotný) | [
"We look for $n = p \\cdot q \\cdot r$, with primes $p \\le q \\le r$ satisfying\n$$(p+1)(q+1)(r+1) = pqr + 963. \\quad (1)$$\nIf $p=2$, the right-hand side of (1) is odd, hence the factors $q+1, r+1$ on the left must be odd too. This implies that $p=q=r=2$, which contradicts to (1). Thus we have proved that $p \\g... | Czech Republic | 63rd Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 2013 | |
09to | Problem:
In een niet-gelijkbenige driehoek $\triangle A B C$ geldt $\angle B A C=60^{\circ}$. Zij $D$ het snijpunt van de bissectrice van $\angle B A C$ met de zijde $B C$, $O$ het middelpunt van de omgeschreven cirkel van $\triangle A B C$ en $E$ het snijpunt van $A O$ met $B C$. Bewijs dat $\angle A E D+\angle A D O... | [
"Solution:\n\nZij $M$ het tweede snijpunt van $A D$ met de omgeschreven cirkel van $\\triangle A B C$. Dan is $M$ het midden van de boog $B C$ waar $A$ niet op ligt. Er geldt nu $\\angle C O M=\\frac{1}{2} \\angle C O B=\\angle C A B=60^{\\circ}$. Verder is natuurlijk $|O C|=|O M|$, dus $\\triangle O C M$ is gelijk... | Netherlands | Selectietoets | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
004e | En el año 2007 murió una tortuga y la cantidad de años que vivió coincide con el producto de los dígitos de su año de nacimiento. Se sabe que la tortuga vivió al menos un año y a lo más 2000 años. ¿En qué año nació la tortuga? | [] | Argentina | XVI Olimpiada Matemática Rioplatense | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Spanish | proof and answer | 1863 | |
0jwy | Problem:
Start by writing the integers $1,2,4,6$ on the blackboard. At each step, write the smallest positive integer $n$ that satisfies both of the following properties on the board.
- $n$ is larger than any integer on the board currently.
- $n$ cannot be written as the sum of 2 distinct integers on the board.
Find t... | [
"Solution:\n\nThe sequence goes\n$$\n1,2,4,6,9,12,17,20,25, \\ldots\n$$\nCommon differences are $5,3,5,3,5,3, \\ldots$, starting from $12$. Therefore, the answer is $12+47 \\times 8=388$."
] | United States | February 2017 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 388 | |
0ejx | Problem:
Poišči vsa realna števila $x$, $y$ in $z$, ki rešijo sistem enačb
$$
\frac{3 x y}{x-y}=2, \quad \frac{2 y z}{y+2 z}=3, \quad \frac{x z}{z-4 x}=3
$$ | [
"Solution:\n\nV enačbah najprej odpravimo ulomke in dobimo\n$$\n\\begin{aligned}\n3 x y & =2 x-2 y \\\\\n2 y z & =3 y+6 z \\\\\nx z & =3 z-12 x\n\\end{aligned}\n$$\nPrvo enačbo preoblikujemo v $(3 y-2) x=-2 y$. Če je $3 y-2=0$ oziroma $y=\\frac{2}{3}$, tedaj sledi $y=0$, kar pa je protislovje. Torej enačbo lahko de... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | x = -1, y = 2, z = -3 | |
0i9k | Problem:
Write down an integer $N$ between $0$ and $10$, inclusive. You will receive $N$ points unless some other team writes down the same $N$, in which case you receive nothing. | [] | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | final answer only | null | |
0740 | Let $n \ge 2$ be an integer. If $n$ divides $3^n + 4^n$, prove that $7$ divides $n$. | [
"Observe that $n$ is odd and $3$ does not divide $n$. Let $p$ be the least prime dividing $n$. Let $c$ be an integer such that $0 < c < p$ and $4c \\equiv 3 \\pmod{p}$. Such a number $c$ exists since $\\gcd(3, 4) = 1$. Thus we have\n$$\n\\begin{align*}\n4^n(c^{2n} - 1) &\\equiv (c^n - 1)((4c)^n + 4^n) \\\\\n&\\equi... | India | Indija TS 2009 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0huq | Problem:
Prove that for every positive integer $n$, there is an integer $x$ such that $x^{2}-17$ is divisible by $2^{n}$. | [
"Solution:\nWe prove this by induction on $n$.\n\nIf $n=1,2$, or $3$, then $x=1$ works.\n\nSuppose that $x^{2}-17$ is divisible by $2^{n}$ and $n \\geq 3$. We seek to find $y$ such that $y^{2}-17$ is divisible by $2^{n+1}$. Let\n$$\nx^{2}-17=k \\cdot 2^{n} \\text{.}\n$$\nIf $k$ is even, we are done since $x^{2}-17$... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
05at | Juku and Miku play the following game on a grid of dimensions $n \times m$: In the beginning, all unit squares are white. Each player on their turn paints one white unit square either red or blue of their choice, but no two unit squares with a common side or a common vertex can be painted the same color. The players ta... | [
"If $n$ and $m$ are even, then there is a middle square on the grid. Let Juku paint the middle square any color on the first move. From now on, each of Juku's moves should mirror Miku's last move relative to the center of the grid. If before Miku's move the position is symmetrical with respect to the center of the ... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) Yes; b) No; c) No | |
053e | In Wonderland there are at least 5 towns. Some towns are connected directly by roads or railways. Every town is connected to at least one other town and for any four towns there exists some direct connection between at least three pairs of towns among those four. When entering the public transportation network of this ... | [
"Let $A$ and $B$ be any two towns. We know that it must be possible to move from $A$ to some other town $X$ and from $B$ to some other town $Y$. From four towns $A, B, X, Y$ we can form three pairs which all have a direct connection between them. Of those at least one way goes from either $A$ or $X$ to either $B$ o... | Estonia | IMO Team Selection Contest | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0iq2 | Let $ABC$ be an acute, scalene triangle, and let $M$, $N$, and $P$ be the midpoints of $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively. Let the perpendicular bisectors of $\overline{AB}$ and $\overline{AC}$ intersect ray $AM$ in points $D$ and $E$ respectively, and let lines $BD$ and $CE$ intersect ... | [
"Assume without loss of generality that $AB > AC$, and we consider the configuration shown below. Our proof can be modified for other configurations.\nLet $O$ be the circumcenter of triangle $ABC$. Point $D$ lies on $PO$, which is the perpendicular bisector of segment $AB$. Thus, $ABO$ is an isosceles triangle with... | United States | USAMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transfor... | null | proof only | null | |
0ayw | Problem:
Given that $a$ and $b$ are real numbers satisfying the equation
$$
\log_{16} 3 + 2 \log_{16}(a-b) = \frac{1}{2} + \log_{16} a + \log_{16} b
$$
find all possible values of $\frac{a}{b}$. | [] | Philippines | 21st PMO Area Stage | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | 3 | |
0308 | Problem:
Lembrando que
$$
\begin{aligned}
(a+b+c)^{3} = a^{3} + b^{3} + c^{3} + 3 a^{2} b + 3 a b^{2} + 3 a c^{2} + 3 a^{2} c + 3 b^{2} c + 3 b c^{2} + 6 a b c
\end{aligned}
$$
Encontre as soluções do sistema de equações
$$
\begin{aligned}
a^{3} + 3 a b^{2} + 3 a c^{2} - 6 a b c & = 1 \\
b^{3} + 3 b a^{2} + 3 b c^{2} ... | [
"Solution:\n\nSejam $A = a^{3} + 3 a b^{2} + 3 a c^{2} - 6 a b c$, $B = b^{3} + 3 b a^{2} + 3 b c^{2} - 6 a b c$ e $C = c^{3} + 3 c a^{2} + 3 c b^{2} - 6 a b c$. Usando a identidade algébrica mencionada no enunciado, temos\n$$\n\\begin{aligned}\n-A + B + C & = (-a + b + c)^{3} \\\\\nA - B + C & = (a - b + c)^{3} \\... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | (1, 1, 1) | |
01tn | Vertices of a regular $n$-gon are marked on a circumference. Integer numbers are placed at the vertices such that the difference of any two adjacent numbers equals $\pm 1$ or $0$. At a moment, simultaneously, the half-sums of each pair of the adjacent numbers are placed at the midpoints of the arcs subtending the corre... | [
"**Answer:** all even $n$.\nThe following example shows that $n = 2k + 1$ does not satisfy the problem condition. We number clockwise the vertices of the $n$-gon as $A_0, A_1, \\dots, A_k, A_{-k}, \\dots, A_{-1}$ and place the number $|i|$ at the vertex $A_i$. Since $n$ is odd, the pairs of antipodal points have th... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | all even n | |
05s3 | Problem:
Soit $u$ un entier naturel non nul.
Démontrer qu'il n'existe qu'un nombre fini de triplets d'entiers naturels $(a, b, n)$ tels que $n! = u^{a} - u^{b}$.
Note : on rappelle que $0! = 1! = 1$. | [
"Solution:\n\nSoit $p$ un nombre premier impair qui ne divise pas $u$, et soit $k$ un entier tel que $p^{k} > u^{p-1} - 1$. Posons $q = p^{k-1}$. On montre tout d'abord que $u$ n'est pas une puissance $q$-ième modulo $p^{k}$. En effet, si $u$ était une puissance $q$-ième modulo $p^{k}$, alors il existerait un entie... | France | Préparation Olympique Française de Mathématiques | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analys... | null | proof only | null | |
02zy | Problem:
Seja $f(x)=\frac{9^{x}}{9^{x}+3}$.
a) Encontre o valor de $f(x)+f(1-x)$.
b) Calcule o valor da soma
$$
f\left(\frac{1}{2020}\right)+f\left(\frac{2}{2020}\right)+f\left(\frac{3}{2020}\right)+\ldots+f\left(\frac{2019}{2020}\right)
$$ | [
"Solution:\n\na) Temos\n$$\n\\begin{aligned}\nf(x)+f(1-x) & =\\frac{9^{x}}{9^{x}+3}+\\frac{9^{1-x}}{9^{1-x}+3} \\\\\n& =\\frac{9^{x}}{9^{x}+3}+\\frac{9}{9+3 \\cdot 9^{x}} \\\\\n& =\\frac{9^{x}}{9^{x}+3}+\\frac{3}{3+9^{x}} \\\\\n& =1\n\\end{aligned}\n$$\n\nb) Em virtude do item anterior, podemos juntar os termos cor... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | a) 1; b) 2019/2 | |
08ej | Problem:
Sia $ABCD$ un rettangolo e sia $E$ un punto arbitrario, diverso da $C$, sul lato $DC$. Sia $H$ la proiezione di $E$ sulla diagonale $AC$ e sia $K$ la proiezione di $C$ sulla semiretta $AE$.
a. Dimostrare che $K$ giace sulla circonferenza circoscritta ad $ABCD$ e che il quadrilatero $CKEH$ è ciclico, cioè ins... | [
"Solution:\n\nAffrontiamo in sequenza i punti.\n\n\n\na.\nLa diagonale $AC$ è diametro della circonferenza $\\Gamma_1$ circoscritta al rettangolo $ABCD$. L'angolo $\\widehat{CKA}$ è retto per costruzione e la circonferenza circoscritta al triangolo rettangolo $CKA$ ha $AC$ come diametro. Qu... | Italy | Italian Mathematical Olympiad - February Round | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
078k | All the squares of a $2024 \times 2024$ board are coloured white. In one move, Mohit can select one row or column whose every square is white, choose exactly $1000$ squares in this row or column, and colour all of them red. Find the maximum number of squares that Mohit can colour red in a finite number of moves. | [
"Let $n = 2024$ and $k = 1000$. We claim that the maximum number of squares that can be coloured in this way is $k(2n - k)$, which evaluates to $3048000$.\n\nIndeed, call a row/column *bad* if it has at least one red square. After the first move, there are exactly $k+1$ bad rows and columns: if a row was picked, th... | India | INMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 3048000 | |
04fl | Prove that $2012^9 + 2016^9$ is divisible by $2014$. (Nikola Adžaga) | [] | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
0667 | Let $D$ be the foot of the internal bisector of the angle $\angle A$ of the triangle $ABC$. The straight line which joins the incenters of the triangles $ABD$ and $ACD$ cuts $AB$ and $AC$ at $M$ and $N$, respectively. Show that $BN$ and $CM$ meet on the bisector $AD$. | [] | Greece | Mediterranean Mathematical Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > ... | English | proof only | null | |
0j3c | Problem:
A function $f(x, y)$ is linear in $x$ and in $y$. $f(x, y) = \frac{1}{x y}$ for $x, y \in \{3, 4\}$. What is $f(5,5)$? | [
"Solution:\nThe main fact that we will use in solving this problem is that $f(x+2, y) - f(x+1, y) = f(x+1, y) - f(x, y)$ whenever $f$ is linear in $x$ and $y$. Suppose that $f(x, y) = a x y + b y + c x + d = x(a y + c) + (b y + d)$ for some constants $a, b, c$, and $d$. Then it is easy to see that\n$$\n\\begin{alig... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 1/36 | |
0b0f | Problem:
Let $A = \{ n \in \mathbb{Z} \mid |n| \leq 24 \}$. In how many ways can two distinct numbers be chosen (simultaneously) from $A$ such that their product is less than their sum? | [] | Philippines | Philippines Mathematical Olympiad | [
"Discrete Mathematics > Other",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 623 | |
08z6 | Let $N$ be the set of positive integers. Determine all functions $f : N \to N$ such that, for all positive integers $m$ and $n$,
$$
f^{f(n)}(m) + mn = f(m)f(n).
$$
Note that $f^k(n) = \underbrace{f(f(\cdots f(n)\cdots))}_{k \text{ times}}$. | [
"Let $\\ell$ be a positive integer. By substituting $(m, n) = (f(\\ell), \\ell)$ and $(\\ell, f(\\ell))$ into the original equation and comparing them, we obtain\n$$\nf^{f(\\ell)+1}(\\ell) + \\ell f(\\ell) = f(\\ell)f(f(\\ell)) = f^{f(f(\\ell))}(\\ell) + \\ell f(\\ell),\n$$\nor $f^{f(\\ell)+1}(\\ell) = f^{f(f(\\ell... | Japan | Japan 2022 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | f(n) = n + 1 | |
04lt | Find the number of nine-digit numbers with digits $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ and $9$, such that no three consecutive digits equal $123$, $246$ or $678$. | [] | Croatia | Mathematical competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 348000 | |
0kbw | Problem:
An $11 \times 11$ grid is labeled with consecutive rows $0,1,2, \ldots, 10$ and columns $0,1,2, \ldots, 10$ so that it is filled with integers from $1$ to $2^{10}$, inclusive, and the sum of all of the numbers in row $n$ and in column $n$ are both divisible by $2^{n}$. Find the number of possible distinct gri... | [
"Solution:\n\nWe begin by filling the $10 \\times 10$ grid formed by rows and columns $1$ through $10$ with any values, which we can do in $\\left(2^{10}\\right)^{100} = 2^{1000}$ ways. Then in column $0$, there is at most $1$ way to fill in the square in row $10$, $2$ ways for the square in row $9$, down to $2^{10... | United States | HMMT February 2020 | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2^{1100} | |
0421 | Let $n \ge 2$ and $a_1, a_2, \dots, a_n, b_1, b_2, \dots, b_n$ be non-negative integers. Prove that
$$
\left(\frac{n}{n-1}\right)^{n-1}\left(\frac{1}{n}\sum_{i=1}^{n} a_i^2\right)+\left(\frac{1}{n}\sum_{i=1}^{n} b_i\right)^2 \geqslant \prod_{i=1}^{n}\left(a_i^2+b_i^2\right)^{\frac{1}{n}}.
$$ | [
"Denote $\\lambda = \\left(\\frac{n}{n-1}\\right)^{n-1}$, $n \\ge 2$. Obviously, $\\lambda > 1$.\nFor given $i \\in \\{1, \\dots, n\\}$, fix $p = a_k^2 + b_k^2$ for $k = 1, 2, \\dots, n$ and fix $a_j$ and $b_j$ ($j \\ne i$). Then the left-hand side of\n$$\n\\textcircled{1} = \\frac{\\lambda}{n} (p - b_i^2 + \\sum_{... | China | China National Team Selection Test | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0cag | Problem:
Fie $M$ o mulţime formată din 13 numere naturale de trei cifre.
Arătaţi că există o submulţime nevidă $S \subset M$ şi o combinaţie de operaţii aritmetice elementare (adunare, scădere, înmulţire, împărţire - fără a utiliza parantezele) între elementele lui $S$, astfel încât valoarea expresiei rezultate să fie... | [] | Romania | Al treilea test de selecţie pentru OBMJ Bucureşti | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
09yt | A set consisting of at least two distinct positive integers is called *centenary* if its greatest element is $100$. We will consider the average of all numbers in a centenary set, which we will call the average of the set. For example, the average of the centenary set $\{1, 2, 20, 100\}$ is $\frac{123}{4}$ and the aver... | [
"If you decrease one of the numbers (unequal to $100$) in a centenary set, the average becomes smaller. Also if you add a number that is smaller than the current average, the average becomes smaller. To find the centenary set with the smallest possible average, we can start with $1, 100$ and keep adjoining numbers ... | Netherlands | Final Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | All integers from 14 to 99 inclusive | |
07gx | Let $ABC$ be an acute-angled triangle and $D$ be the foot of altitude from $A$. Let $K$, $L$ be the touching points of tangent lines from $D$ to the circles with diagonals $AB$ and $AC$, respectively. Point $S$ is given on the plane such that
$$
\angle ABC + \angle ABS = \angle ACB + \angle ACS = 180^\circ.
$$
Prove th... | [
"Suppose that $L$ and $K$ lie on the circumcircle of $ABD$ and $ADC$, respectively.\n\n\nNote that $\\angle LAB = 90^\\circ - \\angle ABL = 90^\\circ - \\angle ADL = 90^\\circ - \\angle ACB$. Similarly we have $\\angle CAK = 90^\\circ - \\angle CBA$. Summing these two implies that $\\angle ... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
05zw | Problem:
Soit $a, b, c > 0$ tels que $a^{2} < 16 b c$, $b^{2} < 16 a c$ et $c^{2} < 16 a b$. Montrer que
$$
a^{2} + b^{2} + c^{2} < 2(a b + b c + a c)
$$ | [
"Solution:\n\nRemarquons tout d'abord que l'on a forcément $\\sqrt{a} < \\sqrt{b} + \\sqrt{c}$. En effet, dans le cas contraire, on aurait\n$$\n16 b c > a^{2} = (\\sqrt{a})^{4} \\geqslant (\\sqrt{b} + \\sqrt{c})^{4} \\geqslant 16 b c\n$$\npar IAG car $\\sqrt{b} + \\sqrt{c} \\geqslant 2 \\sqrt[4]{b c}$, absurde. Ain... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - Envoi 5 : Pot Pourri | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0df8 | Consider the polynomial $f(x) = c x(x - 2)$ where $c$ is a positive real number. For any $n \in \mathbb{Z}^{+}$, the notation $g_n(x)$ is a composite function $n$ times of $f$ and assume that the equation $g_n(x) = 0$ has all of the $2^n$ solutions are real numbers.
1. For $c = 5$, find in terms of $n$, the sum of all ... | [
"1) We will prove by induction on $n$ that the solutions of $g_n(x)$ are all distinct. With $n=1$, we have $g_1(x) = f(x)$ which has two solutions, $x=0, x=2$. Suppose the polynomial $g_n(x)$ has all $2^n$ solutions that are distinct, set as $x_1, x_2, \\dots, x_{2^n}$, we write $g_n(x) = k(x - x_1)(x - x_2) \\dots... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | Part 1: 2^n. Part 2: c ≥ 1. |
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