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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0h0j | Real numbers $a_1, a_2, \dots, a_n$ satisfy the following conditions $a_1 + a_2 + \dots + a_n = n$ and $a_1 \ge a_2 \ge \dots \ge a_n \ge 0$. Prove the inequality:
$$
na_1 \ge a_1^2 + a_2^2 + \dots + a_n^2.
$$ | [
"Proof immediately follows from next transformations: $a_1^2 + a_2^2 + \\dots + a_n^2 \\le a_1^2 + a_2^2 + \\dots + a_n^2 + a_1(a_1 - a_1) + a_2(a_2 - a_2) + a_3(a_3 - a_3) + \\dots + a_n(a_n - a_n) = a_1(a_1 + a_2 + \\dots + a_n) = na_1$ and we are done."
] | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 24, 2010) | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0bpi | Problem:
Fie $f:(0, \infty) \rightarrow (0, \infty)$ o funcţie neconstantă care are proprietatea
$$
f\left(x^{y}\right) = (f(x))^{f(y)}
$$
pentru orice $x, y > 0$. Să se arate că
$$
f(xy) = f(x) f(y) \text{ şi } f(x+y) = f(x) + f(y)
$$
pentru orice $x, y > 0$. | [
"Solution:\n\nFie $a > 0$ astfel ca $f(a) \\neq 1$. Avem, pentru $x, y$ arbitrari,\n$$\nf\\left(a^{xy}\\right) = f(a)^{f(xy)}\n$$\ndar\n$$\nf\\left(a^{xy}\\right) = f\\left(\\left(a^{x}\\right)^{y}\\right) = f\\left(a^{x}\\right)^{f(y)} = \\left(f(a)^{f(x)}\\right)^{f(y)} = f(a)^{f(x) f(y)}\n$$\nde unde $f(xy) = f(... | Romania | Olimpiada Naţională de Matematică Etapa Judeţeană şi a Municipiului Bucureşti | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof only | null | |
0cnr | 100 thimbles are arranged on a circle. A token is placed under one of the thimbles. By one turn, a player can pick up arbitrary four thimbles and check whether a token is under one of them. After that, the thimbles return to their places, and the token moves to one of the two neighboring thimbles. Find the least number... | [
"**Ответ.** За 33 хода.\n\nПосле каждого нашего хода и перемещения монетки будем поворачивать все наперстки (вместе с монеткой) по ходу часовой стрелки на одну позицию. Тогда будем считать, что после каждого хода монетка либо остается на месте, либо перемещается на две позиции по часовой стрелке, а наперстки остают... | Russia | Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English; Russian | proof and answer | 33 | |
0i0r | Problem:
Find all functions $f$ from the set $\mathbf{R}$ of real numbers to itself such that $f(x y+1)=x f(y)+2$ for all $x, y \in \mathbf{R}$. | [
"Solution:\nThe only such function is $f(x)=2 x$. First, we note that this function really is a solution of the equation, since $2(x y+1)=2 x y+2=x(2 y)+2$ for all $x, y$.\n\nNow let $f$ be any function satisfying the equation. First put $x=y=0$; the equation gives $f(0 \\cdot 0+1)=0 f(0)+2$ or $f(1)=2$. Now suppos... | United States | Berkeley Math Circle Monthly Contest #1 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 2x | |
0e64 | Let $K$ be the circumscribed circle of a triangle $ABC$. Let $D$ be the midpoint of the arc $AB$ that does not contain the point $C$, and let $E$ the midpoint of the arc $AC$ that does not contain the point $B$. Let $F$ and $G$ denote the points of tangency of the inscribed circle of the triangle $ABC$ with the sides $... | [
"Let $I$ be the center of the inscribed circle of the triangle $ABC$, and let $K$ and $L$ be the intersection points of the line $DE$ with the lines $AB$ and $AC$, respectively. Because the bisector of an angle of a triangle goes through the center of the inscribed circle and the midpoint of the opposite arc, the p... | Slovenia | National Math Olympiad 2012 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadri... | null | proof only | null | |
0fz4 | Problem:
Sei $I$ der Inkreismittelpunkt und $AD$ der Durchmesser des Umkreises eines Dreiecks $ABC$. Seien $E$ und $F$ Punkte auf den Strahlen $BA$ und $CA$ mit
$$
BE = CF = \frac{AB + BC + CA}{2}
$$
Zeige, dass sich die Geraden $EF$ und $DI$ rechtwinklig schneiden. | [
"Solution:\n\nSeien $B_{0}, C_{0}$ die Berührungspunkte des Inkreises des Dreiecks $ABC$ auf den Seiten $AC, AB$. Wir können nun schreiben:\n$$\nx = AB_{0} = AC_{0}, \\quad y = C_{0}B, \\quad z = B_{0}C\n$$\nwobei $x + y + z = \\frac{AB + BC + CA}{2}$ gilt. Die Bedingung aus der Aufgabenstellung liefert uns jetzt $... | Switzerland | IMO Selektion | [
"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 > Miscellaneous > Angle c... | null | proof only | null | |
07gd | Let $a_1, \dots, a_n; b_1, \dots, b_n; c_1, \dots, c_n$ be real numbers. Prove that
$$
\sqrt{\sum_{i=1}^{n} (3a_i - b_i - c_i)^2} + \sqrt{\sum_{i=1}^{n} (3b_i - a_i - c_i)^2} + \sqrt{\sum_{i=1}^{n} (3c_i - a_i - b_i)^2} \\
\geq \sqrt{\sum_{i=1}^{n} a_i^2} + \sqrt{\sum_{i=1}^{n} b_i^2} + \sqrt{\sum_{i=1}^{n} c_i^2}. \qq... | [
"According to the *Minkowski's* inequality we have\n$$\n\\sqrt{\\sum_{i=1}^{n} (3a_i - b_i - c_i)^2} + \\sqrt{\\sum_{i=1}^{n} b_i^2} + \\sqrt{\\sum_{i=1}^{n} c_i^2} \\\\\n\\geq \\sqrt{\\sum_{i=1}^{n} (3a_i - b_i - c_i + b_i + c_i)^2} = 3\\sqrt{\\sum_{i=1}^{n} a_i^2}.\n$$\nAdding the above inequality with other two ... | Iran | 38th Iranian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
025e | Problem:
Prove that the sum of the lengths of the legs of a right triangle never exceeds $\sqrt{2}$ times the length of the hypotenuse of the triangle. | [
"Solution:\nLet $a$ and $b$ be the legs of a right triangle and $c$ its hypotenuse. We know that\n$$\n\\begin{aligned}\n(a-b)^2 & \\geq 0 \\\\\na^2 + b^2 & \\geq 2ab\n\\end{aligned}\n$$\nBy the Pythagorean Theorem, we have\n$$\n\\begin{aligned}\n(a+b)^2 & = a^2 + 2ab + b^2 \\\\\n& \\leq 2(a^2 + b^2) \\\\\n& = 2c^2\... | Brazil | null | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0jad | Problem:
FemtoPravis is walking on an $8 \times 8$ chessboard that wraps around at its edges (so squares on the left edge of the chessboard are adjacent to squares on the right edge, and similarly for the top and bottom edges). Each femtosecond, FemtoPravis moves in one of the four diagonal directions uniformly at ran... | [
"Solution:\n\nAnswer: $\\left(\\frac{1+2^{1005}}{2^{1007}}\\right)^{2}$\n\nWe note the probability that he ends up in the same row is equal to the probability that he ends up in the same column by symmetry. Clearly these are independent, so we calculate the probability that he ends up in the same row.\n\nNow we num... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | final answer only | ((1+2^1005)/2^1007)^2 | |
0ije | Problem:
In bridge, a standard 52-card deck is dealt in the usual way to 4 players. By convention, each hand is assigned a number of "points" based on the formula
$$
4 \times (\# \{ A \text{'s} \}) + 3 \times (\# \{ K \text{'s} \}) + 2 \times (\# \{ Q \text{'s} \}) + 1 \times (\# \{ J \text{'s} \})
$$
Given that a part... | [
"Solution:\nObviously, we can ignore the cards lower than $J$. Simply enumerate the ways to get at least 13 points: $AAAA$ (1), $AAAK$ (16), $AAAQ$ (16), $AAAJ$ (16), $AAKK$ (36), $AAKQ$ (96), $AKKK$ (16). The numbers in parentheses represent the number of ways to choose the suits, given the choices for the values.... | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | proof and answer | 197/1820 | |
03ok | Suppose a quadratic function $f(x) = a x^2 + b x + c$ ($a, b, c \in \mathbb{R}$, and $a \neq 0$) satisfies the following conditions:
(1) When $x \in \mathbb{R}$, $f(x-4) = f(2-x)$ and $f(x) \ge x$.
(2) When $x \in (0, 2)$, $f(x) \le \left(\frac{x+1}{2}\right)^2$.
(3) The minimum value of $f(x)$ on $\mathbb{R}$ is $0$.
... | [
"Since $f(x-4) = f(2-x)$ for $x \\in \\mathbb{R}$, it is known that the quadratic function $f(x)$ has $x = -1$ as its axis of symmetry. By condition (3), we know that $f(x)$ opens upward, that is, $a > 0$. Hence\n$$\nf(x) = a(x+1)^2 \\quad (a > 0).\n$$\nBy condition (1), we get $f(1) \\ge 1$ and by (2), $f(1) \\le ... | China | China Mathematical Competition | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 9 | |
0hnm | Problem:
Mona has $12$ match sticks of length $1$, and she has to use them to make regular polygons, with each match being a side or a fraction of a side of a polygon, and no two matches overlapping or crossing each other. What is the smallest total area of the polygons Mona can make? | [
"$4 \\frac{\\sqrt{3}}{4} = \\sqrt{3}$."
] | United States | null | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | sqrt(3) | |
0i4o | Problem:
Eight knights are randomly placed on a chessboard (not necessarily on distinct squares). A knight on a given square attacks all the squares that can be reached by moving either (1) two squares up or down followed by one square left or right, or (2) two squares left or right followed by one square up or down. ... | [
"Solution:\n\n$0$. Since every knight attacks at most eight squares, the event can only occur if every knight attacks exactly eight squares. However, each corner square must be attacked, and some experimentation readily finds that it is impossible to place a knight so as to attack a corner and seven other squares a... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 0 | |
0izc | Problem:
Call an $2n$-digit base-10 number special if we can split its digits into two sets of size $n$ such that the sum of the numbers in the two sets is the same. Let $p_{n}$ be the probability that a randomly-chosen $2n$-digit number is special. (We allow leading zeros in $2n$-digit numbers).
a. The sequence $p_{... | [
"Solution:\n\na.\n\nAnswer: $\\frac{1}{2}$\n\nWe first claim that if a $2n$-digit number $x$ has at least eight $0$'s and at least eight $1$'s and the sum of its digits is even, then $x$ is special.\n\nLet $A$ be a set of eight $0$'s and eight $1$'s and let $B$ be the set of all the other digits. We split $B$ arbit... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | c = 1/2; r = 1/4; d = -1 | |
09ay | Parliament has $76$ members and $114$ working groups (each has at least two members). There is no two working groups that have exactly the same members. If a member is elected as a speaker, he have to leave all his working groups and after that if there are two working groups that have exactly the same members, then th... | [
"Let $S$ be the set of all the members. We can represent working groups as nonempty subsets of $S$. Let $A_1, A_2, \\dots, A_{114}$ be the working groups.\n\nTo the contrary, assume that for each $x \\in S$, there are at most $112$ distinct sets among the sets $A_1 - \\{x\\}, A_2 - \\{x\\}, \\dots, A_{114} - \\{x\\... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
01a1 | The set of vertices of the graph $G$ is a set of 2014 points in general position on the plane. The segment $AB$ is an edge of the graph iff each of the two (open) half-planes of line $AB$ contains 1006 points. Prove that $G$ does not contain a Hamiltonian path (i.e. a path that passes through every vertex exactly once)... | [
"Each vertex of the convex hull has degree 1 in the graph $G$. (When we rotate the line that passes through such point the numbers of other points in the half-planes change monotonically.) The convex hull contains at least 3 vertices, so $G$ has at least three “leaves”. Therefore there is no Hamiltonian path in it.... | Baltic Way | Baltic Way 2013 | [
"Discrete Mathematics > Graph Theory",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls"
] | null | proof only | null | |
04rh | Given a parallelogram $ABCD$ with center $S$, denote by $O$ the incenter of triangle $ABD$ and by $T$ the point of contact of the incircle of triangle $ABD$ with the diagonal $BD$. Prove that lines $OS$ and $CT$ are parallel. (Jaromír Šimša) | [
"Denote the lengths of $AB$, $AD$, and $BD$ by $a$, $b$, and $c$, respectively. If $a = b$ then both $OS$ and $CT$ coincide with $AC$ and the conclusion is trivial. Suppose $a > b$ (the case $b > a$ being completely analogous).\nLet $T'$ be the reflection of $T$ in $S$ (Fig. 1). As $CT \\parallel AT'$, it suffices ... | Czech Republic | 62nd Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0iwl | Problem:
Descartes's Blackjack: How many integer lattice points (points of the form $(m, n)$ for integers $m$ and $n$) lie inside or on the boundary of the disk of radius $2009$ centered at the origin?
If your answer is higher than the correct answer, you will receive $0$ points. If your answer is $d$ less than the c... | [] | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics"
] | null | final answer only | 4 * sum_{m=0}^{2009} floor(sqrt(2009^2 - m^2)) + 1 | |
0imw | Problem:
Let $a$ be a positive real number. Find the value of $a$ such that the definite integral
$$
\int_{a}^{a^{2}} \frac{\mathrm{d} x}{x+\sqrt{x}}
$$
achieves its smallest possible value. | [
"Solution:\nAnswer: $\\sqrt{3-2 \\sqrt{2}}$\nLet $F(a)$ denote the given definite integral. Then\n$$\nF'(a) = \\frac{\\mathrm{d}}{\\mathrm{d} a} \\int_{a}^{a^{2}} \\frac{\\mathrm{d} x}{x+\\sqrt{x}} = 2a \\cdot \\frac{1}{a^{2}+\\sqrt{a^{2}}} - \\frac{1}{a+\\sqrt{a}}.\n$$\nSetting $F'(a) = 0$, we find that $2a + 2\\s... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Integral Calculus > Applications",
"Calculus > Differential Calculus > Applications",
"Calculus > Differential Calculus > Derivatives"
] | null | proof and answer | 3 - 2√2 | |
0bgk | Problem:
O funcţie $f:(0, \infty) \rightarrow(0, \infty)$ se numeşte contractibilă dacă, pentru orice numere $x, y \in(0, \infty)$, avem $\lim_{n \rightarrow \infty}\left(f^{n}(x)-f^{n}(y)\right)=0$, unde $f^{n}=\underbrace{f \circ f \circ \ldots \circ f}_{\text{de } n \text{ ori } f}$.
a) Considerăm $f:(0, \infty) \r... | [
"Solution:\n\na) Presupunem prin reducere la absurd că $f$ ar admite un al doilea punct fix $x_{1} \\in(0, \\infty) \\setminus\\{x_{0}\\}$. Atunci $\\lim_{n \\rightarrow \\infty}\\left(f^{n}\\left(x_{0}\\right)-f^{n}\\left(x_{1}\\right)\\right)=x_{0}-x_{1} \\neq 0$, contradicţie. În acest caz, din continuitatea lui... | Romania | Olimpiada Naţională de Matematică Etapa Naţională | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof only | null | |
065r | Let $z_1, z_2, z_3, z_4, z_5, z_6$ be six pairwise different complex numbers which their images $A_1, A_2, A_3, A_4, A_5, A_6$ are consecutive points of the circle with center $O(0,0)$ and radius $r > 0$. If $w$ is a solution of the equation $z^2 + z + 1 = 0$ and
$$
z_1 w^2 + z_3 w + z_5 = 0 \quad (I),
$$
$$
z_2 w^2 + ... | [
"(a) Since $w$ is a root of the equation $z^2 + z + 1 = 0$, we have $w^2 + w + 1 = 0$. Multiplying both parts by $w$:\n$$\nw^3 + w^2 + w = 0 \\Leftrightarrow w^3 + \\underbrace{w^2 + w + 1}_{0} = 1 \\Leftrightarrow w^3 = 1.\n$$\nFrom the last equation we find $|w| = 1$. Substituting in relation (I) $w^2 = -w - 1$, ... | Greece | 26th Hellenic Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0egl | Problem:
Če bi plašč stožca razgrnili v ravnino, bi dobili četrtino kroga s polmerom $8~\mathrm{cm}$. Koliko je visok stožec?
(A) $10~\mathrm{cm}$
(B) $2 \sqrt{15}~\mathrm{cm}$
(C) $\sqrt{20}~\mathrm{cm}$
(D) $3~\mathrm{cm}$
(E) $16~\mathrm{cm}$ | [
"Solution:\nUpoštevanje, da je polmer krožnega izseka enak dolžini stranice stožca $s=8~\\mathrm{cm}$. Dolžina krožnega loka razgrnjenega plašča stožca je enaka obsegu osnovne ploskve stožca $\\frac{1}{4} \\cdot 2 \\pi s = 2 \\pi r$. Izračun polmera stožca $r=\\frac{s}{4}=2~\\mathrm{cm}$. Izračun višine stožca $v^{... | Slovenia | Državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | null | MCQ | B | |
0i9y | Let $\overline{AH_1}$, $\overline{BH_2}$, and $\overline{CH_3}$ be the altitudes of an acute scalene triangle $ABC$. The incircle of triangle $ABC$ is tangent to $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ at $T_1, T_2$, and $T_3$, respectively. For $k = 1, 2, 3$, let $P_i$ be the point on line $H_iH_{i+1}$ (... | [
"**First Solution.** (By Po-Ru Loh) We begin by showing that points $O_3, H_2, T_2$, and $O_3$ lie on a cyclic. We will prove this by establishing $\\angle O_3O_1H_2 = \\angle O_3T_2C = \\angle O_3T_2H_2$. To find $\\angle O_3O_1H_2$, observe that triangles $H_2AH_3$ and $H_2H_1C$ are similar. Indeed, quadrilateral... | United States | USA IMO 2003 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circle... | English | proof only | null | |
0704 | Problem:
Find all possible values for the sum of the digits of a square. | [
"Solution:\n$0^2 = 0$, $(\\pm 1)^2 = 1$, $(\\pm 2)^2 = 4$, $(\\pm 3)^2 = 0$, $(\\pm 4)^2 = 7 \\bmod 9$, so the condition is necessary.\n\nWe exhibit squares which give these values.\n\n$0 \\bmod 9$. Obviously $0^2 = 0$. We have $9^2 = 81$, $99^2 = 9801$ and in general $99\\ldots9^2 = (10^n - 1)^2 = 10^{2n} - 2 \\cd... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Number Theory > Modular Arithmetic"
] | null | proof and answer | Exactly the nonnegative integers whose remainder upon division by nine is 0, 1, 4, or 7. | |
00ga | Let $O$ be the circumcentre and $H$ the orthocentre of an acute triangle $A B C$. Prove that the area of one of the triangles $A O H$, $B O H$ and $C O H$ is equal to the sum of the areas of the other two. | [
"Suppose, without loss of generality, that $B$ and $C$ lie on the same side of line $O H$. Such line is the Euler line of $A B C$, so the centroid $G$ lies on this line.\n\n\n\nLet $M$ be the midpoint of $B C$. Then the distance between $M$ and the line $O H$ is the average of the distances... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Algebra > Linear Algebra > Determinants"
] | English | proof only | null | |
0c08 | In an acute triangle $ABC$ with $AB < BC$ let $BB'$ be an altitude, and let $O$ be the circumcenter. A line through $B'$ parallel to $CO$ meets $BO$ at $X$. Prove that $X$ and the midpoints of $AB$ and $AC$ are collinear.
Caucasus Mathematical Olympiad
 | [
"Let $M$ be the midpoint of the side $AB$. Then $\\angle OBC = \\angle OCB = 90^\\circ - \\angle A$ and $MX$ parallel to $BC$ comes to $\\angle MXB = 90^\\circ - \\angle A$. But $\\angle B'XB = \\angle XOC = 2\\angle OBC = 180^\\circ - 2\\angle A$. In the triangle $ABB'$ we have $MA = MB = MB'$ and $\\angle BMB' = ... | Romania | 69th NMO Selection Tests for JBMO | [
"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"
] | null | proof only | null | |
03a2 | Let $p$ and $q$ be prime numbers and let the sequence $\{a_n\}_{n=1}^{\infty}$ be defined by:
$$
a_0 = 0, a_1 = 1 \text{ and } a_{n+2} = pa_{n+1} - qa_n
$$
for $n \ge 0$. Find $p$ and $q$ if it is known that $a_{3k} = -3$ for some integer $k$. | [
"Let $p$ and $q$ be odd. The recurrence relation gives $a_2 = p$ and $a_3 = p^2 - q$. Therefore $a_0$ and $a_3$ are even. Since\n$$\na_{3k+3} = pa_{3k+2} - qa_{3k+1} = p(pa_{3k+1} - qa_{3k}) - qa_{3k+1} = (p^2 - q)a_{3k+1} - pqa_{3k}\n$$\nand the number $p^2 - q$ is even, we prove by induction that $a_{3k}$ is even... | Bulgaria | Fall Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof and answer | p = 2, q = 7 | |
0334 | Problem:
In triangle $ABC$ with orthocenter $H$ one has that
$$
AH \cdot BH \cdot CH = 3 \text{ and } AH^2 + BH^2 + CH^2 = 7
$$
Find:
a) the circumradius of $\triangle ABC$;
b) the sides of $\triangle ABC$ with maximum possible area. | [
"Solution:\n\na) If $\\triangle ABC$ is acute, then by the Law of cosines for $\\triangle AHB$ we get that\n$$\nAB^2 = AH^2 + BH^2 - 2 AH \\cdot BH \\cos(\\pi - \\gamma)\n$$\nSince $AB = 2R \\sin \\gamma$ and $CH = 2R \\cos \\gamma$ (by the Extended Law of sines), we obtain $AB^2 + CH^2 = 4R^2$. Therefore\n$$\n4R^2... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Ge... | null | proof and answer | a) The circumradius is either 1 or 3/2. b) The maximum area occurs for an acute isosceles triangle with side lengths sqrt(6), sqrt(6), and sqrt(8). | |
05jr | Problem:
Soit $(a_{n})$ une suite définie par $a_{1}, a_{2} \in [0,100]$ et
$$
a_{n+1} = a_{n} + \frac{a_{n-1}}{n^{2}-1} \quad \text{ pour tout entier } n \geqslant 2
$$
Existe-t-il un entier $n$ tel que $a_{n} > 2013$ ? | [
"Solution:\n\nLa réponse est non.\n\nPlus précisément, montrons par récurrence que l'on a $a_{n} \\leqslant 400$, pour tout $n \\geqslant 0$.\n\nL'inégalité est vraie pour $n=1$ et $n=2$, d'après l'énoncé.\n\nSupposons qu'elle soit vraie pour tout $k \\leqslant n$ pour un certain entier $n \\geqslant 2$.\n\nPour to... | France | Olympiades Françaises de Mathématiques, Envoi Numéro 3 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | No | |
028w | Problem:
Mostre que se o produto $N = (n + 6m)(2n + 5m)(3n + 4m)$ é múltiplo de $7$, com $m$ e $n$ números naturais, então $N$ é múltiplo de $7^{3} = 343$. | [
"Solution:\nInicialmente, observemos que:\n$$\n\\begin{aligned}\nN & = (n + 6m)(2n + 5m)(3n + 4m) \\\\\n & = (n + 7m - m)(2n + 7m - 2m)(3n + 7m - 3m) \\\\\n & = (n - m + 7m)[2(n - m) + 7m][3(n - m) + 7m] \\\\\n & = (k + 7m)(2k + 7m)(3k + 7m)\n\\end{aligned}\n$$\nonde $k = n - m$.\nAfirmamos que se $N$ é múltiplo... | Brazil | Nível 3 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
033e | Problem:
Find the maximum possible value of the inradius of a triangle with vertices in the interior or on the boundary of a unit square. | [
"Solution:\nIt is easy to see that if a triangle contains another triangle, then its inradius is greater than the inradius of the second one. So we may consider triangles with vertices on the boundary of the square. Moreover, we may assume that at least one vertex of the triangle is a vertex of the square and the o... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | (sqrt(5)-1)/4 | |
0h0d | In a convex quadrilateral $ABCD$ angles $\angle ABC$ and $\angle BCD$ are not less than $120^\circ$. Prove, that $AC + BD > AB + BC + CD$. | [
"Let $AB = a$, $BC = b$, $CD = c$ (Fig.18).\n\nThen, $AC^2 = a^2 + b^2 - 2ab \\cos \\angle B \\ge a^2 + ab + b^2$.\n\nBy analogy, $BD^2 \\ge b^2 + bc + c^2 \\Rightarrow AC + BD \\ge \\sqrt{a^2 + ab + b^2} + \\sqrt{b^2 + bc + c^2}$.\n\nSince $\\sqrt{a^2 + ab + b^2} > a + \\frac{1}{2}b$ and $\\sqrt{b^2 + bc + c^2} > ... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 24, 2010) | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0hdj | The chessboard was split into domino tiles, meaning it was split into $1 \times 2$ and $2 \times 1$ rectangles. Each tile has a number written on it equal to the number of tiles that it has a common line segment with, without taking into account the tile itself. What is a least possible sum of all numbers that are writ... | [
"It is clear that there are $32$ domino tiles, hence there are exactly $32$ unit intervals that are covered by domino tiles thus they are internal for tiles and therefore they cannot be common for two tiles. The rest of unit intervals that are not at the edge of the chessboard are common for two tiles. Let their nu... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 104 | |
008h | Let a positive integer be called balanced if the difference between any two adjacent digits of it is $0$, $1$ or $-1$. For instance, the numbers $232$, $555$ and $876$ are balanced, but the numbers $244$ and $890$ are not.
How many three-digit balanced numbers are there? | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 75 | |
004c | Sea $k \ge 1$ un entero. En un grupo de $2k+1$ personas algunas son sinceras (siempre dicen la verdad) y las restantes son impredecibles (a veces dicen la verdad y a veces mienten). Se sabe que las impredecibles son a lo sumo $k$. Alguien ajeno al grupo debe determinar quién es sincero y quién impredecible mediante una... | [] | Argentina | Argentina 2006 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Logic"
] | Español | proof only | null | |
0auu | Problem:
The irrational number $0.123456789101112 \ldots$ is formed by concatenating, in increasing order, all the positive integers. Find the sum of the first 2016 digits of this number after the decimal point. | [
"Solution:\n\nFrom $0$ to $99$, there are $10$ occurrences of $0$ to $9$ in the ones place, and ten iterations of $0$ to $9$ in the tens place. Thus, the total digit sum from $0$ to $99$ is $20(45)=900$.\n\nFrom $100$ to $699$, there are $6$ occurrences of $0$ to $99$, plus $100$ iterations each of $0$ to $6$ (in t... | Philippines | 18th PMO National Stage Oral Phase | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Other"
] | null | proof and answer | 8499 | |
00e7 | An integer $n \ge 3$ is said to be a *polygonal pythagorean number* if there are $n$ positive integers, no two of them equal, which can be placed in the vertices of a regular $n$-gon in such a way that the sum of the squares of the numbers in any two consecutive vertices is a perfect square. For instance, 3 is a polygo... | [
"The answer is $n \\ge 3$. We abbreviate PP = polygonal pythagorean.\n\nFirst, assume that $n$ is PP. We will show then that $n + 2$ is also PP. Let $a_1, \\dots, a_n$ be pairwise different positive integers such that $a_i^2 + a_{i+1}^2$ is a perfect square for all $i = 1, \\dots, n$, where $a_{n+1} = a_1$. Choose ... | Argentina | Rioplatense Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | all integers n ≥ 3 | |
0fxk | Problem:
Seien $a$, $b$, $c$, $d$ positive reelle Zahlen. Beweise die Ungleichung
$$
\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b} \geq 0
$$
und bestimme alle Fälle, in denen das Gleichheitszeichen steht. | [
"Solution:\n\nWegen\n$$\n\\frac{a-b}{b+c}=\\frac{a+c}{b+c}-1\n$$\nist die Ungleichung äquivalent zu\n$$\n\\frac{a+c}{b+c}+\\frac{b+d}{c+d}+\\frac{c+a}{d+a}+\\frac{d+b}{a+b} \\geq 4\n$$\nFür die linke Seite erhält man mit AM-HM nun die Abschätzung\n$$\n\\begin{aligned}\nLS & =(a+c)\\left(\\frac{1}{b+c}+\\frac{1}{d+a... | Switzerland | SMO Finalrunde | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | Equality holds if and only if a = c and b = d. | |
03uy | Solve the inequality
$$
\log_2(x^{12} + 3x^{10} + 5x^8 + 3x^6 + 1) < 1 + \log_2(x^4 + 1).
$$ | [
"As\n$$\n1 + \\log_2(x^4 + 1) = \\log_2(2x^4 + 2),\n$$\nand $\\log_2 y$ is monotonically increasing over $(0, +\\infty)$, the given inequality is equivalent to\n$$\nx^{12} + 3x^{10} + 5x^8 + 3x^6 + 1 < 2x^4 + 2\n$$\nor\n$$\nx^{12} + 3x^{10} + 5x^8 + 3x^6 - 2x^4 - 1 < 0.\n$$\nIt can be rewritten as\n$$\n\\begin{alig... | China | China Mathematical Competition | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | (-sqrt((-1+sqrt(5))/2), sqrt((-1+sqrt(5))/2)) | |
05eb | Problem:
An infinite increasing sequence $a_{1} < a_{2} < a_{3} < \dots$ of positive integers is called central if for every positive integer $n$, the arithmetic mean of the first $a_{n}$ terms of the sequence is equal to $a_{n}$.
Show that there exists an infinite sequence $b_{1}, b_{2}, b_{3}, \ldots$ of positive ... | [
"Solution:\n\nWe claim that the sequence $b_{1}, b_{2}, b_{3}, \\ldots$ defined by $b_{i} = 2i - 1$ has this property. \nLet $d_{i} = a_{i} - b_{i} = a_{i} - 2i + 1$. The condition $a_{i} < a_{i + 1}$ now becomes $d_{i} + 2i - 1 < d_{i + 1} + 2i + 1$, which can be rewritten as $d_{i + 1} \\geqslant d_{i} - 1$. Thu... | European Girls' Mathematical Olympiad (EGMO) | EGMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | b_n = 2n - 1 for all n | |
0kpc | Problem:
In right triangle $A B C$, a point $D$ is on hypotenuse $A C$ such that $B D \perp A C$. Let $\omega$ be a circle with center $O$, passing through $C$ and $D$ and tangent to line $A B$ at a point other than $B$. Point $X$ is chosen on $B C$ such that $A X \perp B O$. If $A B=2$ and $B C=5$, then $B X$ can be ... | [
"Solution:\n\nNote that since $A D \\cdot A C = A B^{2}$, we have the tangency point of $\\omega$ and $A B$ is $B'$, the reflection of $B$ across $A$. Let $Y$ be the second intersection of $\\omega$ and $B C$. Note that by power of point, we have $B Y \\cdot B C = B B'^{2} = 4 A B^{2} \\Longrightarrow B Y = \\frac{... | United States | HMMT February | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | null | final answer only | 8041 | |
0gdn | 設 $f$ 為一正整數值函數, 且對於所有正整數 $a, b$, 有 $(a + f(b)) | (a^2 + b f(a))$。證明：存在正整數 $k$ 使得 $f(n) = kn$ 對所有正整數 $n$ 均成立。
Let $f$ be a positive integer valued function that satisfies $(a+f(b)) | (a^2+bf(a))$ for all positive integers $a$ and $b$. Prove that there is a positive integer $k$ such that $f(n) = kn$ for all positive int... | [
"Easy to see that $f(n) \\leq f(1)n$ by substituting $a = 1$. By substituting $a = nb - f(b)$ in the inequality for any large enough $n$, we have\n$$\nnb \\mid (nb - f(b))^2 + b f(nb - f(b))\n$$\nand hence $b \\mid f(b)^2$. In particular, for every prime $p$, $f(p) = k_p p$ for some integer $0 < k_p \\leq f(1)$. Th... | Taiwan | 2020 Taiwan IMO 2J | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(n) = k n for all positive integers n, for some fixed positive integer k. | |
0244 | Problem:
Quatro frações e um inteiro - Quantos números naturais $a, b, c$ e $d$, todos distintos, existem tais que $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}$ seja um inteiro? | [
"Solution:\n\n1"
] | Brazil | Desafios | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | 1 | |
02lj | Problem:
As cinco cartas abaixo estão sobre uma mesa, e cada uma tem um número numa face e uma letra na outra. Simone deve decidir se a seguinte frase é verdadeira: "Se uma carta tem uma vogal numa face, então ela tem um número par na outra." Qual o menor número de cartas que ela precisa virar para decidir corretament... | [
"Solution:\n\n\nEla não precisa virar a carta que tem o número $2$, porque sendo vogal ou consoante, ela cumpre a condição, de igual forma. Ela também não precisa virar a carta com a letra $M$. A carta que tem o número $3$ tem que ser virada, para comprovar que na outra face tem uma consoan... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Logic"
] | null | proof and answer | 3 | |
0dy7 | Prove that for all real numbers $x \in \left(-\frac{3\pi}{2}, \frac{\pi}{2}\right)$ the equality
$$
\frac{2}{1 - \sin x} = \tan^2 \left( \frac{x}{2} + \frac{\pi}{4} \right) + 1
$$
holds. | [
"From the addition formula for tangents we get\n$$\n\\tan^2 \\left( \\frac{x}{2} + \\frac{\\pi}{4} \\right) = \\left( \\frac{\\tan \\frac{x}{2} + \\tan \\frac{\\pi}{4}}{1 - \\tan \\frac{x}{2} \\tan \\frac{\\pi}{4}} \\right)^2 .\n$$\nUsing the fact that $\\tan \\frac{\\pi}{4} = 1$ and expressing $\\tan \\frac{x}{2}$... | Slovenia | Slovenija 2008 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
0ahk | Let $x$, $y$ and $z$ be positive real numbers such that $x^4 + y^4 + z^4 = 3$. Prove that
$$
\frac{9}{x^2 + y^4 + z^6} + \frac{9}{x^4 + y^6 + z^2} + \frac{9}{x^6 + y^2 + z^4} \le x^6 + y^6 + z^6 + 6.
$$
When does equality hold? | [
"If we use the Cauchy-Bunyakovsky-Schwarz inequality for the positive numbers $(x, y^2, z^3)$ and $(x^3, y^2, z)$ we get\n$$ (x^4 + y^4 + z^4)^2 \\leq (x^2 + y^4 + z^6)(x^6 + y^4 + z^2) \\text{ i.e. } $$\n$$ \\frac{1}{x^2 + y^4 + z^6} \\leq \\frac{x^6 + y^4 + z^2}{9}. $$\n(1)\nAnalogously, using the Cauchy-Bunyakov... | North Macedonia | Macedonian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Equality holds when x = y = z = 1. | |
0c32 | Show that a number of the form $n(n + 1)$, where $n$ is a positive integer, is the sum of two numbers of like form, say $k(k + 1)$ and $m(m + 1)$, where $k$ and $m$ are positive integers, if and only if the number $2n^2 + 2n + 1$ is composite. | [
"Rewrite the condition $k(k + 1) + m(m + 1) = n(n + 1)$ in the equivalent form\n$$\n(2k + 1)^2 + (2m + 1)^2 = 2N. \\quad (*)\n$$\nSince the prime factors of $N$ are all congruent to $1$ modulo $4$, the number of solutions of the equation $x^2 + y^2 = 2N$ in integers $x$ and $y$ (necessarily odd) is equal to $4d(N)$... | Romania | 69th NMO Selection Tests for BMO and IMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof only | null | |
03y5 | Given integer $a_1 \ge 2$, for $n \ge 2$, define $a_n$ to be the least positive integer not coprime to $a_{n-1}$ and not equal to $a_1$, $a_2, \dots, a_{n-1}$. Prove that every integer except $1$ appears in the sequence $\{a_n\}$. (Posed by Yu Hongbing) | [
"*Step 1:* We prove that there are infinitely many even numbers in this sequence.\nSuppose on the contrary that there are only finitely many even numbers, and there is an integer $E$, such that all even numbers greater than $E$ do not appear in the sequence. It follows that there exists a positive integer $K$ such ... | China | China National Team Selection Test | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0grb | Let $a$, $b$, $c$ be distinct real numbers and $x$ be a real number.
Given that three numbers among
$$ax^2 + bx + c,\ ax^2 + cx + b,\ bx^2 + cx + a,\ bx^2 + ax + c,\ cx^2 + ax + b,\ cx^2 + bx + a$$
coincide, prove that $x = 1$. | [
"Let\n$$\n\\begin{align*}\nax^2 + bx + c &= x_1, & ax^2 + cx + b &= x_2 \\\\\nbx^2 + cx + a &= x_3, & bx^2 + ax + c &= x_4 \\\\\ncx^2 + ax + b &= x_5, & cx^2 + bx + a &= x_6\n\\end{align*}\n$$\nThen we get\n$$\n\\begin{align*}\n(b-c)(x-1) &= x_1 - x_2, \\quad (c-a)(x-1) = x_3 - x_4, \\quad (a-b)(x-1) = x_5 - x_6 \\... | Turkey | Team Selection Test | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | x = 1 | |
0f1l | Problem:
There are finitely many polygons in the plane. Every two have a common point. Prove that there is a straight line intersecting all the polygons. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Combinatorial Geometry > Helly's theorem"
] | null | proof only | null | |
0f3b | Problem:
A set of square carpets have total area $4$. Show that they can cover a unit square. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
09ud | Problem:
Laat $x$ en $y$ positieve reële getallen zijn.
a) Bewijs: als $x^{3}-y^{3} \geq 4 x$, dan geldt $x^{2}>2 y$.
b) Bewijs: als $x^{5}-y^{3} \geq 2 x$, dan geldt $x^{3} \geq 2 y$. | [
"Solution:\na)\nEr geldt $x^{3}-4 x \\geq y^{3}>0$, dus $x\\left(x^{2}-4\\right)>0$. Omdat $x$ positief is, volgt hieruit dat $x^{2}-4>0$, dus $x^{2}>4$. Dat betekent $x>2$. Verder geldt $x^{3}-y^{3} \\geq 4 x>0$, dus $x>y$. Combineren van deze twee resultaten (wat mag omdat $x$ en $y$ beide positief zijn) geeft $x... | Netherlands | Selectietoets | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0ha3 | Let $a, b$ are such that $\frac{7a^3b^3}{a^6-8b^6} = 1$. Determine what the value of $\frac{a^2-b^2}{a^2+b^2}$ can be. | [
"Rewrite given equation as $a^6 - 7a^3b^3 - 8b^6 = 0$, hence $(a^3 + b^3)(a^3 - 8b^3) = 0$. We can show that $x^2 \\pm xy + y^2 > 0$ holds iff $x, y \\neq 0$, since\n$$\nx^2 \\pm xy + y^2 = (x \\pm \\frac{1}{2}y)^2 + \\frac{3}{4}y^2 = 0 \\Leftrightarrow y = 0 \\text{ and } x \\pm \\frac{1}{2}y = 0 \\Leftrightarrow ... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 0 or 3/5 | |
00cy | Sea $n$ un entero positivo. Se tienen $n$ colores, $n \ge 1$. Cada uno de los números enteros entre $1$ y $1000$ se quiere pintar con uno de los $n$ colores de modo que cada dos números diferentes, si uno divide al otro tengan colores diferentes. Dar el menor número $n$ para que esto sea posible. | [
"Observamos que los $10$ números $2^0=1$, $2^1=2$, $2^2=4$, $2^3=8$, $2^4=16$, $2^5=32$, $2^6=64$, $2^7=128$, $2^8=256$, $2^9=512$ tienen la propiedad que para cualesquiera dos, uno de ellos divide al otro. Por lo tanto no pueden tener el mismo color, lo que implica que $n \\ge 10$. Damos una coloración para $n=10$... | Argentina | Nacional OMA | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Spanish | proof and answer | 10 | |
01pm | Each of twenty two sets contains five elements. An intersection of any two of these sets contains exactly two elements.
Prove that the intersection of all these sets contains exactly two elements. | [
"Assume there are no two such elements. Let $A_1, \\ldots, A_{22}$ be the initial sets, $|A_i| = 5$, $i = 1, \\ldots, 22$.\nPut $S = \\bigcup_{i=1}^{22} A_i$ and $a(x, y) = |\\{i \\mid \\{x, y\\} \\subset A_i\\}|$ for $x, y \\in S$.\nAssume there are five sets that contain $x$ and $y$, without loss of generality, $... | Belarus | BelarusMO 2013_s | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0bcm | Problem:
Fie $n \in \mathbb{N}^*$. Să se arate că numărul
$$
2 \sqrt{2^{n}} \cos \left(n \arccos \frac{\sqrt{2}}{4}\right)
$$
est număr întreg impar. | [] | Romania | Olimpiada Naţională de Matematică Etapa Judeţeană şi a Municipiului Bucureşti | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0aag | Problem:
In Wonderland, the towns are connected by roads, and whenever there is a direct road between two towns there is also a route between these two towns that does not use that road. (There is at most one direct road between any two towns.) The Queen of Hearts ordered the Spades to provide a list of all "even" sub... | [
"Solution:\n\nThe answer is $\\frac{1}{2} n x$.\n\nProof: We reformulate the problem in terms of graph theory with the towns being vertices and the roads being edges of a graph $G = (V, E)$. The given information implies that every edge $e \\in E$ is part of a cycle. The subgraphs to be counted are those with every... | Nordic Mathematical Olympiad | Nordic Mathematical Contest | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | n x / 2 | |
0hof | Problem:
Determine whether the number
$$
\frac{1}{2 \sqrt{1}+1 \sqrt{2}}+\frac{1}{3 \sqrt{2}+2 \sqrt{3}}+\frac{1}{4 \sqrt{3}+3 \sqrt{4}}+\cdots+\frac{1}{100 \sqrt{99}+99 \sqrt{100}}
$$
is rational or irrational. Explain your answer. | [
"Solution:\nNotice that\n$$\n\\begin{aligned}\n\\frac{1}{(n+1) \\sqrt{n}+n \\sqrt{n+1}} & =\\frac{1}{\\sqrt{n(n+1)}} \\cdot \\frac{1}{\\sqrt{n+1}+\\sqrt{n}} \\\\\n& =\\frac{1}{\\sqrt{n(n+1)}} \\cdot \\frac{\\sqrt{n+1}-\\sqrt{n}}{(\\sqrt{n+1}+\\sqrt{n})(\\sqrt{n+1}-\\sqrt{n})} \\\\\n& =\\frac{1}{\\sqrt{n}}-\\frac{1}... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | rational | |
0emd | Let $CH$ be the altitude of an acute angled triangle $ABC$, and let $O$ be the centre of its circumcircle. If $T$ is the foot of the perpendicular drawn from vertex $C$ to the line $AO$, prove that the line $TH$ bisects the side $BC$. | [
"Let $TH$ cut $BC$ at $P$. We want to show that $P$ bisects $BC$. Since $\\angle AHC = \\angle ATC = 90^\\circ$, the points $A$, $H$, $T$ and $C$ lie on the same circle.\n\n\n\nNow $\\angle PHC = \\angle TAC = \\frac{1}{2}(180^\\circ - \\angle AOC) = 90^\\circ - \\angle ABC = 90^\\circ - \\... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0kev | Problem:
If $x, y, z$ are real numbers such that $x y=6$, $x-z=2$, and $x+y+z=9$, compute $\frac{x}{y}-\frac{z}{x}-\frac{z^{2}}{x y}$. | [
"Solution:\nLet $k=\\frac{x}{y}-\\frac{z}{x}-\\frac{z^{2}}{x y}=\\frac{x^{2}-y z-z^{2}}{x y}$. We have\n$$\nk+1=\\frac{x^{2}+x y-y z-z^{2}}{x y}=\\frac{x^{2}-x z+x y-y z+z x-z^{2}}{x y}=\\frac{(x+y+z)(x-z)}{x y}=\\frac{9 \\cdot 2}{6}=3,\n$$\nso $k=2$."
] | United States | HMMO 2020 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 2 | |
030u | Problem:
Fie $p_{i}\left(i \in \mathbb{N}^{*}\right)$ al $i$-ulea număr prim (în ordine crescătoare). Pentru fiecare număr natural nenul $k$, notăm cu $a_{k}$ numărul de numere naturale nenule $i$ cu proprietatea că produsul $p_{i} p_{i+1}$ divide numărul $k$.
Dacă $n$ este un număr natural nenul, arătați că
$$
a_{1}+... | [] | Brazil | Al treilea baraj de selecție pentru OBMJ | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
0107 | Problem:
Let $n \geqslant 4$ be an even integer. A regular $n$-gon and a regular $(n-1)$-gon are inscribed into the unit circle. For each vertex of the $n$-gon consider the distance from this vertex to the nearest vertex of the $(n-1)$-gon, measured along the circumference. Let $S$ be the sum of these $n$ distances. Pr... | [
"Solution:\nFor simplicity, take the length of the circle to be $2n(n-1)$ rather than $2\\pi$. The vertices of the $(n-1)$-gon $A_{0} A_{1} \\ldots A_{n-2}$ divide it into $n-1$ arcs of length $2n$. By the pigeonhole principle, some two of the vertices of the $n$-gon $B_{0} B_{1} \\ldots B_{n-1}$ lie in the same ar... | Baltic Way | Baltic Way 1998 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0d9q | Consider the infinite, strictly increasing sequence of positive integers $a_{n}$ such that
i. All terms of the sequence are pairwise coprime.
ii. The sum $\frac{1}{\sqrt{a_{1} a_{2}}} + \frac{1}{\sqrt{a_{2} a_{3}}} + \frac{1}{\sqrt{a_{3} a_{4}}} + \cdots$ is unbounded.
Prove that this sequence contains infinitely ma... | [
"Suppose on the contrary that there are infinite primes in this sequence, thus there is some index $k$ such that $a_{n}$ is composite for all $n > k$. Denote $S$ as all prime divisors of the first $k$ terms of the given sequence.\n\nBy comparing $a_{k+1}$ with the first prime $p_{1}$ that does not appear in $S$ the... | Saudi Arabia | Team selection tests for IMO 2018 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof only | null | |
0ef9 | Problem:
Koliko je takih 6-mestnih števil, ki se ne začnejo z 0 in vsebujejo število 2017 kot strnjen podniz? Npr. število 820178 vsebuje strnjen podniz 2017, število 820817 pa ne.
(A) 100
(B) 190
(C) 200
(D) 280
(E) 300 | [
"Solution:\n6-mestno število, ki vsebuje strnjen podniz 2017 je ene od treh oblik: $x y 2017$, $x 2017 y$ ali $2017 x y$, kjer sta $x$ in $y$ števki. Hitro opazimo, da nobeno naravno število ne more biti dveh oblik hkrati. Torej moramo le prešteti, koliko števil je posamezne oblike, saj s tem nobenega števila ne bo... | Slovenia | Slovenian Mathematical Olympiad | [
"Discrete Mathematics > Other"
] | null | MCQ | D | |
0cnf | A mushroom containing not less than $10$ worms is called bad. A basket with $90$ bad and $10$ good mushrooms is given. Determine if all the mushrooms can become good after several worms creep from bad to good mushrooms. | [
"Suppose each bad mushroom contains exactly $10$ worms, and each good mushroom contains no worms. Next, let one worm from each bad mushroom crawl into the good mushrooms, $9$ worms into each. As a result, each mushroom will have $9$ worms, and all mushrooms will be good."
] | Russia | Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof and answer | Yes | |
01e4 | Positive integers from $1$ to $n$ are written on the blackboard. The first player chooses a number and erases it. Then the second player chooses two consecutive numbers and erases them. After that the first player chooses three consecutive numbers and erases them. And finally the second player chooses four consecutive ... | [
"Answer: $n = 14$.\n\nAt first, let's show that for $n = 13$ the first player can ensure that after his second move no $4$ consecutive numbers are left. In the first move he can erase number $4$ and in the second move he can ensure that numbers $8$, $9$ and $10$ are erased. No interval of length $4$ is left.\n\nIf ... | Baltic Way | Baltic Way shortlist | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | 14 | |
0kdv | Problem:
A snake of length $k$ is an animal which occupies an ordered $k$-tuple $(s_{1}, \ldots, s_{k})$ of cells in a $n \times n$ grid of square unit cells. These cells must be pairwise distinct, and $s_{i}$ and $s_{i+1}$ must share a side for $i=1, \ldots, k-1$. If the snake is currently occupying $(s_{1}, \ldots, s... | [
"Solution:\nLet $n=30$. The snake can get stuck in only 8 positions, while the total number of positions is about $n^{2} \\times 4 \\times 3 \\times 3=36 n^{2}$. We can estimate the answer as $\\frac{36 n^{2}}{8}=4050$, which is good enough for 13 points.\n\nLet's try to compute the answer as precisely as possible.... | United States | HMMT February 2020 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Linear Algebra > Matrices",
"Discrete Mathematics > Algorithms"
] | null | final answer only | 4050 | |
0gb5 | 給定 $\triangle ABC$ 與三點 $D, E, F$ 使得: $DB = DC$, $EC = EA$, $FA = FB$, 且 $\angle^*BDC = \angle^*CEA = \angle^*AFB$ (這裡 $\angle^*$ 係指有向角)。設 $\Omega_D$ 是以 $D$ 為圓心且經過 $B, C$ 的圓, 並類似定義 $\Omega_E$ 與 $\Omega_F$。證明: $\Omega_D, \Omega_E, \Omega_F$ 的根心落在 $\triangle DEF$ 的尤拉線上。
註：$\triangle DEF$ 的尤拉線是指通過 $\triangle DEF$ 的垂心、重心與外... | [
"設 $T$ 為 $\\Omega_E, \\Omega_F$ 異於 $A$ 的交點，且設 $AT$ 與 $\\triangle BTC$ 的外接圓再交於點 $H_D$。由 $\\angle^*BCH_D = \\angle^*BTA = \\frac{1}{2}\\angle^*BFA = \\frac{1}{2}\\angle^*CDB$，可知 $BH_D \\perp CD$。同理可得 $CH_D \\perp BD$，所以 $H_D$ 為 $\\triangle BDC$ 的垂心。\n\n類似地，可設出 $H_E, H_F$ 分別為 $\\triangle CEA, \\triangle AFB$ 的垂心，則 $BH... | Taiwan | 二〇一七數學奧林匹亞競賽第二階段選訓營 | [
"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 > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Cons... | null | proof only | null | |
0lb6 | Let $n$ be a positive integer. Show that the polynomial
$$
P(x, y) = x^n + x y + y^n
$$
can not be written in the form
$$
P(x, y) = G(x, y) \cdot H(x, y),
$$
where $G(x, y)$ and $H(x, y)$ are non-constant polynomials with real coefficients. | [
"We will show the claim by assuming the contrary.\nAssume that there exist non-constant polynomials $G(x, y)$ and $H(x, y)$, with real coefficients, such that\n$$\nP(x, y) = G(x, y) \\cdot H(x, y), \\qquad (1)\n$$\nwhere $P(x, y) = x^n + x y + y^n$, $n \\in \\mathbb{N}^*$.\nPresent $G(x, y)$ and $H(x, y)$ in the fo... | Vietnam | Vijetnam 2011 | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0ay8 | Problem:
For a positive integer $n$, let $\varphi(n)$ denote the number of positive integers less than and relatively prime to $n$. Let $S_{k}=\sum_{n} \frac{\varphi(n)}{n}$, where $n$ runs through all positive divisors of $42^{k}$. Find the largest positive integer $k<1000$ such that $S_{k}$ is an integer. | [
"Solution:\n\nAnswer: 996\n\nThe function $\\varphi$ is the well-known Euler totient function which satisfies the property\n$$\n\\frac{\\varphi(n)}{n}=\\prod_{\\substack{p \\mid n \\\\ p \\text{ prime }}}\\left(1-\\frac{1}{p}\\right)\n$$\nfor any integer $n>2$. Note that the problem defines $\\varphi(1)=0$.\n\nFor ... | Philippines | 21st PMO Area Stage | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 996 | |
00iq | Let $x$ be the smallest positive integer such that $2x$ is the square of an integer, $3x$ is the cube of an integer and $5x$ is the fifth power of an integer. Find the prime factorization of $x$. | [
"Let the prime factor decomposition of $x$ be given by $2^a 3^b 5^c p_4^{e_4} \\dots p_r^{e_r}$ (with $a, b, c \\ge 0$). We conclude that\n* $a$ is a multiple of $15$ and is odd,\n* $b$ is a multiple of $10$ and $b+1$ is a multiple of $3$,\n* $c$ is a multiple of $6$ and $c + 1$ is a multiple of $5$.\n\nThe smalles... | Austria | AustriaMO2011 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | x = 2^15 * 3^20 * 5^24 | |
083i | Problem:
Due candele hanno la stessa lunghezza. La prima si consuma in 5 ore, la seconda in 3 ore. Le candele vengono accese contemporaneamente. Dopo quanti minuti l'altezza della prima candela sarà uguale a 3 volte l'altezza della seconda? | [] | Italy | UNIONE MATEMATICA ITALIANA Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO TRIENNIO | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 150 | |
00vh | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$$
f(x + y f(x)) + y = x y + f(x + y)
$$
for all real numbers $x, y$. | [
"Let $P(x, y)$ denote the given relation. If there is an $a \\in \\mathbb{R}$ such that $f(a) = 0$, then $P(a, y)$ gives that $y = a y + f(a + y)$, and so $f$ must be linear. Then we can easily check and get that the only linear solutions are $f(x) = x$ and $f(x) = 2 - x$ ($x \\in \\mathbb{R}$).\n\nNow suppose that... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | f(x) = x and f(x) = 2 - x | |
0aq4 | Problem:
If $a$ and $b$ are positive real numbers, what is the minimum value of the expression
$$
\sqrt{a+b}\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right) ?
$$ | [
"Solution:\n$2 \\sqrt{2}$\n\nBy the AM-GM Inequality, we have\n$$\n\\sqrt{a+b} \\geq \\sqrt{2 \\sqrt{a b}} = \\sqrt{2} (a b)^{1 / 4}\n$$\nand\n$$\n\\frac{1}{\\sqrt{a}} + \\frac{1}{\\sqrt{b}} \\geq 2 \\sqrt{\\frac{1}{\\sqrt{a}} \\cdot \\frac{1}{\\sqrt{b}}} = \\frac{2}{(a b)^{1 / 4}}\n$$\nwhere both inequalities beco... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 2√2 | |
04p2 | Find all pairs $(p, q)$ of prime integers such that the solutions of the quadratic equation $x^2 + px + q = 0$ are two distinct integers. | [
"Let the roots of $x^2 + px + q = 0$ be $r$ and $s$, with $r \\neq s$ and both integers.\n\nBy Vieta's formulas:\n$r + s = -p$\n$rs = q$\n\nSince $p$ and $q$ are primes, $-p$ is the sum of two distinct integers, and $q$ is their product.\n\nLet $r$ and $s$ be distinct integers. Since $q$ is prime, $rs = q$ implies ... | Croatia | Croatian Mathematical Society Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | (3, 2) and (-3, 2) | |
07oi | A square $ABCD$ is inscribed in a circle with centre $O$. Let $E$ be the midpoint of $AD$. The line $CE$ meets the circle again at $F$. The lines $FB$ and $AD$ meet at $H$. Prove $|HD| = 2|AH|$. | [
"Because $A, B, C, F$ are on a circle, we have $\\angle BFC = \\angle BAC = 45^\\circ$. Since $\\angle AFC = 90^\\circ$ this means that $FB$ is the bisector of the angle $\\angle AFE$. Because the angle bisector cuts the opposite side in a triangle at a ratio equal to the ratio of the adjacent sides (which follows ... | Ireland | Irska 2014 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry ... | null | proof only | null | |
06k5 | Let $d$ be a nonnegative integer. Determine all functions $f : \mathbb{R}^2 \to \mathbb{R}$ such that, for any real constants $A$, $B$, $C$ and $D$, $f(At+B, Ct+D)$ is a polynomial in $t$ of degree at most $d$. | [
"We claim that $f(x, y)$ is a polynomial in $x$ and $y$ of degree at most $d$.\nIt is obvious that every such polynomial satisfies the desired condition. To prove the converse, let $f$ be a function satisfying the desired condition. Pick $(d+2)$ straight lines $\\ell_1$, $\\ell_2$, $\\dots$, $\\ell_{d+2}$ in $\\mat... | Hong Kong | Pre-IMO 2017 Mock Exam | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | All polynomials in two variables of total degree at most d. | |
0i1k | Problem:
Find $x-y$, given that $x^{4} = y^{4} + 24$, $x^{2} + y^{2} = 6$, and $x + y = 3$. | [
"Solution:\n\n$\\frac{24}{6 \\cdot 3} = \\frac{x^{4} - y^{4}}{(x^{2} + y^{2})(x + y)} = \\frac{(x^{2} + y^{2})(x + y)(x - y)}{(x^{2} + y^{2})(x + y)} = x - y = \\frac{4}{3}$."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | final answer only | 4/3 | |
0eab | Let $AB$ be the longest side of the triangle $ABC$. Let $M$ and $N$ denote the points on the side $AB$, such that $|AM| = |AC|$ and $|BN| = |BC|$. Denote the midpoints of the segments $MC$ and $NC$ by $P$ and $R$. The incircle of the triangle $ABC$ touches the sides $BC$ and $AC$ at $D$ and $E$. Prove that the points $... | [
"Let $I$ be the incentre of the triangle $ABC$.\n\n\n\nTriangle *AMC* is isosceles with the apex at *A*, so the line *AP* is the altitude to the base and at the same time the bisector of the angle $\\angle MAC$. It follows that $I$ lies on $AP$. Similarly, $I$ lies on the line $BR$. This im... | Slovenia | National Math Olympiad in Slovenia | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07v6 | Prove that the sum of the elements in any finite subset of the set
$$
\left\{ \frac{1}{mn(m+n+1)} : m, n = 1, 2, 3, \dots \right\}
$$
is less than 2. | [
"We will use the following lemma three times.\n**Lemma.** For all $k \\ge 1$ and all $1 \\le M \\le N$ we have\n$$\n\\sum_{n=M}^{N} \\frac{1}{n(n+k)} < \\frac{1}{k} \\sum_{n=M}^{M+k-1} \\frac{1}{n} .\n$$\n*Proof.* Using $\\frac{1}{n(n+k)} = \\frac{1}{k} \\left( \\frac{1}{n} - \\frac{1}{n+k} \\right)$ we see that\n$... | Ireland | IRL_ABooklet | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
03iy | Problem:
The geometric mean (G.M.) of a $k$ positive numbers $a_{1}, a_{2}, \ldots, a_{k}$ is defined to be the (positive) $k$-th root of their product. For example, the G.M. of $3,4,18$ is $6$. Show that the G.M. of a set $S$ of $n$ positive numbers is equal to the G.M. of the G.M.'s of all non-empty subsets of $S$. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
09dp | $\{x_n\}$ дараалал нь $x_0 = a$, $x_1 = 2$ ба
$$2x_n - 1 = 2x_{n-1}x_{n-2} - x_{n-1} - x_{n-2} + 1$$
томёогоор өгөгдсөн бол $2x_{3n} - 1$ тоо бүхэл тооны квадрат байх бүх $a$ тоог ол. | [
"$2x_n - 1 = 2(2x_{n-1}x_{n-2} - x_{n-1} - x_{n-2} + 1) - 1$\n$$\n= (2x_{n-1} - 1)(2x_{n-2} - 1)\n$$\n$$\n\\Rightarrow a_n = 2x_n - 1 \\text{ гэе. } a_n = a_{n-1} \\cdot a_{n-2}, a_1 = 3, a_0 = 2a - 1.\n$$\n$$\na_0 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2a - 1 \\Rightarrow a = 2k^2 + 2k + 1\n$$\n$$\na_{3n} = k^2 \\text{ ба... | Mongolia | ММО-48 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | Mongolian | proof and answer | a = 2k^2 + 2k + 1 for all integers k | |
00ca | Facu and Nico play the following game with a $13 \times 13$ grid square. Facu cuts the square into rectangles having a side equal to $1$, in any way he wishes. Then Nico chooses a number $k$ among $1, 2, \ldots, 13$ and takes all the obtained rectangles $1 \times k$. How many grid cells can he take with certainty? | [
"Nico can always ensure $16$ cells. Suppose that there is a partition like in the statement so that no $16$ cells can be taken. Then this partition has at most $1$ rectangle $1 \\times k$ for each $k = 8, 9, 10, 11, 12, 13$; at most $2$ such rectangles for $k = 6, 7$; at most $3$ such rectangles for $k = 4, 5$; at ... | Argentina | Argentina_2018 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | 16 | |
0gnx | Let $H$ be the orthocenter of an acute triangle $ABC$, and let $A_1, B_1, C_1$ be the feet of the altitudes belonging to the vertices $A, B, C$, respectively. Let $K$ be a point on the smaller $AB_1$ arc of the circle with diameter $AB$ satisfying the condition $\angle HKB = \angle C_1KB$. Let $M$ be the point of inter... | [
"Since $A, C_1, L, K$ and $A, C_1, H, B_1$ are concyclic, so is $L, K, B_1, H$. Using these and the fact that $KL$ bisects $\\angle C_1KH$, we get $\\angle C_1AL = \\angle LB_1H$ and hence $\\angle ALC = \\angle LB_1C$. Therefore the triangles $ALC$ and $LB_1C$ are similar. Using this similarity as well as the fact... | Turkey | 17th Turkish 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 > Miscellaneous > Angle c... | English | proof only | null | |
02k8 | Problem:
Na figura estão indicadas em graus as medidas de alguns ângulos em função de $x$. Quanto vale $x$?
A) $6^{\circ}$
B) $12^{\circ}$
C) $18^{\circ}$
D) $20^{\circ}$
E) $24^{\circ}$
 | [
"Solution:\n\nCompletamos a figura marcando os ângulos $\\alpha$ e $\\beta$, lembrando que ângulos opostos pelo vértice são iguais. Como a soma dos ângulos internos de um triângulo é $180^{\\circ}$, podemos escrever as três igualdades abaixo, uma para cada um dos triângulos da figura:\n$$\n\\begin{aligned}\n& \\alp... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | C | |
0apx | Problem:
Let each of the characters $A, B, C, D, E$ denote a single digit, and $A B C D E 4$ and $4 A B C D E$ represent six-digit numbers. If
$$
4 \times A B C D E 4=4 A B C D E
$$
what is $C$ ? | [
"Solution:\nLet $x = A B C D E$. Then $4 \\times A B C D E 4 = 4 A B C D E$ implies\n$$\n4(10x + 4) = 400000 + x\n$$\nwhich gives us $x = 10256$, so that $C = 2$."
] | Philippines | Tenth Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 2 | |
00y0 | Problem:
Solve the system of equations in integers:
$$
\left\{\begin{array}{l}
z^{x}=y^{2 x} \\
2^{z}=4^{x} \\
x+y+z=20 .
\end{array}\right.
$$ | [
"Solution:\nFrom the second and third equation we find $z=2x$ and $x=\\frac{20-y}{3}$. Substituting these into the first equation yields $\\left(\\frac{40-2y}{3}\\right)^{x}=\\left(y^{2}\\right)^{x}$. As $x \\neq 0$ (otherwise we have $0^{0}$ in the first equation which is usually considered undefined) we have $y^{... | Baltic Way | Baltic Way 1993 | [
"Algebra > Intermediate Algebra > Exponential functions",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | x = 8, y = -4, z = 16 | |
04lw | Let $A$, $B$ and $C$ be points on the curve (hyperbola) with equation $xy = 1$. Prove that the orthocentre of the triangle $ABC$ also belongs to that curve. | [] | Croatia | Mathematical competitions in Croatia | [
"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 only | null | |
0gv5 | Solve (in real numbers) the equation
$$
\left|x - \frac{\pi}{6}\right| + \left|x + \frac{\pi}{3}\right| = \arcsin \frac{x^3 - x + 2}{2}.
$$ | [
"Використовуючи властивості модуля числа та властивості функції $\\arcsin$, маємо:\n$$\n\\frac{\\pi}{2} \\leq \\left|x - \\frac{\\pi}{6}\\right| + \\left|x + \\frac{\\pi}{3}\\right| = \\arcsin \\frac{x^3 - x + 2}{2} \\leq \\frac{\\pi}{2}.\n$$\nЗвідси випливає, що\n$$\n\\arcsin \\frac{x^3 - x + 2}{2} = \\frac{\\pi}{... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Algebra > Equations and Inequalities"
] | null | proof and answer | x = -1, 0 | |
02kx | Problem:
O perfume de Rosa - Rosa ganhou um vidro de perfume no formato de um cilindro de $7~\mathrm{cm}$ de raio da base e $10~\mathrm{cm}$ de altura. Depois de duas semanas usando o perfume restou $0{,}45~l$ no vidro. Qual a fração que representa o volume que Rosa já usou? | [
"Solution:\n\nO volume de um cilindro é o produto da área da base pela altura. Como o raio da base é $7~\\mathrm{cm}$, a área da base é: $\\pi \\times 7^{2}$, e então o volume do vidro é\n$$\n\\pi \\times 7^{2} \\times 10~\\mathrm{cm}^{3} = 490 \\pi~\\mathrm{cm}^{3} = \\frac{490 \\pi}{1000}~\\mathrm{dm}^{3} = 0,49 ... | Brazil | Nível 3 | [
"Geometry > Solid Geometry > Volume"
] | null | final answer only | (49π - 45)/(49π) | |
04f6 | Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that $f(1) \ge 0$ and
$$
f(x) - f(y) \ge (x - y)f(x - y),
$$
for all real numbers $x$ and $y$. | [
"If we plug $y \\to x - 1$ into the given inequality we get:\n$$\nf(x) - f(x-1) \\ge 1 \\cdot f(1) \\ge 0,\n$$\ni.e.\n$$\nf(x) \\ge f(x-1), \\quad \\text{for every } x \\in \\mathbb{R}. \\qquad (1)\n$$\n\nIf we plug $y \\to 0$ into the given inequality we get:\n$$\nf(x) - f(0) \\ge x f(x), \\qquad (2)\n$$\n\nand pl... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 0 for all real x | |
0jde | Problem:
Tim the Beaver can make three different types of geometrical figures: squares, regular hexagons, and regular octagons. Tim makes a random sequence $F_{0}, F_{1}, F_{2}, F_{3}, \ldots$ of figures as follows:
- $F_{0}$ is a square.
- For every positive integer $i$, $F_{i}$ is randomly chosen to be one of the 2 ... | [
"Solution:\n\nAnswer: 7\n\nWe write $F_{i}=n$ as shorthand for \"the $i$th figure is an $n$-sided polygon.\"\n\nIf $F_{1}=8$, then $F_{2}=6$ or $F_{2}=4$. If $F_{2}=6$, Tim is making a 6-gon at time 13 (probability contribution $1/4$). If $F_{2}=4$, $F_{3}=6$ or $F_{3}=8$ will take the time 13 mark ($1/8$ contribut... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 7 | |
0jkl | Problem:
The side lengths of a triangle are distinct positive integers. One of the side lengths is a multiple of $42$, and another is a multiple of $72$. What is the minimum possible length of the third side? | [
"Solution:\n\nSuppose that two of the side lengths are $42a$ and $72b$, for some positive integers $a$ and $b$. Let $c$ be the third side length. We know that $42a$ is not equal to $72b$, since the side lengths are distinct. Also, $6 \\mid 42a - 72b$. Therefore, by the triangle inequality, we get $c > |42a - 72b| \... | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 7 | |
0bn0 | Let $k \ge 1$, be a positive integer, let $p_1, p_2, \dots, p_k$ be distinct primes and denote $n = p_1p_2\dots p_k$. For a function $f : \{1, 2, \dots, n\} \to \{1, 2, \dots, n\}$, define $p(f) = f(1)f(2)\dots f(n)$.
a) Determine the number of functions $f$ such that $p(f)$ divides $n$.
b) For $n=6$, determine the n... | [
"a) If $p(f)$ divides $n$, then $p(f) = p_1^{a_1} p_2^{a_2} \\dots p_k^{a_k}$, with $a_i \\in \\{0, 1\\}$. It follows that a prime factor $p_i$ either does not divide $p(f)$, or it appears in the prime decomposition of exactly one of the numbers $f(1), f(2), \\dots, f(n)$. Thus, for each $p_i$ we have $n+1$ ways to... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | a) (n+1)^k; b) 580 | |
0g3n | Problem:
Let $ABC$ be a triangle with $AB = AC$ and $\angle BAC = 20^{\circ}$. Let $D$ be the point on the side $AB$ such that $\angle BCD = 70^{\circ}$. Let $E$ be the point on the side $AC$ such that $\angle CBE = 60^{\circ}$. Determine the value of the angle $\angle CDE$. | [
"Solution:\n\nDefine $E'$ on the side $AB$ such that $\\angle BCE' = 60^{\\circ}$, and $X$ the intersection of $BE$ and $CE'$. Straightforward angle chasing gives that $\\triangle E'EX$ is equilateral, as it has angles of $60^{\\circ}$. Moreover, $\\triangle AE'C$ is isosceles at $E'$, as $\\angle E'AC = 20^{\\circ... | Switzerland | Final round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 20° | |
0gy8 | Does there exist a polynomial $f(x) = x^3 + a x^2 + b x + c$ satisfying simultaneously the following conditions: $|c| \leq 2009$, $f$ has 3 integer roots and $|f(34)|$ is a prime number. | [
"Let $f(x) = (x - \\alpha)(x - \\beta)(x - \\gamma)$ be a polynomial which satisfies the necessary conditions. Then $\\alpha, \\beta, \\gamma$ are integer numbers and $|f(34)| = |(34 - \\alpha)(34 - \\beta)(34 - \\gamma)|$ is a prime number. Without loss of generality we have $|34 - \\alpha| = |34 - \\beta| = 1$ an... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | No | |
010a | Problem:
Given a triangle $A B C$ with $|A B| < |A C|$. The line passing through $B$ and parallel to $A C$ meets the external bisector of angle $B A C$ at $D$. The line passing through $C$ and parallel to $A B$ meets this bisector at $E$. Point $F$ lies on the side $A C$ and satisfies the equality $|F C| = |A B|$. Pro... | [
"Solution:\n\nSince the lines $B D$ and $A C$ are parallel and since $A D$ is the external bisector of $\\angle B A C$, we have $\\angle B A D = \\angle B D A$; denote their common size by $\\alpha$ (see Figure 5). Also $\\angle C A E = \\angle C E A = \\alpha$, implying $|A B| = |B D|$ and $|A C| = |C E|$. Let $B'... | Baltic Way | Baltic Way 1998 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
05ej | Problem:
Prouver qu'il n'existe qu'un nombre fini de nombres premiers s'écrivant sous la forme $n^{3}+2 n+3$ avec $n \in \mathbb{N}$. | [
"Solution:\n\nRemarquons que $n^{3}-n = n(n-1)(n+1)$ est un produit de trois entiers consécutifs. Puisque parmi trois entiers consécutifs il y a toujours un multiple de $3$, on obtient que $n^{3}-n$ est divisible par $3$, c'est-à-dire $n^{3} \\equiv n \\pmod{3}$. (On peut le voir aussi en utilisant le petit théorèm... | France | Envoi 1 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof only | null | |
0lbp | Let $a > 0$ and the sequence $(x_n)$ is defined by
$$ x_1 = a, $$
$$ x_{n+1} = x_n + \frac{\sqrt{x_n}}{n^2}, \forall n \ge 1. $$
Prove that $(x_n)$ has a finite limit when $n$ tends to infinity. | [
"We have $x_{n+1} < x_{n+1} + \\frac{1}{4n^4} = x_n + 2\\frac{\\sqrt{x_n}}{2n^2} + \\frac{1}{4n^4} = \\left(\\sqrt{x_n} + \\frac{1}{2n^2}\\right)^2$.\nTherefore $\\sqrt{x_{n+1}} < \\sqrt{x_n} + \\frac{1}{2n^2}$, $n = 1, 2, 3, \\dots$ thus\n$$\n\\begin{aligned}\n\\sqrt{x_{n+1}} &< \\sqrt{a} + \\sum_{i=1}^{n} \\frac{... | Vietnam | Vietnamese Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
0hb0 | It is given that real roots of a quadratic polynomial $g(x) = x^2 - 3x + a$ are also roots of polynomial $f(x) = x^3 - x^2 + c x + 4$. Analogously, both real roots of a quadratic polynomial $h(x) = x^2 + x + b$ are also roots of $f(x)$. What values can $f(1)$ take? | [
"Since $f(x)$ is a cubic polynomial, it has no more than three real roots, hence, quadratic polynomials $g(x)$ and $h(x)$ have the same root. Let us denote this root by $t$. Then,\n$$\ng(t) = t^2 - 3t + a = 0 \\text{ and } h(t) = t^2 + t + b = 0 \\Rightarrow 4t + b - a = 0 \\Rightarrow t = \\frac{1}{4}(a-b).\n$$\n\... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 0 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.