Let there be two circles. Find all points $M$ such that there exist two points, one on each circle such that $M$ is their midpoint.

Call a positive integer $n$ $k$[i]-pretty[/i] if $n$ has exactly $k$ positive divisors and $n$ is divisible by $k$. For example, $18$ is $6$[i]-pretty[/i]. Let $S$ be the sum of positive integers less than $2019$ that are $20$[i]-pretty[/i]. Find $\tfrac{S}{20}$.

Prove that $p(x)=1+x+x^2 +\ldots+x^n$ is reducible over $\mathbb{F}_{2}$ in case $n+1$ is composite. If $n+1$ is prime, is $p(x)$ irreducible over $\mathbb{F}_{2}$ ?

Triangle $ABC$ has sidelengths $AB=1$, $BC=\sqrt{3}$, and $AC=2$. Points $D,E$, and $F$ are chosen on $AB, BC$, and $AC$ respectively, such that $\angle EDF = \angle DFA = 90^{\circ}$. Given that the maximum possible value of $[DEF]^2$ can be expressed as $\frac{a}{b}$ for positive integers $a, b$ with $\gcd (a, b) = 1$, find $a + b$. (Here $[DEF]$ denotes the area of triangle $DEF$.) [i]Proposed by Vismay Sharan[/i]

Over all real numbers $x$ and $y$ such that $$x^3=3x+y \qquad \text{and} \qquad y^3=3y+x,$$ compute the sum of all possible values of $x^2+y^2.$

$$\sin 30^o + \sin 45^o + \sin 60^o + \sin 90^o + \cos 120^o + \cos 135^o + \cos 150^o + \cos 180^o = ?$$

In a word of more than $10$ letters, any two consecutive letters are different. Prove that one can change places of two consecutive letters so that the resulting word is not [i]periodic[/i], that is, cannot be divided into equal subwords.

The natural numbers $x_1, x_2, \ldots , x_n$ are such that all their $2^n$ partial sums are distinct. Prove that: $$ {x_1}^2 + {x_2}^2 + \ldots + {x_n}^2 \geq \frac{4^n – 1}{3}. $$

A number of linked rings, each 1 cm thick, are hanging on a peg. The top ring has an outside diameter of 20 cm. The outside diameter of each of the outer rings is 1 cm less than that of the ring above it. The bottom ring has an outside diameter of 3 cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring? [asy] size(200); defaultpen(linewidth(3)); real[] inrad = {40,34,28,21}; real[] outrad = {55,49,37,30}; real[] center; path[][] quad = new path[4][4]; center[0] = 0; for(int i=0;i<=3;i=i+1) { if(i != 0) { center[i] = center[i-1] - inrad[i-1] - inrad[i]+3.5; } quad[0][i] = arc((0,center[i]),inrad[i],0,90)--arc((0,center[i]),outrad[i],90,0)--cycle; quad[1][i] = arc((0,center[i]),inrad[i],90,180)--arc((0,center[i]),outrad[i],180,90)--cycle; quad[2][i] = arc((0,center[i]),inrad[i],180,270)--arc((0,center[i]),outrad[i],270,180)--cycle; quad[3][i] = arc((0,center[i]),inrad[i],270,360)--arc((0,center[i]),outrad[i],360,270)--cycle; draw(circle((0,center[i]),inrad[i])^^circle((0,center[i]),outrad[i])); } void fillring(int i,int j) { if ((j % 2) == 0) { fill(quad[i][j],white); } else { filldraw(quad[i][j],black); } } for(int i=0;i<=3;i=i+1) { for(int j=0;j<=3;j=j+1) { fillring(((2-i) % 4),j); } } for(int k=0;k<=2;k=k+1) { filldraw(circle((0,-228 - 25 * k),3),black); } real r = 130, s = -90; draw((0,57)--(r,57)^^(0,-57)--(r,-57),linewidth(0.7)); draw((2*r/3,56)--(2*r/3,-56),linewidth(0.7),Arrows(size=3)); label("$20$",(2*r/3,-10),E); draw((0,39)--(s,39)^^(0,-39)--(s,-39),linewidth(0.7)); draw((9*s/10,38)--(9*s/10,-38),linewidth(0.7),Arrows(size=3)); label("$18$",(9*s/10,0),W); [/asy] $ \textbf{(A) } 171\qquad \textbf{(B) } 173\qquad \textbf{(C) } 182\qquad \textbf{(D) } 188\qquad \textbf{(E) } 210$

Eight tiles are located on an $8\times 8$ board in such a way that no pair of them are in the same row or in the same column. Prove that, among the distances between each pair of tiles, we can find two of them that are equal (the distance between two tiles is the distance between the centers of the squares in which they are located).

Let $A\in \mathcal{M}_n(\mathbb{R}^*)$. If $A\cdot\ ^t A=I_n$, prove that: a)$|\text{Tr}(A)|\le n$; b)If $n$ is odd, then $\det(A^2-I_n)=0$.

Let $f$, $g$, $h : [a, b] \to \mathbb{R}$, three integrable functions such that:$$\int \limits_a^b fgdx=\int \limits_a^bghdx=\int \limits_a^bhfdx=\int \limits_a^bg^2dx\int \limits_a^bh^2dx=1$$Then$$\int \limits_a^bg^2dx=\int \limits_a^bh^2dx=1$$

Paul and Sara are playing a game with integers on a whiteboard, with Paul going first. When it is Paul’s turn, he can pick any two integers on the board and replace them with their product; when it is Sara’s turn, she can pick any two integers on the board and replace them with their sum. Play continues until exactly one integer remains on the board. Paul wins if that integer is odd, and Sara wins if it is even. Initially, there are $2021$ integers on the board, each one sampled uniformly at random from the set $\{0, 1, 2, 3, . . . , 2021\}$. Assuming both players play optimally, the probability that Paul wins is $m/n$ , where $m, n$ are positive integers and $gcd(m, n) = 1$. Find the remainder when $m + n$ is divided by $1000$.

For an integer $m\geq 4,$ let $T_{m}$ denote the number of sequences $a_{1},\dots,a_{m}$ such that the following conditions hold: (1) For all $i=1,2,\dots,m$ we have $a_{i}\in \{1,2,3,4\}$ (2) $a_{1} = a_{m} = 1$ and $a_{2}\neq 1$ (3) For all $i=3,4\cdots, m, a_{i}\neq a_{i-1}, a_{i}\neq a_{i-2}.$ Prove that there exists a geometric sequence of positive integers $\{g_{n}\}$ such that for $n\geq 4$ we have that \[ g_{n} - 2\sqrt{g_{n}} < T_{n} < g_{n} + 2\sqrt{g_{n}}.\]

Show that there are infinitely many positive integers $c$, such that the following equations both have solutions in positive integers: $(x^2 - c)(y^2 -c) = z^2 -c$ and $(x^2 + c)(y^2 - c) = z^2 - c$.

A directed graph has each vertex with outdegree 2. Prove that it is possible to split the vertices into 3 sets so that for each vertex $v$, $v$ is not simultaneously in the same set with both of the vertices that it points to. [i]David Yang.[/i] [hide="Stronger Version"]See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=492100]here[/url].[/hide]

Rays $l_1$ and $l_2$ pass through a point $O$. Segments $OA_1$ and $OB_1$ on $l_1$, and $OA_2$ and $OB_2$ on $l_2$, are drawn so that $\frac{OA_1}{OA_2} \ne \frac{OB_1}{OB_2}$ . Find the set of all intersection points of lines $A_1A_2$ and $B_1B_2$ as $l_2$ rotates around $O$ while $l_1$ is fixed.

Let $ABC$ be a triangle and $P$ be an interior point. Let $\omega_A$ be the circle that is tangent to the circumcircle of $BPC$ at $P$ internally and tangent to the circumcircle of $ABC$ at $A_1$ internally and let $\Gamma_A$ be the circle that is tangent to the circumcircle of $BPC$ at $P$ externally and tangent to the circumcircle of $ABC$ at $A_2$ internally. Define $B_1$, $B_2$, $C_1$, $C_2$ analogously. Let $O$ be the circumcentre of $ABC$. Prove that the lines $A_1A_2$, $B_1B_2$, $C_1C_2$ and $OP$ are concurrent.

Let $ABC$ be an acute-angled triangle with $AB\not= AC$. Let $\Gamma$ be the circumcircle, $H$ the orthocentre and $O$ the centre of $\Gamma$. $M$ is the midpoint of $BC$. The line $AM$ meets $\Gamma$ again at $N$ and the circle with diameter $AM$ crosses $\Gamma$ again at $P$. Prove that the lines $AP,BC,OH$ are concurrent if and only if $AH=HN$.

The map shows sections of three straight roads connecting the three villages, but the villages themselves are located outside the map. In addition, the fire station located at an equal distance from the three villages is not indicated on the map, although its location is within the map. Is it possible to find this place with the help of a compass and a ruler, if the construction is carried out only within the map?

Let $X,Y$ be two points which lies on the line $BC$ of $\triangle ABC(X,B,C,Y\text{lies in sequence})$ such that $BX\cdot AC=CY\cdot AB$, $O_1,O_2$ are the circumcenters of $\triangle ACX,\triangle ABY$, $O_1O_2\cap AB=U,O_1O_2\cap AC=V$. Prove that $\triangle AUV$ is a isosceles triangle.

In a country, there are some cities and the city named [i]Ben Song[/i] is capital. Each cities are connected with others by some two-way roads. One day, the King want to choose $n$ cities to add up with [i]Ben Song[/i] city to establish an [i]expanded capital[/i] such that the two following condition are satisfied: (i) With every two cities in [i]expanded capital[/i], we can always find a road connecting them and this road just belongs to the cities of [i]expanded capital[/i]. (ii) There are exactly $k$ cities which do not belong to [i]expanded capital[/i] have the direct road to at least one city of [i]expanded capital[/i]. Prove that there are at most $\binom{n+k}{k}$ options to expand the capital for the King.

Prove that for all $a,b,c$ positive real numbers $\dfrac{a^4+1}{b^3+b^2+b}+\dfrac{b^4+1}{c^3+c^2+c}+\dfrac{c^4+1}{a^3+a^2+a} \ge 2$

In triangle $ABC$ angle bisectors $AA_{1}$ and $CC_{1}$ intersect at $I$. Line through $B$ parallel to $AC$ intersects rays $AA_{1}$ and $CC_{1}$ at points $A_{2}$ and $C_{2}$ respectively. Let $O_{a}$ and $O_{c}$ be the circumcenters of triangles $AC_{1}C_{2}$ and $CA_{1}A_{2}$ respectively. Prove that $\angle{O_{a}BO_{c}} = \angle{AIC} $

$ABC$ is a right triangle with $\angle C=90$,$CD$ is perpendicular to $AB$,and $D$ is the foot,$\omega$ is the circumcircle of triangle $BCD$,$\omega_{1}$ is a circle inside triangle $ACD$,tangent to $AD$ and $AC$ at $M$ and $N$ respectively,and $\omega_{1}$ is also tangent to $\omega$.prove that: (1)$BD*CN+BC*DM=CD*BM$ (2)$BM=BC$