We are given two mutually tangent circles in the plane, with radii $r_1, r_2$. A line intersects these circles in four points, determining three segments of equal length. Find this length as a function of $r_1$ and $r_2$ and the condition for the solvability of the problem.

In a triangle scalene $ABC$, let $I$ be its incenter and $D$ the intersection of $AI$ and $BC$. Let $M$ and $N$ points where the incircle touches $AB$ and $AC$, respectively. Let $F$ be the second intersection of the circumcircle $(AMN)$ with the circumcircle $(ABC)$. Let $T$ the intersection of $AF$ and $BC$. Let $J$ be the intersection of $TI$ with the line parallel of $FI$ that passes through $D$. Prove that the line $AJ$ is perpendicular to $BC$.

By straightedge and compass, reconstruct a right triangle $ABC$ ($\angle C = 90^o$), given the vertices $A, C$ and a point on the bisector of angle $B$.

The radii of the circumcircle and the incircle of a right triangle are given. Cconstruct that triangle with compass and ruler, describe the construction and justify why it is correct.

If \[N=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}},\] then $N$ equals $\textbf{(A) }1\qquad\textbf{(B) }2\sqrt{2}-1\qquad\textbf{(C) }\frac{\sqrt{5}}{2}\qquad\textbf{(D) }\sqrt{\frac{5}{2}}\qquad \textbf{(E) }\text{none of these}$

A point $P$ lies on side $AB$ of a convex quadrilateral $ABCD$. Let $\omega$ be the inscribed circumference of triangle $CPD$ and $I$ the centre of $\omega$. It is known that $\omega$ is tangent to the inscribed circumferences of triangles $APD$ and $BPC$ at points $K$ and $L$ respectively. Let $E$ be the point where the lines $AC$ and $BD$ intersect, and $F$ the point where the lines $AK$ and $BL$ intersect. Prove that the points $E, I, F$ are collinear.

Let $f(x)=x^2+x+1$. Determine, with proof, all positive integers $n$ such that $f(k)$ divides $f(n)$ whenever $k$ is a positive divisor of $n$.

$ABC$ is a triangle. Take $a = BC$ etc as usual. Take points $T_1, T_2$ on the side $AB$ so that $AT_1 = T_1T_2 = T_2B$. Similarly, take points $T_3, T_4$ on the side BC so that $BT_3 = T_3T_4 = T_4C$, and points $T_5, T_6$ on the side $CA$ so that $CT_5 = T_5T_6 = T_6A$. Show that if $a' = BT_5, b' = CT_1, c'=AT_3$, then there is a triangle $A'B'C'$ with sides $a', b', c'$ ($a' = B'C$' etc). In the same way we take points $T_i'$ on the sides of $A'B'C' $ and put $a'' = B'T_6', b'' = C'T_2', c'' = A'T_4'$. Show that there is a triangle $A'' B'' C'' $ with sides $a'' b'' , c''$ and that it is similar to $ABC$. Find $a'' /a$.

$\vartriangle A_0B_0C_0$ has side lengths $A_0B_0 = 13$, $B_0C_0 = 14$, and $C_0A_0 = 15$. $\vartriangle A_1B_1C_1$ is inscribed in the incircle of $\vartriangle A_0B_0C_0$ such that it is similar to the first triangle. Beginning with $\vartriangle A_1B_1C_1$, the same steps are repeated to construct $\vartriangle A_2B_2C_2$, and so on infinitely many times. What is the value of $\sum_{i=0}^{\infty} A_iB_i$?

Triangle $ABC$ is inscribed in a circle with center $O'$. A circle with center $O$ is inscribed in triangle $ABC$. $AO$ is drawn, and extended to intersect the larger circle in $D$. Then, we must have: $\text{(A)}\ CD=BD=O'D \qquad \text{(B)}\ AO=CO=OD \qquad \text{(C)}\ CD=CO=BD \qquad\\ \text{(D)}\ CD=OD=BD \qquad \text{(E)}\ O'B=O'C=OD $ [asy] size(200); defaultpen(linewidth(0.8)+fontsize(12pt)); pair A=origin,B=(15,0),C=(5,9),O=incenter(A,B,C),Op=circumcenter(A,B,C); path incirc = incircle(A,B,C),circumcirc = circumcircle(A,B,C),line=A--3*O; pair D[]=intersectionpoints(circumcirc,line); draw(A--B--C--A--D[0]^^incirc^^circumcirc); dot(O^^Op,linewidth(4)); label("$A$",A,dir(185)); label("$B$",B,dir(355)); label("$C$",C,dir(95)); label("$D$",D[0],dir(O--D[0])); label("$O$",O,NW); label("$O'$",Op,E);[/asy]

The diagonals $AC$ and $BD$ of a convex cyclic quadrilateral $ABCD$ intersect at point $E$. Given that $AB = 39, AE = 45, AD = 60$ and $BC = 56$, determine the length of $CD.$

In triangle $ABC$ let $M$ and $N$ be midpoints of $BC$ and $AC,$ respectively. Point $P$ is inside $ABC$ such that $\angle BAP = \angle PBC = \angle PCA .$ Prove that if $\angle PNA = \angle AMB,$ then $ABC$ is isosceles triangle.

We consider the non-zero natural numbers $(m, n)$ such that the numbers $$\frac{m^2 + 2n}{n^2 - 2m} \,\,\,\, and \,\,\, \frac{n^2 + 2m}{m^2-2n}$$ are integers. a) Show that $|m - n| \le 2$: b) Find all the pairs $(m, n)$ with the property from hypothesis $a$.

Is there a real number $\alpha$ such that $\cos\alpha$ is irrational but $\cos 2\alpha$, $\cos 3\alpha$, $\cos 4\alpha$, $\cos 5\alpha$ are all rational? (Author: V. Senderov)

Let $n,k \in \mathbb{Z}^{+}$ with $k \leq n$ and let $S$ be a set containing $n$ distinct real numbers. Let $T$ be a set of all real numbers of the form $x_1 + x_2 + \ldots + x_k$ where $x_1, x_2, \ldots, x_k$ are distinct elements of $S.$ Prove that $T$ contains at least $k(n-k)+1$ distinct elements.

Find all functions $f:\mathbb{N}_0\to\mathbb{N}$ such that [b]1)[/b] \(f(a)\) divides \(a\) for every \(a\in\mathbb{N}_0\), and [b]2)[/b] for all \(a,b,k\in\mathbb{N}_0\) we have \[ f\bigl(f(a)+kb\bigr)\;=\;f\bigl(a + k\,f(b)\bigr). \]

Right isosceles triangle $T$ is placed in the first quadrant of the coordinate plane. Suppose that the projection of $T$ onto the $x$-axis has length $6$, while the projection of $T$ onto the $y$-axis has length $8$. What is the sum of all possible areas of the triangle $T$? [asy] import olympiad; size(120); defaultpen(linewidth(0.8)); pair A = (0.9,0.6), B = (1.7, 0.8), C = rotate(270, B)*A; pair PAx = (A.x,0), PBx = (B.x,0), PAy = (0, A.y), PCy = (0, C.y); draw(PAx--A--PAy^^PCy--C^^PBx--B, linetype("4 4")); draw(rightanglemark(A,B,C,3)); draw(A--B--C--cycle); draw((0,2)--(0,0)--(2,0),Arrows(size=8)); label("$6$",(PAx+PBx)/2,S); label("$8$",(PAy+PCy)/2,W); [/asy]

Let $ABC$ be a triangle and the points $M$ and $N$ on the sides $AB$ respectively $BC$, such that $2 \cdot \frac{CN}{BC} = \frac{AM}{AB}$. Let $P$ be a point on the line $AC$. Prove that the lines $MN$ and $NP$ are perpendicular if and only if $PN$ is the interior angle bisector of $\angle MPC$.

Let ${ABC}$ be an acute angled triangle, and ${H}$ a point in its interior. Let the reflections of ${H}$ through the sides ${AB}$ and ${AC}$ be called ${H_{c} }$ and ${H_{b} }$ , respectively, and let the reflections of H through the midpoints of these same sidesbe called ${H_{c}^{'} }$ and ${H_{b}^{'} }$, respectively. Show that the four points ${H_{b}, H_{b}^{'} , H_{c}}$, and ${H_{c}^{'} }$ are concyclic if and only if at least two of them coincide or ${H}$ lies on the altitude from ${A}$ in triangle ${ABC}$.

Let $p$ and $n$ be a natural numbers such that $p$ is a prime and $1+np$ is a perfect square. Prove that the number $n+1$ is sum of $p$ perfect squares.

Find the greatest integer $n$ such that $10^n$ divides $$\frac{2^{10^5} 5^{2^{10}}}{10^{5^2}}$$

Maria buys computer disks at a price of 4 for 5 dollars and sells them at a price of 3 for 5 dollars. How many computer disks must she sell in order to make a profit of 100 dolars? $ \text{(A)}\ 100\qquad\text{(B)}\ 120\qquad\text{(C)}\ 200\qquad\text{(D)}\ 240\qquad\text{(E)}\ 1200 $

The winners of first AGMC in 2019 gifts the person in charge of the organiser, which is a polyhedron formed by $60$ congruent triangles. From the photo, we can see that this polyhedron formed by $60$ quadrilateral spaces. (Note: You can find the photo in 3.4 of [url]https://files.alicdn.com/tpsservice/18c5c7b31a7074edc58abb48175ae4c3.pdf?spm=a1zmmc.index.0.0.51c0719dNAbw3C&file=18c5c7b31a7074edc58abb48175ae4c3.pdf[/url]) A quadrilateral space is the plane figures that we fold the figures following the diagonal on a $n$ sides on a plane (i.e. form an appropriate dihedral angle in where the chosen diagonal is). "Two figure spaces are congruent" means they can coincide completely by isometric transform in $\mathbb{R}^3$. A polyhedron is the bounded space region, whose boundary is formed by the common edge of finite polygon. (a) We know that $2021=43\times 47$. Does there exist a polyhedron, whose surface can be formed by $43$ congruent $47$-gon? (b) Prove your answer in (a) with logical explanation.

Let $n > 1$ be a fixed integer. On an infinite row of squares, there are $n$ stones on square $1$ and no stones on squares $2$, $3$, $4$, $\ldots$. Curious George plays a game in which a [i]move[/i] consists of taking two adjacent piles of sizes $a$ and $b$, where $a-b$ is a nonzero even integer, and transferring stones to equalize the piles (so that both piles have $\frac{a+b}{2}$ stones). The game ends when no more moves can be made. George wants to analyze the number of moves it takes to end the game. (a) Suppose George wants to end the game as quickly as possible. How many moves will it take him? (b) Suppose George wants to end the game as slowly as possible. Show that for all $n > 2$, the game will end after at most $\frac{3}{16}n^2$ moves. [i]Scoring note:[/i] For part (b), partial credit will be awarded for correct proofs of weaker bounds, eg. $\frac{1}{4}n^2$, $n^k$, or $k^n$ (for some $k \geq 2$).

Let $A$ and $E$ be opposite vertices of an octagon. A frog starts at vertex $A.$ From any vertex except $E$ it jumps to one of the two adjacent vertices. When it reaches $E$ it stops. Let $a_n$ be the number of distinct paths of exactly $n$ jumps ending at $E$. Prove that: \[ a_{2n-1}=0, \quad a_{2n}={(2+\sqrt2)^{n-1} - (2-\sqrt2)^{n-1} \over\sqrt2}. \]