The sides of a triangle are 1 and 2, and the angle between them is $120^{\circ}$.

A circle with its center on the third side of the triangle touches the other two sides. A second circle touches these sides and the first circle. Find the radii of the circles.
To find the radius of the first circle, express the area of triangle $ABC$ in two ways.

## Solution

Let the circle of radius $R$ with center $O$ on side $BC$ of triangle $ABC$ touch sides $AB$ and $AC$ at points $M$ and $N$ respectively, with $AB=1$, $AC=2$, and $\angle A=120^{\circ}$. Denote $S=S_{ABC}$. Then $2S=AB \cdot AC \sin 120^{\circ}=\sqrt{3}$ and $2S=AB \cdot R + AC \cdot R = 3R$.

From this, $R=\frac{\sqrt{3}}{3}$.

Let $Q$ be the center of the second circle of radius $r$. Points $O$ and $Q$ lie on the bisector of angle $A$.

Drop a perpendicular $QF$ to the radius $ON$. From the right triangle $OQF$ we get

$R-r=(R+r) \sin 60^{\circ}$, hence

$$
r=\frac{(2-\sqrt{3}) R}{2+\sqrt{3}}=(7-4 \sqrt{3}) R=\frac{\sqrt{3}(7-4 \sqrt{3})}{3}
$$

## Answer

$\frac{\sqrt{3}}{3}, \frac{\sqrt{3}(7-4 \sqrt{3})}{3}$

Given a triangle with sides $a, b$, and $c$. A line parallel to the side equal to $a$ touches the inscribed circle of the triangle and intersects the other two sides at points $M$ and $N$. Find $MN$.

## Hint

Let points $M$ and $N$ lie on sides $AB$ and $AC$ of triangle $ABC$, and the inscribed circle of triangle $ABC$ touches side $AB$ at point $K$. Then the segment $AK$ is equal to the semiperimeter of triangle $AMN$ and, at the same time, the difference between the semiperimeter of triangle $ABC$ and side $BC$.

## Solution

Let points $M$ and $N$ lie on sides $AB$ and $AC$ of triangle $ABC$ and $BC=a, AB=c, AC=b$, and the inscribed circle of triangle $ABC$ touches side $AB$ at point $K$. Then $AK=\frac{1}{2}(b+c-a)$.

On the other hand, the segment $AK$ is equal to the semiperimeter of triangle $AMN$.

Since $MN \parallel BC$, triangle $AMN$ is similar to triangle $ABC$, and the similarity ratio is

the ratio of the perimeters of these triangles, that is

$$
\frac{b+c-a}{b+c+a} \text{. Therefore, } MN=BC \cdot \frac{b+c-a}{b+c+a}=\frac{a(b+c-a)}{b+c+a} \text{. }
$$

Chords $AB$ and $CD$ intersect at point $P$. It is known that $AB=CD=12, \angle APC=60^{\circ}$, and $AC=2BD$. Find the sides of triangle $BPD$.

## Hint

Triangle $BPD$ is equilateral.

## Solution

Let $DP=x, BP=y$. Triangles $DPB$ and $APC$ are similar by two angles, so $AP=2x$ and $CP=2y$. By the condition $x+2y=12=2x+y$, from which

$x=y=4$, and since $\angle BPD=\angle APC=60^{\circ}$, triangle $BPD$ is equilateral.

## Answer