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Inscribed and Circumscribed Circles
Two tangents drawn from one point
[The area of a figure is equal to the sum of the areas of the figures into which it is divided]

On the sides $B C, C A$, and $A B$ of a triangle, points $A_{1}, B_{1}, C_{1}$ are taken respectively, such that the radii of the circles inscribed in triangles $A_{1} B C_{1}, A B_{1} C_{1}$, and $A_{1} B_{1} C$ are equal to each other and equal to $r$. The radius of the circle inscribed in triangle $A_{1} B_{1} C_{1}$ is $r_{1}$. Find the radius of the circle inscribed in triangle $A B C$.

#
Using the equality of tangent segments drawn from one point to a circle, prove that the perimeter of triangle $A_{1} B_{1} C_{1}$ is equal to the perimeter of the triangle with vertices at the centers of the inscribed circles of triangles $A_{1} B C_{1}, A B_{1} C_{1}$, and $A_{1} B_{1} C$.

## Solution

Let $O_{1}, O_{2}, O_{3}$ be the centers of the circles inscribed in triangles $A_{1} B C_{1}, A B_{1} C_{1}$, and $A_{1} B_{1} C$ respectively, and let the circle with center $O_{1}$ touch segments $A B$ and $C_{1} B_{1}$ at points $L$ and $N$ respectively, the circle with center $O_{2}$ touch segments $A B$ and $A_{1} C_{1}$ at points $K$ and $M$ respectively, and the circle with center $O_{3}$ touch $A_{1} B_{1}$ at point $P$. Then

$$
O_{1} O_{2}=L K=L C_{1}+K C_{1}=C_{1} N+C_{1} M
$$

Similarly, we can prove that

$$
O_{1} O_{3}=B_{1} N+B_{1} P, O_{2} O_{3}=A_{1} M+A_{1} P
$$

Thus, the perimeter of triangle $O_{1} O_{2} O_{3}$ is equal to the perimeter of triangle $A_{1} B_{1} C_{1}$.

Note that the radius of the circle inscribed in triangle $ABC$ is equal to the sum of the radius of the inscribed circle of triangle $O_{1} O_{2} O_{3}$ and $r$. We will prove that the radius of the inscribed circle of triangle $O_{1} O_{2} O_{3}$ is $r_{1}$.

Since the perimeters of triangles $O_{1} O_{2} O_{3}$ and $A_{1} B_{1} C_{1}$ are equal, and the area of a triangle is the product of its semiperimeter and the radius of the inscribed circle, it is sufficient to prove the equality of the areas of these triangles.

Consider the sum of the areas of triangles $A_{1} B C_{1}, A B_{1} C_{1}$, and $A_{1} B_{1} C$:

$$
\begin{gathered}
S_{\Delta A_{1} B C_{1}}+S_{\Delta A B_{1} C_{1}}+S_{\Delta A_{1} B_{1} C}= \\
=\frac{1}{2}\left(B C_{1}+A_{1} C_{1}+B A_{1}\right) r+\frac{1}{2}\left(A B_{1}+B_{1} C_{1}+A C_{1}\right) r+\frac{1}{2}\left(C A_{1}+A_{1} B_{1}+C B_{1}\right) r= \\
=\frac{1}{2}\left(B C_{1}+A_{1} C_{1}+B A_{1}+A B_{1}+B_{1} C_{1}+A C_{1}+C A_{1}+A_{1} B_{1}+C B_{1}\right) r= \\
=\frac{1}{2}\left(\left(B C_{1}+A C_{1}\right)+\left(B A_{1}+C A_{1}\right)+\left(A B_{1}+C B_{1}\right)+\left(A_{1} C_{1}+B_{1} C_{1}+A_{1} B_{1}\right)\right) r= \\
\left.=\frac{1}{2}(A B+B C+A C)+\left(A_{1} C_{1}+B_{1} C_{1}+A_{1} B_{1}\right)\right) r=\left(p+p_{1}\right) r
\end{gathered}
$$

where $p$ and $p_{1}$ are the semiperimeters of triangles $ABC$ and $A_{1} B_{1} C_{1}$. Thus,

$$
S_{\triangle A B C}=S_{\Delta A_{1} B_{1} C_{1}}+\left(p+p_{1}\right) r .
$$

On the other hand, the area of triangle $ABC$ is equal to the sum of the area of triangle $O_{1} O_{2} O_{2}$ and three trapezoids: $A O_{1} O_{2} B, A O_{1} O_{3} C$, and $B O_{2} O_{3} C$, i.e.,

$$
\begin{gathered}
S_{\triangle A B C}=S_{\Delta O_{1} O_{2} O_{2}}+S_{A O_{1} O_{2} B}+S_{A O_{1} O_{3} C}+S_{B O_{2} O_{3} C}= \\
=S_{\Delta O_{1} O_{2} O_{2}}+\frac{1}{2}\left(A B+O_{1} O_{2}\right) r+\frac{1}{2}\left(A C+O_{1} O_{3}\right) r+\frac{1}{2}\left(B C+O_{2} O_{3}\right) r= \\
=S_{\Delta O_{1} O_{2} O_{2}}+\frac{1}{2}\left(A B+O_{1} O_{2}+A C+O_{1} O_{3}+B C+O_{2} O_{3}\right) r= \\
=p_{1} r_{2}+p r+p_{1} r=p_{1} r_{2}+\left(p+p_{1}\right) r
\end{gathered}
$$

where $r_{2}$ is the radius of the inscribed circle of triangle $O_{1} O_{2} O_{2}$. Thus,

$$
p_{1} r_{1}+\left(p+p_{1}\right) r=p_{1} r_{2}+\left(p+p_{1}\right) r .
$$

Therefore, $r_{2}=r_{1}$, i.e., the radii of the circles inscribed in triangles $O_{1} O_{2} O_{2}$ and $A_{1} B_{1} C_{1}$ are equal.

Thus, the radius of the circle inscribed in triangle $ABC$ is $r_{1}+r$.

## Answer

$r+r_{1}$

## Problem