\left.\begin{array}{l}\text { Tangent circles } \\ {[\text { Area of a circle, sector, and segment }]}\end{array}\right]

In triangle $ABC$, three circles of equal radii are placed such that each circle touches two sides of the triangle. One of the circles (with center $O_{1}$) touches the other two (with centers $O_{2}$ and $O_{3}$, respectively), and $\angle \mathrm{O}_{2} \mathrm{O}_{1} O_{3}=90^{\circ}$. Determine which is greater: the area of the circle bounded by the circle with center $O_{1}$, or one-fifth of the area of triangle $ABC$?
$A B C$ - is an isosceles right triangle.

## Solution

Let $R$ be the radius of the circles. Suppose the circle with center $O_{1}$ touches sides $A B$ and $B C$ at points $M$ and $N$ respectively, and the circle with center $O_{2}$ touches side $A B$ at point $P$. From the tangency condition and the equality of the circles, the following relationships follow:

$$
O_{1} O_{2}=M P=2 R, O_{1} O_{3}=2 R, P O_{2}=M O_{1}=R
$$

as well as the parallelism of the sides of triangles $O_{1} O_{2} O_{3}$ and $A B C$ ( $B C \| O_{1} O_{3}$ and $\left.A C \| O_{2} O_{3}\right)$.

Since $\angle O_{2} O_{1} O_{3}=90^{\circ}$, then $\angle A B C=90^{\circ}$ and triangle $A B C$ is isosceles and right. Therefore,

$$
\begin{gathered}
B C=A B=B M+M P+P A=R+2 R+R(\sqrt{2}+1)=R(4+\sqrt{2}) \\
S_{\triangle \mathrm{ABC}}=\frac{1}{2} A B \cdot B C=\frac{1}{2} R(4+\sqrt{2})^{2}=\frac{R^{2}(18+8 \sqrt{2})}{2}
\end{gathered}
$$

Therefore,

$$
\frac{\frac{1}{5} S_{\triangle A B C}}{\pi R^{2}}=\frac{18+\sqrt{2}}{10 \pi}<\frac{18+12}{10 \pi}=\frac{3}{\pi}<1
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_ae1b0d3bbbf9aa5c38e8g-55.jpg?height=489&width=834&top_left_y=2282&top_left_x=612)

## Answer

## Problem