In triangle $A B C$, three equal circles are placed, each touching two sides of the triangle. All three circles have one common point. Find the radii of these circles if the radii of the inscribed and circumscribed circles of triangle $A B C$ are $r$ and $R$.

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The triangle with vertices at the centers of equal circles is similar to the given one, and the radius of its circumscribed circle is equal to the radius of these circles.

## Solution

Let $x$ be the desired radius. Denote $B C=a, \angle A=\alpha, O_{1}, O_{2}, O_{3}$ as the centers of the equal circles inscribed in the angles $A, B$ and $C$ respectively, $M$ - the common point of these circles.

Since $x$ is the radius of the circumscribed circle of triangle $O_{1} \mathrm{O}_{2} \mathrm{O}_{3}$, then $\mathrm{O}_{2} \mathrm{O}_{3}=2 x \sin \angle \mathrm{O}_{2} \mathrm{O}_{1} \mathrm{O}_{3}=2 x \sin \alpha, a=$ $2 R \sin \alpha, \sin \alpha={ }^{a} / 2 R, O_{2} O_{3}=a x / R$.

Rays $\mathrm{BO}_{2}$ and $\mathrm{CO}_{3}$ intersect at the center $O$ of the inscribed circle of triangle $A B C$. The heights of similar triangles $\mathrm{OO}_{2} \mathrm{O}_{3}$ and $\mathrm{OBC}$, drawn from the common vertex $O$, are in the same ratio as the bases of these triangles. Therefore, ${ }^{r-x} / r=\mathrm{O}_{2} \mathrm{O}_{3 / a}={ }^{x / / R}$, from which $x=r{ }^{r R} / R+r$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9e7ee8b57660b9721937g-38.jpg?height=317&width=548&top_left_y=1465&top_left_x=755)

## Answer

$r R /{ }_{R+r}$.