10.245. Three circles are inscribed in a corner - a small, a medium, and a large one. The large circle passes through the center of the medium one, and the medium one through the center of the small one. Determine the radii of the medium and large circles if the radius of the smaller one is $r$ and the distance from its center to the vertex of the angle is $a$.
Solution.

Let $O_{1}, O_{2}, O_{3}$ be the centers of the circles mentioned in the problem, inscribed in the angle $B A C$ (Fig. 10.52), $O_{1} A=a . E, F, K$ - the points of tangency of the small, medium, and large circles with the side $A C$ of the angle, respectively. Then $O_{1} E=r, O_{1} E \perp A C, O_{2} F \perp A C, O_{3} K \perp A C$ and $\triangle A O_{1} E \sim \triangle A O_{2} F \sim \Delta A O_{3} K$. Let $x$ be the radius of the medium circle, and $y$ be the radius of the large circle. Then $O_{2} O_{1}=O_{2} F=x, O_{2} O_{3}=O_{3} K=y$. From the similarity of $\triangle A O_{1} E$ and

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0744.jpg?height=486&width=773&top_left_y=76&top_left_x=298)

Fig. 10.52

$\triangle A O_{2} F$ it follows that $\frac{A O_{1}}{O_{1} E}=\frac{A O_{2}}{O_{2} F} ; \frac{a}{r}=\frac{a+x}{x} ; x=\frac{a r}{a-r}$. From the similarity of $\triangle A O_{1} E$ and $\triangle A O_{3} K$ it follows that:

$$
\frac{A O_{1}}{O_{1} E}=\frac{A O_{3}}{O_{3} K} ; \frac{a}{r}=\frac{a+x+y}{y} ; y=\frac{r(a+x)}{a-x}=\frac{r\left(a+\frac{a r}{a-r}\right)}{a-r}=\frac{a^{2} r}{(a-r)^{2}}
$$

Answer: $\frac{a r}{a-r} ; \frac{a^{2} r}{(a-r)^{2}}$