In an isosceles triangle with base $a$ and angle at the base $\alpha$, a circle is inscribed. Additionally, a second circle is constructed that touches the lateral sides of the triangle and the inscribed circle.

Find the radius of the second circle.
Let $O$ and $Q$ be the radii of the first and second circles, $F$ be the point of tangency of the first circle with the side of the given triangle, and $P$ be the projection of point $Q$ onto $O F$. Consider triangle $O P Q$.

## Solution

Let $\mathrm{O}$ be the center of the circle with radius $r$, inscribed in triangle $A B C (A B = B C, A B = a, \angle A = \angle B = \alpha)$, and $M$ and $F$ be the points of tangency with sides $A B$ and $A C$, respectively. From the right triangle $O M A$, we find that

$$
r = A M \operatorname{tg} \angle O A M = \frac{1}{2} a \operatorname{tg} \frac{\alpha}{2}
$$

Let $P$ be the projection of point $Q$ onto $O F$. From the right triangle $Q P O$, we find that

$$
O P = O Q \cos \angle P O Q, \text{ or } r - x = (r + x) \cos \alpha.
$$

Therefore,

$$
x = \frac{r(1 - \cos \alpha)}{1 + \cos \alpha} = \frac{\frac{1}{2} a \operatorname{tg} \frac{\alpha}{2} \cdot 2 \sin^2 \frac{\alpha}{2}}{2 \cos^2 \frac{\alpha}{2}} = \frac{1}{2} a \operatorname{tg}^3 \frac{\alpha}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_0e468145bb05e34a17aag-36.jpg?height=426&width=688&top_left_y=1742&top_left_x=685)

Answer

$\frac{1}{2} a \operatorname{tg}^3 \frac{\alpha}{2}$