A circle is inscribed with an isosceles triangle with base $a$ and angle at the base $\alpha$. Additionally, a second circle is constructed that is tangent to the first circle and the base of the triangle, with the point of tangency being the midpoint of the base. Find the radius of the second circle. If the solution is not unique, consider all cases.
The diameter of one of the sought circles is the height of the given triangle, and the other is the difference between the diameter of the circumscribed circle and the diameter of the first sought circle.

## Solution

Let $C K$ be the diameter of the circle circumscribed around the isosceles triangle $A B C (A C = B C, A B = a, \angle A = \angle B = \alpha)$. Then the midpoint $M$ of the base $A B$ lies on this diameter, and $C M$ and $M K$ are the diameters of the sought circles.

Let $r$ and $x$ be the radii of the sought circles. Then

$$
\begin{gathered}
r = \frac{1}{2} C M = \frac{1}{2} A M \cdot \operatorname{tg} \alpha = \frac{a}{4} \operatorname{tg} \alpha \\
x = \frac{1}{2} M K = \frac{1}{2} A M \cdot \operatorname{ctg} \angle A K M = \frac{a}{4} \operatorname{ctg} \alpha
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_c2dae46d614e2e953886g-36.jpg?height=475&width=464&top_left_y=2412&top_left_x=797)

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