10.2. An isosceles triangle with base $a$ and angle at the base $\alpha$ is inscribed in a circle. 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.
Answer: $\frac{a}{4} \operatorname{tg} \alpha, \frac{a}{4} \operatorname{ctg} \alpha$.

Solution. The diameter of one of the desired circles is the height of the given triangle, and

![](https://cdn.mathpix.com/cropped/2024_05_06_7bf17e53486882b92a25g-1.jpg?height=320&width=305&top_left_y=2179&top_left_x=173)
the diameter of the other is the difference between the diameter of the circumscribed circle and the diameter of the first desired circle. Let $CK$ be the diameter of the circle circumscribed around the isosceles triangle $ABC (AC = BC, AB = a, \angle A = \angle B = \alpha)$. Then the midpoint $M$ of the base $AB$ lies on this diameter, and $CM$ and $MK$ are the diameters of the desired circles. Let $r$ and $x$ be the radii of the desired circles.

Then $r = \frac{1}{2} CM = \frac{1}{2} AM \cdot \operatorname{tg} \alpha = \frac{a}{4} \operatorname{tg} \alpha, x = \frac{1}{2} MK = \frac{1}{2} AM \cdot \operatorname{ctg} \angle AKM = \frac{a}{4} \operatorname{ctg} \alpha$.