12.176. A circle is inscribed in an isosceles triangle with base $a$ and angle $\alpha$ at the base. Find the radius of the circle that is tangent to the inscribed circle and the lateral sides of the triangle.
Solution.

In $\triangle ABC$ (Fig. 12.46) $AB = BC, AC = a, \angle BAC = \alpha, O$ is the center of the inscribed circle, $R$ is its radius, $O_1$ is the center of the circle that is tangent to the inscribed circle and the lateral sides of the triangle, $r$ is its radius. Points $O$ and $O_1$ lie on the height $BM$ of the given triangle.

In $\triangle AMB \left(\angle AMB = 90^\circ\right): BM = AM \tan \angle BAC = \frac{a}{2} \tan \alpha$.

In $\triangle AMO \left(\angle AMO = 90^\circ\right): R = OM = AM \tan \angle OAM = \frac{a}{2} \tan \frac{\alpha}{2}$.

Let $L$ be the point of tangency of the circles with centers $O$ and $O_1$, and $NK$ be their common tangent passing through point $L$.

$BL = BM - LM = BM - 2R = \frac{a}{2} \tan \alpha - a \tan \frac{\alpha}{2} = \frac{a}{2} \left(\tan \alpha - 2 \tan \frac{\alpha}{2}\right) = \frac{a}{2} \left(\frac{2 \tan \frac{\alpha}{2}}{1 - \tan^2 \frac{\alpha}{2}} - 2 \tan \frac{\alpha}{2}\right) = \frac{a}{2} \cdot \frac{2 \tan^3 \frac{\alpha}{2}}{1 - \tan^2 \frac{\alpha}{2}} = \frac{a}{2} \tan \alpha \tan^2 \frac{\alpha}{2}$.

$\triangle ABC \sim \triangle NBK$.

Then $\frac{r}{R} = \frac{BL}{BM}; r = R \cdot \frac{BL}{BM} = \frac{a}{2} \tan \frac{\alpha}{2} \cdot \frac{\frac{a}{2} \tan \alpha \tan^2 \frac{\alpha}{2}}{\frac{a}{2} \tan \alpha} = \frac{a}{2} \tan^3 \frac{\alpha}{2}$.

Answer: $\frac{a}{2} \tan^3 \frac{\alpha}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0929.jpg?height=402&width=613&top_left_y=885&top_left_x=81)

Fig. 12.47