3B. In an isosceles triangle, the angle at the base is equal to $\alpha$. Calculate the length of the base of the triangle if the height dropped to the base is $m$ greater than the radius of the inscribed circle.
Solution. Let $ABC$ be a triangle with base $AB$ where $\measuredangle BAC = \alpha$. From vertex $C$, we drop a perpendicular $CD, \overline{CD} = h$, which is also the angle bisector of $\measuredangle BCA$. The angle bisector of $\measuredangle BAC$ and the perpendicular $CD$ intersect at point $O$, which is the center of the inscribed circle. The points where the inscribed circle $k$ touches the sides $AB, BC$, and $CA$ will be denoted by $D, M, N$ respectively. The radius of the circle is $\overline{ON} = r = \overline{OD}$ and $\measuredangle OAN = \measuredangle OAB = \frac{\alpha}{2}$. Triangles $CON$ and $CAD$ are similar (they have one right angle and angles with mutually perpendicular sides). In triangle $ONC$, we have $\measuredangle NOC = \alpha, \overline{CO} = h - r = m$, and $\overline{ON} = r$, so according to this, $r = \overline{ON} = \overline{OC} \cos \alpha = m \cos \alpha$.

Triangle $ADO$ is a right triangle and one of its acute angles is $\frac{\alpha}{2}$, and one of its sides is $\overline{OD} = m \cos \alpha$. Since $\frac{\overline{AD}}{\overline{OD}} = \operatorname{ctg} \frac{\alpha}{2}$, we get

$$
\frac{a}{2} = \overline{AD} = \overline{OD} \operatorname{ctg} \frac{\alpha}{2} = m \cos \alpha \operatorname{ctg} \frac{\alpha}{2}
$$

Finally, the length of the base is $a = 2 m \cos \alpha \operatorname{ctg} \frac{\alpha}{2}$.