3.1. The sum of two unequal heights of an isosceles triangle is equal to $l$, the angle at the vertex is $\alpha$. Find the lateral side.
3.1. Given $A B=A C$, $A A_{1} \perp B C$, $B B_{1} \perp A C$, $\angle B A C=\alpha$, $A A_{1} + B B_{1}=l$ (Fig. 3.6). Let $B C=a$. From $\triangle A A_{1} C$ we find $A A_{1}=\frac{a}{2} \operatorname{ctg} \frac{\alpha}{2}$, $A C=\frac{a}{2 \sin \frac{\alpha}{2}}$, and from $\triangle B B_{1} C$ we get $B B_{1}=a \sin \left(\frac{\pi}{2}-\frac{\alpha}{2}\right)=a \cos \frac{\alpha}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-500.jpg?height=583&width=468&top_left_y=797&top_left_x=685)

Fig. 3.6

Using the given condition, we have

$$
\frac{a}{2} \operatorname{ctg} \frac{\alpha}{2} + a \cos \frac{\alpha}{2} = l, \text{ or } \frac{a}{2} \operatorname{ctg} \frac{\alpha}{2} \left(1 + 2 \sin \frac{\alpha}{2}\right) = l,
$$

i.e.,

$$
a = \frac{2 l \operatorname{tg} \frac{\alpha}{2}}{1 + 2 \sin \frac{\alpha}{2}}
$$

$$
\begin{gathered}
A C = \frac{l \operatorname{tg} \frac{\alpha}{2}}{\left(1 + 2 \sin \frac{\alpha}{2}\right) \sin \frac{\alpha}{2}} = \frac{l}{\cos \frac{\alpha}{2} \left(1 + 2 \sin \frac{\alpha}{2}\right)} = \frac{l}{\cos \frac{\alpha}{2} + \sin \alpha} = \\
= \frac{l}{\sin \left(\frac{\pi}{2} - \frac{\alpha}{2}\right) + \sin \alpha} = \frac{l}{2 \sin \frac{\pi + \alpha}{4} \cos \frac{\pi - 3 \alpha}{4}}
\end{gathered}
$$