12.001. The sum of two unequal heights of an isosceles triangle is $l$, the angle at the vertex is $\alpha$. Find the lateral side.
Solution.

By the condition $A B=A C, A A_{1} \perp B C, B B_{1} \perp A C, \angle B A C=\alpha, A A_{1}+B B_{1}=l$ (Fig. 12.4). Let $B C=a$. From $\triangle A A_{1} C$ we have $A A_{1}=\frac{a}{2} \operatorname{ctg} \frac{\alpha}{2}, A C=\frac{a}{2 \sin \frac{\alpha}{2}}$, and from $\triangle B B_{1} C$ we find $B B_{1}=a \sin \left(\frac{\pi}{2}-\frac{\alpha}{2}\right)=a \cos \frac{\alpha}{2}$. By the condition $\frac{a}{2} \operatorname{ctg} \frac{\alpha}{2}+a \cos \frac{\alpha}{2}=l, \frac{a}{2} \operatorname{ctg} \frac{\alpha}{2}\left(1+2 \sin \frac{\alpha}{2}\right)=l, a=\frac{2 l \operatorname{tg} \frac{\alpha}{2}}{1+2 \sin \frac{\alpha}{2}}$. We have obtained

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-716.jpg?height=476&width=404&top_left_y=80&top_left_x=188)

Fig. 12.4

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-716.jpg?height=370&width=340&top_left_y=187&top_left_x=782)

Fig. 12.5

\[
\begin{aligned}
& A C=\frac{l \operatorname{tg} \frac{\alpha}{2}}{\left(1+2 \sin \frac{\alpha}{2}\right) \sin \frac{\alpha}{2}}=\frac{l}{\cos \frac{\alpha}{2}\left(1+2 \sin \frac{\alpha}{2}\right)}=\frac{l}{\cos \frac{\alpha}{2}+\sin \alpha}= \\
& =\frac{l}{\sin \left(\frac{\pi}{2}-\frac{\alpha}{2}\right)+\sin \alpha}=\frac{l}{2 \sin \frac{\pi+\alpha}{4} \cos \frac{\pi-3 \alpha}{4}} . \\
& \text { Answer: } \frac{l}{2 \sin \frac{\pi+\alpha}{4} \cos \frac{\pi-3 \alpha}{4}}
\end{aligned}
\]