3.5. The area of an isosceles triangle is $S$, and the angle between the medians drawn to its lateral sides, opposite the base, is $\alpha$. Find the base.
3.5. By the condition, $A B=A C, S_{\triangle A B C}=S$, $A B_{1}=B_{1} C, A C_{1}=C_{1} B, B B_{1} \cap C C_{1}=O$, $\angle B O C=\alpha$ (Fig.3.9). We have $S_{\triangle B O C}=\frac{1}{3} S_{\triangle A B C}=\frac{1}{3} S$, since the height of $\triangle B O C$, drawn from $O$, is $\frac{1}{3}$ of the height of $\triangle A B C$, drawn from $A$. To find the area of $\triangle B O C$, we will use formula (1.6):

$$
\begin{gathered}
S_{\triangle B O C}=\frac{B C^{2} \sin \left(\frac{\pi}{2}-\frac{\alpha}{2}\right) \sin \left(\frac{\pi}{2}-\frac{\alpha}{2}\right)}{2 \sin \alpha}= \\
=\frac{B C^{2} \cos ^{2} \frac{\alpha}{2}}{2 \sin \alpha}=\frac{B C^{2} \cos ^{2} \frac{\alpha}{2}}{4 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}=\frac{B C^{2} \operatorname{ctg} \frac{\alpha}{2}}{4}
\end{gathered}
$$

Thus, $\frac{B C^{2} \operatorname{ctg} \frac{\alpha}{2}}{4}=\frac{1}{3} S$, from which $B C^{2}=\frac{4 S \operatorname{tg} \frac{\alpha}{2}}{3}$; finally, we get

$$
B C=\frac{2 \sqrt{S \operatorname{tg} \frac{\alpha}{2}}}{\sqrt{3}}=\frac{2 \sqrt{3 S \operatorname{tg} \frac{\alpha}{2}}}{3} .
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-503.jpg?height=390&width=679&top_left_y=628&top_left_x=45)

Fig. 3.10