12.179. Find the sine of the angle at the vertex of an isosceles triangle, given that the perimeter of any inscribed rectangle, two vertices of which lie on the base, has a constant value.
Solution.

Let $K L M N$ and $D E F G$ be two arbitrary rectangles inscribed in $\triangle A B C, A B=B C$ (Fig. 12.49), $L K=x_{1}, L M=y_{1}, E D=x_{2}$, $D G=y_{2}, \angle A B C=\alpha$. By the condition $x_{1}+y_{1}=x_{2}+y_{2}$.

$$
\text { Then } x_{2}-x_{1}=y_{1}-y_{2}
$$

If $B S$ is the altitude of $\triangle A B C, P$ is the intersection point of $L M$ and $E D$, then

$$
L P=\frac{y_{1}-y_{2}}{2}, E P=x_{2}-x_{1} \text {. }
$$

$\triangle L P E \sim \triangle A S B$.

From this, $\frac{A S}{B S}=\frac{L P}{E P}=\frac{y_{1}-y_{2}}{2\left(x_{2}-x_{1}\right)}=\frac{1}{2}$.

Since $\frac{A S}{B S}=\operatorname{tg} \frac{\alpha}{2}$, then $\sin \alpha=\frac{2 \operatorname{tg} \frac{\alpha}{2}}{1+\operatorname{tg}^{2} \frac{\alpha}{2}}=\frac{1}{1+\frac{1}{4}}=\frac{4}{5}$.

Answer: $\frac{4}{5}$.