An isosceles triangle's legs form an angle $\alpha$ with each other. For what $\alpha$ can a line parallel to the base be found that bisects both the area and the perimeter of the triangle?
Let the sides of the triangle be denoted by $b$, and its base by $a$. Suppose that a line $e$ parallel to the base divides both the area and the perimeter of the triangle.

# $1993-05-216-1 . e p s$ 

The triangle cut off by the line $e$ is similar to the original triangle. Since the ratio of their areas is $1:2$, the ratio of their sides is $\frac{x}{b}=\frac{1}{\sqrt{2}}$, hence $x=\frac{b \sqrt{2}}{2}$. The division of the perimeter means that $2 x=a+2(b-x)$.

From these relationships, it follows that $\frac{a}{2 b}=\sqrt{2}-1$. Since $\frac{a}{2 b}=\sin \frac{\alpha}{2}$, it follows that $\alpha \approx 48.94^{\circ}$.

Tahócz László (Tamási, Béri Balogh Á. Gymn., III. o. t.)