Problem 737

Let $a$, $b$, and $h$ be the length of the congruent sides, the base, and the height to the base of the obtuse isosceles triangle, respectively. Then the area of the triangle is $\frac{1}{2} b h$, which by the stated condition equals $\frac{1}{4} a^2$. The area is also $\frac{1}{2} a^2 \sin \theta$, where $\theta$ is the vertex angle. Equating these expressions shows that $\sin \theta = \frac{1}{2}$. Because the triangle is obtuse, this implies that $\theta = 150^{\circ}$.