Year: 1965
Topic: Geometry

Problem:
Let $ABC$ be a triangle with angle $A < \angle C < 90^\circ < \angle B$. Consider the bisectors of the external angles at $A$ and $B$, each measured from the vertex to the opposite side (extended). Suppose both of these line-segments are equal to $AB$. Compute the angle $A$.

Solution:
Suppose the bisector of the exterior angle at $A$ intersects line $BC$ at $X$ and the bisector of the exterior angle at $B$ meets the line $AC$ at $Y$. The assumption that $C$ is between $B$ and $X$ contradicts the fact that $\angle B > \angle C$, so we may assume that $B$ is between $X$ and $C$. Similarly, we conclude that $C$ is between $A$ and $Y$ because $\angle A < \angle C$.

If $Z$ is a point on line $AB$ with $B$ between $A$ and $Z$, we have from triangle $ABY$ that $\angle ZBY = 2A$. Hence, $\angle BXA = \angle ABX = \angle ZBC = 2 \angle ZBY = 4A$, and the angle sum of triangle $ABX$ is $90^\circ - \frac{1}{2}A + 8A$. Thus, $A = \boxed{12}^\circ$.