2015
Problem - 2034
Let $ABCDE$ be a convex pentagon such that $\angle ABC = \angle BCD = 108^{\circ}$, $\angle CDE = 168^{\circ}$ and $AB =BC = CD = DE$. Find the measure of $\angle AEB$.
Construct a point $F$ such that $ABCDF$ is a regular pentagon. Note that $\triangle{FDE}$ is equilateral. In addition, observe that $\angle{AEB} = \angle{FEB} − \angle{FEA}$. By symmetry, we know that $\angle{FEB}$ is half of $\angle{FED}$ or $30^\circ$. Because $\triangle{AFE}$ is isosceles, $\angle{FEA} = (180^\circ - \angle{AFE})/2 = (180^\circ−168^\circ)/2=6^\circ$. This allows us to compute $\angle{AEB}$ as $30^\circ − 6^\circ = \boxed{24^\circ}$.