# CUNY Math Challenge Blog

Solutions, Resources, Further Information & Discussion

## Round 4: Problem 2 Solution

Here’s a geometric construction proof:

• Extend $AF$ and $BC$ so that they intersect at an exterior point $\text{X}$.
• Extend $FE$ and $CD$ in a similar fashion so they intersect at another exterior point, $\text{Y}$, forming quadrilateral $XFYC$.
• $\angle XAB \cong \angle XBA \cong \angle YED \cong \angle YDE = 60^{\circ}$ since each external angle is supplementary with the 60 degree internal angle.
• $\angle X \cong \angle Y = 60^{\circ}$, so $\triangle XAB, \triangle YDE$ are equilateral triangles.
• $AB = XB, DE = YE$ in equilateral triangles.
• $XFYC$ is a parallelogram (both pairs of opposite angles are congruent), so $FY = XC$.
• $FE + EY = AB + BC$, so $FE + ED = AB + BC$ by equilateral triangles as shown above.