## Solution 30 In hexagon $A B C D E F$, let $A B=B C=C D=3$ and let $D E=E F=F A=5$. Let $P$ be the intersection of $\overline{B E}$ and $\overline{C F}, Q$ that of $\overline{B E}$ and $\overline{A D}$, and $R$ that of $\overline{C F}$ and $\overline{A D}$. Since [[Chord|chords]] of the same length correspond with [[Arc|arcs]] of the same length, we can show that arcs $\overparen{BAF}$ and $\overparen{CDE}$ are both $120^\circ$ in measure $\begin{aligned} g & =m \overparen{A B}=m \overparen{B C}=m \overparen{C D} \\ h & =m \overparen{D E}=m \overparen{E F}=m \overparen{F A} \\ 3 g+3 h & =360^\circ \\ g+h & =m \overparen{BAF}=m \overparen{CDE} \\ \frac{1}{3}=\frac{g+h}{3 g+3 h} & =\frac{m \overparen{BAF}}{360^\circ}=\frac{m \overparen{CDE}}{360^\circ} \\ 120^{\circ}=\frac{1}{3}\left(360^{\circ}\right) & =m \overparen{BAF}=m \overparen{CDE}\end{aligned}$ These arcs are subtended by [[Inscribed angle of a Circle|inscribed angles]] $\angle B C F=\angle B E F=60^{\circ}$ and $\angle C B E=\angle C F E=60^{\circ}$. Triangles $E F P$ and $B C P$ are equilateral, and by symmetry, triangle $P Q R$ is isosceles and thus also equilateral. ![[ahsme-1996-p30-diagram.svg]] ![[ahsme-1996-p30-similarity.svg]]