Regular Shapes

In the regular hexagon, we use Law of Cosines on $\Delta AEF$ to find that $\overline{AE} = \sqrt{3}$ . Then use Pythagorean Theorem on $\Delta ADE$ to find that $\overline{AD}=2$ . Now we must find the diagonal of the regular pentagon. Calling the length of the diagonal $x$ , we use Ptolemy's Theorem on isosceles trapezoid $GHIK$ to find that $x+1=x^2$ . Solving this gives us $x = \dfrac{1+ \sqrt{5}}{2}$ . (Interestingly, this is the Golden Ratio ). Now we must find $\dfrac{2}{\dfrac{1+\sqrt{5}}{2}}$ . This yields $\dfrac{4}{1 + \sqrt{5}} = \dfrac{4(\sqrt{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)} = \dfrac{4\sqrt{5} - 4}{4} = \sqrt{5} - 1$ , which is the answer.