A regular pentagon with area $\sqrt{5}+1$ is printed on paper and cut out. The five vertices of the pentagon are folded into the center of the pentagon, creating a smaller pentagon. What is the area of the new pentagon? $\textbf{(A)}~4-\sqrt{5}\qquad$ $\textbf{(B)}~\sqrt{5}-1\qquad$ $\textbf{(C)}~8-3\sqrt{5}\qquad$ $\textbf{(D)}~\frac{\sqrt{5}+1}{2}\qquad$ $\textbf{(E)}~\frac{2+\sqrt{5}}{3}$
Solution
Let the pentagon have circumradius $R$ and center $O$. Folding a vertex $V$ onto $O$ creates the crease that is the perpendicular bisector of $VO$, at distance $\frac R2$ from $O$ and perpendicular to $OV$. The new pentagon is the intersection of the five half-planes bounded by these creases, a regular pentagon with apothem $\frac R2$.
The original pentagon has apothem $R\cos36^\circ$, so the ratio of the areas is $\left(\dfrac{R/2}{R\cos36^\circ}\right)^2=\dfrac1{4\cos^236^\circ}$.
With $\cos36^\circ=\frac{1+\sqrt5}4$, $\cos^236^\circ=\frac{3+\sqrt5}8$, so the ratio is $\dfrac2{3+\sqrt5}=\dfrac{3-\sqrt5}2$.
The new area is $(\sqrt5+1)\cdot\dfrac{3-\sqrt5}2=\sqrt5-1$.
A regular pentagon with area $2\sqrt{5}+2$ is printed on paper and cut out. The five vertices of the pentagon are folded into the center, creating a smaller pentagon. What is the area of the new pentagon?
