3. Let $A B C D E$ be a regular pentagon with side length 1, and let $P$ be the intersection of the diagonals $A C$ and $B E$. What is the length of the segment $P C$?
(A) 1
(B) $\frac{\sqrt{5}}{2}$
(C) $\sqrt{5}-1$
(D) $4(\sqrt{5}-2)$
(E) None of the above.
## 3. The answer is (A).

## FIRST SOLUTION

The internal angles of a regular $n$-gon measure $\frac{n-2}{n} 180^{\circ}$. Therefore, the internal angles of a regular pentagon measure $\frac{3}{5} 180^{\circ}=108^{\circ}$. The triangles $E A B$ and $C B A$ are isosceles, so $\widehat{A E B}=\widehat{E B A}=\widehat{B C A}=\widehat{C A B}=\frac{180^{\circ}-108^{\circ}}{2}=36^{\circ}$. Then $\widehat{A P B}=180^{\circ}-\widehat{P A B}-\widehat{P B A}=108^{\circ}, \widehat{C P B}=180^{\circ}-\widehat{A P B}=$ $72^{\circ}$ and $\widehat{C B P}=\widehat{C B A}-\widehat{P B A}=72^{\circ}$. Therefore, the triangle $C B P$ is isosceles and $C P=C B=1$.

SECOND SOLUTION

![](https://cdn.mathpix.com/cropped/2024_04_17_8e3f264af5d5452f36c1g-08.jpg?height=383&width=399&top_left_y=1319&top_left_x=1431)

By symmetry, the diagonal $A C$ is parallel to the side $D E$ and the diagonal $B E$ is parallel to the side $C D$; it follows that the quadrilateral $E P C D$ is a parallelogram and thus $P C=E D=1$.