Problem 1185

## Solution

Let the pentagon be ABCDE. Triangles BCD and ECD have the same area, so $\mathrm{B}$ and $\mathrm{E}$ are the same perpendicular distance from $\mathrm{CD}$, so $\mathrm{BE}$ is parallel to $\mathrm{CD}$. The same applies to the other diagonals (each is parallel to the side with which it has no endpoints in common). Let BD and CE meet at X. Then ABXE is a parallelogram, so area $\mathrm{BXE}=$ area $\mathrm{EAB}=1$. Also area $\mathrm{CDX}+$ area $\mathrm{EDX}=$ area $\mathrm{CDX}+$ area $\mathrm{BCX}=1$. Put
area $\mathrm{EDX}=\mathrm{x}$. Then $\mathrm{DX} / \mathrm{XB}=$ area $E D X /$ area $\mathrm{BXE}=\mathrm{x} / 1$ and also $=$ area $\mathrm{CDX} / \operatorname{area} \mathrm{BCX}=(1-\mathrm{x}) / \mathrm{x}$. So $\mathrm{x}^{2}+$ $x-1=0, x=\sqrt{5}-1) / 2$ (we know $x<0$, so it cannot be the other root). Hence area $\operatorname{ABCDE}=3+x=(\sqrt{ } 5+$ $5) / 2$.

Take any triangle XCD of area $(3-\sqrt{5}) / 2$ and extend DX to $B$, so that $B C D$ has area 1 , and extend CX to E so that CDE has area 1. Then take BA parallel to CE and EA parallel to BD. It is easy to check that the pentagon has the required property.

## 2nd USAMO 1973