An Inscribed Pentagon

Let $x$ be $\angle BOC$ and $r$ be the radius of circle $O$ . Since the sides of $\triangle AOC$ are all equal to $r$ , it is an equilateral triangle, and so $\angle AOC = 60°$ and $\angle AOB = \angle AOC - \angle BOC = 60° - x$ . Likewise, $\angle BOD = 60°$ and $\angle COD = 60° - x$ . Since $\angle AOE$ is a straight angle, $\angle AOE = 180°$ , and $\angle DOE = \angle AOE - \angle AOB - \angle BOC - \angle COD$ $=$ $180° - (60° - x) - x - (60° - x)$ $=$ $60° + x$ .

The red area is the combined area of triangles $\triangle BOC$ and $\triangle COD$ , which is $A_r = \frac{1}{2}r^2 \sin x + \frac{1}{2}r^2 \sin (60° - x)$ . The blue area is the area of $\triangle DOE$ which is $A_b = \frac{1}{2}r^2 \sin (60° + x)$ . Then:

$A_r$

$= \frac{1}{2}r^2 \sin x + \frac{1}{2}r^2 \sin (60° - x)$

$= \frac{1}{2}r^2 (\sin x + \sin (60° - x))$

$= \frac{1}{2}r^2 (\sin x + \sin 60° \cos x - \cos 60° \sin x))$

$= \frac{1}{2}r^2 (\sin x + \sin 60° \cos x - \frac{1}{2} \sin x))$

$= \frac{1}{2}r^2 (\sin 60° \cos x + \frac{1}{2} \sin x))$

$= \frac{1}{2}r^2 (\sin 60° \cos x + \cos 60° \sin x))$

$= \frac{1}{2}r^2 \sin (60° + x)$

$= A_b$

Therefore, no matter what the value of $x$ is, the red and blue areas are always equal.