Fanning Fans

The figure is symmetrical about the straight line $OB$ . Therefore $OB$ bisects both $\angle AOC$ and $\angle FBG$ . Since $\angle AOC = \angle FBG= \theta$ , $\angle AOB = \angle GBO = \frac 12 \theta$ and $AO \parallel BG$ . Let the radii of sector $FBG$ and sector $AOC$ be $r$ and $1$ respectively. Then we note that \BG=r) and $OB=OG = 1$ . Therefore $\triangle OBG$ is isosceles and $\angle BGO = \angle GBO = \frac 12 \theta$ . Also $BG = r = 2\cos \frac \theta 2$ . We also note that $BD = r = \sin \frac \theta 2$ . So we have:

$\begin{aligned} \sin \frac \theta 2 & = 2 \cos \frac \theta 2 \\ \implies \tan \frac \theta 2 & = 2 \\ r & = \sin \frac \theta 2 = \frac 2{\sqrt 5} \end{aligned}$

The ratio of areas of the two sectors $\dfrac {[FBG]}{[AOC]} = \dfrac {r^2}{1^2} = \dfrac 45$ and the required answer $a+b = 4+5 = \boxed 9$ .