Dynamic Geometry: P8

The double angle formula for tan is $\tan (α)=\frac{2\tan(α/2)}{1-\tan^2(α/2)}$ $1-\tan^2(α/2)=1519/1600$ $\tan(α/2)=9/40$ $y=x\tan(α/2)=1$ $d=\sqrt{x^2+y^2}=\sqrt{\frac{40^2+9^2}{9^2}}y=\frac{41}{9}$

Given that

$\begin{aligned} \frac {\tan \alpha}{\tan \frac \alpha 2} & = \frac {3200}{1519} \\ 1519 \tan \alpha & = 3200 \tan \frac \alpha 2 \\ 1519 \cdot \frac {2 \tan \frac \alpha 2}{1-\tan^2 \frac \alpha 2} & = 3200 \tan \frac \alpha 2 \\ 1519 & = 1600 \left(1 - \tan^2 \frac \alpha 2 \right) \\ \tan^2 \frac \alpha 2 & = \frac {1600-1519}{1600} = \frac {81}{1600} \\ \implies \tan \frac \alpha 2 & = \frac 9{40} \\ \implies \sin \frac \alpha 2 & = \frac 9{\sqrt{40^2+9^2}} = \frac 9{41} \end{aligned}$

Note that the distance between the purple point and cyan point is $\dfrac 1{\sin \frac \alpha 2} = \dfrac {41}9$ . Therefore $a+b = 41+9 = \boxed{50}$ .