Dynamic Geometry: P31

Let the center of the large and cyan semicircle be $O(0,0)$ , the origin of the $xy$ -plane, $Q(x_q,y_q)$ be the center of the right pink circle, and $P(x_p,y_p)$ , the tangent point of the cyan semicircle and the right pink circle. Let the radii of the large semicircle, the cyan semicircle, and the pink circle be $1$ , $r_1$ , and $r_2$ respectively.

We note that $r_1+2r_2 = 1$ . Let $k = \dfrac {r_2}{r_1} \implies r_2 = kr_1$ . Then $r_1 + 2kr_1 = 1 \implies r_1 = \dfrac 1{1+2k}$ .

We also note that $y_q = r_2$ and by Pythagorean theorem :

$\begin{aligned} x_q^2+y_q^2 & = (r_1+r_2)^2 \\ x_q^2 + r_2^2 & = r_1^2(1+k)^2 \\ x_q^2 + k^2r_1^2 & = r_1^2(1+2k + k^2) \\ x_q^2 & = r_1 \blue{(r_1 + 2kr_1)} & = r_1 & \small \blue{\text{Note that }r_1 + 2kr_1 = 1} \\ \implies x_q & = \sqrt{r_1} \end{aligned}$

By similar triangles , $\dfrac {x_p}{x_q} = \dfrac {r_1}{r_1+r_2} = \dfrac 1{1+k} \implies x_p = \dfrac {\sqrt{r_1}}{1+k}$ and $y_p = \dfrac {kr_1}{1+k}$ .

The area of the trapezoid is given by:

$\begin{aligned} \frac {2x_q+2x_p}2 \cdot (y_q - y_p) & = \frac {63}{400} \\ \left(\sqrt{r_1} + \frac {\sqrt{r_1}}{1+k} \right) \left(kr_1 - \frac {kr_1}{1+k} \right) & = \frac {63}{400} \\ \frac {k^2r_1^\frac 32(2+k)}{(1+k)^2} & = \frac {63}{400} \\ \frac {k^2(2+k)}{(1+2k)^\frac 32(1+k)^2} & = \frac {63}{400} \\ \implies k & = \frac 32 \end{aligned}$

Therefore $a+b = 3 + 2 = \boxed{5}$ .