Dynamic Geometry: P125

From the previous similar problem Dynamic Geometry: P120 , we found that the purple locus of the centers of the two semicircles is an ellipse with the following equation with the origin of the $xy$ -plane being at the center of the circle:

$8x^2 + 16 \left(y - \frac 54\right)^2 = 313$

If we shift the origin $O(0,0)$ to be at the center of the ellipse, then the equation becomes:

$8x^2 + 16 y^2 = 313$

Since the centers of the two semicircles are moving at the same rate horizontally but in the opposite directions and an ellipse is symmetrical about its center, if the center of the blue semicircle is $P(x,y)$ , then the center of the green semicircle is $Q(-x.-y)$ . We note that the radius of the blue semicircle $r_1 = y - \frac 54$ and the radius of the green semicircle is $r_2 = y + \frac 54$ . Then we have:

$\begin{aligned} \frac {r_2}{r_1} & = \frac {21 + \sqrt{185}}{16} \\ \frac {y+\frac 54}{y-\frac 54} & = \frac {21 + \sqrt{185}}{16} \\ 64y + 80 & = (84+4\sqrt{185})y - 105 - 5\sqrt{185} \\ (20+4\sqrt{185}) y & = 185 + 5\sqrt{185} \\ \implies y & = \frac {\sqrt{185}}4 \end{aligned}$

Then

$\begin{aligned} 8x^2 + 16 \cdot \frac {185}{16} & = 313 \\ \implies x & = 4 \end{aligned}$

And $PQ = 2\sqrt{x^2+y^2} = 2\sqrt{16+\dfrac {185}{16}} = \dfrac {21}2$ . Therefore $p+q = 21+2 = \boxed{23}$ .