Dynamic Geometry: P89

. Let the center of the large semicircle be $O(0,0)$ , the origin of the $xy$ -plane and its radius be $1$ . Since the system is symmetrical about the $y$ -axis, I have flipped the system horizontally so the $x$ increases to the right. Let an arbitrary point on the locus be $P(x,y)$ , the centers and radii of the right yellow circle and the cyan semicircle be $Q(u,v)$ and $R$ , and $r_2$ and $r_1$ respectively, the radius of the green semicircle be $r$ , the red line be $KM$ , $LQ$ be perpendicular to $KM$ , and $QN$ be perpendicular to the $x$ -axis. Then $r+r_1 = 1$ and $OR=1-r_1 = r$ .

By Pythagorean theorem ,

$\begin{aligned} QN^2 & = QR^2 - NR^2 = (QP+PR)^2 + (MR-MN)^2 \\ v^2 & = (r_1+r_2)^2 - (r_1 - r_2)^2 = 4r_1r_2 \end{aligned}$

Also

$\begin{aligned} QN^2 & = OQ^2 - ON^2 = OQ^2 - (OM+MN)^2 \\ v^2 & = (1-r_2)^2 - (1-2r_1+r_2)^2 \\ 4r_1r_2 & = (1-r_2)^2 - (1+r_2)^2 + 4r_1(1+r_2)-4r_1^2 \\ r_2 & = r_1(1-r_1) = rr_1 \\ v^2 & = 4r_1r_2 = 4rr_1^2 \\ \implies v & = 2 r_1 \sqrt r = 2(1-r)\sqrt r \end{aligned}$

From similar triangles, we note that

$\frac yv = \frac {r_1}{r_1+r_2} = \frac {r_1}{r_1+rr_1} = \frac 1{1+r} \implies y = \frac v{1+r} = \frac {2(1-r)\sqrt r}{1+r}$

Since $u = ON = OM + MN = 1-2r_1+r_2$ . From similar triangles,

$x = u + \frac {r_2}{r_1+r_2} \cdot NR = 1-2r_1+r_2 + \frac {r_2}{r_1+r_2} (r_1-r_2)= \frac {3r-1}{1+r}$

The area under the locus is given by:

$\begin{aligned} A & = \int_{-1}^1 y \ dx \\ & = \int_{-1}^1 \frac {2(1-r)\sqrt r}{1+r} \ \ce d \left(\frac {3r-1}{1+r} \right) \\ & = \int_0^1 \frac {8(1-r)\sqrt r}{(1+r)^3} \ \ce dr & \small \blue{\text{Let }r = u^2 \implies \ce dr = 2u \ \ce du} \\ & = 16 \int_0^1 \frac {(1-u^2)u^2}{(1+u^2)^3} \ce du & \small \blue{\text{Let }u = \tan \theta \implies \ce du = \sec^2 \theta \ \ce d\theta} \\ & = 16 \int_0^\frac \pi 4 \frac {(1-\tan^2 \theta)\tan^2 \theta}{\sec^6 \theta} \cdot \sec^2 \theta \ \ce d\theta \\ & = 16 \int_0^\frac \pi 4 (\cos^2 \theta - \sin^2 \theta) \sin^2 \theta \ \ce d \theta \\ & = 8 \int_0^\frac \pi 4 \cos 2\theta (1 - \cos 2 \theta) \ \ce d \theta \\ & = \int_0^\frac \pi 4 (8\cos 2\theta - 4 - 4 \cos 4 \theta) \ \ce d\theta \\ & = 4 \sin 2 \theta - 4 \theta - \sin 4\theta \ \bigg|_0^\frac \pi 4 \\ & = 4 - \pi \end{aligned}$

Therefore the ratio of the area under the locus to the area of the unit semicircle is $\dfrac {4-\pi}{\frac \pi 2} = \dfrac 8\pi - 2$ , and $p+q = 8-2 = \boxed 6$ .