Dynamic Geometry: P83

Let the center of the unit semicircle be $O(0,0)$ , the origin of the $xy$ -plane. 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 $Q$ and $R$ , and $r_2$ and $r_1$ respectively, the radius of the green semicircle be $r$ , the red line be $PM$ , and $QN$ be perpendicular to the $x$ -axis. Then $r+r_1 = 1$ , $x=OM=1 - 2r_1= 2r -1$ , and $y = PM = QN$ .

By Pythagorean theorem ,

$\begin{aligned} QN^2 & = QR^2 - NR^2 \\ y^2 & = (r_1+r_2)^2 - (r_1 - r_2)^2 = 4r_1r_2 \end{aligned}$

Also

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

Then the area under the locus is given by:

$\begin{aligned} A & = \int_{-1}^1 y \ \ce d x \\ & = \int_0^1 2(1-r)\sqrt r \cdot \ce d (2r -1) \\ & = \int_0^1 4(1-r)\sqrt r \cdot \ce d r & \small \blue{\text{Let }u^2 = r \implies 2u \ \ce du = \ce d r} \\ & = \int_0^1 8u^2(1-u^2) \ \ce du \\ & = \frac 83u^3 - \frac 85u^5 \ \bigg|_0^1 = \frac {16}{15} \end{aligned}$

Therefore $p-q = 16-15 = \boxed 1$ .