Dynamic Geometry: P62

Let the center of the big semicircle be $O(0,0)$ , the origin of the $xy$ -plane, an arbitrary point of the right locus as $P(x,y)$ , $PN$ be perpendicular to the $x$ -axis, and the radii of the cyan semicircle and pink circle be $r$ and $r_1$ respectively. Then we have:

$\begin{cases} x = ON = 1 - r \\ y = PN = r+r_1 \end{cases}$

By Pythagorean theorem , we have:

$\begin{aligned} ON^2 + PN^2 & = OP^2 \\ x^2 + y^2 & = (1-r_1)^2 \\ & = \left(1 - (\blue{1-r} + \red{r + r_1} -1)\right)^2 \\ & = \left(1 - (\blue x+ \red y -1)\right)^2 \\ & = (2-x-y)^2 \\ & = x^2 + 2xy + y^2 - 4(x+y) + 4 \\ xy - 2x - 2y + 2 & = 0 \\ \implies y & = \frac {2(1-x)}{2-x} \end{aligned}$

The above is the equation of the locus. The area under the two loci is

$\begin{aligned} A & = 2 \int_0^1 y \ dx = 2 \int_0^1 \frac {2(1-x)}{2-x} \ dx = 2\int_0^1 \left(2 - \frac 2{2-x} \right) dx \\ & = 4x + 4 \ln (2-x) \ \bigg|_0^1 = 4 - 4 \ln 2 = 4 - 2 \ln 4 \end{aligned}$

Therefore $p+q = 4+2 = \boxed 6$ .

Let the $r_p$ be the radius of the pink circle and $r_c$ be the radius of the cyan semicircle, and label the diagram as follows:

Then $OA = OB - AB = 1 - r_c$ , $AC = AE + EC = r_c + r_p$ , and $OC = OD - CD = 1 - r_p$ .

The blue curve has $x = OA = 1 - r_c$ and $y = AC = r_c + r_p$ , so that $r_c = 1 - x$ and $r_p = y - r_c = y - (1 - x)$ , which makes $OC = 1 - r_p = 1 - (y - (1 - x)) = 2 - x - y$ .

By the Pythagorean Theorem on $\triangle OAC$ , $OA^2 + AC^2 = OC^2$ or $x^2 + y^2 = (2 - x - y)^2$ , which rearranges to $y = 2 + \cfrac{2}{x - 2}$ .

The area bounded by the blue curves and the orange semicircle's diameter is then $\displaystyle 2 \int_{0}^{1} \bigg(2 + \cfrac{2}{x - 2}\bigg)\,dx = 4 - 2 \ln 4$ , so $p = 4$ , $q = 2$ , and $p + q = \boxed{6}$ .