Dynamic Geometry: P87

Let the center of the unit semicircle be $O(0,0)$ , the origin of the $xy$ -plane, and an arbitrary point on the locus or the center of the square be $P(x,y)$ when the square has a side length of $a$ . We note that

$\begin{cases} y = \dfrac a2 \\ x = \sqrt{1-a^2} - \dfrac a2 = \sqrt{1-4y^2} - y\end{cases}$

The area under the locus is given by:

$\begin{aligned} A & = 2 \int_0^\blue{\frac 1{\sqrt 5}} x \ dy & \small \blue{\text{Note that when }x=0, y = \frac 1{\sqrt 5}} \\ & = 2 \int_0^{\frac 1{\sqrt 5}} \left(\sqrt{1-4 y^2} - y \right) dy \\ & = 2 \int_0^{\frac 1{\sqrt 5}} \sqrt{1-4\blue y^2} \ dy - 2\int_0^{\frac 1{\sqrt 5}} y \ dy & \small \blue{\text{Let }y = \frac {\sin \theta}2 \implies dy = \frac {\cos \theta}2 \ d\theta} \\ & = \int_0^{\sin^{-1} \frac 2{\sqrt 5}} \cos^2 \theta \ d\theta - y^2 \ \bigg|_0^{\frac 1{\sqrt 5}} \\ & = \int_0^{\tan^{-1} 2} \frac {1+ \cos 2\theta}2 d\theta - \frac 15 \\ & = \frac \theta 2 + \frac {\sin 2\theta}4 \ \bigg|_0^{\tan^{-1}2} - \frac 15 \\ & = \frac {\tan^{-1}2}2 + \frac 15 - \frac 15 \\ & = \frac {\tan^{-1} 2}2 \end{aligned}$

Therefore $p+q = 1+2 = \boxed 3$ .

Consider the purple curve in the first quadrant. Let $\left( x,y \right)$ be the coordinates of the center $P$ of the square. Then, referring to the figure, $OD=x+y$ and $CD=2y$ .

By Pythagorean theorem on $\triangle OCD$ , $O{{A}^{2}}=O{{B}^{2}}+A{{B}^{2}}\Rightarrow 1={{\left( x+y \right)}^{2}}+{{\left( 2y \right)}^{2}}$ which solves, for $y\ge 0$ , to $y=\frac{1}{5}\left( \sqrt{5-4{{x}^{2}}}-x \right)$ The area ${{A}_{1}}$ bounded by the semicircle's diameter and the purple curve in the first quadrant is $\begin{aligned} {{A}_{1}} & =\frac{1}{5}\int\limits_{0}^{1}{\left( \sqrt{5-4{{x}^{2}}}-x \right)} \\ & =\frac{1}{5}\int\limits_{0}^{1}{\sqrt{5-4{{x}^{2}}}dx}-\frac{1}{5}\int\limits_{0}^{1}{xdx} \\ & =\frac{2}{5}\int\limits_{0}^{1}{\sqrt{\frac{5}{4}-{{x}^{2}}}dx}-\frac{1}{10} \\ & =\frac{2}{5}\left( \left[ \frac{1}{2}x\sqrt{\frac{5}{4}-{{x}^{2}}}+\frac{1}{2}\cdot \frac{5}{4}{{\tan }^{-1}}\frac{x}{\sqrt{\frac{5}{4}-{{x}^{2}}}} \right]_{0}^{1} \right)-\frac{1}{10} \\ & =\frac{1}{4}{{\tan }^{-1}}2 \\ \end{aligned}$ Due to symmetry, the total area bounded by the semicircle's diameter and the curves is $A=2{{A}_{1}}=\frac{1}{2}{{\tan }^{-1}}2$ For the answer, $p=1$ , $q=2$ , thus, $p+q=\boxed{3}$ .