Dynamic Geometry: P45

Let the black semi-circle have a radius of $1$ and the blue semi-circle have a radius of $r$ , extend the black semi-circle to a full circle, and draw the perpendicular bisector of the diameter of the blue semi-circle and label the diagram as follows:

Let $k = OE$ . By the intersecting chord theorem , $AE \cdot EB = CE \cdot ED$ , so $r \cdot r = (1 + k) \cdot (1 - k)$ , which solves to $k^2 = 1 - r^2$ .

By the Pythagorean Theorem on $\triangle OFE$ , $OF^2 = OE^2 - EF^2 = k^2 - r^2 = (1 - r^2) - r^2 = 1 - 2r^2$ .

The pink curve can be defined by $x^2 = OF^2 = 1 - 2r^2$ and $y = r$ , which combines to $x^2 + 2y^2 = 1$ , an ellipse with a semi-major axis of $a = 1$ and a semi-minor axis of $b =\frac{\sqrt{2}}{2}$ .

The area of the cyan semi-circle is a maximum when its center is directly above the center of the black semi-circle when $r = b =\frac{\sqrt{2}}{2}$ , and has an area of $A_b = \frac{1}{2}\pi r^2 = \frac{1}{2}\pi (\frac{\sqrt{2}}{2})^2 = \frac{1}{4}\pi$ .

The area between the pink curve and the black semicircle's diameter is half the area of the ellipse, so $A_p = \frac{1}{2} \pi ab = \frac{1}{2} \cdot 1 \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4}\pi$ .

The ratio is then $\cfrac{A_b}{A_p} = \cfrac{\frac{1}{4}\pi}{\frac{\sqrt{2}}{4}\pi} = \cfrac{1}{\sqrt{2}}$ , so $q = 1$ , $p = 2$ , and $q + p = \boxed{3}$ .

Let the center of big semicircle be $O(0,0)$ , the origin of the $xy$ -plane and its radius be $1$ . Then the equation of the big semicircle is $x^2 + y^2 = 1$ . Let radius of the cyan semicircle be $r$ and its center $P$ (green point). Since the $y$ -coordinate of $P$ is always $r$ , let its coordinates be $P(s,r)$ . Then the equation of the cyan semicircle is $(x-s)^2 + (y-r)^2 = r^2$ or

$x^2 - 2sx + s^2 + y^2 - 2ry + r^2 = r^2 \implies x^2 - 2sx + s^2 + y^2 - 2ry = 0$

For the intersection points of the two semicircles, we have $x^2+y^2 = 1$ , and the equation reduces to:

$1 + s^2 - 2sx - 2ry = 0$

This is the equation of the diameter of the cyan semicircle. And a point on the diameter is $P(s,r)$ , putting $x=s$ and $y=r$ , we have:

$\begin{aligned} 1 + s^2 - 2s^2 - 2r^2 & = 0 \\ s^2 + 2r^2 & = 1 & \small \blue{\text{Replace }s \text{ with }x \text{ and }r \text{ with }y} \\ x^2 + \frac {y^2}{\frac 12} & = 1 \end{aligned}$

Therefore the locus is half of an ellipse centered at $(0,0)$ , with major-axis $a=1$ , minor-axis $b =\dfrac 1{\sqrt 2}$ , and an area of $\dfrac 12 \pi ab = \dfrac \pi{2\sqrt 2}$ . The cyan semicircle has a maximum area when $x=0$ and $y = r = b = \dfrac 1{\sqrt 2}$ and the area is $\dfrac 12 \pi \left(\dfrac 1{\sqrt 2}\right)^2 = \dfrac \pi 4$ . The ratio of areas is:

$\frac \pi 4 \cdot \frac {2\sqrt 2}\pi = \frac {\sqrt 2}2 = \frac 1{\sqrt 2}$

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