Dynamic Geometry: P64

The locus of the center of the cyan semicircle is part of an ellipse. This solution proves it.

Let the center of the big semicircle be $O(0,0)$ , the origin of the $xy$ -plane. Tilt the figure by $45^\circ$ counterclockwise. Let the center and radius of the cyan semicircle be $P(k,h)$ and $r_1$ , and those of the green semicircle $Q$ and $r_0$ .

Then $Q = \left( \frac {1-r_0}{\sqrt 2}, \frac {1-r_0}{\sqrt 2} \right)$ . Note that $PQ = r_0+r_1$ . Therefore the coordinates of $P$ are

$\begin{cases} k = \dfrac {1-2r_0-r_1}{\sqrt 2} \\ h = \dfrac {1+r_1}{\sqrt 2} \end{cases}$

The equation of the cyan semicircle is $(x-k)^2+(y-h)^2 = r_1^2$ and that of the big semicircle is $x^2 + y^2 = 1$ . The two points of intersection satisfy both equations and we have:

$1 - 2kx + k^2 - 2hy + h^2 = r_1^2$

which is the diameter of the cyan semicircle. Since the center $P(k,h)$ is on the diameter, putting $x=k$ and $y=h$ ,

$\begin{aligned} 1 - 2k^2 + k^2 - 2h^2 + h^2 & = r_1^2 \\ k^2 + h^2 + \blue{r_1^2} & = 1 & \small \blue{\text{Since }h = \frac {1+r_1^2}{\sqrt 2}} \\ k^2 + h^2 + \blue{(\sqrt 2h-1)^2} & = 1 \\ k^2 + 3h^2 - 2\sqrt 2 h & = 0 \\ k^2 + 3\left(h - \frac {\sqrt 2}3 \right)^2 & = \frac 23 \\ \frac {k^2}{\frac 23} + \frac {\left(h-\frac {\sqrt 2}3\right)^2}{\frac 29} & = 1 & \small \blue{\text{Replace }k \text{ with }x \text{ and }h \text{ with }y.} \\ \frac {x^2}{\frac 23} + \frac {\left(y-\frac {\sqrt 2}3\right)^2}{\frac 29} & = 1 \end{aligned}$

Therefore, the locus is part of an ellipse with center at $C \left(0, \frac {\sqrt 2}3 \right)$ , major-axis of $a=\sqrt{\frac 23}$ and minor-axis of $b=\frac {\sqrt 2}3$ . Solving the ellipse equation with $x^2 + y^2 = 1$ , we can find the two tangent points of the ellipse and the big semicircle are $A\left(-\frac 1{\sqrt 2}, \frac 1{\sqrt 2}\right)$ and $B\left(\frac 1{\sqrt 2}, \frac 1{\sqrt 2}\right)$ . The area bounded by the semicircle and the ellipse is given by:

$\begin{aligned} A & = 90^\circ \text{ circular segment } - 120^\circ \text{ elliptical segment } \\ & = \frac \pi 4 - \frac 12 - \frac ba \left(\frac {\pi a^2}3 - \frac 12 a^2 \sin 120^\circ \right) \\ & = \frac \pi4 - \frac 12 - ab \left(\frac \pi 3 - \frac {\sqrt 3}4 \right) \\ & = \frac \pi4 - \frac 12 - \frac 2{3\sqrt 3} \left(\frac \pi 3 - \frac {\sqrt 3}4 \right) \\ & = \frac \pi 4 - \frac 13 - \frac {2\pi}{9\sqrt 3} \end{aligned}$

Therefore the required answer is $4+3 = \boxed 7$ .