Dynamic Geometry: P48

The the center of the semicircle be $O(0,0)$ , the origin of the $xy$ -plane. Let $AB$ be its diameter such that $A(-1,0)$ and $B(1,0)$ , the point on the arc be $P(u,v)$ and the center of the square be $C(x,y)$ . Let the side length of the square be $s$ and the height of $\triangle PDE$ be $\alpha s$ ; then $(1+\alpha)s = v$ .

Let the top-left vertex of the square be $D(x_d,s)$ . Draw $DB'$ parallel to $PB$ . Then we note that $\triangle PDE$ and $\triangle DAB'$ are similar, therefore

$\begin{aligned} \frac {DE}{AB'} & = \frac {\alpha s}s = \alpha \\ \frac s{2-s} & = \alpha = \frac vs - 1 & \small \blue{\text{Since }(1+\alpha) s = v} \\ s^2 & = (v-s)(2-s) = 2v - 2s - sv + s^2 \\ \implies s & = \frac {2v}{2+v} \\ \implies y & = \frac s2 = \frac v{2+v} \end{aligned}$

By similar triangles again,

$\begin{aligned} \frac {u-x_d}{x_d+1} & = \alpha = \frac vs - 1 = \frac v2 \\ 2u - 2x_d & = vx_d + v \\ x_d & = \frac {2u-v}{2+v} \\ \implies x & = x_d + \frac s2 = \frac {2u-v}{2+v} + \frac v{2+v} = \frac {2u}{2+v} \end{aligned}$

Then we have:

$\begin{aligned} x^2 + 4y^2 & = \frac {4(u^2+v^2)}{(2+v)^2} & \small \blue{\text{Note that }u^2 + v^2 = 1} \\ & = \frac 4{\left(2+\frac {2y}{1-y}\right)^2} & \small \blue{\text{and }y = \frac v{2+v}} \\ & = (1-y)^2 \\ x^2 + 3y^2 + 2y & = 1 \\ x^2 + 3 \left(y+\frac 13\right)^2 & = \frac 43 \\ \frac {x^2}{\frac 43} + \frac {\left(y+\frac 13\right)^2}{\frac 49} & = 1 \end{aligned}$

The locus is part of an ellipse with major-axis $a = \frac 2{\sqrt 3}$ , minor-axis $b=\frac 23$ , and center $\left(0, -\frac 13 \right)$ . The area bounded by the locus and the diameter is equivalent to a $120^\circ$ circular segment of radius $a = \frac 2{\sqrt 3}$ tilted along the minor-axis at $\cos^{-1}\frac ba = \cos^{-1} \frac 1{\sqrt 3}$ . The area is given by:

$A = \frac ba \left(\frac {\pi a^2}3 - \frac {a^2\sin 120^\circ}2 \right) = \frac 1{\sqrt 3} \left(\frac {4\pi}9 - \frac 1{\sqrt 3} \right) = \frac {4\pi}{9\sqrt 3} - \frac 13$

Therefore $m+n+l+p = 4 + 9 + 1 + 3 = \boxed{17}$ .