Dynamic Geometry: P60

Let the center of the semicircle be $O(0,0)$ , the origin of the $xy$ -plane, and the blue triangle be $ABC$ , where $AB$ is the diameter of the semicircle. Consider the locus on in the positive quadrant and let a point on the locus be $P(x,y)$ and $PN$ be perpendicular to $AB$ ; then $PN=y$ and $ON=x$ . Let $\angle COB = 2 \theta$ ; then $\angle OCB = \angle OBC = \frac \pi 2 - \theta$ . Note that $OP$ bisects $\angle COB$ . Therefore, $\angle POB = \theta$ and $\angle PBO = \frac \pi 4 - \frac \theta 3$ . From

$\begin{aligned} ON + NB & = OB \\ y \cot \theta + y \cot \left(\frac \pi 4 - \frac \theta 2 \right) & = 1 & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ y \cdot \frac {1-t^2}{2t} + y \cdot \frac {1+t}{1-t} & = 1 \\ \frac {(1+t)(1+t^2)}{2t(1-t)} \cdot y & = 1 \\ \implies y & = \blue{\frac {2t}{1+t^2}} \cdot \frac {1-t}{1+t} \\ & = \blue{\sin \theta} \cdot \frac {(1-t)^2}{(1+t)(1-t)} \\ & = \sin \theta \cdot \frac {1-2t+t^2}{1-t^2} \\ & = \sin \theta \left(\frac {1+t^2}{1-t^2} - \frac {2t}{1-t^2} \right) \\ & = \sin \theta (\sec \theta - \tan \theta) \\ \implies y & = \tan \theta (1-\sin \theta) & \small \blue{\text{Note that }\frac yx = \tan \theta} \\ \implies x & = 1 - \sin \theta \end{aligned}$

Now the area under the locus is given by:

$\begin{aligned} A & = \int_0^1 y \ \text dx \\ & = \int_\frac \pi 2^0 \tan \theta (1-\sin \theta) \ \text d(1-\sin \theta) \\ & = \int_0^\frac \pi 2 \sin \theta (1-\sin \theta) \ \text d \theta \\ & = \int_0^\frac \pi 2 \left(\sin \theta - \frac {1 - \cos 2 \theta}2 \right) \text d \theta \\ & = - \cos \theta - \frac \theta 2 + \frac {\sin 2 \theta}4 \ \bigg|_0^\frac \pi 2 \\ & = 1 - \frac \pi 4 \end{aligned}$

Therefore $p+q = 1 + 4 = \boxed 5$ .