Dynamic Geometry: P40

Consider a triangle with height $h$ and width $w$ inscribing a rectangle. Let the height of the triangle on top of the rectangle be $\alpha h$ , where $0 \le \alpha \le 1$ . Then the height of the rectangle is $(1-\alpha)h$ . As the triangle on top of the rectangle is similar to the large triangle, the width of the rectangle is $\alpha w$ and the area of the rectangle

$A = \alpha w (1 - \alpha) h = (\alpha - \alpha^2) hw = \left(\frac 14 - \left(\alpha - \frac 12\right)^2 \right) hw$

Therefore, the inscribed rectangle has a maximum area of $\frac 14 hw$ , half that of the triangle, when $\alpha = \frac 12$ or a height and width a half of those of the inscribing triangle.

Now let the center of the semicircle be $O(0,0)$ , the origin of the $xy$ -plane. Since the radius of the semicircle is $1$ , we can assign the top vertex of the triangle as $P(\cos \theta, \sin \theta$ as shown. As the height of the rectangle is a half of the triangle, its top two vertices are $A \left(\frac 12 (\cos \theta -1), \frac 12 \sin \theta \right)$ and $B \left(\frac 12 (\cos \theta +1), \frac 12 \sin \theta \right)$ . And the coordinates of the center of the rectangle are:

$\begin{cases} x = \dfrac 12 \left(\dfrac {\cos \theta -1}2 + \dfrac {\cos \theta +1}2 \right) = \dfrac {\cos \theta}2 & \implies \cos \theta = \dfrac x{\frac 12} \\ y = \dfrac 12 \cdot \dfrac {\sin \theta}2 = \dfrac {\sin \theta}4 & \implies \sin \theta = \dfrac y{\frac 14}\end{cases} \implies \dfrac {x^2}{\frac 14} + \dfrac {y^2}{\frac 1{16}} = 1$

This means that the blue curve is a half of an ellipse with major axis of $a=\frac 12$ and minor axis of $b=\frac 14$ and an area of $\dfrac {\pi ab}2 = \dfrac \pi{16}$ . Therefore the required answer is $1+6 = \boxed 7$ .