Dynamic Geometry: P26

Let the half the width of the rectangle be $x$ . Then the height of the rectangle $h = \sqrt{1-x^2}$ and the perimeter of the rectangle $p = 4x + 2\sqrt{1-x^2}$ .

Note that the center of the semicircle, the center of the purple circle, and the tangent point of the purple circle and semicircle are colinear. By Pythagorean theorem , we have:

$\begin{aligned} (x+r)^2 + r^2 & = (1-r)^2 \\ x^2 + 2rx + r^2 + r^2 & = r^2 - 2r + 1 \\ r^2 + 2(x+1)r + x^2 - 1 & = 0 \\ \implies r & = \sqrt{2(x+1}-x-1 \end{aligned}$

Then the ratio of the diameter of the semicircle to the perimeter of the rectangle:

$\begin{aligned} \frac {\sqrt{2(x+1)}-x-1}{2x + \sqrt{1-x^2}} & = \frac {2\sqrt 5 -4}5 \end{aligned}$

Solving the equation, we get $x=\frac 35$ . Then $h = \sqrt{1-\frac 9{25}} = \frac 45$ and the area of the rectangle $\frac 65 \times \frac 45 = \frac {24}{25}$ . Therefore $\sqrt{a+b} = \boxed 7$ .