Dynamic Geometry: P56

• Let the center of the blue semicircle be at (0,0)
• Let $\sqrt{(1-a)^2-(x+a)^2}$ describe the pink semicircle with center at (-a,0) and 0<a<1
• Let $-\frac{x^{2}-1}{2}$ describe the locus of the center of the yellow circle. In problem P28 Chew-Seong and Thanos Petropoulos prove this setup creates a parabola.
• Let $r_{1}=1-a$ be the radius of the pink semicircle, $r_{2}$ be the radius of the yellow circle and $r_{3}$ be the radius of green circle.
• Let (x,y) be the center of the yellow circle and (w,z) be the center of the green circle.

We have:

1. $(w+a)^{2}+z^{2}=(r_{1}+r_{3})^{2}$
2. $(x+a)^{2}+y^{2}=(r_{1}+r_{2})^{2}$
3. $(x-w)^{2}+(y-z)^{2}=(r_{2}+r_{3})^{2}$
4. $w^{2}+z^{2}=(1-r_{3})^{2}$

Substituting $r_{1}$ with $(1-a)$ and also $y$ and $r_{2}$ with $-\frac{x^2-1}{2}$ , will give the simultaneous equations enough information to get $r_{2}=-\frac{4(a^2-a)}{(a-2)^2}$ and $r_{3}=-\frac{4(a^2-a)}{9a^2-4a+4}$

The final step is noting that $\frac{Area}{Perimeter} = \frac{\sqrt{(r_1+r_2+r_2)(r_1r_2r_3)}}{2(r_1+r_2+r_3)} = \frac{1}{2}\sqrt{\frac{r_1r_2r_3}{r_1+r_2+r_3}}$

With everything written as a function of $a$ , $\frac{A}{P}=\frac{12}{127} ⟹ a=\frac{3}{5} ⟹ \frac{r_2}{r_3}=\frac{121}{49} ⟹ \sqrt{p}+\sqrt{q} = 11 + 7$