Dynamic Geometry: P63

Let the center of the big semicircle be $O(0,0)$ , the radii of the pink and cyan semicircles be $r_0$ and $r_1$ respectively, and center of the right cyan semicircle be $P(k,h)$ . Then $k = 1- r_0$ and $h = r_0 + r_1$ . The equation of the big semicircle is $x^2 + y^2 = 1$ and that of the cyan semicircle is $(x-k)^2 + (y-h)^2 = r_1^2$ . The two intersection points $A$ and $B$ satisfy:

$\begin{aligned} \blue{x^2} - 2kx + k^2 + \blue{y^2} - 2hy + h^2 & = r_1^2 & \small \blue{\text{Note that }x^2 + y^2 =1} \\ \blue 1 - 2kx + k^2 - 2hy + h^2 & = r_1^2 \end{aligned}$

This is the equation of the diameter $AB$ of the cyan semicircle. Since the center $P(k,h)$ is on $AB$ , putting $x=k$ and $y=h$ , we have:

$\begin{aligned} 1 - 2k^2 + k^2 - 2h^2 + h^2 & = r_1^2 & \small \blue{\text{Rearranging}} \\ k^2 + h^2 + \blue{r_1}^2 & = 1 & \small \blue{\text{Note that }k+h = 1 + r_1} \\ k^2 + h^2 + \blue{(k+h-1)^2} & = 1 & \small \blue{\implies k+h-1 = r_1} \\ k^2 + h^2 + k^2 + 2kh + h^2 - 2k - 2h + 1 & = 1 \\ k^2 + h^2 + kh - k - h & = 0 \\ (k+h)^2 - 2kh + kh - (k+h) & = 0 \\ kh & = (k+h)^2 - (k+h) \\ & = (1+r_1)^2 - 1 - r_1 \\ \implies kh & = r_1(1+r_1) & ...(1) \\ (1-r_0)(r_0+r_1) & = r_1^2+r_1 \\ r_0 + r_1 - r_0^2 - r_0r_1 & = r_1^2 + r_1 \\ r_1^2 +r_0r_1 & = r_0 - r_0^2 & ...(2) \end{aligned}$

Note that the area of the rectangle $A = 2kh$ . $A$ is maximum when $kh$ is maximum. Equation $(1): \ kh = r_1(1+r_1)$ shows that $kh$ is maximum, when $r_1$ is maximum. From $(2)$ :

$\begin{aligned} r_1^2 +r_0r_1 + r_0^2 - r_0 & = 0 & \small \blue{\text{Differentiate both sides w.r.t. }r_0} \\ 2r_1 \cdot \frac {dr_1}{dr_0} + r_1 + r_0 \cdot \frac {dr_1}{dr_0} & = 1 - 2r_0 & \small \blue{\text{Putting }\frac {dr_1}{dr_0}=0} \\ \implies r _ 1 & = 1 - 2r_0 \end{aligned}$

This means that $A$ is maximum, when $r_1 = 1 - 2r_0$ . From $(2)$ :

$\begin{aligned} (1-2r_0)^2 + r_0(1-2r_0) & = r_0 - r_0^2 \\ 1 - 4r_0 + 4r_0^2 + r_0 - 2r_0^2 & = r_0 - r_0^2 \\ 3r_0^2 - 4r_0 + 1 & = 0 \\ (3r_0-1)(r_0-1) & = 0 \end{aligned}$

This means that when $r_0 = \dfrac 13$ and $r_1 = 1-2r_0 = \dfrac 13$ , $A=2kh = 2r_1(1+r_1) = \dfrac 89$ is maximum (when $r_0 = 1$ , $r_1 = -1$ which is not acceptable). Then the perimeter of the rectangle $p = 4(1-r_0)+ 2(r_0+r_1) = 4$ . Then $\dfrac pA = \dfrac 4{\frac 89} = \dfrac 92$ and $p+q = 9+2 = \boxed{11}$ .