Dynamic Geometry: P111

The radius of the circle is $R=\dfrac{4+9}{2}=6.5$ . Let $AB$ be the yellow cord and $M$ its midpoint. Let $EM$ and $FM$ be the heights of the circular segments, as seen in the figure.
By Intersecting Cords Theorem ,

$AM\cdot MB=EM\cdot MF\Rightarrow {{\left( \dfrac{AB}{2} \right)}^{2}}=4\times 9\Rightarrow AB=12$ We place the configuration on a coordinate system, so that the center of the circle is point $K\left( 0,-2.5 \right)$ . Then, the equation of the circle is ${{x}^{2}}+{{\left( y+2.5 \right)}^{2}}={{6.5}^{2}}$ The common cord $AB$ of the circular segments lies on the $x$ -axis with $A\left( -6,0 \right)$ and $B\left( 6,0 \right)$ . Let the upper triangle be $\triangle CAB$ and the lower $\triangle DAB$ , as seen on the figure. Since the biggest rectangle inscribed in each triangle has half the height and half the width of the corresponding triangle, its area is one fourth the area of the triangle.

Consequently, denoting each green angle by $\theta$ , we have $\dfrac{11}{61}=\dfrac{\text{area of upper rectangle}}{\text{area of lower rectangle}}=\dfrac{\cancel{\dfrac{1}{2}}\left[ CAB \right]}{\cancel{\dfrac{1}{2}}\left[ CDB \right]}=\dfrac{\dfrac{1}{2}AB\cdot AC\cdot \sin \theta }{\dfrac{1}{2}AB\cdot AD\cdot \sin \theta }=\dfrac{AC}{AD}$ $\Rightarrow AD=\frac{61}{11}AC \ \ \ \ \ (1)$ Because the green angles are equal, the reflection ${C}'$ of point $C$ w.r.t. the $x$ -axis lies on $AD$ , thus, vectors $\overrightarrow{A{C}'}$ and $\overrightarrow{AD}$ have the same direction.
Using points $A\left( -6,0 \right)$ , hence, taking into account $(1)$ and the fact that $A{C}'=AC$ , we have

$\begin{aligned} \overrightarrow{AD}=\dfrac{61}{11}\overrightarrow{A{C}'} & \Rightarrow \left( {{x}_{D}}-{{x}_{A}},{{y}_{D}}-{{y}_{A}} \right)=\dfrac{61}{11}\left( {{x}_{{{C}'}}}-{{x}_{A}},{{y}_{{{C}'}}}-{{y}_{A}} \right) \\ & \Rightarrow \left( {{x}_{D}}+6,{{y}_{D}} \right)=\dfrac{61}{11}\left( {{x}_{C}}+6,-{{y}_{C}} \right) \\ \phantom{0} \\ & \Rightarrow \left\{ \begin{matrix} {{x}_{D}}=\dfrac{61{{x}_{C}}+300}{11} \\ \phantom{0} \\ {{y}_{D}}=-\dfrac{61}{11}{{y}_{C}} \\ \end{matrix} \right. \ \ \ \ \ \ (2) \\ \end{aligned}$ Since both points $C$ and $D$ belong to the circle we get the system of equations

$\left\{ \begin{matrix} {{x}_{C}}^{2}+{{\left( {{y}_{C}}+2.5 \right)}^{2}}={{6.5}^{2}} \\ \phantom{0} \\ {{x}_{D}}^{2}+{{\left( {{y}_{D}}+2.5 \right)}^{2}}={{6.5}^{2}} \\ \end{matrix} \right.\Leftrightarrow \left\{ \begin{matrix} {{x}_{C}}^{2}+{{\left( {{y}_{C}}+2.5 \right)}^{2}}={{6.5}^{2}} \\ \phantom{0} \\ {{\left( \dfrac{61{{x}_{C}}+300}{11} \right)}^{2}}+{{\left( -\dfrac{61}{11}{{y}_{C}} \right)}^{2}}={{6.5}^{2}} \\ \end{matrix} \right.$ which solves to ${{x}_{C}}=-\dfrac{171}{34} \ \ \ \ \ \ {{y}_{C}}=\dfrac{55}{34}$ Consequently,
$\left( 2 \right)\Rightarrow \left( {{x}_{D}},{{y}_{D}} \right)=\left( -\dfrac{21}{34},-\dfrac{305}{34} \right)$ Now we can calculate the length of $CD$ :

$CD=\sqrt{{{\left( {{x}_{D}}-{{x}_{C}} \right)}^{2}}+{{\left( {{y}_{D}}-{{y}_{C}} \right)}^{2}}}=\sqrt{{{\left( -\dfrac{21}{34}+\dfrac{171}{34} \right)}^{2}}+{{\left( -\dfrac{305}{34}-\dfrac{55}{34} \right)}^{2}}}=\dfrac{195}{17}$ For the answer, $p=195$ , $q=17$ , thus, $p+q=\boxed{212}$ .

$\red{\text{[Updated to provide a better explanation.]}}$ Let the dividing chord of the circle be $AB$ , the upper and lower variable triangles be $\triangle ABC$ and $\triangle ABC'$ respectively, and the measure of the green angle be $\theta$ . Referring the calculations in Dynamic Geometry: P96 Series , the radius of the circle is $6.5$ , $AB=12$ , $\angle ACB = \pi - \tan^{-1} \frac {12}5$ , $\angle AC'B = \tan^{-1} \frac {12}5$ , and the biggest rectangle inscribed by a triangle is one with half the height and width of those of the triangle.

$\red{\text{[Updated; much simpler solution using cyclic quadrilateral.]}}$ Since $\triangle ABC$ and $\triangle ABC'$ have the same base $AB = 12$ , the width of both upper and lower rectangles is $6$ . The heights of the two triangles are $h = AC \sin \theta$ and $h' = AC' \sin \theta$ . Then the ratio of their areas ' $\frac {6 \cdot \frac h2}{6 \cdot \frac {h'}2} = \frac {6 \cdot \frac 12 AC \sin \theta}{6 \cdot \frac 12 AC' \sin \theta} = \frac {AC}{AC'} = \frac {11}{61}$

If $AC = a$ , $AC' = \dfrac {61}{11} a = ka$ , where $k = \dfrac {61}{11}$ .

Since $ACBC'$ is a cyclic quadrilateral , $\angle BCC' = \angle BAC' = \theta$ and $\angle BC'C = \angle BAC = \theta$ . Therefore $\triangle BCC'$ is isosceles and $BC = BC' = b$ . By cosine rule ,

$\begin{cases} a^2 + b^2 - 2ab \cos \left(\pi - \tan^{-1} \dfrac {12}5 \right) = a^2 + b^2 + \dfrac {10}{13}ab = 144 & ...(1) \\ k^2a^2 + b^2 - 2kab \cos \left(\tan^{-1} \dfrac {12}5 \right) = k^2a^2 + b^2 - \dfrac {10}{13}kab = 144 & ...(2) \end{cases} \\ \begin{aligned} (2) - (1): \quad (k^2 -1)a^2 - \frac {10ab(k+1)}{13} & = 0 \\ \implies b & = \frac {13a}{10} (k-1) = \frac {13a}{10}\left(\frac {61}{11}-1 \right) = \frac {65}{11}a \\ (1): \quad a^2 + \frac {65^2}{11^2}a^2 + \frac {10}{13} \cdot \frac {65}{11} a^2 & = 144 \\ \implies a & = \frac {11}{\sqrt{34}} \\ b & = \frac {65}{\sqrt{34}} \end{aligned}$

Let the distance between the two moving points $CC' = d$ . By Ptolemy's theorem ,

$\begin{aligned} AB \cdot CC' & = AC \cdot BC' + AC' \cdot BC \\ 12d & = ab + \frac {61}{11}ab = \frac {72}{11}ab = \frac {72}{11} \cdot \frac {11 \cdot 65}{34} \\ \implies d & = \frac {195}{17} \end{aligned}$

Therefore $p+q = 195+17 = \boxed{212}$ .