Dynamic Geometry: P114

Let the dividing yellow chord be $AB$ , the upper and lower moving points on the circumference be $C$ and $C'$ respectively, and $\angle CAB = \theta$ . From the compiled calculations in Dynamic Geometry: P96 Series , we have $AN = 12$ , $\angle ACB = \pi - \tan^{-1} \frac {12}5$ and $\angle AC'B = \tan^{-1} \frac {12}5$ . By sine rule :

$\frac {BC}{AB} = \frac {\sin \angle CAB}{\sin \angle ACB} \implies BC = \frac {12\sin \theta}{\frac {12}{13}} = 13\sin \theta$

Similarly,

$\begin{aligned} AC & = 13 \sin \left(\tan^{-1} \frac {12}5 - \theta \right) = 12 \cos \theta - 5 \sin \theta \\ AC' & = 13 \sin \left(\frac \pi 2- \theta \right) = 13 \cos \theta \\ BC' & = 13 \sin \left(\frac \pi 2 - \tan^{-1} \frac {12}5 + \theta \right) = 5 \cos \theta + 12 \sin \theta \end{aligned}$

Then the perimeter of $\triangle ABC$ , $p = 8 \sin \theta + 12 \cos \theta + 12$ and that of $\triangle ABC'$ , $p' = 12 \sin \theta + 18 \cos \theta + 12$ . When $\dfrac {p'}p = \dfrac {483}{380}$ ,

$\begin{aligned} \frac {12 \sin \theta + 18 \cos \theta + 12}{8 \sin \theta + 12 \cos \theta + 12} & = \frac {483}{380} \\ \frac {2 \sin \theta + 3 \cos \theta + 2}{2 \sin \theta + 3 \cos \theta + 3} & = \frac {161}{190} \\ 1 - \frac 1{2 \sin \theta + 3 \cos \theta + 3} & = \frac {161}{190} \\ 2 \sin \theta + 3 \cos \theta & = \frac {103}{29} \\ \sqrt{13} \sin \left(\theta + \tan^{-1} \frac 32 \right) & = \frac {103}{29} \\ \tan \left(\theta + \tan^{-1} \frac 32 \right) & = \frac {103}{18} \\ \frac {\tan \theta + \frac 32}{1-\frac 32 \tan \theta} & = \frac {103}{18} \\ 345 \tan \theta & = 152 \\ \implies \theta & = \tan^{-1} \frac {152}{345} \end{aligned}$

From the compilation of calculations, we know that the largest rectangle inscribed by a triangle has a half of the height and a half of the width the triangle. Since the upper and lower rectangles have the same width of $6$ the area of the rectangles are directly proportional to their heights. Then we have:

$\begin{aligned} \frac {A_\red{\rm red}}{A_\blue{\rm blue}} & = \frac {AC \cdot \sin \theta}{AC' \cdot \sin \left(\frac \pi 2 - \tan^{-1} \frac {12}5 + \theta \right)} \\ & = \tan \theta \tan \left(\tan^{-1} \frac {12}5 - \theta \right) \\ & = \frac {152}{345} \cdot \frac {\frac {12}5 - \frac {152}{345}}{1+\frac {12}5 \cdot \frac {152}{345}} \\ & = \frac {608}{1449} \end{aligned}$

Therefore $\sqrt{q-p} = \sqrt{1449-608} = \sqrt{841} = \boxed{29}$ .