Dynamic Geometry: P108

Let the variable triangle be $ABC$ , where $AB=5$ is the fixed side, and the measure of a cyan angle be $\theta$ . The inradius of $\triangle ABC$ is given by:

$\begin{aligned} r \cot \frac A2 + r \cot \frac B2 & = 5 \\ r \cot \left(\frac {\tan^{-1} \frac 43 - \theta}2 \right) + r \cot \left(\frac {\tan^{-1} \frac 34 - \theta}2 \right) & = 5 & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ r \left(\frac {1+\frac t2}{\frac 12 - t} + \frac {1+\frac t3}{\frac 13 - t} \right) & = 5 \\ r \left(\frac {2+t}{1 - 2t} + \frac {3+t}{1 - 3t} \right) & = 5 \\ \implies r & = \frac {(1-2t)(1-3t)}{1-2t-t^2} \end{aligned}$

We note that $\angle C = \frac \pi 2 - 2\theta$ . Then the circumradius of $\triangle ABC$ is given by:

$\begin{aligned} 2R & = \frac 5{\sin C} = \frac 5{\sin \left(\frac \pi 2 - 2\theta\right)} = \frac 5{\cos 2\theta} \\ \implies R & = \frac 5{4\left(\frac {1-t^2}{1+t^2}\right)^2-2} = \frac {5(1+t^2)^2}{2(1-6t^2+t^4)} \end{aligned}$

When $\dfrac rR = \dfrac {124}{625}$ , then

$\begin{aligned} \frac {(1-2t)(1-3t)(1-6t^2+t^4)}{(1-2t-t^2)(1+t^2)^2} & = \frac {62}{125} \\ \frac {(1-2t)(1-3t)(1+2t-t^2)}{(1+t^2)^2} & = \frac {62}{125} \\ 812t^4-2125t^3+749t^2+375t - 63 & = 0 \\ (4t-3)(7t-1)(29t^2-50t-21) & = 0 \\ \implies t & = \frac 17 & \small \blue{\text{For }\theta < \tan^{-1} \frac 34} \\ \implies \theta & = \tan^{-1} \frac {\frac 27}{1-\frac 1{49}} = \frac 7{24} \end{aligned}$

The area of $\triangle ABC$ :

$\begin{aligned} [ABC] & = \frac 12 AB \cdot CA \sin A & \small \blue{\text{By cosine rule}} \\ & = \frac {AB^2 \cdot \sin A \sin B}{2\sin C} \\ & = \frac {25 \sin \left(\tan^{-1} \frac 43-\theta\right)\sin \left(\tan^{-1} \frac 34-\theta\right)}{2 \blue{\sin \left(\frac \pi 2 -2\theta \right)}} & \small \blue{\text{Note that }\sin \left(\frac \pi 2 - \phi\right) = \cos \phi} \\ & = \frac {25\left(\frac 45 \cdot \frac {24}{25} - \frac 35 \cdot \frac 7{25} \right)\left(\frac 35 \cdot \frac {24}{25} - \frac 45 \cdot \frac 7{25} \right)}{2 \blue{\left(\frac {576}{625} - \frac {49}{625} \right)}} \\ & = \frac {75 \cdot 44}{2\cdot 527} = \frac {1650}{527} \end{aligned}$

Therefore $p+q = 1650 + 527 = \boxed{2177}$ .