Dynamic Geometry :P17

Label the triangle $ABC$ . Let the variable $\angle A = \theta$ (orange), the $\angle B = \phi$ (black), the height of the triangle be $h$ , and the inradius and circumradius of the triangle be $r$ and $R$ . Then the inradius $r$ is given by:

$\begin{aligned} r \cot \frac \theta 2 + r \cot \frac \phi 2 & = 4 & \small \blue{\text{Let }\tan \frac \theta 2 = t} \\ \frac rt + 3 r & = 4 & \small \blue{\text{Note that }\tan \frac \phi 2 = \frac {1-\cos \phi}{\sin \phi} = \frac 13} \\ \implies r & = \frac 4{\frac 1t + 3} = \frac {4t}{1+3t} \end{aligned}$

The height of triangle:

$\begin{aligned} h \cot \theta + h \cot \phi & = 4 \\ \implies h & = \frac 4{\cot \theta + \cot \phi} = \frac 4{\frac {1-t^2}{2t} + \frac 43} = \frac {24t}{3+8t-3t^2} = \frac {24t}{(3-t)(1+3t)} \end{aligned}$

The circumradius is $2R = \dfrac {CA}{\sin \phi} \implies R = \dfrac {5h}{6\sin \theta}$ . Then the ratio of radii:

$\begin{aligned} \frac rR & = \frac {4t}{1+3t} \cdot \frac {6\sin \theta}{5h} = \frac {4t}{1+3t} \cdot \frac {12t}{1+t^2} \cdot \frac {(3-t)(1+3t)}{5\cdot 24t} = \frac {2t(3-t)}{5(1+t^2)} \\ \frac d{dt} \left(\frac rR\right) & = \frac {2\left((3-2t)(1+t^2)-(3t-t^2)(2t)\right)}{5(1+t^2)^2} = \frac {2(3-2t-3t^2)}{5(1+t^2)} \end{aligned}$

Therefore $\dfrac rR$ is maximum, when $3-2t-3t^2 = 0$ , because $\dfrac {d^2}{dt^2} \left(\dfrac rR\right) < 0$ . when $3-2t-3t^2 = 0$ . Then the area of the triangle:

$\begin{aligned} A_\triangle & = \dfrac {4h}2 = 2h = \frac {2 \cdot 24t}{3+8t-3t^2} = \frac {2 \cdot 24t}{\blue{\cancel{3-2t-3t^2}^0} + 10t} = \frac {24}5 \end{aligned}$

Therefore $a+b = 24+5 = \boxed{29}$ .