Maximum area triangle

Let the center of the three circles be $O$ and the radius $AO=a=385$ , $BO=b=119$ , and $CO=c=65$ , and the central angle opposite the respective radii be $\alpha$ , $\beta$ , and $\gamma$ . Then the area of $\triangle ABC$ is given by:

$\begin{aligned} A & = \frac 12 ab \sin \gamma + \frac 12 bc \sin \alpha + \frac 12 ca \sin \beta & \small \blue{\text{Since } \gamma = 2\pi - \alpha - \beta} \\ 2A & = -ab\sin (\alpha + \beta) + bc \sin \alpha + ca \sin \beta \end{aligned}$

$A$ is maximum when $\dfrac {\partial A}{\partial \alpha} = 0$ and $\dfrac {\partial A}{\partial \beta} = 0$ . $\begin{cases} 2 \dfrac {\partial A}{\partial \alpha} = -ab\cos (\alpha + \beta) + bc \cos \alpha \\ 2 \dfrac {\partial A}{\partial \beta} = -ab\cos (\alpha + \beta) + ca \cos \beta \end{cases}$ .

Equating to zero, we have $bc \cos \alpha = ca \cos \beta = ab \cos (\alpha + \beta) = ab \cos \gamma$ $\implies \dfrac {\cos \alpha}a = \dfrac {\cos \beta}b = \dfrac {\cos \gamma}c$ .

From $c \cos \alpha = a\cos (\alpha + \beta)$ , we have $2bc \cos^3 \alpha - (a^2+b^2+c^2) \cos \alpha + a^2 = 0$ . Solving the equation, we have

$\begin{cases} \cos \alpha = - \dfrac {77}{85} & \implies \sin \alpha = \dfrac {36}{85} \\ \cos \beta = - \dfrac 7{25} & \implies \sin \beta = \dfrac {24}{25} \\ \cos \gamma = - \dfrac {13}{85} & \implies \sin \gamma = \dfrac {84}{85} \end{cases}$

And the maximum area of $\triangle ABC$ , $A_{\max} = \dfrac 12 \left(119 \times 65 \times \dfrac {36}{85} + 65 \times 385 \times \dfrac {24}{25} + 385 \times 119 \times \dfrac {84}{85} \right) = \boxed{36288}$ .