Maximum area 3-4-5

Let the central angles at $O$ be $\alpha$ , $\beta$ , and $\gamma$ as shown. Then the area of $\triangle ABC$ is

$\begin{aligned} S & = \frac 12 (4 \cdot 5 \sin \alpha + 5 \cdot 3 \sin \beta + 3 \cdot 4 \sin \gamma) & \small \blue{\text{Since }\alpha + \beta + \gamma = 2\pi} \\ & = 10 \sin \alpha + 7.5 \sin \beta - 6 \sin (\alpha + \beta) \end{aligned}$

Let us fix the value of $\beta$ and find the value of $\alpha$ which maximize $S$ . (Note that $S$ has a minimum value of $0$ , when $\alpha = 0$ .)

$\begin{aligned} \frac {\partial S}{\partial \alpha} & = 10 \cos \alpha - 6 \cos (\alpha + \beta) & \small \blue{\text{Putting } \frac {\partial S}{\partial \alpha} = 0} \\ \implies 10 \cos \alpha & = 6 \cos (\alpha + \beta) \end{aligned}$

Similarly, fixing the value of $\alpha$ , the value of $\beta$ which maximizes $S$ is given by $7.5 \cos \beta = 6 \cos (\alpha + \beta)$ , which in turn equals to $10 \cos \alpha$ . Then $S$ is maximized when $10 \cos \alpha = 7.5 \cos \beta \implies \cos \beta = \frac 43 \cos \alpha$ and

$\begin{aligned} 10 \cos \alpha & = 6 \cos (\alpha + \beta) \\ 5 \cos \alpha & = 3 (\cos \alpha \cos \beta - \sin \alpha \sin \beta) \\ & = 4 \cos^2 \alpha - \sqrt{(1-\cos^2 \alpha)(9 - 16\cos^2 \alpha)} \end{aligned}$

Solving the equation above, we have $\cos \alpha \approx -0.372402014 \implies \cos \beta \approx -0.496536018$ , $\sin \alpha \approx 0.928071517$ , $\sin \beta \approx 0.868016119$ , $\sin (\alpha+\beta) \approx -0.784071886$ , and

$S = 10 \sin \alpha + 7.5 \sin \beta - 6 \sin (\alpha + \beta) \approx 20.49526738$

The required answer $\lfloor 100S \rceil = \boxed{2050}$ .