Given triangle ABC and its inscribed circle O, AO = 3, BO = 4, CO = 5, find the perimeter of ABC

Not sure if there is a more straightforward solution but in meantime,

$3 \sin\frac{A}{2} = 4 \sin\frac{B}{2} = 5 \sin\frac{C}{2} = r$

$\frac{C}{2} = 90^0 - \frac{A+B}{2}$

So, $\cos \frac{A+B}{2} = \frac{r}{5}$

Expanding and writing values in terms of $r$ and squaring both sides, we simplify it to a cubic

$\frac{\sqrt {9-r^2} \times \sqrt {16-r^2}}{12} - \frac{r^2}{12} = \frac{r}{5}$

$25(9-r^2)(16-r^2) = (5r^2 + 12r)^2$

$120r^3 + 769r^2 - 3600 = 0$.

As $0 \lt r \lt 3$, we get $r \approx 1.9$ (with help from WolframAlpha)

From here on, we get sides as $\approx 5.842, 6.947, 8.145$.