Three tangent circles in a triangle

Let the radii of the three circles be $r_A,r_B,r_C$ . Label the points of tangency along $BC$ as $P,Q$ as below:

Since the centre of the circle nearest $B$ lies on the bisector of $\angle B$ , we have $BP=r_B \cot \frac{B}{2}$ . Similarly, $CQ=r_C \cot \frac{C}{2}$ .

By Pythagoras, $PQ^2 = \left(r_B+r_C \right)^2 - \left(r_B-r_C \right)^2$ , so that $PQ=2\sqrt{r_B r_C}$ .

Putting all these together, $BC=r_B \cot \frac{B}{2} + 2\sqrt{r_B r_C} + r_C \cot \frac{C}{2}$

We can use the same approach on the other two sides to find $CA=r_C \cot \frac{C}{2} + 2\sqrt{r_C r_A} + r_A \cot \frac{A}{2}$ and $AB=r_A \cot \frac{A}{2} + 2\sqrt{r_A r_B} + r_B \cot \frac{B}{2}$

Since we know the sides of $\Delta ABC$ , we can work out the required angles as well. I didn't find a nice algebraic solution so plugged the equations into Wolfram|Alpha to find $r_A=1.33596\ldots,\;\;\;\;r_B=1.81827\ldots,\;\;\;\;r_C=2.12937\ldots$

giving an answer of $\boxed{528360}$ .