# Competition math geometry question

The perimeter of triangle ABC is $36$, and its area is $36$. Compute $\tan\frac{A}2 \tan\frac{B}2 \tan\frac{C}2$.

I found that the answer is $1/9$, but I was not able to find a reason for this. Could someone please give me a good explanation as to why it is this?

Let $r$ be the inradius of a triangle, classical geometry tell us the perimeter $\mathcal{P}$ and area $\mathcal{A}$ is related to $r$ through the relations:

$$\mathcal{P} = 2r\left(\cot\frac{A}{2} + \cot\frac{B}{2} + \cot\frac{C}{2}\right) \quad\text{ and }\quad \mathcal{A} = \frac{r\mathcal{P}}{2}$$ Eliminating $r$ gives us $$\cot\frac{A}{2} + \cot\frac{B}{2} + \cot\frac{C}{2} = \frac{\mathcal{P}^2}{4\mathcal{A}}$$ Since $A + B + C = \pi$, we have

$$0 = \cot\frac{\pi}{2} = \cot\left(\frac{A}{2} + \frac{B}{2} + \frac{C}{2}\right) = \frac{\cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot \cot\frac{C}{2} - \left(\cot\frac{A}{2} + \cot\frac{B}{2} + \cot\frac{C}{2}\right) }{ \cot\frac{A}{2}\cdot\cot\frac{B}{2} + \cot\frac{B}{2}\cdot\cot\frac{C}{2} + \cot\frac{C}{2}\cdot\cot\frac{A}{2} - 1}$$

This implies (the triple cotangent identity) $$\tan\frac{A}{2}\cdot\tan\frac{B}{2}\cdot\tan\frac{C}{2} = \frac{1}{\cot\frac{A}{2} + \cot\frac{B}{2} + \cot\frac{C}{2}} = \frac{4\mathcal{A}}{\mathcal{P}^2} = \frac{4\times36}{36^2} = \frac19$$