1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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) $$\tan\frac{A}{2}\cdot\tan\frac{B}{2}\cdot\tan\frac{C}{2} = \frac{1}{\cot\frac{A}{2} + \cot\frac{B}{2} + \cot\frac{C}{2}} = \frac{4\mathcal{A}}{\mathcal{P}^2} = \frac{4\times36}{36^2} = \frac19$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.