Question

Prove that a $\triangle$ ABC is equilateral if and only if
tan A + tan B + tan C = 3 $ \sqrt 3 $

Answer 1

If the triangle is equilateral, then
A = B = C = 60^\circ$
$\Rightarrow $ tan A + tan B + tan C = 3 tan 60^\circ = 3 \sqrt 3 $
Conversely assume that,
tan A + tan B + tan C$ 3 \sqrt 3 $
But in $\triangle ABC, A + B = 180^\circ$ - C
Taking tan on both sides, we get
tan $ (A + B) = tan \, (180^\circ - C)$
$\Rightarrow \frac{ tan \, A + tan \, B }{ 1 - tan \, A \, tan \, B } = - tan \, C $
$\Rightarrow $ tan A + tan B = - tan C + tan A tan B tan C
$\Rightarrow $ tan A + tan B + tan C = tan A tan B tan C = $3 \sqrt 3$
$\Rightarrow $ None of the tan A, tan B, tan C can be negative
So, $\triangle$ ABC cannot be obtuse angle triangle
Also, $ AM \ge GM$
$\Rightarrow \frac{1}{3} [tan A + tan B + tan C] \ge [tan A tan B tan C]^{1/3}$
$\Rightarrow \frac{1}{3} (3 \sqrt 3) \ge (3 \sqrt 3)^{ 1/ 3} \Rightarrow \sqrt 3 \ge \sqrt 3 $.
So, equality can hold if and only if
tan A = tan B = tan C
or A = B = C or when the triangle is equilateral.