Question

Prove that a $△$ 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$<br/>$\Rightarrow $tanA+tanB+tanC=3tan60^{\circ }=3\sqrt{3}$
Conversely assume that,
tan A + tan B + tan C$3\sqrt{3}$
But in $△ABC,A+B=180^{\circ }$ - C
Taking tan on both sides, we get
tan $(A+B)=tan(180^{\circ }-C)$
$\Rightarrow \frac{tanA+tanB}{1-tanAtanB}=-tanC$
$\Rightarrow$ 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, $△$ ABC cannot be obtuse angle triangle
Also, $AM\ge GM$
$\Rightarrow \frac{1}{3}[tanA+tanB+tanC]\ge [tanAtanBtanC{]}^{1/3}$
$\Rightarrow \frac{1}{3}(3\sqrt{3})\ge (3\sqrt{3}{)}^{1/3}\Rightarrow \sqrt{3}\ge \sqrt{3}$.
So, equality can hold if and only if
tan A = tan B = tan C
or A = B = C or when the triangle is equilateral.