Question

Prove that a triangle ABC is equilateral if and only if tanA + tanB + tanC = 3√3.

Answer 1

If the triangle is equilateral, then

A = B = C = 60º

⇒ tanA + tanB + tanC = 3tan60º = 3√3

Conversely assume that

tanA + tanB + tanC = 3√3

But in triangle ABC, A + B = 180º – C

Taking tan on both sides, we get

tan(A+B) = tan (180º–C)

⇒ (tanA + tanB)/(1 - tanAtanB) = - tanC

⇒ tanA + tanB = – tanC + tanA tanB tanC

⇒ tanA + tanB + tanC + tanA tanB tanC =3√3 

⇒ none of the tanA, tanB, tanC can be negative 

⇒ ∆ ABC cannot be obtuse angle triangle 

Also, AM ≥ GM

= 1/3[tanA + tanB + tanC ] ≥ [tanA tanB tanC ]^1/3

⇒ 1/3(3√3) ≥ (3√3)1/3⇒ √3 ≥ √3

So, the equality can hold if and only if 

tanA = tanB = tanC or A = B = C or 

when the triangle is equilateral.