Question

For a triangle ABC it is given that cosA + cosB + cosC = 3/2. Prove that the triangle is equilateral. 

Answer 1

Let a, b, c are the sides of a ∆ ABC.

Given, cos A + cos B + cos C = 3/2

⇒ (b² + c² - a²)/2bc + (a² + c² - b²)/2ac + (a² + b² - c²)/2ab = 3/2

⇒ ab² + ac² – a³ + ba² + bc² – b³ + ca² + cb² – c³ = 3abc 

⇒ a(b – c)² + b(c – a)² + c(a – b)²

= (a + b + c)/2[(a - b)² + (b - c)² + (c - a)²]

⇒ (a + b – c) (a – b)² + (b + c – a) (b – c)² + (c + a – b)

(c – a)² = 0

(as we know, a+b–c > 0, b+c–a > 0, c+a–b > 0)

∴ Each term on the left of equation has positive coefficient multiplied by perfect square, each term must be separately zero.

⇒ a = b = c

∴ Triangle is an equilateral.