In a ΔABC if cosA+cosB+cosC=3/2 then prove that the Δ is equilateral.
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2 Answers
We need to consider only cases when the angles are acute.This is because if one of the angles is obtuse(say A).cosB≤0.
B+C = Ï€/2. and cosB+cosC ≤ 2cos(Ï€4)=√2<32
Hence A,B,C ≤ π/2
cos(θ) is concave for θ ≤ π/2
(i.e on differentiating it twice we get a value which is ≤0 for all θ between 0 to π/2).
Hence we can apply Jensen's Inequality,
cos(A)+cos(B)+cos(C) \leq 3cos\left(\frac{A+B+C}{3}\right)=3cos\left(\frac{\pi}{3}\right)=\frac{3}{2}
Equality holds when cosA=cosB=cosC
This is only possible when the Δ is equilateral.
- Shaswata Roy First line should be read as cosA <=0 and not cosB....sorry!Upvote·0· Reply ·2013-03-20 09:02:25
- Akshay Ginodia What is this Jenson's Inequality? Shashwata plz explain
@Akshay: The Jensen's Inequality states that if the double derivative of a function(say f(x)) is ≥ 0.
i.e. f"(x)≥0
Then,
f(x_{1})+f(x_{2})+\dots +f(x_{n})\geq n\times f\left(\frac{x_{1}+x_{2}+x_{3}+\dots +x_{n}}{n}\right)
and if the double derivative of a function is ≤ 0.
i.e. f"(x)≤0
Then,
f(x_{1})+f(x_{2})+\dots +f(x_{n})\leq n\times f\left(\frac{x_{1}+x_{2}+x_{3}+\dots +x_{n}}{n}\right)