Sunday, February 11, 2007

Return Of The Triangle. Topic: Geometry/Inequalities/Trigonometry. Level: AIME.

Problem: (1961 IMO - #2) Let $ a, b, c $ be the sides of a triangle, and $ T $ its area. Prove:

$ a^2+b^2+c^2 \ge 4T\sqrt{3} $.

In what case does equality hold?

Solution: We begin with the trivial inequality, $ (a-b)^2 \ge 0 $, which has equality at $ a = b $. Rearrange to get

$ a^2+b^2 \ge 2ab $.

Let $ \theta $ be the angle between the sides with lengths $ a, b $. Since $ 2 \ge \cos{\theta}+\sqrt{3}\sin{\theta} $ (can be proved by combining RHS) with equality at $ \theta = \frac{\pi}{3} $, we know

$ a^2+b^2 \ge ab(\cos{\theta}+\sqrt{3}\sin{\theta}) $

$ 2(a^2+b^2) \ge 2ab(\cos{\theta}+\sqrt{3}\sin{\theta}) $

$ a^2+b^2+(a^2+b^2-2ab\cos{\theta}) \ge 2\sqrt{3} \cdot ab\sin{\theta} $.

Recalling the Law of Cosines, we know $ c^2 = a^2+b^2-2ab\cos{\theta} $. Also, $ T = \frac{1}{2}ab\sin{\theta} $, so substituting we obtain

$ a^2+b^2+c^2 \ge 4T\sqrt{3} $

as desired. Equality holds when $ a = b $ and $ \theta = \frac{\pi}{3} $, which means the triangle must be equilateral. QED.


Comment: There are lots of ways to prove this, but this is one of the more elementary ones, requiring only basic knowledge of inequalities and trigonometry. Which is always good because I don't know any geometry. We see that this inequality is in general pretty weak, with equality only when the triangle is equilateral - there is a stronger version that states

$ a^2+b^2+c^2 \ge 4T\sqrt{3}+(a-b)^2+(b-c)^2+(c-a)^2 $.

See if you can prove that...


Practice Problem: Let $ a, b, c $ be the sides of a triangle, and $ T $ its area. Prove:

$ a^2+b^2+c^2 \ge 4T\sqrt{3}+(a-b)^2+(b-c)^2+(c-a)^2 $.


  1. This is in Engel and they give an IDENTITY (if I recall correctly) from which this follows by trivial inequality. It's so hardcore.

  2. Your solution seems completely unmotivated.

  3. That's cuz it's written in reverse =). Try reading it backwards, then some of the steps may seem more obvious.