## 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.

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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...

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

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