Equilateralism. Topic: Geometry/Inequalities. Level: AIME.

Problem: Let $a,b,c$ be the side lengths of a triangle and $S$ be its area. Given that

$48S^2 = (a^2+b^2+c^2)^2$,

the triangle must be equilateral.

Solution: Well we have an annoying $S$ on the LHS, so how can we get rid of it... it's a good idea to get a few main formulas down for the area of a triangle, including Heron's:

$ [ABC] = \sqrt{s(s-a)(s-b)(s-c)} $

where $s = \frac{a+b+c}{2}$. So if we just plug that into our condition, we get

$ 3(a+b+c)(a+b-c)(a-b+c)(-a+b+c) = (a^2+b^2+c^2)^2$.

This is where some experience with inequalities comes in hand. Try applying AM-GM to get

$ (a+b-c)(a-b+c)(-a+b+c) \le \left(\frac{a+b+c}{3}\right)^3 $.

Multiply that by a factor of $3(a+b+c)$ to get

$ 3(a+b+c)(a+b-c)(a-b+c)(-a+b+c) \le \frac{1}{9}(a+b+c)^4 $.

By Cauchy or AM-QM, we have $ (a+b+c)^2 \le 3(a^2+b^2+c^2)^2 $, so

$ \frac{1}{9}(a+b+c)^4 \le (a^2+b^2+c^2)^2 $.

Then

$ 3(a+b+c)(a+b-c)(a-b+c)(-a+b+c) \le (a^2+b^2+c^2)^2$

with equality iff $ a = b = c $, but equality is the given condition, so the triangle is equilateral. QED.

--------------------

Practice Problem #1: Prove Heron's Formula.

Practice Problem #2: Given a quadrilateral $ABCD$ with an inscribed circle of radius $r$, prove that $AB+BC+CD+DA \ge 8r$ and find the equality condition.