## Monday, January 9, 2006

### Symmetrific! Topic: Geometry/Inequalities. Level: Olympiad.

Problem: (2004 USAMO - #1) Let $ABCD$ be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least $60$ degrees. Prove that

$\frac{1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|$.

When does equality hold?

Solution: First of all, we notice conveniently enough that if the left inequality is true, the right is true by symmetry (just applying the same inequality cycled around the quadrilateral). Hence it is only necessary to prove the first inequality.

We establish two inequalities that we will use throughout the problem:

For $a,b, 60^{\circ} \le \theta \le 120^{\circ}$, we have $-\frac{1}{2} \le \cos{\theta} \le \frac{1}{2} \Rightarrow -ab \le 2ab\cos{\theta} \le ab$ (1).

Also note that the interior angles and exterior angles $\ge 60^{\circ}$ means all the angles of the quadrilateral will qualify as $\theta$ under the set conditions.

So we begin with the Trivial Inequality,

$0 \le (AB-AD)^2$

$(AB)(AD) \le AB^2+AD^2-(AB)(AD)$

$(AB)(AD) \le AB^2+AD^2-2(AB)(AD)\cos{\angle BAD} = BD^2$ by (1) and Law of Cosines.

Then we have

$\frac{1}{3}(BD^2+2(AB)(AD)) \le BD^2$

$\frac{1}{3}(AB^2+AD^2-2(AB)(AD)\cos{\angle BAD}+2(AB)(AD)) \le BD^2$ by Law of Cosines.

$\frac{1}{3}(AB^2+AD^2+(AB)(AD)) \le BD^2$ by (1).

So then we manipulate the RHS to find

$\frac{1}{3}(AB^2+AD^2+(AB)(AD)) \le BC^2+CD^2-2(BC)(CD)\cos{\angle BCD} \le BC^2+CD^2+(BC)(CD)$ by Law of Cosines and (1).

Noticing that $|AB-AD| = |BC-CD| = |y-w|$ in the diagram (true because of the inscribed circle), we multiply on both sides to get

$\frac{1}{3}|(AB-AD)(AB^2+AD^2+(AB)(AD))| \le |(BC-CD)(BC^2+CD^2+(BC)(CD))|$

$\frac{1}{3}|AB^3-AD^3| \le |BC^3-CD^3|$

as desired. And the equality occurs whenever we have $AB = AD, \angle BAD = 60^{\circ}, \angle BCD = 120^{\circ}$ for the first inequality or $BC = CD, \angle BCD = 60^{\circ}, \angle BAD = 120^{\circ}$ for the second. QED.

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Comment: A decent USAMO #1. Basically if you saw the factoring and noticed the similarity to the Law of Cosines, it just took a bit to work your way through the inequalities.