# 1999 USAMO Problems/Problem 2

## Problem

Let $ABCD$ be a cyclic quadrilateral. Prove that $$|AB - CD| + |AD - BC| \geq 2|AC - BD|.$$

## Solution

Let arc $AB$ of the circumscribed circle (which we assume WLOG has radius 0.5) have value $2x$, $BC$ have $2y$, $CD$ have $2z$, and $DA$ have $2w$. Then our inequality reduces to, for $x+y+z+w = 180^\circ$: $$|\sin x - \sin z| + |\sin y - \sin w| \ge 2|\sin (x+y) - \sin (y+z)|.$$

This is equivalent to by sum-to-product and use of $\cos x = \sin (90^\circ - x)$: $$|\sin \frac{x-z}{2} \sin \frac{y+w}{2}| + |\sin \frac{y-w}{2} \sin \frac{x+z}{2}| \ge 2|\sin \frac{x-z}{2} \sin \frac{y-w}{2}|.$$

Clearly $90^\circ \ge \frac{x+z}{2} > \frac{x-z}{2} \ge 0^\circ$. As sine is increasing over $[0, \pi/2]$, $|\sin \frac{x+z}{2}| > |\sin \frac{x-z}{2}|$.

Similarly, $|\sin \frac{y+w}{2}| > |\sin \frac{y-w}{2}|$. The result now follows after multiplying the first inequality by $|\sin \frac{x-z}{2}|$, the second by $|\sin \frac{y-w}{2}|$, and adding. (Equality holds if and only if $x=z$ and $y=w$, ie. $ABCD$ is a parallelogram.)

--Suli 11:23, 5 October 2014 (EDT)

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