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)

1999 USAMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
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1999 USAMO Problems/Problem 2

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