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Difference between revisions of "1999 USAMO Problems/Problem 2"
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== Solution ==
 
== Solution ==
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{{solution}}
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Let arc <math>AB</math> of the circumscribed circle (which we assume WLOG has radius 0.5) have value <math>2x</math>, <math>BC</math> have <math>2y</math>, <math>CD</math> have <math>2z</math>, and <math>DA</math> have <math>2w</math>.
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Then our inequality reduces to, for <math>x+y+z+w = 180^\circ</math>: <cmath> |\sin x - \sin z| + |\sin y - \sin w| \ge 2|\sin (x+y) - \sin (y+z)|.</cmath>
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This is equivalent to by sum-to-product and use of <math>\cos x = \sin (90^\circ - x)</math>:
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<cmath>|\sin \frac{x-z}{2} \cos \frac{x+z}{2}| + |\sin \frac{y-w}{2} \cos \frac{y+w}{2}| \ge 2|\sin \frac{x-z}{2} \sin \frac{y-w}{2}|.</cmath>
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Clearly <math>y + w + x - z < 180^\circ</math>, so <math>\frac{y+w}{2} < 90^\circ - \frac{x-z}{2}</math>. Because <math>\frac{y+w}{2} \le 90^\circ</math> and <math>\frac{x-z}{2} \le 90^\circ</math>, their cosine and sine are respectively non-negative, and thus, as cosine is decreasing over <math>(0, \pi/2)</math>, <math>|\cos \frac{y+w}{2}| > |\sin \frac{x-z}{2}|</math>.
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Similarly, we have as <math>x + z + y - w < 180^\circ</math> that <math>|\cos \frac{x+z}{2}| > |\sin \frac{y-w}{2}|</math>. The result now follows after multiplying the first inequality by <math>|\sin \frac{x-z}{2}|</math>, the second by <math>|\sin \frac{y-w}{2}|</math>, and adding. (Equality holds if and only if <math>x=z</math> and <math>y=w</math>.)
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--[[User:Suli|Suli]] 11:23, 5 October 2014 (EDT)
  
 
== See Also ==
 
== See Also ==

Revision as of 10:23, 5 October 2014

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 $\cos x = \sin (90^\circ - x)$:

\[|\sin \frac{x-z}{2} \cos \frac{x+z}{2}| + |\sin \frac{y-w}{2} \cos \frac{y+w}{2}| \ge 2|\sin \frac{x-z}{2} \sin \frac{y-w}{2}|.\]

Clearly $y + w + x - z < 180^\circ$, so $\frac{y+w}{2} < 90^\circ - \frac{x-z}{2}$. Because $\frac{y+w}{2} \le 90^\circ$ and $\frac{x-z}{2} \le 90^\circ$, their cosine and sine are respectively non-negative, and thus, as cosine is decreasing over $(0, \pi/2)$, $|\cos \frac{y+w}{2}| > |\sin \frac{x-z}{2}|$.

Similarly, we have as $x + z + y - w < 180^\circ$ that $|\cos \frac{x+z}{2}| > |\sin \frac{y-w}{2}|$. 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 $y=w$.)

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

See Also

1999 USAMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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