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Difference between revisions of "1999 USAMO Problems/Problem 2"

(Created page with "== Problem == Let <math>ABCD</math> be a cyclic quadrilateral. Prove that <cmath> |AB - CD| + |AD - BC| \geq 2|AC - BD|. </cmath> == Solution == {{solution}} == See Also == {{...")
 
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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} \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}|.</cmath>
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Clearly <math>90^\circ \ge \frac{x+z}{2} > \frac{x-z}{2} \ge 0^\circ</math>. As sine is increasing over <math>[0, \pi/2]</math>, <math>|\sin \frac{x+z}{2}| > |\sin \frac{x-z}{2}|</math>.
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Similarly, <math>|\sin \frac{y+w}{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>, ie. <math>ABCD</math> is a parallelogram.)
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--[[User:Suli|Suli]] 11:23, 5 October 2014 (EDT)
  
 
== See Also ==
 
== See Also ==
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[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 10:11, 18 March 2023

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)

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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