Difference between revisions of "1999 USAMO Problems/Problem 2"
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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>. | 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>. | ||
| − | 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>.) | + | 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.) |
--[[User:Suli|Suli]] 11:23, 5 October 2014 (EDT) | --[[User:Suli|Suli]] 11:23, 5 October 2014 (EDT) | ||
Latest revision as of 10:11, 18 March 2023
Problem
Let 
be a cyclic quadrilateral. Prove that ![\[|AB - CD| + |AD - BC| \geq 2|AC - BD|.\]](http://latex.artofproblemsolving.com/6/a/7/6a70d0405d1612efa759ecc19ba17ee9ef428354.png)
Solution
Let arc 
of the circumscribed circle (which we assume WLOG has radius 0.5) have value 
, 
have 
, 
have 
, and 
have 
.
Then our inequality reduces to, for 
: ![\[|\sin x - \sin z| + |\sin y - \sin w| \ge 2|\sin (x+y) - \sin (y+z)|.\]](http://latex.artofproblemsolving.com/c/5/5/c55c209aa2df1094cfd01367fc3886d0a4e6510b.png)
This is equivalent to by sum-to-product and use of 
:
![\[|\sin \frac{x-z}{2} \sin \frac{y+w}{2}| + |\sin \frac{y-w}{2} \sin \frac{x+z}{2}| \ge 2|\sin \frac{x-z}{2} \sin \frac{y-w}{2}|.\]](http://latex.artofproblemsolving.com/b/4/0/b40b2fb657b6a6a555fd19baee18f1f6a03c0a71.png)
Clearly 
. As sine is increasing over ![$[0, \pi/2]$](http://latex.artofproblemsolving.com/3/7/2/37278732e6da75cdf01fab8f86657b7b3feb2006.png)
, 
.
Similarly, 
. The result now follows after multiplying the first inequality by 
, the second by 
, and adding. (Equality holds if and only if 
and 
, ie. 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
