Difference between revisions of "1999 USAMO Problems/Problem 2"
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== Solution == | == Solution == | ||
− | {{ | + | 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>. |
+ | 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> | ||
+ | |||
+ | This is equivalent to by sum-to-product and use of <math>\cos x = \sin (90^\circ - x)</math>: | ||
+ | |||
+ | <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> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>.) | ||
+ | |||
+ | --[[User:Suli|Suli]] 11:23, 5 October 2014 (EDT) | ||
== See Also == | == See Also == |
Revision as of 11:23, 5 October 2014
Problem
Let be a cyclic quadrilateral. Prove that
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
:
This is equivalent to by sum-to-product and use of :
Clearly , so
. Because
and
, their cosine and sine are respectively non-negative, and thus, as cosine is decreasing over
,
.
Similarly, we have as that
. The result now follows after multiplying the first inequality by
, the second by
, and adding. (Equality holds if and only if
and
.)
--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.