Difference between revisions of "1999 USAMO Problems/Problem 2"
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== Problem == | == Problem == | ||
Let <math>ABCD</math> be a cyclic quadrilateral. Prove that <cmath> |AB - CD| + |AD - BC| \geq 2|AC - BD|. </cmath> | Let <math>ABCD</math> be a cyclic quadrilateral. Prove that <cmath> |AB - CD| + |AD - BC| \geq 2|AC - BD|. </cmath> | ||
+ | |||
+ | ==Hint== | ||
+ | Use trigonometry. Bashing is helpful for this problem. | ||
+ | |||
== Solution == | == Solution == |
Revision as of 10:28, 5 October 2014
Contents
[hide]Problem
Let be a cyclic quadrilateral. Prove that
Hint
Use trigonometry. Bashing is helpful for this problem.
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 . As sine is increasing over , .
Similarly, . 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 |
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