Difference between revisions of "2000 USAMO Problems/Problem 2"
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− | holds true, with equality [[iff]] <math>4x = y = z</math>. Then <math>s = x+y+z = 9x</math>, and <math>s-a = x, s-b = 4x, s-c = 4x</math> yields <math>\boxed{a:b:c = 5: | + | holds true, with equality [[iff]] <math>4x = y = z</math>. Then <math>s = x+y+z = 9x</math>, and <math>s-a = x, s-b = 4x, s-c = 4x</math> yields <math>\boxed{a:b:c = 8:5:5}</math>. |
Revision as of 19:30, 8 April 2013
Problem
Let be the set of all triangles for which
where is the inradius and are the points of tangency of the incircle with sides respectively. Prove that all triangles in are isosceles and similar to one another.
Solution
We let , and without loss of generality let . Then , so . Thus,
Squaring yields
We claim that the inequality
holds true, with equality iff . Then , and yields .
Note that is homogeneous in , so without loss of generality, scale so that . Then
which is a quadratic in . As , it suffices to show that the quadratic cannot have more than one root, or the discriminant . Then,
as desired. Equality comes when ; since is symmetric in and , it follows that is also necessary for equality. Reversing our scaling, it follows that .
See Also
2000 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |