2000 USAMO Problems/Problem 2
Contents
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 and simplifying yields (after much grueling work)
We claim that the inequality
holds true, with equality iff .
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 .
Then , and yields . Thus, we have proved that all possible ABC are isosceles (as b = c), and that they are similar to a 5-5-8 triangle.
Solution 2
Let , , be the three angles of the triangle, then , , and . Without loss of generality, assume , then . Denote , , , then we have, Meanwhile, Substitute in , we get Treating (2) as a quadratic equation for , the discriminant is: For , to be both real numbers, must not be negative, so , which yields . Solving for x, we get . Using the formula , we have . Therefore . This proves that all possible triangle are isosceles with the same set of angles, i.e., they are similar to each other.
It's inconsequential to the proof, but it can be easily found that, since , the altitude to the base of the isosceles triangle divides the triangle into two 3-4-5 right triangles with on the base; therefore, the entire triangle is 5-5-8. .
Solution by .
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 |
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