Difference between revisions of "2000 USAMO Problems/Problem 2"
(→Solution) |
(→Solution) |
||
Line 19: | Line 19: | ||
<cmath> | <cmath> | ||
− | holds true, with equality [[iff]] <math>4x = y = z | + | holds true, with equality [[iff]] <math>4x = y = z</math>. |
Line 30: | Line 30: | ||
<cmath>\Delta = z^2(14-16z)^2 - 4(4z^2 - 16z + 25)(25z^2) = -144(z-4)^2 \le 0</cmath> | <cmath>\Delta = z^2(14-16z)^2 - 4(4z^2 - 16z + 25)(25z^2) = -144(z-4)^2 \le 0</cmath> | ||
− | as desired. Equality comes when <math>z = 4</math>; since <math>(*)</math> is symmetric in <math>y</math> and <math>z</math>, it follows that <math>y = 4</math> is also necessary for equality. Reversing our scaling, it follows that <math>x:y:z = 1:4:4</math>. <math>\blacksquare</math> | + | as desired. Equality comes when <math>z = 4</math>; since <math>(*)</math> is symmetric in <math>y</math> and <math>z</math>, it follows that <math>y = 4</math> is also necessary for equality. Reversing our scaling, it follows that <math>x:y:z = 1:4:4</math>. |
+ | |||
+ | Then <math>s = x+y+z = 9x</math>, and <math>s-a = x, s-b = 4x, s-c = 4x</math> yields <math>a:b:c = 8:5:5</math>. Thus, we have proved that all possible ABC are isosceles (as b = c), and that they are similar to a 5-5-8 triangle. | ||
+ | <math>\blacksquare</math> | ||
== See Also == | == See Also == |
Revision as of 20:21, 18 April 2014
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 .
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.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.