Difference between revisions of "2000 USAMO Problems/Problem 2"
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We let <math>x = AP = s - a, y = BQ = s-b, z = CR = s-c</math>, and [[without loss of generality]] let <math>x \le y \le z</math>. Then <math>x + y + z = 3s - (a+b+c) = s</math>, so <math>r = \frac {A}{s} = \frac{\sqrt{(x+y+z)xyz}}{x+y+z} = \sqrt{\frac{xyz}{x+y+z}}</math>. Thus, | We let <math>x = AP = s - a, y = BQ = s-b, z = CR = s-c</math>, and [[without loss of generality]] let <math>x \le y \le z</math>. Then <math>x + y + z = 3s - (a+b+c) = s</math>, so <math>r = \frac {A}{s} = \frac{\sqrt{(x+y+z)xyz}}{x+y+z} = \sqrt{\frac{xyz}{x+y+z}}</math>. Thus, | ||
− | <cmath>6\sqrt{\frac | + | <cmath>6\sqrt{\frac{x+y+z}{xyz}} = \frac{2yz + 5xy + 5xz}{xyz}</cmath> |
Squaring yields | Squaring yields |
Revision as of 20:17, 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 . 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.