Difference between revisions of "2000 USAMO Problems/Problem 2"
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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>. <math>\blacksquare</math> | ||
− | == See | + | == See Also == |
{{USAMO newbox|year=2000|num-b=1|num-a=3}} | {{USAMO newbox|year=2000|num-b=1|num-a=3}} | ||
[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Revision as of 07:24, 16 September 2012
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 |