Difference between revisions of "2002 USAMO Problems/Problem 2"
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
[[Category:Olympiad Inequality Problems]] | [[Category:Olympiad Inequality Problems]] | ||
Revision as of 20:48, 6 April 2013
Problem
Let 
be a triangle such that
,
where 
and 
denote its semiperimeter and inradius, respectively. Prove that triangle is similar to a triangle 
whose side lengths are all positive integers with no common divisor and determine those integers.
Solution
Let 
denote 
, respectively. Since the line from a triangle's incenter to one of its vertices bisects the angle at the triangle's vertex, the condition of the problem is equivalent to the
,
or
.
But by the Cauchy-Schwarz Inequality, we know
![$\begin{matrix} (36 + 9 + 4) \left[ \frac{(s-a)^2}{36} + \frac{(s-b)^2}{9} + \frac{(s-c)^2}{4} \right] & \ge &\left[ (s-a) + (s-b) + (s-c) \right]^2\\ & = & s^2 \\ \qquad\qquad \quad \quad \frac{(s-a)^2}{36} + \frac{(s-b)^2}{9} + \frac{(s-c)^2}{4} & \ge &\frac{s^2}{36 + 9 + 4} \; , \end{matrix}$](http://latex.artofproblemsolving.com/8/f/f/8ff3702c4f3caf1928f0f6804aa07aebea95134a.png)
with equality only when 
are directly proportional to 36, 9, 4, respectively. Therefore (clearing denominators and taking square roots) our problem requires that 
be directly proportional to 36, 9, 4, and since 
etc., this is equivalent to the condition that be in proportion with 13, 40, 45, Q.E.D.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See also
| 2002 USAMO (Problems • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAMO Problems and Solutions | ||
