Difference between revisions of "2002 USAMO Problems/Problem 2"
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with equality only when <math> \frac{(s-a)^2}{36}, \frac{(s-b)^2}{9}, \frac{(s-c)^2}{4}</math> are directly proportional to 36, 9, 4, respectively. Therefore (clearing denominators and taking square roots) our problem requires that <math>(s-a), (s-b), (s-c) </math> be directly proportional to 36, 9, 4, and since <math>a = (s-b) + (s-c) </math> etc., this is equivalent to the condition that <math>a,b,c </math> be in proportion with 13, 40, 45, Q.E.D. | with equality only when <math> \frac{(s-a)^2}{36}, \frac{(s-b)^2}{9}, \frac{(s-c)^2}{4}</math> are directly proportional to 36, 9, 4, respectively. Therefore (clearing denominators and taking square roots) our problem requires that <math>(s-a), (s-b), (s-c) </math> be directly proportional to 36, 9, 4, and since <math>a = (s-b) + (s-c) </math> etc., this is equivalent to the condition that <math>a,b,c </math> be in proportion with 13, 40, 45, Q.E.D. | ||
− | Sidenote: A 13, 40, 45 obtuse triangle has an integer area of 252 and an inradius of 36/ | + | Sidenote: A 13, 40, 45 obtuse triangle has an integer area of 252 and an inradius of <math>\frac{36}{7}</math>. |
Latest revision as of 22:32, 6 April 2016
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
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.
Sidenote: A 13, 40, 45 obtuse triangle has an integer area of 252 and an inradius of .
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 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.