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Difference between revisions of "2002 USAMO Problems/Problem 2"

 
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== Problem ==
 
== Problem ==
  
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Let <math> \displaystyle ABC </math> be a triangle such that
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Let <math>ABC </math> be a triangle such that
 
<center>
 
<center>
 
<math>
 
<math>
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</math>,
 
</math>,
 
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where <math> \displaystyle s </math> and <math> \displaystyle r </math> denote its [[semiperimeter]] and [[inradius]], respectively.  Prove that triangle <math> \displaystyle ABC </math> is similar to a triangle <math> \displaystyle T </math> whose side lengths are all positive integers with no common divisor and determine those integers.
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where <math>s </math> and <math>r </math> denote its [[semiperimeter]] and [[inradius]], respectively.  Prove that triangle <math>ABC </math> is similar to a triangle <math>T </math> whose side lengths are all positive integers with no common divisor and determine those integers.
  
 
== Solution ==
 
== Solution ==
  
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Let <math> \displaystyle a,b,c </math> denote <math> \displaystyle BC, CA, AB </math>, 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
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Let <math>a,b,c </math> denote <math>BC, CA, AB </math>, 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
 
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<math>
 
<math>
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<math>
 
<math>
 
\begin{matrix}
 
\begin{matrix}
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\displaystyle (36 + 9 + 4) \left[ \frac{(s-a)^2}{36} + \frac{(s-b)^2}{9} + \frac{(s-c)^2}{4} \right] & \ge & \displaystyle \left[ (s-a) + (s-b) + (s-c) \right]^2\\
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(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 \\
 
& = & s^2 \\
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\displaystyle \qquad\qquad \quad \quad \frac{(s-a)^2}{36} + \frac{(s-b)^2}{9} + \frac{(s-c)^2}{4} & \ge & \displaystyle \frac{s^2}{36 + 9 + 4} \; ,
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\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}
 
\end{matrix}
 
</math>
 
</math>
 
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</center>
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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> \displaystyle (s-a), (s-b), (s-c) </math> be directly proportional to 36, 9, 4, and since <math> \displaystyle a = (s-b) + (s-c) </math> etc., this is equivalent to the condition that <math> \displaystyle a,b,c </math> be in proportion with 13, 40, 45, Q.E.D.
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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.
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Sidenote: A 13, 40, 45 obtuse triangle has an integer area of 252 and an inradius of <math>\frac{36}{7}</math>.
  
  
 
{{alternate solutions}}
 
{{alternate solutions}}
  
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== Resources ==
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== See also ==
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{{USAMO newbox|year=2002|num-b=1|num-a=3}}
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* [[2002 USAMO Problems]]
 
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* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=337847#p337847 Discussion AoPS/MathLinks]
 
  
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]
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[[Category:Olympiad Inequality Problems]]
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{{MAA Notice}}

Latest revision as of 22:32, 6 April 2016

Problem

Let $ABC$ be a triangle such that

$\left( \cot \frac{A}{2} \right)^2 + \left( 2 \cot \frac{B}{2} \right)^2 + \left( 3 \cot \frac{C}{2} \right)^2 = \left( \frac{6s}{7r} \right)^2$,

where $s$ and $r$ denote its semiperimeter and inradius, respectively. Prove that triangle $ABC$ is similar to a triangle $T$ whose side lengths are all positive integers with no common divisor and determine those integers.

Solution

Let $a,b,c$ denote $BC, CA, AB$, 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

$\left( \frac{s-a}{r} \right)^2 + 4\left( \frac{s-b}{r} \right)^2 + 9\left( \frac{s-c}{r} \right)^2 = \left( \frac{6s}{7r} \right)^2$,

or

$\frac{(s-a)^2}{36} + \frac{(s-b)^2}{9} + \frac{(s-c)^2}{4} = \frac{s^2}{36 + 9 + 4}$.

But by the Cauchy-Schwarz Inequality, we know

$\begin{matrix} (36 + 9 + 4) \left[ \frac{(s-a)^2}{36} + \frac{(s-b)^2}{9} + \frac{(s-c)^2}{4} \right] & \ge &\left[ (s-a) + (s-b) + (s-c) \right]^2\\ & = & s^2 \\ \qquad\qquad \quad \quad \frac{(s-a)^2}{36} + \frac{(s-b)^2}{9} + \frac{(s-c)^2}{4} & \ge &\frac{s^2}{36 + 9 + 4} \; , \end{matrix}$

with equality only when $\frac{(s-a)^2}{36}, \frac{(s-b)^2}{9}, \frac{(s-c)^2}{4}$ are directly proportional to 36, 9, 4, respectively. Therefore (clearing denominators and taking square roots) our problem requires that $(s-a), (s-b), (s-c)$ be directly proportional to 36, 9, 4, and since $a = (s-b) + (s-c)$ etc., this is equivalent to the condition that $a,b,c$ 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 $\frac{36}{7}$.


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

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