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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.
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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>.
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==Solution 2 (Lagrange Multipliers)==
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Let <math>D, E, F</math> be the projections of the incenter onto the sides <math>BC</math>, <math>AC</math>, and <math>AB</math>, respectively, and let <math>x=AF . y=BD, z=CE</math>. We thus have
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<cmath> \cot \frac{A}{2} = \frac{x}{r} , \cot \frac{B}{2} = \frac{y}{r} , \cot \frac{C}{2} = \frac{z}{r}</cmath>
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and additionally,
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<cmath>x+y+z = s</cmath>
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<cmath>x^2 + 4 y^2 + 9 z^2 = (\frac{6s}{7})^2</cmath>
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We first determine the side lengths of the triangle similar to this triangle with semiperimeter <math>1</math>. It is asked of us to show that such a triangle is unique, but we note that this is true if and only if the plane <math>x+y+z=1</math> and the ellipsoid <math>x^2 + 4 y^2 + 9 z^2 = (\frac{6}{7})^2</math> is tangent. This can be shown if we use the method of Lagrange Multipliers on the constraint <math>x+y+z=1</math> and function <math>x^2 + 4 y^2 + 9 z^2</math> to prove that <math>(\frac{6}{7})^2</math> is a extrema.
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The Method of Lagrange Multipliers gives the system of
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<cmath>2x= \lambda </cmath>
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<cmath>8y= \lambda</cmath>
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<cmath>18z= \lambda</cmath>
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<cmath>x+y+z=1</cmath>
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from which we easily obtain <math>(x,y,z)=(\frac{36}{49} , \frac{9}{49} , \frac{4}{49} )</math>. We calculate that this indeed satisfy <math>x^2 + 4 y^2 + 9 z^2 = (\frac{6}{7})^2</math>, so <math>(\frac{6}{7})^2</math> is indeed an relative extrema. To check that it is a global extrema, one notes that as x, y, or z goes towards plus or minus infinity, the function <math>x^2 + 4 y^2 + 9 z^2</math> goes to positive infinity, so that <math>(\frac{6}{7})^2</math> is the global minima. Now, <math>AB=x+y=\frac{45}{49}</math>, <math>AC=x+z=\frac{40}{49}</math>, and <math>BC=y+z=\frac{13}{49}</math>, so we see that the desired side lengths are <math>AB=45, AC=40, BC=13</math>. <math>\square</math>.
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~[[Ddk001]]
  
  
 
{{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]]
 
[[Category:Olympiad Inequality Problems]]
 
[[Category:Olympiad Inequality Problems]]
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{{MAA Notice}}

Latest revision as of 20:25, 6 June 2025

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 $\frac{36}{7}$.

Solution 2 (Lagrange Multipliers)

Let $D, E, F$ be the projections of the incenter onto the sides $BC$, $AC$, and $AB$, respectively, and let $x=AF . y=BD, z=CE$. We thus have

\[\cot \frac{A}{2} = \frac{x}{r} , \cot \frac{B}{2} = \frac{y}{r} , \cot \frac{C}{2} = \frac{z}{r}\]

and additionally,

\[x+y+z = s\] \[x^2 + 4 y^2 + 9 z^2 = (\frac{6s}{7})^2\]

We first determine the side lengths of the triangle similar to this triangle with semiperimeter $1$. It is asked of us to show that such a triangle is unique, but we note that this is true if and only if the plane $x+y+z=1$ and the ellipsoid $x^2 + 4 y^2 + 9 z^2 = (\frac{6}{7})^2$ is tangent. This can be shown if we use the method of Lagrange Multipliers on the constraint $x+y+z=1$ and function $x^2 + 4 y^2 + 9 z^2$ to prove that $(\frac{6}{7})^2$ is a extrema.

The Method of Lagrange Multipliers gives the system of

\[2x= \lambda\] \[8y= \lambda\] \[18z= \lambda\] \[x+y+z=1\]

from which we easily obtain $(x,y,z)=(\frac{36}{49} , \frac{9}{49} , \frac{4}{49} )$. We calculate that this indeed satisfy $x^2 + 4 y^2 + 9 z^2 = (\frac{6}{7})^2$, so $(\frac{6}{7})^2$ is indeed an relative extrema. To check that it is a global extrema, one notes that as x, y, or z goes towards plus or minus infinity, the function $x^2 + 4 y^2 + 9 z^2$ goes to positive infinity, so that $(\frac{6}{7})^2$ is the global minima. Now, $AB=x+y=\frac{45}{49}$, $AC=x+z=\frac{40}{49}$, and $BC=y+z=\frac{13}{49}$, so we see that the desired side lengths are $AB=45, AC=40, BC=13$. $\square$.

~Ddk001


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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