Difference between revisions of "2003 USAMO Problems/Problem 2"

 
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== Solution ==
 
== Solution ==
  
When <math> \mathcal{P} </math> is a triangle, the problem is trivial.  Otherwise, it is sufficient to prove that any two diagonals of the polygon cut each other into rational lengths.  Let two diagonals which intersect at a point <math> \displaystyle P </math> within the polygon be <math> \displaystyle AC </math> and <math> \displaystyle BD </math>.  Since <math> \displaystyle ABCD </math> is a convex quadrilateral with sides and diagonals of rational length, we consider it in isolation.
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When <math> \mathcal{P} </math> is a triangle, the problem is trivial.  Otherwise, it is sufficient to prove that any two diagonals of the polygon cut each other into rational lengths.  Let two diagonals which intersect at a point <math>P </math> within the polygon be <math>AC </math> and <math>BD </math>.  Since <math>ABCD </math> is a convex quadrilateral with sides and diagonals of rational length, we consider it in isolation.
  
By the [[Law of Cosines]], <math> \cos BAC = \frac{AB^2 + CA^2 - BC^2}{2AB \cdot CA} </math>, which is rational.  Similarly, <math> \displaystyle \cos CAD </math> is rational, as well as <math> \displaystyle \cos BAD = \cos (BAC + CAD) = \cos BAC \cos CAD - \sin BAC \sin CAD </math>.  It follows that <math> \displaystyle \sin BAC \sin CAD </math> is rational.  Since <math> \displaystyle \sin^2 CAD = 1 - \cos^2 CAD </math> is rational, this means that <math> \frac{\sin BAC \sin CAD}{\sin^2 CAD} = \frac{\sin BAC}{\sin CAD} = \frac{\sin BAP}{\sin PAD} </math> is rational.  This implies that <math> \frac{ AB \sin BAP}{ AD \sin PAD} = \frac{\frac{1}{2} AB \cdot AP \sin BAP}{\frac{1}{2} AP \cdot AD \sin PAD} = \frac{[BAP]}{[PAD]} = \frac{BP}{PD} </math> is rational.  This means that for some rational number <math> \displaystyle r </math><math> \displaystyle (1+r)BP = BP + PD = BD </math>, which is, of course, rational.  It follows that <math> \displaystyle BP </math> and <math> \displaystyle PD </math> both have rational length, as desired.
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By the [[Law of Cosines]], <math> \cos BAC = \frac{AB^2 + CA^2 - BC^2}{2AB \cdot CA} </math>, which is rational.  Similarly, <math>\cos CAD </math> is rational, as well as <math>\cos BAD = \cos (BAC + CAD) = \cos BAC \cos CAD - \sin BAC \sin CAD </math>.  It follows that <math>\sin BAC \sin CAD </math> is rational.  Since <math>\sin^2 CAD = 1 - \cos^2 CAD </math> is rational, this means that <math> \frac{\sin BAC \sin CAD}{\sin^2 CAD} = \frac{\sin BAC}{\sin CAD} = \frac{\sin BAP}{\sin PAD} </math> is rational.  This implies that <math> \frac{ AB \sin BAP}{ AD \sin PAD} = \frac{\frac{1}{2} AB \cdot AP \sin BAP}{\frac{1}{2} AP \cdot AD \sin PAD} = \frac{[BAP]}{[PAD]} = \frac{BP}{PD} </math> is rational.  Define <math>r</math> to be equal to <math>\frac{BP}{PD} </math>. We know that <math>\frac{BP}{PD} </math> is rational; hence <math>r</math> is rational. We also have <math>(1+r)PD = (1+\frac{BP}{PD} )PD = PD + BP = BD </math>, which is, of course, rational.  It follows that <math>BP </math> and <math>PD </math> both have rational length, as desired.
  
== Resources ==
 
  
* [[2003 USAMO Problems]]
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{{alternate solutions}}
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=336193#p336193 Discussion on AoPS/MathLinks]
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== See also ==
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{{USAMO newbox|year=2003|num-b=1|num-a=3}}
  
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 20:50, 29 April 2014

Problem

A convex polygon $\mathcal{P}$ in the plane is dissected into smaller convex polygons by drawing all of its diagonals. The lengths of all sides and all diagonals of the polygon $\mathcal{P}$ are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers.

Solution

When $\mathcal{P}$ is a triangle, the problem is trivial. Otherwise, it is sufficient to prove that any two diagonals of the polygon cut each other into rational lengths. Let two diagonals which intersect at a point $P$ within the polygon be $AC$ and $BD$. Since $ABCD$ is a convex quadrilateral with sides and diagonals of rational length, we consider it in isolation.

By the Law of Cosines, $\cos BAC = \frac{AB^2 + CA^2 - BC^2}{2AB \cdot CA}$, which is rational. Similarly, $\cos CAD$ is rational, as well as $\cos BAD = \cos (BAC + CAD) = \cos BAC \cos CAD - \sin BAC \sin CAD$. It follows that $\sin BAC \sin CAD$ is rational. Since $\sin^2 CAD = 1 - \cos^2 CAD$ is rational, this means that $\frac{\sin BAC \sin CAD}{\sin^2 CAD} = \frac{\sin BAC}{\sin CAD} = \frac{\sin BAP}{\sin PAD}$ is rational. This implies that $\frac{ AB \sin BAP}{ AD \sin PAD} = \frac{\frac{1}{2} AB \cdot AP \sin BAP}{\frac{1}{2} AP \cdot AD \sin PAD} = \frac{[BAP]}{[PAD]} = \frac{BP}{PD}$ is rational. Define $r$ to be equal to $\frac{BP}{PD}$. We know that $\frac{BP}{PD}$ is rational; hence $r$ is rational. We also have $(1+r)PD = (1+\frac{BP}{PD} )PD = PD + BP = BD$, which is, of course, rational. It follows that $BP$ and $PD$ both have rational length, as desired.


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See also

2003 USAMO (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6
All USAMO Problems and Solutions

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