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

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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> \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. | 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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== Resources == | == Resources == |

## Revision as of 06:33, 22 April 2007

## Problem

A convex polygon 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 are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers.

## Solution

When 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 within the polygon be and . Since is a convex quadrilateral with sides and diagonals of rational length, we consider it in isolation.

By the Law of Cosines, , which is rational. Similarly, is rational, as well as . It follows that is rational. Since is rational, this means that is rational. This implies that is rational. This means that for some rational number , , which is, of course, rational. It follows that and 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.*