2003 USAMO Problems/Problem 2

Revision as of 20:47, 29 April 2014 by Mathcool2009 (talk | contribs) (Solution)


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.


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. This means that for some rational number $r$, $(1+r)PD = BP + PD = BD$, 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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