2003 USAMO Problems/Problem 2

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.

2003 USAMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6
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2003 USAMO Problems/Problem 2

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