2003 USAMO Problems/Problem 2

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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 $\displaystyle P$ within the polygon be $\displaystyle AC$ and $\displaystyle BD$. Since $\displaystyle ABCD$ 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, $\displaystyle \cos CAD$ is rational, as well as $\displaystyle \cos BAD = \cos (BAC + CAD) = \cos BAC \cos CAD - \sin BAC \sin CAD$. It follows that $\displaystyle \sin BAC \sin CAD$ is rational. Since $\displaystyle \sin^2 CAD = 1 - \cos^2 CAD$ 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 $\displaystyle r$, $\displaystyle (1+r)BP = BP + PD = BD$, which is, of course, rational. It follows that $\displaystyle BP$ and $\displaystyle PD$ 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.


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