USAMO 2003 Problem 2
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.
Re: USAMO 2003 Problem 2
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 $AC$ and $BD$ of the polygon intersect at a point $P$ within the polygon.Since $ABCD$ is a convex quadrilateral with sides and diagonals of rational length, we consider it differently.
By Cosine law,$\displaystyle{\cos BAC = \frac{AB^2 + CA^2 - BC^2}{2AB \cdot CA}}$, 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 $\displaystyle{\frac{\sin BAC \sin CAD}{\sin^2 CAD}} =\displaystyle{\frac{\sin BAC}{\sin CAD}} = \displaystyle{\frac{\sin BAP}{\sin PAD}}$ is rational. This implies that $\displaystyle{\frac{ AB \sin BAP}{ AD \sin PAD}} = \displaystyle{\frac{\frac{1}{2} AB \cdot AP \sin BAP}{\frac{1}{2} AP \cdot AD \sin PAD}} = \displaystyle{\frac{[BAP]}{[PAD]} = \frac{BP}{PD}}$ is rational. Define $r$ to be equal to $\displaystyle{\frac{BP}{PD}}$. We know that $\displaystyle{\frac{BP}{PD}}$ is rational; hence $r$ is rational. We also have $(1+r)PD = (1+\displaystyle{\frac{BP}{PD}} )PD = PD + BP = BD$, which is, of course, rational. It follows that $BP$ and $PD$ both have rational length, as desired.
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By Cosine law,$\displaystyle{\cos BAC = \frac{AB^2 + CA^2 - BC^2}{2AB \cdot CA}}$, 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 $\displaystyle{\frac{\sin BAC \sin CAD}{\sin^2 CAD}} =\displaystyle{\frac{\sin BAC}{\sin CAD}} = \displaystyle{\frac{\sin BAP}{\sin PAD}}$ is rational. This implies that $\displaystyle{\frac{ AB \sin BAP}{ AD \sin PAD}} = \displaystyle{\frac{\frac{1}{2} AB \cdot AP \sin BAP}{\frac{1}{2} AP \cdot AD \sin PAD}} = \displaystyle{\frac{[BAP]}{[PAD]} = \frac{BP}{PD}}$ is rational. Define $r$ to be equal to $\displaystyle{\frac{BP}{PD}}$. We know that $\displaystyle{\frac{BP}{PD}}$ is rational; hence $r$ is rational. We also have $(1+r)PD = (1+\displaystyle{\frac{BP}{PD}} )PD = PD + BP = BD$, which is, of course, rational. It follows that $BP$ and $PD$ both have rational length, as desired.
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