2003 USAMO Problems/Problem 2
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.
See also
2003 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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