Difference between revisions of "2003 USAMO Problems/Problem 2"
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== Solution == | == Solution == | ||
− | When <math> \mathcal{P} </math> 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 <math> | + | When <math> \mathcal{P} </math> 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 <math>P </math> within the polygon be <math>AC </math> and <math>BD </math>. Since <math>ABCD </math> is a convex quadrilateral with sides and diagonals of rational length, we consider it in isolation. |
− | By the [[Law of Cosines]], <math> \cos BAC = \frac{AB^2 + CA^2 - BC^2}{2AB \cdot CA} </math>, which is rational. Similarly, <math> | + | By the [[Law of Cosines]], <math> \cos BAC = \frac{AB^2 + CA^2 - BC^2}{2AB \cdot CA} </math>, which is rational. Similarly, <math>\cos CAD </math> is rational, as well as <math>\cos BAD = \cos (BAC + CAD) = \cos BAC \cos CAD - \sin BAC \sin CAD </math>. It follows that <math>\sin BAC \sin CAD </math> is rational. Since <math>\sin^2 CAD = 1 - \cos^2 CAD </math> is rational, this means that <math> \frac{\sin BAC \sin CAD}{\sin^2 CAD} = \frac{\sin BAC}{\sin CAD} = \frac{\sin BAP}{\sin PAD} </math> is rational. This implies that <math> \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} </math> is rational. Define <math>r</math> to be equal to <math>\frac{BP}{PD} </math>. We know that <math>\frac{BP}{PD} </math> is rational; hence <math>r</math> is rational. We also have <math>(1+r)PD = (1+\frac{BP}{PD} )PD = PD + BP = BD </math>, which is, of course, rational. It follows that <math>BP </math> and <math>PD </math> both have rational length, as desired. |
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− | + | {{alternate solutions}} | |
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+ | == See also == | ||
+ | {{USAMO newbox|year=2003|num-b=1|num-a=3}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:50, 29 April 2014
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. Define to be equal to . We know that is rational; hence is rational. We also have , 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.