Difference between revisions of "1975 IMO Problems/Problem 5"
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− | Determine, with proof, whether or not one can find 1975 points on the circumference | + | ==Problem== |
− | of a circle with unit radius such that the distance between any | + | Determine, with proof, whether or not one can find <math>1975</math> points on the circumference of a circle with unit radius such that the distance between any two of them is a rational number. |
− | two of them is a rational number. | + | |
+ | ==Solution== | ||
+ | Since there are infinitely many primitive Pythagorean triples, there are infinitely many angles <math>\theta</math> s.t. <math>\sin\theta, \cos\theta</math> are both rational. Call such angles good. By angle-sum formulas, if <math>a,b</math> are good, then <math>a+b,a-b</math> are also good. | ||
+ | |||
+ | For points <math>A,B</math> on the circle <math>\omega</math>, let <math>\angle AB</math> be the angle subtended by <math>AB</math>. Now inductively construct points on <math>\omega</math> s.t. all angles formed by them are good; for 1,2 take any good angle. If there are <math>n</math> points chosen, pick a good angle <math>\theta</math> and a marked point <math>A</math> s.t. the point <math>B</math> on <math>\omega</math> with <math>\angle AB=\theta</math> is distinct from the <math>n</math> points. Since there are infinitely many good angles but finitely many marked points, such <math>\theta</math> exists. For a previously marked point <math>P</math> we have <math>\angle BP=\pm \angle AP\pm \angle AB</math> for suitable choices for the two <math>\pm</math>. Since <math>\angle AP ,\angle AB</math> are both good, it follows that <math>\angle BP</math> is good, which finishes induction by adding <math>B</math>. | ||
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+ | Observe that these points for <math>n=1975</math> work: since <math>AB=27sin\angle AB</math> for <math>A,B</math> on the circle, it follows that <math>AB</math> is rational, and so we're done. | ||
+ | |||
+ | The above solution was posted and copyrighted by tobash_co. The original thread for this problem can be found here: [https://aops.com/community/p2603158] | ||
+ | |||
+ | == See Also == {{IMO box|year=1975|num-b=4|num-a=6}} |
Latest revision as of 15:17, 29 January 2021
Problem
Determine, with proof, whether or not one can find points on the circumference of a circle with unit radius such that the distance between any two of them is a rational number.
Solution
Since there are infinitely many primitive Pythagorean triples, there are infinitely many angles s.t. are both rational. Call such angles good. By angle-sum formulas, if are good, then are also good.
For points on the circle , let be the angle subtended by . Now inductively construct points on s.t. all angles formed by them are good; for 1,2 take any good angle. If there are points chosen, pick a good angle and a marked point s.t. the point on with is distinct from the points. Since there are infinitely many good angles but finitely many marked points, such exists. For a previously marked point we have for suitable choices for the two . Since are both good, it follows that is good, which finishes induction by adding .
Observe that these points for work: since for on the circle, it follows that is rational, and so we're done.
The above solution was posted and copyrighted by tobash_co. The original thread for this problem can be found here: [1]
See Also
1975 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |