1975 IMO Problems/Problem 5: Difference between revisions

Latest revision as of 15:17, 29 January 2021

Problem

Determine, with proof, whether or not one can find $1975$ 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 $\theta$ s.t. $\sin\theta, \cos\theta$ are both rational. Call such angles good. By angle-sum formulas, if $a,b$ are good, then $a+b,a-b$ are also good.

For points $A,B$ on the circle $\omega$ , let $\angle AB$ be the angle subtended by $AB$ . Now inductively construct points on $\omega$ s.t. all angles formed by them are good; for 1,2 take any good angle. If there are $n$ points chosen, pick a good angle $\theta$ and a marked point $A$ s.t. the point $B$ on $\omega$ with $\angle AB=\theta$ is distinct from the $n$ points. Since there are infinitely many good angles but finitely many marked points, such $\theta$ exists. For a previously marked point $P$ we have $\angle BP=\pm \angle AP\pm \angle AB$ for suitable choices for the two $\pm$ . Since $\angle AP ,\angle AB$ are both good, it follows that $\angle BP$ is good, which finishes induction by adding $B$ .

Observe that these points for $n=1975$ work: since $AB=27sin\angle AB$ for $A,B$ on the circle, it follows that $AB$ 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]

@@ Line 1: / Line 1: @@
-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>.
+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}}

1975 IMO (Problems) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
All IMO Problems and Solutions

Page

Toolbox

Search

1975 IMO Problems/Problem 5: Difference between revisions

Latest revision as of 15:17, 29 January 2021

Problem

Solution

See Also