1975 IMO Problems/Problem 5


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.


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]

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