Problem:
Let be an irrational number with , and draw a circle in the plane whose circumference has length . Given any integer , define a sequence of points , as follows. First select any point on the circle, and for define as the point on the circle for which the length of arc is , when travelling counterclockwise around the circle from to . Suppose that and are the nearest adjacent points on either side of . Prove that .
Solution:
Observe that since is irrational no two of the points will coincide. It will be useful to define the auxiliary point such that the length of arc is , when travelling counter-clockwise around the circle from to . We begin by noting that for any , if then lies on the arc from to containing . For if we travel back (clockwise) around the circle through a distance of from then we reach . The same translation must map to , and since is situated between and , we deduce that must be also.
The claim is clearly true for . Now suppose to the contrary that for some value of we have and consider the minimal such counterexample. If in fact , then we may translate the three points , and clockwise around the circle through a distance to find points and adjacent to on either side. But then we would have for this trio of points, which contradicts our assumption that was the minimal counterexample.
Therefore we must have . Again we translate points , and clockwise around the circle through a distance to obtain points and adjacent to on either side with . By our earlier observation this implies that lies on the arc from to containing . But now translating forward again, we conclude that lies on the arc from to containing , contradicting the fact that and were the nearest adjacent points to on either side. This completes the proof.
This problem and solution were suggested by Sam Vandervelde.
The problems on this page are the property of the MAA's American Mathematics Competitions