Existence of Antipodes on a Circle

The following problem was offered at the LXI Moscow Mathematical Olympiad for grade 9 (which is likely to be equivalent to the US grade 11):


200 points have been cosen on a circle, all with integer number of degrees. Prove that the points there are at least one pair of antipodes, i.e., the points 180° apart.


Solution

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny

Solution

200 points have been cosen on a circle, all with integer number of degrees. Prove that the points there are at least one pair of antipodes, i.e., the points 180° apart.


Denote the given set of points A and its generic member a. Let B be the set of antipodes of the points in A, with b as a generic member.

Assume to the contrary that set A contains no antipodes. This assumptions leaves, for 200 members of B, 160 available locations. Since all b's are different (as are the a's), some of b's will spill out of the 160 left-over points, meaning that necessarily c = b for some c∈A and b∈B. But, by the definition, b is an antipode of some a∈A. It follows that both a and c are members of A and are each other's antipodes.


[an error occurred while processing this directive]

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny

[an error occurred while processing this directive]
[an error occurred while processing this directive]