Any partition corresponds to the partition of a regular polygon into triangles , each of which has angles (measured in degrees) . We may number the vertices in such a way so that corresponds to in some permutation. WLOG, we may assume that in that order. One of the remaining triples must contain two numbers in the interval : it follows that there exists a triplet .

Each other triple contain one number from each interval , . Hence the mapping yields an appropriate partition of . This is clearly impossible for , so a simple induction implies that it is impossible for all odd , but if , then the triples and , for all , satisfy the requirements.