MathDB

Let

n>1

be an integer. Suppose we are given

2n

points in the plane such that no three of them are collinear. The points are to be labelled

A_1, A_2, \dots , A_{2n}

in some order. We then consider the

2n

angles

\angle A_1A_2A_3, \angle A_2A_3A_4, \dots , \angle A_{2n-2}A_{2n-1}A_{2n}, \angle A_{2n-1}A_{2n}A_1, \angle A_{2n}A_1A_2

. We measure each angle in the way that gives the smallest positive value (i.e. between

0^{\circ}

and

180^{\circ}

). Prove that there exists an ordering of the given points such that the resulting

2n

angles can be separated into two groups with the sum of one group of angles equal to the sum of the other group.

Problem Statement