62nd Belarusian Mathematical Olympiad

Question

N boys (N 3), no two of them having the same height, are arranged along a circle. A boy in the given arrangement is said to be *middle* if he is taller than one of his neighbors and shorter than the other one. Prove that if N is odd number then there exists at least one middle boy in any arrangement of the boys along the circle. FIGURE_0

Accepted Answer

Consider arbitrary arrangement of the boys along the circle. We put FIGURE_0 the signs "+" or "-" before any boy in accordance with the following rule: we move clockwise along the circle and put the sign "+" before the boy if he is taller than the previous boy and we put the sign "-" if he is shorter than the previous one (see the fig.). It is evident that the boy is middle if and only if both the signs before and after him are either "+" or "-". If there is no middle boy in the arrangement, then the signs "+" and "-" are alternate, so the total number of the signs (equaled the number of the boys) is even, contrary to the condition. Therefore, there is at least one middle boy in any arrangement.