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Ukrainian Mathematical Olympiad, Final Round

Geometry Difficulty 5.6 AIME, harder Prove it Ukraine

We are given $n$ ( $n \ge 3$ ) points in the plane not all lying on the same line. For an arbitrary point $M$ we define $f(M)$ to be the sum of the distances from these $n$ points to $A$ . It is known that there exists a point $M_1$ such that for every point $M$ of the plane the inequality $f(M_1) \le f(M)$ holds. Let $M_2$ be a point such that $f(M_1) = f(M_2)$ . Prove that the points $M_1$ and $M_2$ coincide.

Solution

It is easy to prove that if a point $M$ is the midpoint of a segment $BC$ , then for any point $A$ the inequality $AM \le \frac{1}{2}(AB + AC)$ holds. Suppose that the points $M_1$ and $M_2$ mentioned in the problem statement do not coincide. Let $M_3$ be the midpoint of the segment $M_1M_2$ . Then $A_kM_3 \le \frac{1}{2}(A_kM_1 + A_kM_2)$ , $1 \le k \le n$ , and at least one of these inequalities is strict (since the points $A_1, A_2, \dots, A_n$ do not all lie on the same line). Adding these $n$ inequalities, we obtain
$f(M_3) < \frac{1}{2}(f(M_1) + f(M_2)) = f(M_1).$
This is a contradiction.

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