AoPS - Problem

Source: Ukraine 2005 grade 9

Tags: inequalities, triangle inequality, geometry unsolved, geometry

moldovan

26.07.2009 18:30

On the plane are given $n \ge 3$ points, not all on the same line. For any point $M$ on the same plane, $f(M)$ is defined to be the sum of the distances from $M$ to these $n$ points. Suppose that there is a point $M_1$ such that $f(M_1)\le f(M)$ for any point $M$ on the plane. Prove that if a point $M_2$ satisfies $f(M_1)=f(M_2),$ then $M_1 \equiv M_2.$

Erken

26.07.2009 19:08

as a very brutal force you may use the lemma from here http://www.mathlinks.ro/viewtopic.php?p=1306273#1306273

Mithril

08.08.2009 19:22

Let $P_1,P_2,...,P_n$ the points given. Suppose that $M_1 \neq M_2$ . Let $M$ be the midpoint of $M_1M_2$ . For all $P_i$ , consider the point $X_i$ such that $P_iM_1X_iM_2$ is a parallelogram. This gives $MP_i = \frac {P_iX}2 \leq \frac {M_1P_i + M_2P_i}2$ , by triangle inequality, with equality iff $P_i$ , $M_1$ and $M_2$ are collinear. Then $f(M)\leq\frac{M_1P_1 + M_2P_2}2+\frac{M_1P_2+M_2P_2}2+ ... +\frac { M_1P_n + M_2P_n}2$ . That gives $f(M) \leq \frac {f(M_1) + f(M_2)}2 = f(M_1) \leq f(M)$ . Then all the inequalities are equalities, and this implies that all the points given are collinear, absurd.