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.$

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.