Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.1 National olympiad Prove it

59. Position in space. Let us be given a plane $E$ and three non-collinear points $A, B, C$ , located on one side of $E$ and lying in a plane intersecting with $E$ . Take three arbitrary points $A^{\prime}, B^{\prime}$ and $C^{\prime}$ on the plane $E$ . Denote by $L, M, N$ the midpoints of the segments $A A^{\prime}, B B^{\prime}, C C^{\prime}$ , and by $S$ the centroid of the triangle $L M N$ . Suppose now that the points $A^{\prime}, B^{\prime}$ and $C^{\prime}$ move independently of each other on the plane $E$ . Where can the point $S$ end up in this case?

Solution

59. Place a particle of unit mass at each of the points $A, B, C, A^{\prime}, B^{\prime}, C^{\prime}$ . Let $R$ be the center of mass of the particles located at points $A, B, C$ , and $T$ be the analogous center of mass for the particles at points $A^{\prime}, B^{\prime}, C^{\prime}$ . We can then consider that $S$

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represents the center of mass of a system of three particles (each with a mass of 2) placed at points $L, M$ , and $N$ . However, we can equally consider that $S$ is the center of mass of a system of two particles (each with a mass of 3) located at points $R$ and $T$ . Therefore, $S$ is the midpoint of the segment $R T$ ; but since $R$ is fixed and $T$ can move arbitrarily in the plane $E$ , the point $S$ will trace out a certain plane parallel to $E$ .

[D. Pedoe, A. M. M., 71, 670 (June 1964).]

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.