Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.1 National olympiad Find the answer

18. C6 (FRA 2) Let $O$ be a point of three-dimensional space and let $l_{1}, l_{2}, l_{3}$ be mutually perpendicular straight lines passing through $O$ . Let $S$ denote the sphere with center $O$ and radius $R$ , and for every point $M$ of $S$ , let $S_{M}$ denote the sphere with center $M$ and radius $R$ . We denote by $P_{1}, P_{2}, P_{3}$ the intersection of $S_{M}$ with the straight lines $l_{1}, l_{2}, l_{3}$ , respectively, where we put $P_{i} \neq O$ if $l_{i}$ meets $S_{M}$ at two distinct points and $P_{i}=O$ otherwise ( $i=1,2,3$ ). What is the set of centers of gravity of the (possibly degenerate) triangles $P_{1} P_{2} P_{3}$ as $M$ runs through the points of $S$ ?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

18. Set the coordinate system with the axes $x, y, z$ along the lines $l_{1}, l_{2}, l_{3}$ respectively. The coordinates $(a, b, c)$ of $M$ satisfy $a^{2}+b^{2}+c^{2}=R^{2}$ , and so $S_{M}$ is given by the equation $(x-a)^{2}+(y-b)^{2}+(z-c)^{2}=R^{2}$ . Hence the coordinates of $P_{1}$ are $(x, 0,0)$ with $(x-a)^{2}+b^{2}+c^{2}=R^{2}$ , implying that either $x=2 a$ or $x=0$ . Thus by the definition we obtain $x=2 a$ . Similarly, the coordinates of $P_{2}$ and $P_{3}$ are $(0,2 b, 0)$ and $(0,0,2 c)$ respectively. Now, the centroid of $\triangle P_{1} P_{2} P_{3}$ has the coordinates $(2 a / 3,2 b / 3,2 c / 3)$ . Therefore the required locus of points is the sphere with center $O$ and radius $2 R / 3$ .

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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.