Difference between revisions of "1961 IMO Problems/Problem 6"

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{{IMO box|year=1961|num-b=5|after=Last question}}
{{IMO box|year=1961|num-b=5|after=Last question}}
[[Category:Olympiad Geometry Problems]]
[[Category:3D Geometry Problems]]

Latest revision as of 22:29, 18 July 2016


Consider a plane $\epsilon$ and three non-collinear points $A,B,C$ on the same side of $\epsilon$; suppose the plane determined by these three points is not parallel to $\epsilon$. In plane $\epsilon$ take three arbitrary points $A',B',C'$. Let $L,M,N$ be the midpoints of segments $AA', BB', CC'$; Let $G$ be the centroid of the triangle $LMN$. (We will not consider positions of the points $A', B', C'$ such that the points $L,M,N$ do not form a triangle.) What is the locus of point $G$ as $A', B', C'$ range independently over the plane $\epsilon$?


We will consider the various points in terms of their coordinates in space. We have $L=\frac{A+A^\prime}{2},M=\frac{B+B^\prime}{2},N=\frac{C+C^\prime}{2}$. Since the centroid of a triangle is the average of the triangle's vertices, we have $G=\frac{1}{3}\left(L+M+N\right)=\frac{1}{6}\left(A+B+C+A^\prime+B^\prime+C^\prime\right)$. It is clear now that $G$ is midpoint of the line segment connecting the centroid of $ABC$ and the centroid of $A^\prime B^\prime C^\prime$. It is obvious that the centroid of $A^\prime B^\prime C^\prime$ can be any point on plane $\epsilon$. Thus, the locus of $G$ is the plane parallel to $\epsilon$ and halfway between the centroid of $ABC$ and $\epsilon$.

1961 IMO (Problems) • Resources
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