Difference between revisions of "1961 IMO Problems/Problem 6"
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− | {{ | + | We will consider the various points in terms of their coordinates in space. We have <math>L=\frac{A+A^\prime}{2},M=\frac{B+B^\prime}{2},N=\frac{C+C^\prime}{2}</math>. Since the centroid of a triangle is the average of the triangle's vertices, we have <math>G=\frac{1}{3}\left(L+M+N\right)=\frac{1}{6}\left(A+B+C+A^\prime+B^\prime+C^\prime\right)</math>. It is clear now that <math>G</math> is midpoint of the line segment connecting the centroid of <math>ABC</math> and the centroid of <math>A^\prime B^\prime C^\prime</math>. It is obvious that the centroid of <math>A^\prime B^\prime C^\prime</math> can be any point on plane <math>\epsilon</math>. Thus, the locus of <math>G</math> is the plane parallel to <math>\epsilon</math> and halfway between the centroid of <math>ABC</math> and <math>\epsilon</math>. |
{{IMO box|year=1961|num-b=5|after=Last question}} | {{IMO box|year=1961|num-b=5|after=Last question}} |
Revision as of 11:06, 6 September 2008
Problem
Consider a plane and three non-collinear points on the same side of ; suppose the plane determined by these three points is not parallel to . In plane take three arbitrary points . Let be the midpoints of segments ; Let be the centroid of the triangle . (We will not consider positions of the points such that the points do not form a triangle.) What is the locus of point as range independently over the plane ?
Solution
We will consider the various points in terms of their coordinates in space. We have . Since the centroid of a triangle is the average of the triangle's vertices, we have . It is clear now that is midpoint of the line segment connecting the centroid of and the centroid of . It is obvious that the centroid of can be any point on plane . Thus, the locus of is the plane parallel to and halfway between the centroid of and .
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