Prove that the distance of a centroid to a plane is the arithmetic mean of the distances of the vertexes of a triangle to that plane

Question

I tried to use the fact that the centroid divides the median in a ratio of $2:1$ and using similar triangles, but I got stuck.

Do you mean "the distance of a centroid of a given triangle to a plane"? — user247327, Nov 25 '17 at 12:59

score 1 · Accepted Answer · answered Nov 25 '17 at 13:11

First set up a coordinate system so that the triangle has one vertex at (0, 0, 0), a second vertex at (a, 0, 0), and the third vertex at (b, c, 0). Take the given plane to be px+ qy+ rz= s. The centroid of that triangle is ((a+b)/3, c/3, 0). A normal vector to the plane is . Determine the line through the centroid in the direction of the normal vector and find where that line intersects the given plane. Find the length of that line segment.

Then do the same with each of the three vertices to determine their distances from the plane and average them.

score 0 · Answer 2 · answered Nov 25 '17 at 13:18

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Let the plane be the $x,y$ plane. Let the three vertices have $z$ coordinates $u,v,w$. Then the centroid has $z$ coordinate $\frac{u+v+w}{3}$, as required.

answered Nov 25 '17 at 13:18

Prove that the distance of a centroid to a plane is the arithmetic mean of the distances of the vertexes of a triangle to that plane

2 Answers2