# Finding the center of gravity of a homogeneous tetrahedron

I am reading 'How to solve it' by George Polya and am stuck on a problem concerning finding the center of gravity of a homogeneous tetrahedron.

The approach described by the author is to first solve for finding the center of gravity of a homogeneous triangle, then by analogy, use that to solve for the homogeneous tetrahedron.

I understand the solution for the triangle, which is to find the intersection of all (3) of the triangle's medians.

On the solution to the tetrahedron part, the author states:

In the case of the tetrahedron we have six median planes like MCD, joining the midpoint of some edge to the opposite edge, each of which has to contain the center of gravity of the tetrahedron. Therefore, these six median planes must meet in one point which is precisely the center of gravity

My problem is that I don't know how the author counted six median planes. In my opinion, the tetrahedron has 4 triangles (the faces) and each triangle has 3 medians, therefore the total median planes should be 12 and not 6.

I think there is some overlap between the 12 median planes I counted hence there is some double counting. I tried to draw each median plane on paper but was unsuccessful as it got messy quickly.

What is a systematic way to count and visualize the 6 median planes in a homogeneous tetrahedron?

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It might aid your visualization to imagine the tetrahedron being embedded in a parallelepiped... –  Ｊ. Ｍ. Aug 7 '11 at 17:20

A tetrahedron has $6$ edges, which should be instructive. For any edge, draw the one median line on each of the two opposite triangles that meet the edge at one of its two vertex endpoints. That gives a triangle, and the plane it lies in constitutes a median plane because it splits the tetrahedron symmetrically in half.

Here's an illustration of the two medians corresponding to the edge on the $y$-axis: $\hskip 1.5 in$

(I shifted the medians over to the right a bit so you could see them both.) Image source is here, though I obviously added in the red medians using MSPaint.

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You are not double-counting; you are just counting a different kind of thing. A median of a triangular face is a line, and does not determine a median plane of the tetrahedron.

Actually, as it happens, if you take two adjacent faces $\triangle_1$ and $\triangle_2$, there is one median line of $\triangle_1$ and one median line of $\triangle_2$ that are both incident on the midpoint of their shared edge. These two lines together determine a plane, which is a median plane of the tetrahedron. As anon's answer describes, the median plane intersects the tetrahedron's surface at the median lines and the edge connecting the two far vertices. So exactly two median lines make a median plane, and a median plane passes through exactly two median lines, which is why you got 12 instead of 6.

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