I need to prove that the medians of a tetrahedron are concurrent, and to find the ratio at which they intersect each other. I cannot use coordinate geometry, and I must use barycentric arguments in my solution.

Thus far, I have the following: Consider tetrahedron ABCD (with triangle ABC being the "bottom" triangle) with masses of 1 at each of the four vertices Consider face ABC, which has medians AD, BE, and CF, which intersects at the centroid of ABC, M. Each median is divided, by M, into segments of ratio 2:1. Further, M has a mass of 3.

Construct DM, the median of the tetrahedron from vertex D to centroid M of ABC> Also, construct CN, the median of the tetrahedron from vertex C to centroid N on triangle ABD.

I'm confident that CN and DM intesect, but I'm not sure why (I know they're on the same plane but I'm having trouble "seeing" it). In addition, I'm not certain how to go about showing that the other two medians are concurrent with these two.

I'm also pretty sure that the ratio that I'm looking for is 3:1, but I'm not sure how to go about showing it; the center of mass of the tetrahedron would have a mass of 4 since there are masses of 1 at each vertex, but I'm not sure how to go from there either.

  • $\begingroup$ Can you argue like this: $M=(0,1/3,1/3,1/3)$ and $A=(1,0,0,0)$, so $AM$ contains the point $G=(1/4,1/4,1/4,1/4)$ and the ratio is certainly $3:1$. $\endgroup$ – Quang Hoang Mar 2 '16 at 16:48
  • $\begingroup$ We've never used notation like that before, so I don't think so; I'm not quite sure I understand it. The n-tuples represent weights of each point relative to the other points? I'm a little confused.... We've weighted various points with numerical values (normally 1) and then found weights of other points, but never as an ordered n-tuple. $\endgroup$ – analysischallenged Mar 2 '16 at 17:09

Try calculating the center of mass of the tetrahedron from different perspectives.

  1. Center of mass of triangle $ABC$ = centroid of the triangle $ABC$ (Let's call this $D'$). The center of mass of the tetrahedron can be considered to be the center of mass of two objects. One with mass $1$ placed at $D$ and another with mass $3$ placed at $D'$. This means that the center of mass of the tetrahedron is placed on the median from $D$, and divides the median in the ration $3:1$.

Similarly, instead of $D$, now calculate the center of mass of the tetrahedron with respect to $A, B, C$. As the center of mass of the tetrahedron is just one single point, all these points must coincide. Thus, it lies on all the $3$ medians. Thus, all the medians intersect and the center of mass divides them in the ratio $3:1$.

  • $\begingroup$ I understand the first argument. I'm a little lost as to why the medians must all intersect: Is that because the first argument can be made again for triangles ABD, BCD, and ACD, and thus all four medians must contain a center of mass, and by definition the center of mass is a unique point so, if all four segments contain it, they must intersect at that common point? Is that reasoning valid? $\endgroup$ – analysischallenged Mar 2 '16 at 17:13
  • $\begingroup$ Yes, that's the argument I was trying to make. $\endgroup$ – Ojas Mar 2 '16 at 17:16
  • $\begingroup$ Thank you so very much! I was trying to mathematically show the various medians were on the same plane and then attempting to show that the common point of all of them was the center of mass, but was having such trouble. This is so much simpler! $\endgroup$ – analysischallenged Mar 2 '16 at 17:17
  • $\begingroup$ @analysischallenged You're welcome. Glad I could help. $\endgroup$ – Ojas Mar 2 '16 at 17:18

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