Relationship between the side lengths of a tetrahedron and an inscribed tetrahedron with vertices at the centroids 
Suppose that $OABC$ is a regular tetrahedron with sides having centroids $\lbrace E,F,G,H\rbrace$ also forming a regular tetrahedron. What is the relationship between the side lengths of $OABC$ and $EFGH$? 

I suspect that tetrahedron $EFGH$ has side lengths $1/3$ of that of $OABC$ but I'm unsure as to how to prove it.
 A: The centroid of the face $ABC$ is $\frac13(A+B+C)$.  The centroid of the whole large tetrahedron is $\frac14(O+A+B+C)$.  Note that
$$ \tfrac14(O+A+B+C) = \tfrac14 O + \tfrac34 (\tfrac13(A+B+C)) $$
This shows that the centroid of the large tetrahedron is $\frac34$ of the way along the line segment from the vertex $O$ to the centroid of the opposite face.  Thus the distance from the centroid of the large tetrahedron to the centroid of the face is $\frac13$ of the distance to the vertex $O$.  Now, $O$ was arbitrary, so the analogous statement is true for each vertex and the centroid of its opposite face.  Thus the little tetrahedron can be obtained from the large one by scaling (with centre the centroid of the big tetrahedron) by $-\frac13$.
(Compare the fact that the medians of a triangle concur at its centroid, which lies $\frac23$ of the way along each median.  In fact this argument generalizes to $n$ dimensions quite readily.)
A: The ratio of the edge-lengths is equal to the ratio of the circumradii. Since the insphere of $ABCD$ is tangent to the faces at the centroids, it is also the circumsphere of $EFGH$. As Wikipedia notes, 
$$\text{inradius of regular tetrahedron} = \frac13 \cdot \text{circumradius of regular tetrahedron}$$
so your suspicion is correct.
