Distance between Centroids of the Faces of a Regular Tetrahedron I was recently given this question to solve: 
In regular tetrahedron $ABCD$, $AB = 1$. What is the distance between the centroids of triangles $ABC$ and $ABD$?
This is how I solved it:
If we set $A = (0, 0, 0)$, $B = (1, 0, 0)$, and correspondingly $C = (\frac{1}{2}, \frac{\sqrt{3}}{2},0)$, then the centroid of $\triangle{ABC}$ is $\frac{A + B + C}{3}$, or $(\frac{1}{2}, \frac{\sqrt{3}}{6},0)$.
In  $\triangle{ABD}$, the length of the median from $D$ is equal to $\frac{\sqrt{3}}{2}$, so the height of the tetrahedron from $D$ to base ${ABC}$ can be found to be $\frac{\sqrt{6}}{3}$ using the Pythagorean Theorem. 
Knowing that the centroid of $\triangle{ABD}$ lies $\frac{1}{3}$ of the way from the midpoint of $AB$ to $D$, the coordinates of the centroid can be determined as $(\frac{1}{2}, \frac{\sqrt{3}}{18},\frac{\sqrt{6}}{9})$. Then, using the distance formula the desired length can be found as $\frac{1}{3}$.
However, I wasn't satisfied with this solution because I'm convinced there is a cleaner way to solve this, ideally using Euclidean geometry (or anything without coordinate bashing). Can anyone provide such a solution?
 A: Construct a cube with side length $\frac {\sqrt 2}{2}$
choose one vertex.  There are 3 vertexes that are diagonally across each face.
Join these 4 vertexes.
You have constructed a regular tetrahedron with side length equal to 1.
See figure.
 
I have marked the centroid of two faces with the little dots.
Now lets move to the top view.... The centroid is $\frac 23$ the distance from the vertex to the opposite edge.  Which means that the distance between the centroid of two faces is $\frac 13$ the lenght of the diagonal of the face.  And we have already discussed that that length is 1.
$\frac 13$
A: The plane perpendicular to $\overline{AB}$ at its midpoint ($M$) contains edge $\overline{CD}$, creating isosceles $\triangle MBC$ whose legs coincide with altitudes of adjacent faces $\triangle ABC$ and $\triangle ABD$.  

It is "known" that the centroid of a triangle lies one-third of the way from a side to its opposite vertex. Consequently, centroids $P$ and $Q$ of the adjacent faces faces trisect $\overline{MC}$ and $\overline{MD}$; by proportionality, $|PQ|=\frac13|CD|$: The distance between centroids is one-third the side of the tetrahedron. $\square$
A: You're right, there is a much simpler approach.
Define $D'$ as the centroid of $\triangle ABC $ and $C'$ as the centroid of $\triangle ABD $.  Draw ray $AD'$ which intersects side $BC $ at point  $D''$.  Likewise draw ray $AC'$ which intersects side $BD $ at point  $C''$.  These rays, of course, include medians of the respective triangular faces in which they lie.
$C''$ and  $D''$ are midpoints of two sides of face $BCD $ whose sides measure one unit, thus $|C''D''|=1/2$.  $\triangle AC'D'$ is similar to $\triangle AC''D''$ with corresponding side ratio $|AC'|/|AC''|=2/3$, thus $|C'D'|=(2/3)(1/2)=1/3$.
