# Does the centroid of a tetrahedron divide the height in two equal parts?

I am trying to calculate a/r ratio for a diamond cubic crystalline structure, but I cannot remember the relationship between the two parts of the height of tetrahedron, which are divided by the centroid.

• Hint: in vector terms the centroid is $\frac14(a+b+c+d)$ where $a$ etc. are the position vectors of the vertices. – Lord Shark the Unknown Oct 13 '17 at 6:37
• So it divides the height in a 3:1 ratio? @LordSharktheUnknown – Ingenium98 Oct 13 '17 at 6:40
• Yes, it does! It's nearer the face than the vertex opposite it. There's more of the tetrahedron nearer the face than the opposite vertex! – Lord Shark the Unknown Oct 13 '17 at 6:41
• Intuitively NOT. A tetrahedron from the centre to the vertex is lighter than the part from the centre of the height to the base. Centroid is the centre of gravity if the tetrahedron is homogeneous... – Raffaele Oct 13 '17 at 7:32
• @Raffaele I think that the critallographic context of the OP refers to the case where the weights are placed on the vertices. – Jean Marie Oct 13 '17 at 8:09