# Linear Algebra Tetrahedron Centroid 3/4 Down Segment Proof

Here is the question:

Show that the centroid of the four vertices of a tetrahedron (a solid with four vertices joined by six lines that bound the tetrahedron’s four triangular faces) is the intersection of the following seven lines:

a) It is $3/4$ of the way from the vertex of the tetrahedron along the line segment joining the vertex to the centroid of the opposite face.

Basically what it means is if there is a line segment starting from one of the tetrahedron's points to the centroid of the point's opposite triangle face, the centroid would be placed 75% into the length of the line (if line segment is $4$ units long, centroid would be at $3$ units).

I spent the day thinking about it so it's not that there's no work, but the work doesn't feel very close to the answer at all. I apologize for that. Here it is:

Properties of a tetrahedron:

• Equilateral triangles compose the 3D shape. Faces = same shape

• Due to symmetry, a height line intersects the centroid of the house triangle (basically tetrahedron $PQRS$ with $G$ as the centroid of its base triangle, $PG$ would equal to its height, $A$ being the top vertex)

• Base of the vertical triangle composed of $P,\, G$ and another vertex is equal to $1/2$ the length of the median of the base triangle.

• Height and base of the vertical triangle are not equal because the vertical triangle is isosceles whereas the base triangle is equilateral (although they are both related to bisectors)

Now that's over with, onto the solutions.

$3/4$ down $PG$ is the centroid, and also the midpoint of the midpoint. i.e. the midpoint, $x$, of $PG$ is $\frac{1}{2}(P + G)$, the midpoint of $x$ and $G$, call it $u$, is $\frac{1}{2}\left(\frac{1}{2}(P + G) + G\right) = \frac14 P + \frac34 G$. Remember, $u$ is the location of the supposed centroid.

$3/4$ down the line between $Q$ to the centroid of the face of its opposite triangle, you measure it the same way and get $v = \frac14 Q + \frac34 a$.

Now that I've got the location of the midpoint of the midpoint, I think I will need to prove that it is a centroid by proving that the two dots are equal.

I did manage to prove vectors $QA = PG$, but that process is so tiring and I'm not sure if two equal vectors have any meaning for my solution, so I'm refraining from posting the entire proof until needed.

It's late and I'm dead tired. Please someone help me a little.

Thank you.

• Hmm the topic must be bad...even I can feel it. – Jackie Doesque Sep 24 '15 at 4:06
• It's a good topic, and you've done well! It just is indeed late :) – pjs36 Sep 24 '15 at 4:21

Let $O$ be the tetrahedron centroid and $G$ be the centroid of face $QRS$, opposite to $P$. Then: $$G={Q+R+S\over 3},\quad O={P+Q+R+S\over 4},$$ so that: $$P-O=P-{P+Q+R+S\over 4}={3\over4}P-{Q+R+S\over 4}= {3\over4}P-{3\over4}{Q+R+S\over 3}={3\over4}(P-G).$$