MHB Vector algebra- centroid of tetrahedron

[mathisfun1]

How to find out the position vector of the centroid of tetrahedron , the position vectors of whose vertices are a,b,c,d respectively.
I am familiar with the result, namely a+b+c+d/4 but want to know how to derive it without using the 3:1 ratio property.
Any help would be appreciated. Thank you.

 

[Evgeny.Makarov]

Which definition of centroid are you using? According to Wikipedia, "the centroid of a plane figure or two-dimensional shape is the arithmetic mean ("average") position of all the points in the shape", so the fact that the position vector of the centroid is $(a+b+c+d)/4$ holds by definition in that case.

 

[mathisfun1]

Which definition of centroid are you using? According to Wikipedia, "the centroid of a plane figure or two-dimensional shape is the arithmetic mean ("average") position of all the points in the shape", so the fact that the position vector of the centroid is $(a+b+c+d)/4$ holds by definition in that case.

The centroid of a tetrahedron is the intersection of all line segments that connect each vertex to the centroid of the opposite face.

 

[Evgeny.Makarov]

The centroid of a tetrahedron is the intersection of all line segments that connect each vertex to the centroid of the opposite face.

Not the easiest definition to work with, but OK. I assume then that the centroid of a triangle is the intersection of its medians. I prefer to work from weighted points and prove that the center of mass, or barycenter, is the intersection of medians.

A weighted point is an ordered pair $(A,x)$ where $A$ is a point and $x$ is a real number. The barycenter of a set $\{(A_1,x_1),\dots,(A_n,x_n)\}$ of weighted points is a point $A$ such that
\[
\overrightarrow{OA}=\frac{1}{S}\left(x_1\overrightarrow{OA}_1+\dots+x_n\overrightarrow{OA}_n\right)
\]
where $S=x_1+\dots+x_n$ and $O$ is any point. One can show that the resulting point $A$ does not depend on the choice of $O$. Another property is that the barycenter of two points lies on the line passing through these points.

According to this definition, the barycenter of $\{(A,1),(B,1),(C,1)\}$ is a point $M$ such that
\[
\overrightarrow{OM}=\frac{1}{3}\left(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right).
\]
But the right-hand side equals
\[
\frac{1}{3}\left(\overrightarrow{OA}+2\frac{1}{2}\left(\overrightarrow{OB}+\overrightarrow{OC}\right)\right).
\]
Thus, $M$ is the barycenter of $(A,1)$ and $(A',2)$ where $A'$ is the barycenter of $(B,1)$ and $(C,1)$, i.e., the center of $BC$. Therefore, $M$ lies on the median $AA'$. Similarly, $A$ lies on the other two medians, so it is the centroid of $\triangle ABC$ according to your definition.

The barycenter of a tetrahedron $ABCD$ is a point $N$ such that
\[
\overrightarrow{ON}=\frac{1}{4}\left(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}\right).
\]
But the right-hand side equals
\[
\frac{1}{4}\left(\overrightarrow{OA}+3\frac{1}{3}\left(\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}\right)\right)
\]
Thus, $N$ is the barycenter of $(A,1)$ and $(A',3)$ where $A'$ is the barycenter and the centroid of $(B,1)$, $(C,1)$ and $(D,1)$. Therefore, $N$ lies on $AA'$. Similarly, $A$ lies on the other three segments connecting vertices to the centroids of the opposite faces, so it is the centroid of the tetrahedron $ABCD$ according to your definition.