Tetrahedron Centroid?

Geometry Level pending

Given any tetrahedron, you connect each of its four vertices with the centroid of the opposing triangular face with a line segment. Do these four line segments always intersect in one point? And is that point the centroid of the tetrahedron defined by $G = \dfrac{1}{4} (A + B + C + D ) ?$

Yes, they meet in one point, and that point is

G

No, not necessarily

1 solution

Chris Lewis
May 5, 2021

Vectors and symmetry give a quick proof of this: let $\mathbf{a}=\overrightarrow{OA}$ be the position vector of point $A$ , etc. Also define $\mathbf{g}=\frac14(\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d})$ .

The centroid of face $BCD$ has position vector $\frac13(\mathbf{b}+\mathbf{c}+\mathbf{d})$ . The equation of the line joining this point and $A$ is $\mathbf{r}=\lambda \left[ \frac13 (\mathbf{b}+\mathbf{c}+\mathbf{d})\right]+(1-\lambda)\mathbf{a}$ where $\lambda$ is a scalar parameter.

Now note that setting $\lambda=\frac34$ shows that point $G$ lies on this line. By symmetry, it also lies on the line joining $B$ and the centroid of its opposite face, and so on.

Since the same point lies on all four lines joining vertices to the centroids of their opposite faces, they are concurrent; there can be at most one such point, and we've shown it's $G$ .

A nice bonus here is that this proof works in any dimension - the centroid of a simplex is always the arithmetic mean of its vertices.

Bonus question: if the centroid of face $BCD$ is $G_{BCD}$ , find the value of $\frac{AG}{GG_{BCD}}$

How does this compare to other dimensions?

You've answered your bonus question in your solution. Since $\lambda = \dfrac{3}{4}$ , then

$G = \dfrac{3}{4} G_{BCD} + \dfrac{1}{4} A$

hence,

$AG = G - A = \dfrac{3}{4} ( G_{BCD} - A ) = \dfrac{3}{4} A G_{BCD}$

but then,

$AG_{BDC} = A G + G G_{BCD}$

$\dfrac{4}{3} A G = A G + G G_{BCD}$

and finally,

$\dfrac{1}{3} AG = G G_{BCD}$

Thus,

$\dfrac{AG}{GG_{BCD} } = 3$

Hosam Hajjir - 1 month, 1 week ago

True (though I expect you knew this anyway ;-)) - but I think the interesting part is how it works in other dimensions. Perhaps it's not that interesting, though!

Chris Lewis - 1 month, 1 week ago

Extension to higher dimensions is surely interesting, perhaps a possible sequel problem to this one.

Hosam Hajjir - 1 month, 1 week ago

Tetrahedron Centroid?

1 solution

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