Find locus of center of gravity

Question

A variable plane makes with the coordinate planes a tetrahedron of constant volume =96k^3. Find the locus of center of gravity of the tetrahedron.

I dont know how to find the center of gravity of a tetrahedron. How to proceed?

Answer 1

Let

$$A=\begin{pmatrix}a\\0\\0\end{pmatrix}, \ B=\begin{pmatrix}0\\b\\0\end{pmatrix}, \ C=\begin{pmatrix}0\\0\\c\end{pmatrix}$$

be the intersection points of the plane with the coordinate axes.

The volume of tetrahedron $OABC$ (see (Volume of a pyramid as a determinant?)) is

$$\tag{1}\frac16 \det(\vec{OA}, \vec{OB}, \vec{OC})=\frac16 abc = 96 k^3$$

(a sixth of the volume of the box having diagonal OD with $D=\begin{pmatrix}a\\b\\c\end{pmatrix} $).

The center of gravity $G$ of tetrahedron $OABC$ is defined by:

$$\tag{2}\vec{OG}=\dfrac14(\vec{OO}+\vec{OA}+\vec{OB}+\vec{OC})=\dfrac14\vec{OD}=\begin{pmatrix}\dfrac{a}{4}\\\dfrac{b}{4}\\\dfrac{c}{4}\end{pmatrix} \ \text{such that} \ abc = 6 \times 96 k^3.$$

(by using (1)).

Therefore, setting in (2) : $x=\dfrac{a}{4}, y=\dfrac{b}{4},z=\dfrac{c}{4}$, the locus is the set of points:

$$\begin{pmatrix}x \\ y \\ z \end{pmatrix} \ \ \text{such that} \ \ 64xyz = 6 \times 96 k^3$$

Said otherwise, the locus is the surface with equation

$$xyz=9k^3.$$

Graphical representation : the third figure in the french site (https://www.mathcurve.com/surfaces/astroidal/astroidal.shtml)

Answer 2

HINT

Let’s indicate with (a,0,0) (0,b,0) (0,0,c) the vertices of the tetrahedron, thus the volume is simply:

$$V=\frac13 Ah = \left|\frac13 \frac12 ab\cdot c \right| = \left|\frac16 abc\right|$$

In general the gravity center coordinates of tetrahedron can be find by $$O(\frac{x_1+x_2+x_3+x_4}{4},\frac{y_1+y_2+y_3+y_4}{4},\frac{z_1+z_2+z_3+z_4}{4})$$ where $$P_1,P_2 P_3,P_4 $$ are the corner points of the tetrahedron.