# To find the center of gravity of a homogeneous tetrahedron

The center of gravity coordinates of a triangle can be calculated

$O(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3},\frac{z_1+z_2+z_3}{3})$ where $P_1,P_2, P_3$ are the corner points of a homogeneous triangle and we know that the areas of triangles $(P_1,P_2,O),(P_1,P_3,O),(P_2,P_3,O)$ in the big triangle equal to each other?

I have not found the formulas about the gravity center of a homogeneous tetrahedron.

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

2. Are the volumes of tetrahedrons $(P_1,P_2,P_3,O),(P_1,P_2,P_4,O),(P_2,P_3,P_4,O),(P_1,P_3,P_4,O)$ in the big tetrahedron equal to each other?

How can be proved the Lemma 1 above?

Note: I confirmed that if Lemma (1) is true ,Lemma (2) is true too.

The volume of the big tetrahedron can be computed by 4x4 matrix

$$V=\frac{1}{6} |det(\begin{bmatrix}x_1 & y_1 & z_1& 1\\x_2 & y_2 & z_2& 1\\x_3 & y_3 & z_3& 1\\x_4 & y_4 & z_4& 1\end{bmatrix})|$$

If the center is $O(x_0,y_0,z_0)=O(\frac{x_1+x_2+x_3+x_4}{4},\frac{y_1+y_2+y_3+y_4}{4},\frac{z_1+z_2+z_3+z_4}{4})$ then

$$V_{123}=\frac{1}{6} |det(\begin{bmatrix}x_0 & y_0 & z_0& 1\\x_1 & y_1 & z_1& 1\\x_2 & y_2 & z_2& 1\\x_3 & y_3 & z_3& 1\end{bmatrix})|=\frac{V}{4}$$

$$V_{124}=\frac{1}{6} |det(\begin{bmatrix}x_0 & y_0 & z_0& 1\\x_1 & y_1 & z_1& 1\\x_2 & y_2 & z_2& 1\\x_4 & y_4 & z_4& 1\end{bmatrix})|=\frac{V}{4}$$

$$V_{234}=\frac{1}{6} |det(\begin{bmatrix}x_0 & y_0 & z_0& 1\\x_2 & y_2 & z_2& 1\\x_3 & y_3 & z_3& 1\\x_4 & y_4 & z_4& 1\end{bmatrix})|=\frac{V}{4}$$

$$V_{134}=\frac{1}{6} |det(\begin{bmatrix}x_0 & y_0 & z_0& 1\\x_1 & y_1 & z_1& 1\\x_3 & y_3 & z_3& 1\\x_4 & y_4 & z_4& 1\end{bmatrix})|=\frac{V}{4}$$

The answer is yes to both questions. Start with a regular (all faces equilateral triangles) tetrahedron $P_1'P_2'P_3'P_4'$, where the answer is clearly yes to both questions. Then apply an affine transformation $A$ (a combination of stretching, shearing, rotating, and translation) to bring $P_1'P_2'P_3'P_4'$ to $P_1P_2P_3P_4$. For notation, let $A=T\circ L$, where $L$ is linear and $T$ is a translation by some vector $u$.
The transformation will have modified the volume of each of the sub-tetrahedra in the same way: multiplying it by $\det L$. So the sub-tetrahedra of $P_1P_2P_3P_4$ all have the same volume.
The relation that $\frac{P_1'+P_2'+P_3'+P_4'}{4}=O'$ will continue to hold once the transformation is applied to each of these points: $$P_i=AP_i'=LP_i'+u$$
So \begin{align} \frac{P_1+P_2+P_3+P_4}{4}&=\frac{LP_1'+LP_2'+LP_3'+LP_4'+4u}{4}\\ &=L\left(\frac{P_1'+P_2'+P_3'+P_4'}{4}\right)+u\\ &=LO'+u\\ &=AO'\\ \frac{P_1+P_2+P_3+P_4}{4}&=O \end{align}