2
$\begingroup$

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?

Many thanks for answer and advice

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

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

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

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

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

$\endgroup$
2
$\begingroup$

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 $$\begin{align} \frac{P_1+P_2+P_3+P_4}{4}&=\frac{LP_1'+LP_2'+LP_3'+LP_4'+4u}{4}\\ &=L\left(\frac{P_1'+P_2'+P_3'+P_4'}{4}\right)+u\\ &=LO'+u\\ &=AO'\\ \frac{P_1+P_2+P_3+P_4}{4}&=O \end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.