The 3-D Coordinate System — Question 9

Question 9

Four points are $A=(1,1,1)$, $B=(5,1,1)$, $C=(3,1+2\sqrt 3,1)$, and $D=(3,1+2/\sqrt 3,1+4\sqrt{2/3})$. Prove that they are the vertices of a regular tetrahedron and find its center.

Question 9 – Solution

Strategy A tetrahedron is regular if all six pairwise distances agree. Symmetry suggests that its center is the average of its four vertices.

Edge lengths Computation gives $AB=AC=BC=4$. Also, $$AD^2=BD^2=4+\frac 43+\frac{32}{3}=16, \qquad CD^2=\frac{16}{3}+\frac{32}{3}=16.$$ Thus all six edges have length 4, proving the tetrahedron is regular.

Center Averaging the vertices gives $$\boxed{G=\left(3,\,1+\frac{2\sqrt 3}{3},\,1+\sqrt{\frac 23}\right)}.$$

Verification By symmetry—or by direct substitution—$G$ is equidistant from all four vertices.