Prove that the tetrahedron with vertices at the points $O (0, 0, 0), A (0, 1, 1), B (1, 0, 1)$ and $C (1, 1, 0)$ is a regular one.

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Solution

STEP 1 : Assumptions
Let us assume a tetrahedron with vertices at the points $O (0, 0, 0), A (0, 1, 1), B (1, 0, 1)$ and $C (1, 1, 0)$ as shown in figure.
The distance between any two points $P (x_{1}, y_{1}, z_{1})$ and $Q (x_{2}, y_{2}, z_{2})$ is given by,
$P Q = \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2} + {(z_{1} - z_{2})}^{2}}$
STEP 2 : Finding the distance between all the vertices of tetrahedron
Distance between $O (0, 0, 0)$ and $A (0, 1, 1)$
$O A = \sqrt{{(0 - 0)}^{2} + {(0 - 1)}^{2} + {(0 - 1)}^{2}}$
$O A = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$
Distance between $O (0, 0, 0)$ and $B (1, 0, 1)$
$O B = \sqrt{{(0 - 1)}^{2} + {(0 - 0)}^{2} + {(0 - 1)}^{2}}$
$O B = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$
Distance between $O (0, 0, 0)$ and $C (1, 1, 0)$
$O C = \sqrt{{(0 - 1)}^{2} + {(0 - 1)}^{2} + {(0 - 0)}^{2}}$
$O C = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$
Distance between $A (0, 1, 1)$ and $B (1, 0, 1)$
$A B = \sqrt{{(0 - 1)}^{2} + {(1 - 0)}^{2} + {(1 - 1)}^{2}}$
$A B = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$
Distance between $A (0, 1, 1)$ and $C (1, 1, 0)$
$A C = \sqrt{{(0 - 1)}^{2} + {(1 - 1)}^{2} + {(1 - 0)}^{2}}$
$A C = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$
Distance between $B (1, 0, 1)$ and $C (1, 1, 0)$
$B C = \sqrt{{(1 - 1)}^{2} + {(0 - 1)}^{2} + {(1 - 0)}^{2}}$
$B C = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$
Since, $O A = O B = O C = A B = A C = B C$
Therefore, the tetrahedron with vertices at the points $O (0, 0, 0), A (0, 1, 1), B (1, 0, 1)$ and $C (1, 1, 0)$ is a regular one.

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