regular tetrahedron coordinate

We can use coordinates $(0, 0, 0)$ , $(a, b, c)$ , $(a, c, b)$ , $(c, b, a)$ , so we solve the equation for the two different side lengths in this symmetric but not necessarily regular tetrahedron

$a^2+b^2+c^2=(a-a)^2+(b-c)^2+(c-b)^2$

$c =a + b + 2 \sqrt{ab}$

So, we let $(a,b. c)=(1,1,4)$ , and end up with the volume of $9$ for the regular tetrahedron of side length $\sqrt{18}$ .

The next smallest tetrahedron has a side length of $\sqrt{98}$ , with $(a,b,c)=(1,4,9)$

$\overline{PQ}^2=\overline{PO}^2=(a-c)^2+(c-b)^2+(b-a)^2=a^2+b^2+c^2$

So, $a^2+b^2+c^2-2(ab+bc+ca)=(\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c})(\sqrt{a}-\sqrt{b}+\sqrt{c})(-\sqrt{a}+\sqrt{b}+\sqrt{c})=0$

So if $a<b<c$ ,then $\sqrt{a}+\sqrt{b}=\sqrt{c}$

Minimum value is when $a=b=1,c=4,\overline{PO}=\sqrt{18}$

The volume of a regular tetrahedron is $\dfrac{a^3\sqrt{2}}{12}$ ( $a$ is the side length),so $\dfrac{\sqrt{18}^3\times\sqrt{2}}{12}=9$

The volume of $OPQR$ is $V(a,b,c) \; = \; \left| \tfrac16 \left(\begin{array}{c} a \\ b \\ c \end{array}\right) \cdot \left( \left(\begin{array}{c} b \\ c \\ a \end{array}\right) \times \left(\begin{array}{c} c \\ a \\ b \end{array}\right)\right)\right| \; = \; \tfrac16(a^3 + b^3 + c^3 - 3abc)$ for any positive integers $a,b,c$ where, for simplicity, we shall assume that $a \le b \le c$ . It is easy to show that $V(a,b,c) \le V(a+1,b+1,c+1) \hspace{1cm} V(a,b,c) \le V(a,b+1,c+1) \hspace{1cm} V(a,b,c) \le V(a,b,c+1)$ for all positive integers $a \le b \le c$ .

For $OPQR$ to be regular, we must have $a^2 + b^2 + c^2 = (a-b)^2 + (b-c)^2 + (c-a)^2$ . One solution to this equation is $a=b=1,c=4$ . By the above paragraph, $V(1,1,2)$ and $V(1,1,3)$ are the only two tetrahedron volumes which are smaller than $V(1,1,4)$ . Thus $V(1,1,4)= \boxed{9}$ is certainly the smallest possible regular tetrahedron volume (and the third smallest tetrahedron volume overall).