Tetrahedroid

If the blue tetrahedron has side length $1$ , and has centroid at the origin $O$ , it can be given by coordinates $A\;:\;\big(-\tfrac12,0,-\tfrac{1}{2\sqrt{2}}\big) \hspace{1cm} B\;:\; \big(\tfrac12,0,-\tfrac{1}{2\sqrt{2}}\big) \hspace{1cm} C\;:\; \big(0,-\tfrac12,\tfrac{1}{2\sqrt{2}}\big) \hspace{1cm} D\;:\; \big(0,\tfrac12,\tfrac{1}{2\sqrt{2}}\big)$ The midpoint $M$ of the face $ABC$ has coordinates $\big(0,-\tfrac16,-\tfrac{1}{6\sqrt{2}}\big)$ Since the altitude of a regular tetrahedron of side length $1$ is $\sqrt{\tfrac23}$ , one vertex of the red tetrahedron is the point $P$ , where $\overrightarrow{OP} \; = \; \overrightarrow{OM} + \sqrt{\tfrac23} \frac{\overrightarrow{OM}}{OM} \; = \; 5\overrightarrow{OM}$ and so $OP = 5OM = \frac{5}{2\sqrt{6}}$ , while $OA = \frac{\sqrt{3}}{2\sqrt{2}}$ . Note that, by symmetry, $O$ will also be the centroid of the red tetrahedron

Since the ratio of the sides of the tetrahedral is equal to the ratio of the distances from the common centroid to the vertices (in other words, the ratio of the circumradii), we deduce that the ratio of the sides of the two tetrahedra is $\tfrac{OP}{OA} = \tfrac53$ , making the answer $\boxed{8}$ .