Volume of Tetrahedron $ABCD$ is a regular tetrahedron of volume $1$. Maria glues regular tetrahedra $A'BCD$, $AB'CD$, $ABC'D$, and $ABCD'$
to the faces of $ABCD$. What is the volume of the tetrahedron $A'B'C'D'$?
 A: Hint: Note that the distance from a vertex of a regular tetrahedron to its center of mass is $3$ times the distance from the center of mass to a face. Let the distance from the center of mass to a face be $\lambda$, then the distance to one of the new vertices from the center of mass would be $5\lambda$ ($\lambda$ to a face them $\lambda+3\lambda$ to the new vertex). Thus the linear scale has increased by a factor of $\frac53$.
A: Let the edge length of (original) regular tetrahedron $ABCD$ (centered at $O$) be $a$ then the volume of the regular tetrahedron $ABCD$ $$=\color{red}{\frac{a^3}{6\sqrt{2}}}$$ but the volume is $1$ hence, we have $$\frac{a^3}{6\sqrt{2}}=1$$ $$\implies \color{blue}{a=(6\sqrt{2})^{1/3}}$$ Distance of each face of regular tetrahedron $ABCD$ from the center $O$ is $$=\frac{a}{2\sqrt{6}}$$  Normal height of regular tetrahedron $ABCD$ is $$=a\sqrt{\frac{2}{3}}$$ Now, the distance of each of vertices $A'$, $B'$, $C'$ & $D'$ from the center $O$ $$=\frac{a}{2\sqrt{6}}+a\sqrt{\frac{2}{3}}=\frac{5a}{2\sqrt{6}}$$Now, let $a'$ be the edge length of new regular tetrahedron $A'B'C'D'$ then the distance of its each vertex from the center $O$ $$=\frac{a'}{2}\sqrt{\frac{3}{2}}$$ $$\implies \frac{a'}{2}\sqrt{\frac{3}{2}}=\frac{5a}{2\sqrt{6}}$$ $$\implies a'=\frac{5a}{3}=\color{red}{\frac{5(6\sqrt{2})^{1/3}}{3}}$$
Hence, the volume of new regular tetrahedron $A'B'C'D'$ $$=\frac{(a')^3}{6\sqrt{2}}=\frac{\left(\frac{5(6\sqrt{2})^{1/3}}{3}\right)^3}{6\sqrt{2}}$$ $$=\frac{125(6\sqrt{2})}{27(6\sqrt{2})}$$ $$=\color{blue}{\frac{125}{27}\approx 4.62962963\dots}$$
Note: All the formula of regular tetrahedron have been taken from Table of platonic solids
