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Let the lengths of side of the large and small tetrahedrons be $a$ and $b$ respectively, the top vertex be $A$ and the bottom ones be $B$ , $C$ and $D$ . Let also the midpoints of $BC$ , $CD$ and $DB$ be $M$ , $N$ and $O$ and the centers of $\triangle ABC$ , $\triangle ACD$ and $\triangle ADB$ be $P$ , $Q$ and $R$ respectively.

We note that the ratio of the volumes of the small to the large tetrahedrons $\dfrac {m}{n} = \dfrac {b^3}{a^3}$ .

We also note that because of parallelism, the ratio of the side length of $\triangle PQR$ to that of $\triangle MNO$ is the ratio of their perpendicular distances to $A$ . This ratio is also the ratio of $AP:AM$ .

The side length of $\triangle PQR = b$

The side length of $\triangle MNO = \frac {1}{2}a$

Therefore, we have $\dfrac {b}{\frac{1}{2}a} = \dfrac {AP}{AM}$

Let $AP = BP = r$ , $\quad \Rightarrow PM = r \cos {30^\circ} = \frac {1}{2} r$

$\Rightarrow \dfrac {b}{\frac {1}{2} a} = \dfrac {AP}{AM} = \dfrac {AP}{AP+PM} = \dfrac {r}{r + \frac {1}{2} r} = \dfrac {1} {\frac{3}{2}} = \dfrac {2}{3} \quad \Rightarrow \dfrac {b}{a} = \dfrac {1}{3}$

$\dfrac {m}{n} = \dfrac {b^3}{a^3} = \left( \dfrac {b}{a} \right)^3 = \left( \dfrac {1}{3} \right)^3 = \dfrac {1}{27} \quad \Rightarrow m+n = 1 + 27 = \boxed {28}$ .