, 2007

Consider a division of cube $S = ABCD - EFGH$ into $T_1, \cdots, T_k$ with properties in the problem. Then one of the below holds.
* $\triangle ABC$ and $\triangle ACD$ is a face of a tetrahedron.
* $\triangle ABD$ and $\triangle BCD$ is a face of a tetrahedron.
By symmetry we can get the answer by counting the former case and double it. Assume that $T_1$ has $\triangle ABC$ as its face. Then $T_1$ is $ABCE$, $ABCF$, $ABCG$ or $ABCH$.

$$(1) \quad T_1 = ABCE.$$
There must be another tetrahedron $T_2$ with face $\triangle ACE$. $T_2$ is $ACDE$ or $ACEH$.

$$(a) \quad T_2 = ACDE.$$
There must be another tetrahedron $T_3$ with face $\triangle CDE$. $T_3$ is $CDEG$ or $CDEH$.

i. $T_3 = CDEG$.
There must be $DEGH$. Then we need to count the ways of dividing the square pyramid $BCGF - E$. There are 2 ways: $\{BCEG, BEFG\}$ and $\{BCEF, CEFG\}$.

ii. $T_3 = CDEH$.
We need to count the ways of dividing the triangle prism $BEF - CHG$ such that each of $\triangle BCE$ and $\triangle CEH$ is a face of a tetrahedron. There are 3 ways: $\{BCEF, CEFG, CEGH\}$, $\{BCEF, CEFH, CFGH\}$ and $\{BCEG, BEFG, CEFH\}$.
So there are 5 ways in the case $T_2 = ACDE$.

$$(b) \quad T_2 = ACEH.$$
There must be $ACDH$. Then we need to count the ways of dividing the triangle prism $BEF - CHG$ such that each of $\triangle BCE$ and $\triangle CEH$. There are 3 ways, as we counted in (1)(a)ii.
Therefore, we have 8 ways in the case $T_1 = ABCE$.

(2) $T_1 = ABCF$.
There must be another tetrahedron $T_2$ with face $\Delta ACF$. $T_2$ is $ACFH$, $ACDF$, $ACEF$ or $ACFG$.

(a) $T_2 = AFCH$.
There is only 1 way: the remaining tetrahedron must be $ACDH$, $AEFH$ and $CFGH$.

(b) $T_2 = ACDF, ACEF, ACFG$.
These 3 cases are all congruent, so we only have to consider the case $T_2 = ACDF$. Since if we remove $ABCF$ and $ACDF$ from $S$, there remains a shape congruent to the case $ABCE$ and $ACDE$ removed, there are 5 ways to divide the remaining shape, as we counted in (1)(a). So there are 15 ways in this case.
Therefore we have 16 ways in the case $T_1 = ABCF$.

(3) $T_1 = ABCG$.
Same as $T_1 = ABCE$. We have 8 ways.

(4) $T_1 = ABCH$.
There must be $ACDH$. Since if we remove $ABCH$ and $ACDH$ from $S$, there remains a shape congruent to the case $ABCE$ and $ACDE$ removed, there are 5 ways to divide the remaining shape, as we counted in (1)(a).

By these, we get the answer $(8 + 16 + 8 + 5) \times 2 = 74$.