Japan 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. $△ABC$ and $△ACD$ is a face of a tetrahedron. $△ABD$ and $△BCD$ 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 $△ABC$ as its face. Then $T_1$ is $ABCE$, $ABCF$, $ABCG$ or $ABCH$.

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

$$(a)T_2=ACDE.$$ There must be another tetrahedron $T_3$ with face $△CDE$. $T_3$ 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 $△BCE$ and $△CEH$ 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 $T_2=ACDE$.

$$(b)T_2=ACEH.$$ There must be $ACDH$. Then we need to count the ways of dividing the triangle prism $BEF-CHG$ such that each of $△BCE$ and $△CEH$. 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 $(8+16+8+5)\times 2=74$.