We use vectors, anf from relations of the type $MN=kAB+(1-k)A'B', PQ=kCD+(1-k)C'D'$ we find $MN+PQ=0$ (two-letter combinations are all vectors). This shows that $MNPQ$ is a parallelogram. We use similar methods to show that $OP=kPO'$, where $O,O',P$ are the centers of $ABCD,A'B'C'D',MNPQ$ respectively. The locus is the segment $OO'$. [Edit: it's not a solution, just an approach; this is to make orl happy ]

Please modify your post. Look here.

Here is my solution (sorry, Grobber, my notations conflict with yours, but actually I find your notations a bit confusing, since you have two different points labeled with P ). Since the points M, N, P, Q divide the segments AA', BB', CC', DD' in equal ratios, we can find a real number u such that u = AM : AA' = BN : BB' = CP : CC' = DQ : DD'. Then, if O is an arbitrary point in space, we have the vectorial identity $\overrightarrow{OM}=\overrightarrow{OA}+\overrightarrow{AM}=\overrightarrow{OA}+u\cdot \overrightarrow{AA^{\prime }}=\overrightarrow{OA}+u\cdot \left( \overrightarrow{OA^{\prime }}-\overrightarrow{OA}\right) $ $=\left( 1-u\right) \cdot \overrightarrow{OA}+u\cdot \overrightarrow{OA^{\prime }}$, and similarly $\overrightarrow{ON}=\left( 1-u\right) \cdot \overrightarrow{OB}+u\cdot \overrightarrow{OB^{\prime }}$; $\overrightarrow{OP}=\left( 1-u\right) \cdot \overrightarrow{OC}+u\cdot \overrightarrow{OC^{\prime }}$; $\overrightarrow{OQ}=\left( 1-u\right) \cdot \overrightarrow{OD}+u\cdot \overrightarrow{OD^{\prime }}$. Since the quadrilaterals ABCD and A'B'C'D' are parallelograms, we have $\overrightarrow{AB}=\overrightarrow{DC}$ and $\overrightarrow{A^{\prime }B^{\prime }}=\overrightarrow{D^{\prime }C^{\prime }}$. But $\overrightarrow{MN}=\overrightarrow{ON}-\overrightarrow{OM}=\left( \left( 1-u\right) \cdot \overrightarrow{OB}+u\cdot \overrightarrow{OB^{\prime }}\right) -\left( \left( 1-u\right) \cdot \overrightarrow{OA}+u\cdot \overrightarrow{OA^{\prime }}\right) $ $=\left( 1-u\right) \cdot \left( \overrightarrow{OB}-\overrightarrow{OA}\right) +u\cdot \left( \overrightarrow{OB^{\prime }}-\overrightarrow{OA^{\prime }}\right) $ $=\left( 1-u\right) \cdot \overrightarrow{AB}+u\cdot \overrightarrow{A^{\prime }B^{\prime }}$, and similarly $\overrightarrow{QP}=\left( 1-u\right) \cdot \overrightarrow{DC}+u\cdot \overrightarrow{D^{\prime }C^{\prime }}$. Since $\overrightarrow{AB}=\overrightarrow{DC}$ and $\overrightarrow{A^{\prime }B^{\prime }}=\overrightarrow{D^{\prime }C^{\prime }}$, we thus get $\overrightarrow{MN}=\overrightarrow{QP}$, and it follows that the quadrilateral MNPQ is a parallelogram. This completes the solution of part a.). Now, if U, V and W are the centers of the parallelograms ABCD, A'B'C'D' and MNPQ, respectively, then these points U, V, W are the midpoints of the segments AC, A'C', MP (since the center of a parallelogram is the midpoint of any of its diagonals), and thus we have $\overrightarrow{OU}=\frac{1}{2}\cdot \left( \overrightarrow{OA}+\overrightarrow{OC}\right) $; $\overrightarrow{OV}=\frac{1}{2}\cdot \left( \overrightarrow{OA^{\prime }}+\overrightarrow{OC^{\prime }}\right) $; $\overrightarrow{OW}=\frac{1}{2}\cdot \left( \overrightarrow{OM}+\overrightarrow{OP}\right) $. The last equation yields $\overrightarrow{OW}=\frac{1}{2}\cdot \left( \overrightarrow{OM}+\overrightarrow{OP}\right) $ $=\frac{1}{2}\cdot \left( \left( \left( 1-u\right) \cdot \overrightarrow{OA}+u\cdot \overrightarrow{OA^{\prime }}\right) +\left( \left( 1-u\right) \cdot \overrightarrow{OC}+u\cdot \overrightarrow{OC^{\prime }}\right) \right) $ $=\frac{1}{2}\cdot \left( \left( 1-u\right) \cdot \left( \overrightarrow{OA}+\overrightarrow{OC}\right) +u\cdot \left( \overrightarrow{OA^{\prime }}+\overrightarrow{OC^{\prime }}\right) \right) $ $=\left( 1-u\right) \cdot \frac{1}{2}\cdot \left( \overrightarrow{OA}+\overrightarrow{OC}\right) +u\cdot \frac{1}{2}\cdot \left( \overrightarrow{OA^{\prime }}+\overrightarrow{OC^{\prime }}\right) $ $=\left( 1-u\right) \cdot \overrightarrow{OU}+u\cdot \overrightarrow{OV}$. This shows that the point W lies on the segment UV and that we have UW : UV = u. When the point M moves on the segment AA', the ratio u = AM : AA' varies from 0 to 1 (in fact, we have u = 0 for M = A, and u = 1 for M = A'). Thus, while the point M moves on the segment AA', the locus of the point W is the segment UV (hereby, every point on this segment can be considered as the point W for some value of u). Thus we have shown that the locus of the center W of the parallelogram MNPQ is the segment UV. Note that this locus degenerates to a single point if and only if U = V, i. e. if and only if the parallelograms ABCD and A'B'C'D' have the same center. Thus the solution of part b.) is complete. Darij