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Let $C$ be a hypercube in $\mathbb{R}^D$ with edge length of $L$. Let $\mathcal{P}_1,\ldots,\mathcal{P}_K$ be $K$ convex polytopes that partition $C$. Let $S_k$ be the surface area of the polytope $\mathcal{P}\_k$, where $k\in \{1,\ldots, K \}$. Define $S=\sum_{k=1}^K S_k$, what is the upper bound and lower bound of $S$? In particular, I'm interested in the bounds which can be expressed as function of $K,D$ and $L$.

  1. In case it is not clear, the surface area of a convex polytope $\mathcal{P}$ is the sum of $(D-1)$-dimensional Lebesgue measure of the facets of $\mathcal{P}$.

  2. There may have some nice results when $K\rightarrow\infty$ and reformulating this problem as a tessellation induced by some random processes. However, I'm interested in small $K$, say $1< K<50$.

  3. I guess this problem has been solved, but I'm struggling to find good literature.

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The lower bound is the surface area of the cube. (This is the trivial part). Denote by $\bar S_K$ this upper bound. Let $A$ be area of the cube and $B$ be the maximal area of the intersection of the cube with a hyper plane. then $\bar S_1=A$, $\bar S_2=A+2{\cdot}B$ and I guess that $$\bar S_K=A+2{\cdot}(K-1){\cdot}B$$ for all $K\ge 3$ – Anton Petrunin Sep 25 '11 at 17:14
@Anton: Surely you are right. One can draw many parallel sectios of the cube close to that providig the maximal section area. THis area should be known. – Ilya Bogdanov Sep 26 '11 at 5:21
Thanks Anton Petrunin and Ilya Bogdanov for the nice answer and explanation. To complete the answer, the maximal area of the intersection of a $D$-cube with a $(D-1)$-dimensional hyper plane is $\sqrt{2}L^{D-1}$, where $L$ is the edge length of hypercube. – han Sep 26 '11 at 15:18

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