Hardness of geometric area minimization problem
Never mind. I misremembered the definition of 3-partition.

I'll leave what I wrote here here, but since division into 3 subsets with equal sums is not strongly NP-complete, this doesn't work.

By 3-subset sum, I mean a: given a set of integers, find a partition into 3 subsets whose sums are all equal.

The problem is roughly equivalent to: let $\{p_i\}$ be the multiset of prime factors of $C$; find the best way of partitioning the multiset $\{r_i\}$ into three parts with equal sums, where $r_i = \log p_i$, for $p_i$.

Here's the wrong argument I had before.

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Find a bunch of primes $p_1$, $p_2$, $\ldots$, $p_{\alpha(n)}$ so that $\log p_i \approx M\ i$ for some real $M$.  Now, take a 3-subset sum problem on integers $1 \ldots \alpha(n)$, and replace the integer $i$ with prime $p_i$. Let $C$  be the product of all the primes $p_i$ corresponding to the integers $i$ in the 3-subset sum problem. Then, any $x$, $y$, $z$ with $x \approx y \approx z$ and $xyz = C$ gives a good partition of the integers in the 3-subset sum problem. Let $p_x$, $p_y$, and $p_z$ be the multisets of primes making up the prime factorization of $x$, $y$, and $z$, respectively. If $x$, $y$, and $z$ are close, $\log x \approx \log y \approx \log z$ means $ \sum_{p_i \in p_x} \log p_i \approx \sum_{p_i \in p_y} \log p_i \approx \sum_{p_i \in p_z} \log p_i$, and so you have a good solution to the 3-subset sum problem (and if you've chosen the primes right, you can tell from the optimal $x$, $y$ and $z$ whether or not the original 3-subset-sum problem is solvable). Since you know what the primes you used were, factoring $x$, $y$ and $z$ can be done by trial division to recover the partition.