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Embedded in Anton's answer are the following two aspects of the problem: You first need to decide (or be given) the combinatorial structure of the graph you want to embed, i.e., assign the n distanc …

I used qepcad to compute that the intersection of the set of possible area ratios with the interval [1/2, 3/4] is (1/2, 3/4). Since the set of possible area ratios is the image of a connected space u …

Yes, this is true. The main point is that the "first thing that goes wrong" cannot be two vertices coming together. Let $a_0$, $b_0$, $c_0$, $d_0$, $e_0$, $f_0$ denote the edge lengths of the original …

The paper Vaidya, Pravin M., An $O(n \log n)$ algorithm for the all-nearest-neighbors problem, Discrete Comput. Geom. 4, No. 2, 101-115 (1989), ZBL0663.68058 gives an $O(n \log n)$ algorithm for the " …

The regular dodecahedron is not a local minimum for total edge length either. Consider the five vertices Vi, i = 1, ..., 5 of a face together with the "center" C of the face. The spherical triangle …

If you call the random variable you described X, then it's much easier to compute the expected value of X2. Indeed, for any point P inside triangle ABC, the expected value of X2 where X is the length …

I think this problem is hard--the known algorithms are both slow and nontrivial to implement. See Exact Volume Computation for Polytopes for a survey. An interesting feature is that there are variou …