Right, assuming the proof holds up. But I'm more inclined to believe the validity of the claim in the Math Intelligencer article that you cite. In fact, they write: ----- Incidentally, the Minkowski sum (½)M_V ⊕ (½)M_F, which one obtains half way in the process of morphing M_V into M_F, would render a body with tetrahedal symmetry. It actually has the same constant width as MV and MF . Its volume, however, is larger than that of the Meissner bodies, due to the Brunn-Minkowski inequality. ----- Here M_V is the Meissner tetrahedron where starting from the Reuleaux tetrahedron, 3 edges sharing a vertex have been modified in the same way, to get a body of constant width. Likewise for M_F, but in this case the 3 edges bound one face of the Reuleaux tetrahedron. I believe the example in the example Scott Huddleston linked to below is identical to the example (½)M_V ⊕ (½)M_F mentioned in the Intelligencer article. --Dan P.S. Fred, what do you mean by "elsewhere" in your posted quoted below? On 2013-01-10, at 4:20 PM, Fred lunnon wrote:
On 1/10/13, Huddleston, Scott <scott.huddleston@intel.com> wrote:
See http://www.xtalgrafix.com/Spheroform.htm and its links, including http://www.xtalgrafix.com/Reuleaux/Spheroform%20Tetrahedron.pdf
Looks like that probably wraps Dan's question up, albeit at some length!
Also, p.4 of http://www.mi.uni-koeln.de/mi/Forschung/Kawohl/kawohl/pub100.pdf says that a Meissner-like tetrahedron with congruent "edge" parts is realizable as a Minkowski sum of the two Meissner tetrahedra.
That was obviously true --- but there seems no reason to expect that the width would be constant elsewhere.
WFL
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