Despite initially having been hazy about the details of Minkowski sums, it turns out that my intuition was not quite so wonky as earlier appeared! Lemma: The Minkowski sum of a surface and a sphere of radius r is (congruent to) the parallel surface at distance r ; proof elementary. Consider the mean M_T = (M_V (+) M_F)/2 , the summands positioned so that all three coincide in the spherical caps over each tetrahedral face, and the mean has the symmetry of a regular tetrahedron. Section it along a plane of symmetry: over each edge, the mean composes a circular arc of small radius e from one summand, with a pair of arcs of unit radius from the other meeting in nodal line; at their free ends, the small and unit arcs touch tangentially. * * * * * * * * * * * * * * * * * * * * * * * * * By the lemma, the sum is parallel to the unit arcs at distance e : rescaling, the mean comprises two arcs of radius (1 + e)/2 , touching the summands where they meet, and capped by a third arc of radius e/2 . Since e = 1 - rt3 / 2 is nonzero, the mean section is non-circular; whereas the corresponding portion of the Roberts section is a single arc of radius d = (2 - rt2) / 8 . Finally, the Minkowski and Roberts bodies are both minimal constant-width, with tetrahedral symmetry, but incongruent. The screed is already in need of updating ... comments are invited. While it suffers from an absence of diagrams, that on page 2 of Roberts report provides a partial substitute. In that connection I have already spotted one misprint: Stage (A): for "d = 0" read "d = c" Fred Lunnon On 1/12/13, Fred lunnon <fred.lunnon@gmail.com> wrote:
... Hence it looks a safe bet that Roberts = Minkowski holds, as conjectured by Dan Asimov, and earlier dismissed by myself with gratuitous contempt. A more detailed investigation along the lines above would not come amiss ... But now more generally, does minimal constant-width with tetrahedral symmetry uniquely determine the body modulo similarity --- why should no others exist?