Scratch that idea! Okay, second try: Build an equilateral triangle on the longest side of the scalene triangle, so that it contains the vertex opposite that side. Then the Reuleaux triangle in which this equilateral one is inscribed seems to be the natural shape of constant width containing the original scalene triangle and having the same diameter. I'm sure this is the least area of any such shape. I suspect it's a subset of any such shape, but am chastened by my earlier mistake, so will not try to assert this as more than a guess. --Dan On 2013-01-09, at 10:12 PM, Dan Asimov wrote:
I think the Meissner tetrahedron has constant width, but without full tetrahedral symmetry.
Let L be the longest side length of a scalene triangle.
Then isn't the intersection of the 3 disks of radius L, centered at the triangle's vertices, a shape of constant width?
--Dan
Fred wrote:
Part of the problem is the misnomer "Reuleaux tetrahedron" attached to the figure you described, which is as you observe irrelevant to the discussion. The "Meissner" tetrahedron --- a special case of the construction I posed earlier --- may well have constant width, but my own previous attempts to decide this question have been inconclusive.
|. . .
So what does the constant-width superset (with the same diameter) of a scalene triangle (including interior) look like, anyway? And is it unique, given the triangle?
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