Daniel Asimov wrote:

Another proof, I believe, can be cobbled from the idea of a cube, each of whose faces has been bisected into two rectangles so that none of 6 new lines meet each other. Now each of these 6 new lines can be symmetrically shrunk toward its midpoint, while allowing the cube to hinge identically along all 6 shrinking lines. By continuity there will be a moment that each of the 12 resulting pentagons is regular, and it can be shown that at this point all dihedral angles are equal.

You're constructing a dodecahedron, not the icosahedron I was talking about. But I don't think this quite works. (I mean the proof "by continuity", of course! -- the construction is quite sound.) Your method clearly constructs a figure made of 12 pentagons with tetrahedral symmetry, which is certainly a nice thing to have. As you adjust the "shrinking lines", the pentagons move through a 1-parameter family. It's clear from the construction that these pentagons have only two different edge lengths -- the edge that is a "shrinking line" is one length, and the other four edges are all congruent to one another. Also the pentagon is bilaterally symmetric, along the perpendicular bisector of the "shrinking line". The continuity point of view then tells you that there is a pentagon in the family where all edges have the same length. But it doesn't tell you that it magically happens to be regular. In particular, you don't automatically know that all five of its vertices lie on a sphere -- three of them do by the construction's symmetry, but the ones lying under the vertices of the original cube, a priori, might not. Of course, fixing all this up is precisely how Euclid constructs the dodecahedron, but a bunch of actual calculation is required. On the other hand, consider instead the *convex hull* of those same shrinking line segments. When the lines run the entire length of the cube's face, you get a cuboctahedron. As they shrink, each square of the cuboctahedron breaks into two triangles, a priori isosceles -- and at some point equilateral, by continuity. These together with the 8 always-equilateral triangles are indeed the faces of the icosahedron, constructed with tetrahedral symmetry. This is a "by continuity" version of the "tthree mutually perpendicular golden rectangle" construction. (Triangles are so much easier than pentagons...) --Michael Kleber kleber@brandeis.edu