On 6/30/15, Dan Asimov <asimov@msri.org> wrote:
I think this is just gorgeous.
Is Fred's solution the unique* set of points in R^3 having the combinatorial type of the Fano plane (with unit circles for the lines) and the symmetry group of D_3 == S_3 ???
—Dan _____________________________________ * Up to isometries of R^3, of course.
An equilateral triangle in 3-space has freedom 7 , a second one with the same axis line freedom 3 , a single point on the axis freedom 1 . Via C_3 symmetry the 7 unit radii impose only 3 constraints; so modulo isometry, a 3-fold symmetric Fano plane has freedom 11 - 3 - 6 = 2 . Since the orbits remain unchanged under D_6 , that would yield the same freedom. Or so I reckon ... WFL
On Jun 30, 2015, at 11:12 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Two views of FWH solid circular Fano plane at https://www.dropbox.com/s/iofsjqyln0n929d/hele_solid0.gif?dl=0 https://www.dropbox.com/s/zs7pi91p0f99m5o/hele_solid1.gif?dl=0 the first showing 6-fold symmetry about z-axis, the second with z-axis up page showing one of 3 adventitious double points below right of centre.
No coordinate box this time. For nothing is simple. FWH's circle #4, parallel to xy-plane with centre on z-axis, defeated all attempts to plot it until in desperation I rotated everything thru' an entire radian around all three axes ... Maple solve() moves --- or in this case, fails to go anywhere --- in mysterious ways!