Yes, Adam made a beeline to a proof, including the needed lemma. That paper was numerically amazing, in that for each coordinate of the 14 vertices of the Heawood graph that they determined, it was the solution of a polynomial in one variable of degree 79 that involved all 80 possible coefficients! (Or at least the one they displayed as an example did.) In fact, that polynomial was displayed 2 coefficients {c_j} to a line, which were mostly gigantic integers, and the ragged edge on the display's right-hand side came close to following a smooth curve — at least for its 40 distinct "points", which were essentially each determined by length(line_k) = floor( log_10( c_(2k) * c_(2k+1) ) ). —Dan
On Jun 22, 2015, at 6:15 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
<< Does one circle really contain 4 vertices, or does it just look that way? >> DA
I hadn't noticed that rather ugly feature --- one point is rather close to a circle to which it does not belong. Another example avoiding that infelicity: https://www.dropbox.com/s/yl819xpiy716qwh/fano7pt7rg_1.gif?dl=0 <https://www.dropbox.com/s/yl819xpiy716qwh/fano7pt7rg_1.gif?dl=0>
Adam seems to have nailed the whole problem very neatly --- I haven't checked his reference yet, but it presumably gives an explicit construction. Incidentally, I also tried relaxation based on 14 points and centres, but it was a total failure!