On 6/23/2015 6:13 PM, Fred Lunnon wrote:
[ DWW has raised the question of whether they might be "shaken out" (as it were) so as to properly occupy 3-space --- which appears likely, though I do not currently know for sure. ]
I worked this out yesterday, but I was hoping someone else would save me the trouble of writing it up (and perhaps find something simpler to describe). There is indeed a 3-dimensional arrangement of 7 unit circles with the correct intersections, and it can be chosen to be quite symmetrical (my example has 3-fold rotational symmetry about the z-axis and bilateral symmetry in the xz-plane, so the six-element dihedral group). Start with the unit circle in the xz-plane centered at (1, 0, 0) and add two more by rotating it 120 and 240 degrees about the z-axis (I'll call these circles 1, 2 and 3). They meet at the origin, which is Fano point A. The three lowest points on the circles are points B, C and D, and lie on unit circle 4. To complete the Fano plane, each of the points B, C and D needs to be on a new circle (5, 6 or 7) that meets whichever two of circles 1, 2 and 3 that this point isn't already on; these intersections will be Fano points E, F and G. To locate F and G, for example, start with the circle in the x = 1 plane centered at (1, 0, 0). It passes through B on circle 1, but is far from circles 2 and 3. Fix this by tilting the circle toward the z-axis, keeping the bottom point at B. When it meets circles 2 and 3, those points are F and G. Their 120-degree rotational counterpart is E on circle 1, completing the set; the tilted circle and its rotated versions are circles 5, 6 and 7. (I'm sure that wasn't the easiest thing to follow. If someone who does follow would like to make a picture, that may help.) If I've done the numerical geometry correctly, point E has coordinates (.22893146917093, 0, .63675216588960), approximately. x and z should be algebraic, and in fact they appear to be the real roots of the cubic equations 13x^3 + 3x^2 + 3x - 1 = 0 and 13z^3 + 11z^2 + 5z - 11 = 0. -- Fred W. Helenius fredh@ix.netcom.com