[math-fun] Re: cyclic quadrilaterals
John Conway writes:
Given P, there's a 1-parameter family. You can get them all by dropping "isoclines" from P to the sides - ie., lines such that angles AFP, BDP, CEP are all equal. Alternatively, let AEPF be any circle through A and P, and then BFPD and CEPD be the circumcircles of BFP and CEP (which will automatically intersect at a point DE on BC).
Thanks, John. Since this gives us 3 real degrees of freedom in total, I wonder if we get a nice parametrization of the full set of configurations (for A,B,C fixed) by specifying the three lengths AP, BP, and CP (or perhaps the three lengths DP, EP, and FP, or perhaps the lengths AD, BE, and CF, or ...). What I ideally want is a set of three parameters such that the distances between all the points can be expressed as rational functions of the parameters and the original lengths AB, AC, BC. Is this too much to hope for? Jim
On Mon, 27 Oct 2003, James Propp wrote:
John Conway writes:
Given P, there's a 1-parameter family. You can get them all by dropping "isoclines" from P to the sides
I wonder if we get a nice parametrization...
What I ideally want is a set of three parameters such that the distances between all the points can be expressed as rational functions of the parameters and the original lengths AB, AC, BC. Is this too much to hope for?
Well, I don't exactly know. But I do know that the interesting thing here is that there's a 1-1 correspondence between the point P and the shape of the triangle DEF. (P is called the "pivot point" or "Miquel point" corresponding to that shape.) What happens is that as you rotate the three lines PD,PE,PF bodily, then their intersections with the respective sides are the vertices of the generic triangle that has the same shape as DEF. The smallest DEF is the pedal triangle of P, and if you rotate from this through angle theta it gets scaled up by sec(theta). So there's a generally-two-to-one correspondence between the set of three lengths EF,FD,DE and the configuration here - except that of course the three lengths must saisfy the triangle inequality. No - that's not quite correct - it's really two-to-two because the inverse of P in the circumcircle is the pivot for the everted triangle, which has the same edge-lengths. There's another correspondence that's two-to-two too. That makes two two-to-two correspondences, of course. I hope that doesn't sound too two-to-two to two of you. JHC
participants (2)
-
James Propp -
John Conway