[math-fun] cyclic quadrilaterals
Given a generic triangle ABC, in "how many ways" can you choose an interior point P and points D, E, and F on sides BC, AC, and AB (respectively) so that quadrilaterals AEPF, BFPD, and CDPE are cyclic? That is to say, how many degrees of freedom do you have? Is there a nice way to parametrize the set of such configurations (with A,B,C fixed), so as to express the lengths of segments AP, BP, CP, DP, EP, and FP in terms of the lengths of segments AB, AC, and BC together with the extra parameter(s)? B / \ / \ F D / P \ / \ A-----E-----C Jim Propp University of Wisconsin Madison, WI
On Mon, 27 Oct 2003, James Propp wrote:
Given a generic triangle ABC, in "how many ways" can you choose an interior point P and points D, E, and F on sides BC, AC, and AB (respectively) so that quadrilaterals AEPF, BFPD, and CDPE are cyclic? That is to say, how many degrees of freedom do you have? Is there a nice way to parametrize the set of such configurations (with A,B,C fixed), so as to express the lengths of segments AP, BP, CP, DP, EP, and FP in terms of the lengths of segments AB, AC, and BC together with the extra parameter(s)?
B / \ / \ F D / P \ / \ A-----E-----C
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). This is all part of "the Miquel theory" of the triangle. JHC
participants (2)
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James Propp -
John Conway