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

there are three degrees of freedom. the internal point P can be anywhere inside the triangle. if P is fixed, there is still another degree of freedom, as JHC notes. a nice way to find all such configurations is as follows. draw the triangle ABC on a sheet of paper. draw a point P , with three rays emanating from it, on a transparency. the three rays, PQ , PR and PS should have angles between them so that angle QPR [resp. RPS , SPQ ] is the supplement of angle A [resp. B , C ] of the triangle. (we also need that the orientation of these three rays is the same (clockwise v. counterclockwise) as the orientation of the vertices A , B , C .) the configurations are obtained by placing the transparency over the paper such that P is inside the triangle, and setting D [resp. E , F ] to be the points where ray PQ intersects BC [resp. PR intersects CA , PS intersects AB ]. two degrees of freedom are moving P around inside the triangle, and the third is rotation about P . to see that the three quadrilaterals are cyclic, note that the opposite angles are supplementary, by construction of the angles between the rays. i'll leave the task of finding an explicit parametrization and formulae for the lengths for others. mike