> 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
