--- In Hyacinthos@yahoogroups.com
, "Francois Rideau"
> Dear friend
> As I am in my dotagen I see choo-choo everywhere!
> First, your problem only makes sense when triangle ABC is
isosceles in A.
> So in the sequel, I suppose that AB <> AC.
> Line HM is on the point A '' symmetric of A wrt the ABC-
> Now I give you a solution based on choo-choo theory.
> Given any point P on line HM , we project him orthogonally at Q on
side AC and
> on R on side AB.
> When P moves on line HM (as a choo-choo that is to say with a
> motion!), points Q moves on line AC and R moves on line AB, (also
> hoos!). Then the correspondence g: Q --> R is an affine
> between lines AC and AB, that is to say preserves affine ratio.
> Look now at special pairs of homologous points in this affine
> 1° When P is on A'', its projections on line AC and AB are just C
and B, so
> the pair (C,B) is a pair of homologous points.
> 2) When P is on H, its projections are B' on line AC and C' on line
> remind lines BB' and CC' are altitudes of triangle ABC. So the pair
> is another pair of homologous points in this affine correspondence.
Is it known that the envelop of line (QR) when P moves on line (HM)
is a parabol
If P = k H + (1-k) M
Q = (1+k) S_c A +(b²+S_a - k S_c) C
and R = (1+k)S_b A + (S²+S_a - k S_b) B
it is easy to calculate equation of line (QR)
How can we find equation of the envelop?