--- In

Hyacinthos@yahoogroups.com, "Francois Rideau"

<francois.rideau@...> wrote:

>

> 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-

circumcenter.

>

> 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

uniform

> motion!), points Q moves on line AC and R moves on line AB, (also

as choo-c

> hoos!). Then the correspondence g: Q --> R is an affine

correspondence

> between lines AC and AB, that is to say preserves affine ratio.

>

> Look now at special pairs of homologous points in this affine

> correspondence.

>

>

>

> 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

AB and

> remind lines BB' and CC' are altitudes of triangle ABC. So the pair

(B', C')

> 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?

Michel