Dear Antreas and other Hyacinthists,

> [APH]

> > Let ABC be a triangle, P a point, and A'B'C', A"B"C" the cevian,

> pedal

> > triangles of P, resp.

> >

> > Find the P's inside the triangle such that:

> >

> > PB" PC" PC" PA" PA" PB"

> > 1. ---- + --- = --- + --- = --- + ---

> > AC' AB' BA' BC' CB' CA'

> >

> >

> > PB" PC" 1 PC" PA" 1 PA" PB" 1

> > 2. (---- + ---) * --- = (---- + ----) * --- = (---- + ----) * ---

> > AC' AB' BC BA' BC' CA CB' CA' AB

> >

> > where the line segments are not signed.

[JPE]

> If P lies inside ABC, the three numbers in 2. are barycentric

> coordinates of the homothetic of P in (G,1/4).

> Thus 2. is true only for P = G (with common value 8*area/(3abc))

> 1. is true only for P = homthetic in (G,4) of the isotomic

conjugate

> of the incenter if this point lies inside ABC - with common value

> 8*area/(ab+bc+ca) -

If P is barycentric (x,y,z) and inside ABC, then

AB' = bz/(x+z),..

PA" = 2dx/(a(x+y+z)),... where d = area(ABC)

Hence (PB"/AC' + PC"/AB')/BC = k(2x+y+z) with k = 2d/(abc(x+y+z))

This a four lines proof of the fact that :

The three numbers in 2. are barycentric coordinates of the homothetic

of P in (G,1/4).

Note that 2. has exactly 4 solutions: G and the homthetics of A,B,C

in (G,4).

May be, it could be interesting to find at what condition 1. has a

solution inside ABC (ie the homothetic in(G,4) of the isotomic

conjugate of the incenter lies inside ABC)

A related problem could be the following :

Find P such as A'A" = B'B" = C'C" (3)

Apart the orthocenter, if (3) is signed, we have 3 solutions lying on

the trilinear polar of X333 - if (3) is not signed, we have 9

supplementary solutions corresponding to extra-versions -

The 3 solutions (if signed) are not constructible; may be it could be

interesting to find an easy conic-construction of these points and to

study the complete configuration of the 12 solutions.

Friendly. Jean-Pierre