## 4259Re: Points

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• Nov 2, 2001
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
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