16519Re: Simson lines

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• Jul 1, 2008
Dear Alxei
I wrote :
> Here is a better presentation of a generalization :
> Starting with a point J in the plane of ABC, the perpendicular
lines
> through J to JA, JB, JC intersect BC, CA, AB respectively at A',
B', C'.
> It is well known that A',B',C' are on a same line L(J).
> For any point M,
> Ma is the point of the line AM such as <MJA = <AJMa
> Mb is the point of the line BM such as <MJB = <BJMb
> Mc is the point of the line CM such as <MJC = <CJMc
> Then Ma,Mb,Mc are collinear if and only if M lies on L(J)
> In this case, the four lines A'Ma, B'Mb, C'Mc, MaMbMc touch the
> inscribed conic with focus J
> This can be proved with projective tools using the fact that M->Ma
is
> the harmonic homology with center A and axis the line JA'
> * The harmonic homology (I don't know if this English name is
correct)
> with center O, axis L maps M to his harmonic conjugate wrt O and OM
> inter L
>
>
> I think that, if we consider three harmonic homologies with centers
> A,B,C, the locus of M with three colinear images is generally a
cubic,
> even when the three axis concur. In the particular case of three
axis
> La,Lb,Lc going through a point J and such as there exists an
involution
> swapping (JA,La), (JB,Lb),(JC,Lc), then the cubic degenerates in
three
> lines :
> the fixed lines of the involution and the line through La inter BC,
Lb
> inter CA, Lc inter AB

Now, if we consider a line L with trilinear pole P and intersecting
the sidelines of ABC at A',B',C', we get :
Suppose that M is a point of L; the harmonic homology with center A,
axis JA' maps M to Ma,.., similarly Mb,Mc
Then
Ma, Mb, Mc are collinear
the four lines A'Ma, B'Mb, C'Mc, MaMbMc touch the inscribed conic
with perspector the cevian product J*P.
Looking at that, I've realized something which should be interesting
for Bernard :
For any point J, the cevian product J*o(J), where o(J) is the
orthocorrespondant of J, is the perspector of the inscribed conic
with focus J.
Friendly. Jean-Pierre
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