16516Re: Simson lines
- Jul 1 2:18 AMDear Alexei
> The polar transformation and the properties of Simson lines allow toprove next interesting fact.
> Let I be the incenter of ABC. For an arbitrary point P we denote asP_A the point on the line PA such that IA is the bisector (internal or
external) of angle PIP_A. The points P_B, P_C are defined similarly.
> 1. If P_A, P_B, P_C are collinear then this line touches the incircleof ABC.
> 2. The locus of points P with this property is the line.Here is a better presentation of a generalization :
> Can anybody prove this without using the polar transformation?
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
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
the fixed lines of the involution and the line through La inter BC, Lb
inter CA, Lc inter AB
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