Re: [EMHL] Re: Circular macros
- Dear friends
Now at last, I have come to construct the fixed points of a direct circular
map <f> and today, I begin with the case where <f> is a circular
transposition (of order 2).
If P and Q are the fixed points, we know that (P, Q, M, f(M)) is a harmonic
quadrangle and that the middle O of segment PQ is the central point of <f>:
f(O) = Infty and f(Infty) = O.
At first, I suppose that the central point O is given with a pair (A, A' =
f(A)) of homologous points.
Then line PQ must be the internal bissector of angle (OA, OA').
As the 4 points (P, Q, A, A') must be on a same circle with center O', then
the point O' is the intersection of the perpendicular bissector of segment
AA' with the perpendicular bissector of PQ which is just the external
bissector of the angle (OA, OA').
So the construction of the fixed points P and Q is easy:
1° Draw the perpendicular bissector (L) of segment AA'.
2° Draw internal and external bissectors (Di) and (De) of angle (OA, OA')
3° Draw the intersection O' of the lines (L) and (De).
4° The circle of center O' through A and A' cuts line (Di) in the fixed
points P and Q of <f>.
But in general a circular transposition <f> is given by 2 pairs (A, A') and
(B, B') of homologous points and we go back to the previous case by
constructing the central point O of <f>.
But we have seen in some previous post that O is just the fixed point of the
direct similitude sending A to B' and B to A'.
And so we are done..
Here I want to raise a problem caused by the difference between limit points
and fixed points in circular geometry.
Limit points are not objects of the circular geometry, that is to say are
not invariant by the circular group when in the contrary the fixed points
are circular invariant.
More precisely let f and g, 2 direct circular maps conjugate in the
circular group, that is to say we have:
g = h. f . h' , where <h' > is the inverse map of the circular map <h>.
Let (I, J) be the limit points object and image of <f> and P and Q its fixed
Then the limit points of <g> are not h(I) and h(J) when its fixed points are
always h(P) and h(Q).
So the previous construction of the fixed points P and Q of the
transposition <f> is not circular invariant for we have used the central
Even the great Morley was aware of that for he gives in his book: Inversive
geometry, page 54-55, a circular invariant construction of these fixed
points. As usual, his proof is difficult to understand for he uses his own
And in fact even today, I still don't understand his proof.
So I found a proof of my own and for the moment , I only give you a hint,
try to look at the configuration obtained in the case where one of the fixed
points of the transposition <f> is the point at infinity.
I have called this configuration: the 2 lunes and it is very beautiful to
In the next post, I will give the general construction of the fixed points
of any direct circular map.
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- --- In Hyacinthos@yahoogroups.com, "jpehrmfr" <jean-pierre.ehrmann@...>
> Dear friends,
> thanks to François and his brilliant explanations, I'm now very
> interested by circular geometry.
> So, I suggest a little problem for the amateurs like me :
> L is a line intersecting the circumcircle (O) of ABC, P lies on (O);
> there exists a circular transposition with pole P mapping A,B,C to
> three points A', B', C' of L.
> Find the locus of the fixed points of the transposition when L is
> and P moves on the circumcircle.
> In the same kind of idea :
> P and L are given; construct A', B', C' (easy) and the fixed points