- 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

points

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

point.

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

lingo!

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

look at.

In the next post, I will give the general construction of the fixed points

of any direct circular map.

Friendly

Francois

[Non-text portions of this message have been removed] - --- In Hyacinthos@yahoogroups.com, "jpehrmfr" <jean-pierre.ehrmann@...>

wrote:>

fixed

> 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

> Jean-Pierre

>