Dear Antreas,

You wrote:

> I think that the following search in ETC would be

> interesting but I guess

> it is not easy!

Yes. It is not easy.

>

> Let (A), (B), (C) be three circles defined with known

> points listed in ETC

> such that they do not have a common KNOWN point.

>

> Not like these circles, for example:

>

> 1) 1 2 11 392

> 2) 1 2 23 36

> 3) 1 3 21 759

>

> Are there such circles which are concurrent?

Yes. There are.

In order to reduce the time of calculation by the

computer one of the three circles can be the

circumcircle (O).

Hence one example is the circles

(O) (11, 13, 14), (36, 65, 79)

with common point P not in ETC and a second is

(O), (11, 13, 14), (46, 57, 79)

with the same common point P.

Hence we have found 4 circles concurrent at P.

(11, 13, 14) means the circle passing through

the two Fermat points X(13), X(14)

and the Feuerbach point X(11).

>

> If there are, then the point of concurrence is a NEW

> interesting point !

I think that this point has not simple coordinates.

Best regards

Nikos