Dear Jean-Pierre

>looking at the beginning of the very interesting from EH Lemoyne:

You mean Lemoine (as in the subject line)

>Suite de theoremes et de resultats concernant la geometrie du

>triangle avalaible at http://www.hti.umich.edu/u/umhistmath/

>I see that he discovered the following centers :

>1) Ma is the unique point on the segment BC such as

>MaB + d(Ma,AB) = MaC + d(Ma, AC) with d for distance; define Mb, Mc

>cyclically. Then MaMbMc and ABC are perspective at the point of the

>line IG with trilinear 1+2R/a:...

The perpendicular to BC at Ma intersects AB at Ab, and AC at Ac.

Define

A' = BAc /\ CAb. Similarly B', C'.

If my midnight calculations were correct, then these three

points are collinear.

And probably the same is true for the other case below

but didn't try it.

Good night from Athens

Antreas

>2) Na is the unique point on the segment BC such as

>NaB + d(Na,AC) = NaC + d(Na, AB) with d for distance; define Nb, Nc

>cyclically. Then NaNbNc and ABC are perspective at the point of the

>line IG with trilinear (2R/a)-1:...

>I don't think - but, may be, I'm wrong - that these points are in

>the current ETC.

>Friendly. Jean-Pierre

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