Dear Darij,
[DG]:
[DG]:
second Napoleon triangles give a concurrence in the same points. I have
not yet tried any calculation.
<...>
Kind regards,
Sincerely,
Floorvan Lamoen
[DG]:
>>>> On the sides of a triangle ABC, describe equilateral[FvL]:
>>>> triangles BCA+, CAB+, ABC+ (all outwards). Then we
>>>> know that the circles BCA+, CAB+, ABC+ concur at the
>>>> Fermat point F+ of triangle ABC (and the lines AA+,
>>>> BB+, CC+ concur at F+, too).
>>>>
>>>> Now it is easy to prove that the circles B+C+A,
>>>> C+A+B, A+B+C concur.
>>>
>>> Of course we may take for A+B+C+ also:[DG]:
>>>
>>> * The third vertices of equilateral triangles pointed
>>> inwardly
>>> * The vertices of either of the Napoleon triangles.
> And this is also part of my Schaal triangles theory, which I hope toYes, since condition (1) clearly is symmetric.
> re-examine later. It starts with the following well-known fact:
>
> Let ABC be a triangle. If three points X, Y, Z are given so that
>
> (1) angle BXC + angle CYA + angle AZB = 0°,
>
> with directed angles modulo 180°, then the circles BCX, CAY and ABZ
> concur at a point which also lies on the lines AX, BY and CZ.
>
> What is rather new (but easy to prove): The circles YZA, ZXB and XYC
> concur, too.
[DG]:
> A triangle XYZ satisfying condition (1) is called GENERAL SCHAALYes, I saw this in my sketches, too, and also that indeed the first and
> TRIANGLE.
>
> [Note. Such triangles were studied long before Hermann Schaal with
> his note in the Mathematikunterricht 4/1967, who cites them as "well-
> known", but my naming is rather due to the circumstance that the term
> sorely needs a name!]
> Note that for the Napoleon triangle XYZ, my dynamic sketch shows that
> the circles YZA, ZXB and XYC concur at the point X(110), which lies
> on the circumcircle of triangle ABC. The point X(110) is not fissile,
> and hence if XYZ is the second Napoleon triangle, the circles YZA,
> ZXB and XYC should also pass through the same point X(110). Since
> this would be rather unusual, please verify it.
second Napoleon triangles give a concurrence in the same points. I have
not yet tried any calculation.
<...>
Kind regards,
Sincerely,
Floorvan Lamoen
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