- View SourceDear Jean-Pierre Ehrmann,

LOTS OF THANKS FOR THE BARYCENTRICS !!

I have tested them in the 6-9-13 case, they are correct. The search

number (in actual TRILINEARS) is then

1st Wernau point -13.209317003851505465.

I call the point W the 1st Wernau point, since an analogous 2nd

Wernau point can be found with equilateral triangles constructed

inwards. Also, two Wernau-Napoleon points can be found for Napoleon

triangles instead of Fermat triangles, as Floor van Lamoen has

noticed.

In order to get the 2nd Wernau point, we must undertake fissile

extraversion. We then get the barycentrics

x = a^2(x1 - 2d x2/root(3)), etc.

and the search number (in actual TRILINEARS)

2nd Wernau point -6.9391333766227036231.

Sincerely,

Darij Grinberg - View SourceDear Darij and Jean-Pierre,

I just found the same barycentrics as did Jean-Pierre.

The centers of the three circles form a triangle perspective to ABC

through X_61 (X_62).

Kind regards,

Sincerely,

Floor.

Darij Grinberg wrote:>

> Dear Jean-Pierre Ehrmann,

>

> LOTS OF THANKS FOR THE BARYCENTRICS !!

>

> I have tested them in the 6-9-13 case, they are correct. The search

> number (in actual TRILINEARS) is then

>

> 1st Wernau point -13.209317003851505465.

>

> I call the point W the 1st Wernau point, since an analogous 2nd

> Wernau point can be found with equilateral triangles constructed

> inwards. Also, two Wernau-Napoleon points can be found for Napoleon

> triangles instead of Fermat triangles, as Floor van Lamoen has

> noticed.

>

> In order to get the 2nd Wernau point, we must undertake fissile

> extraversion. We then get the barycentrics

>

> x = a^2(x1 - 2d x2/root(3)), etc.

>

> and the search number (in actual TRILINEARS)

>

> 2nd Wernau point -6.9391333766227036231.

>

> Sincerely,

> Darij Grinberg