- Dear Bernard,

I know nothing about angular coordinates, so hesitate to say much

Your proof below suggests that, if F = X13, then it is also X13 of

its cevian and anticevian triangle. This isn't always so, it can

be X14 for one or other.

> > We have the following fact :

coordinates

> >

> > If F is a Fermat Point of triangle ABC, then F is also a

> > Fermat Point of its Cevian, and of its Anticevian Triangle.

>

> this is direct consequence of the fact that the angular

> of a point wrt ABC, its cevian and anticevian triangles are the

same.

Is this true?

I'm guessing at the definition, but I think there may be

supplements involved. Probably I just guessed wrongly.

> for X13 : 120°, 120°, 120°

Yours

>

> [snip]

>

>

Wilson - Ah, that is what I wanted to know. But it seems to me that their

angular coordinates would not change, so both points should be the

same. It puzzles me that the second points are different. Intriguing

if they are.

Oh I know, it must be true that if we get the pedal and antipedal of

Fs, then the three Fs's are the same. This makes 12 points from these

six triangles, which I bet have an interesting relationship, Can't

wait until I have some time with Sketchpad.

Hmm, algebra at work.

Steve

On Dec 8, 2005, at 9:55 AM, Wilson Stothers wrote:

> the three triangles will have different second

> Fermat Points, so the lines FnFs are not parallel.

[Non-text portions of this message have been removed] - Bernard,

The two Fermat points each have a cevian and anticevian triangle. The

4 Euler lines of these triangles seem to be parallel to that of ABC

(observed on Sketchpad, not proved). Is there a curve, all of whose

points have Euler line parallel to that of ABC?

Steve - Dear Steve,

> The two Fermat points each have a cevian and anticevian triangle. The

It is true.

> 4 Euler lines of these triangles seem to be parallel to that of ABC

> (observed on Sketchpad, not proved).

> Is there a curve, all of whose

this was answered by Wilson yesterday under the thread [EMHL] The

> points have Euler line [of their cevian and anticevian triangles, I

> suppose] parallel to that of ABC?

Cubic K278!

Best regards

Bernard

[Non-text portions of this message have been removed]