## Re: [EMHL] Fermat Points

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• 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
Message 1 of 10 , Dec 8, 2005
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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 :
> >
> > 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
coordinates
> 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°
>
> [snip]
>
>

Yours

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
Message 2 of 10 , Dec 8, 2005
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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
Message 3 of 10 , Dec 9, 2005
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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, ... It is true. ... this was answered by Wilson yesterday under the thread [EMHL] The Cubic K278! Best regards Bernard [Non-text portions of this
Message 4 of 10 , Dec 9, 2005
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Dear Steve,

> 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).

It is true.

> Is there a curve, all of whose
> points have Euler line [of their cevian and anticevian triangles, I
> suppose] parallel to that of ABC?