## Re: Arnold theorem

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• ... call them ABC and A B C ... ... and searched for another point collinear with the two orthologic centers P and P and the perspector Q. If triangle A B C
Message 1 of 13 , Mar 5, 2003
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I wrote:

>> If two orthologic triangles
call them ABC and A'B'C'
>> are perspective, then the perspector is
>> collinear with the two orthologic centers,
>> and the join of the three points is
>> orthogonal to their perspectrix!

... and searched for another point collinear with the two orthologic
centers P and P' and the perspector Q.

If triangle A'B'C' is inscribed in triangle ABC, then the Longchamps
point L of triangle ABC is collinear with P, P' and Q.

If triangle A'B'C' is inscribed in the circumcircle of triangle ABC,
then (I conjecture that) the circumcenter O is collinear with P, P'
and Q.

THERE MUST BE A GENERALIZATION!
----

Sincerely,
Darij Grinberg
• ... Correct. ... I call a calculation trivial if I can do it without needing a computer to handle the polynomials. In many problems the polynomials tend to
Message 2 of 13 , Mar 5, 2003
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Darij Grinberg wrote:

> I don't know of which computational verification is Barry Wolk

> speaking, but I think he uses barycentric coordinates.

Correct.

> Perhaps a reason why
> theorem is that I am still too lazy to learn representing
> perpendicular lines in barycentrics.

I call a calculation "trivial" if I can do it without needing a
computer to handle the polynomials. In many problems the
polynomials tend to get ridiculously large.

It is very easy to handle perpendicular lines. Take two points
at infinity, P(x:y:z) and P'(x':y':z'), so x+y+z = x'+y'+z' = 0.
Then lines through P are perpendicular to lines through P' iff
x x' SA + y y' SB + z z' SC = 0

So given P, we can find P' by letting x' = y SB - z SC, y' and
z' similarly.

And the line through H and P' has coordinates
[x SA : y SB : z SC]
--
Barry Wolk

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• Dear Barry Wolk, ... Many thanks... the next step would be (perhaps) generalizing to Lamoen s Pl-perpendicularity, but I think this is simply replacing SA, SB,
Message 3 of 13 , Mar 5, 2003
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Dear Barry Wolk,

In Hyacinthos message #6661, you wrote:

>> Darij Grinberg wrote:
>>
>> > I don't know of which computational verification is Barry Wolk
>> > speaking, but I think he uses barycentric coordinates.
>>
>> Correct.
>>
>> > Perhaps a reason why
>> > theorem is that I am still too lazy to learn representing
>> > perpendicular lines in barycentrics.
>>
>> I call a calculation "trivial" if I can do it without needing a
>> computer to handle the polynomials. In many problems the
>> polynomials tend to get ridiculously large.
>>
>> It is very easy to handle perpendicular lines. Take two points
>> at infinity, P(x:y:z) and P'(x':y':z'), so x+y+z = x'+y'+z' = 0.
>> Then lines through P are perpendicular to lines through P' iff
>> x x' SA + y y' SB + z z' SC = 0
>>
>> So given P, we can find P' by letting x' = y SB - z SC, y' and
>> z' similarly.
>>
>> And the line through H and P' has coordinates
>> [x SA : y SB : z SC]
>> --
>> Barry Wolk

Many thanks... the next step would be (perhaps) generalizing to
Lamoen's Pl-perpendicularity, but I think this is simply replacing
SA, SB, SC by some constants.

Sincerely,
Darij Grinberg
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