- I wrote:

>> If two orthologic triangles

call them ABC and A'B'C'

>> are perspective, then the perspector is

... and searched for another point collinear with the two orthologic

>> collinear with the two orthologic centers,

>> and the join of the three points is

>> orthogonal to their perspectrix!

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 - Darij Grinberg wrote:

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

Correct.

> speaking, but I think he uses barycentric coordinates.

> Perhaps a reason why

I call a calculation "trivial" if I can do it without needing a

> I had not found a trivial coordinate proof of the Arnold

> theorem is that I am still too lazy to learn representing

> perpendicular lines in barycentrics.

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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http://taxes.yahoo.com/ - Dear Barry Wolk,

In Hyacinthos message #6661, you wrote:

>> Darij Grinberg wrote:

Many thanks... the next step would be (perhaps) generalizing to

>>

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

>> > speaking, but I think he uses barycentric coordinates.

>>

>> Correct.

>>

>> > Perhaps a reason why

>> > I had not found a trivial coordinate proof of the Arnold

>> > 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

Lamoen's Pl-perpendicularity, but I think this is simply replacing

SA, SB, SC by some constants.

Sincerely,

Darij Grinberg