- On Sunday, August 31, 2003, at 02:48 AM, Darij Grinberg wrote:

>

Dear Darij,

I did receive this and it has not yet made it to the group, so I will

reply to you and the group. I divided your long message into several

parts and will reply in several messages.

Your results complement mine perfectly. This is an interesting schema.

> In Hyacinthos message #7702, you wrote:

I did not know that this result is so recent. I hope the authors

>

>>> The Euler lines of IoBC, AIoC, ABIo, where Io

>>> is the original incenter, concur at the Schiffler

>>> point Ho, which is on the Euler line of ABC.

>>> Successively replacing Io by the other members

>>> of its family Ia, Ib, and Ic generates the other

>>> Schiffler points Ha, Hb, Hc. All told this is 12

>>> lines which meet three at a time at 4 points on

>>> the Euler line of ABC.

>

> Nice that we are discussing these points again. A

> property given by the original solvers of Schiffler's

> problem and soon forgotten is that the Schiffler

> point Ho divides the segment between the centroid G

> and the circumcenter O in the proportion

>

> GHo 2r

> --- = --,

> HoO 3R

>

> where r is the inradius and R is the circumradius.

> By the way, the original solution in

>

> K. Schiffler, G. R. Veldkamp and W. A. van der

> Spek, Problem 1018 and solution, Crux

> Mathematicorum 12 (1986) 150–152

>

> was entirely synthetic, using only the Menelaos

> theorem. It was a surprise for me when I learned

> about it.

>

subscribe to this group. Nor did I know the formula that you quoted.

Very useful.

This formula can be extraverted as in

GHa 2r_a

--- = --,

HaO 3R

so we should get the distribution of all four points relative to G and

O. The signs are tricky however. Under extraversion the radii behave

as r_o --> -r_a, and R ---> R, so that the extra-Schifflers have a

negative ratio GHa/HaO, but we have to be careful about signs. I will

think more about this and post again. I would like to fully understand

the distribution of these points.

Friendly,

Steve - Dear Antreas

> There was paper by a tetrad of hyacinthos listmembers

You've proved in your paper that the locus of P such as the Brocard

> on concurrent Euler lines (and other lines through O)

> of the triangles PBC,PCA,PAB,

> which can be found here:

>

> http://forumgeom.fau.edu/FG2001volume1/FG200109index.html

>

> If I remember correctly JPE mentioned also cases with tripolars

> of points. I think that if the Lemoine axes of PBC,PCA,PAB

> are concurrent, then all four Lemoine axes of ABC, PBC,PCA,PAB

> are concurrent.

> I don't remember if there was a discussion on the locus

> of P with this property. (JP may correct my memory!)

axis of PBC, PCA, PAB concur is again the Neuberg cubic and that, if

they concur, they concur on the Brocard axis of ABC.

If the Lemoine axis of PBC, PCA, PAB concur, they concur on the

Lemoine axis of ABC but the locus of such P is a tricircular

circumsextic and I'm unable to find any interesting point on this

sextic.

Friendly. Jean-Pierre