## Re: relations between some points

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• ... 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
Message 1 of 11 , Aug 31, 2003
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:
>
>>> 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
>
> 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
>
I did not know that this result is so recent. I hope the authors
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
the distribution of these points.

Friendly,

Steve
• part 3 of 3 ... Here are how the lines work: We are looking at Euler lines of IxBC, AIxC, ABIx where x has 4 possible o, a, b, c. I denote each of the 12 Euler
Message 2 of 11 , Sep 1, 2003
part 3 of 3

>>> The 12 Euler lines meet 2 at a time at the 6
>>> midpoints of the lines connecting Io, Ia, Ib, Ic,
>>> all of which are on the circumcircle.

...
>
>>> The 12 Brocard lines meet 3 at a time at
>>> : bb/(c+a) : and its 3 extraversions, (known as
>>> the isogonic Spieker points) all of which are on
>>> the Brocard line of ABC.
>
>>> The 12 Brocard lines also meet 3 at a time at
>>> : bb/(c-a) : [X(101)] and its 3 extraversions,
>>> all of which are on the circumcircle.
>
>>> The 12 Brocard lines meet 2 at a time at the 6
>>> midpoints of the lines connecting Io, Ia, Ib, Ic,
>>> all of which are on the circumcircle.
>>>

...
>
>>> The 12 Euler lines also have a property not
>>> matched by the Brocard lines. They meet 2 at a
>>> time at 12 points on the edges of ABC (four on
>>> each edge), each line going between two such
>>> points.
>
> Sorry, could you please explain which ones concur
> at one point on BC ? There are many of them and I
> don't find the exact ones.

Here are how the lines work:
We are looking at Euler lines of IxBC, AIxC, ABIx where x has 4
possible o, a, b, c.
I denote each of the 12 Euler lines by two letters xy, where x = o, a,
b, c tells us which of the 4 incenters is being used and y tells which
triangle. Hence oa represents the Euler line of IoBC and bc represents
the Euler line of ABIb.

Here is how the concurrences occur.

Ho, the original Schiffler point, is the concurrence of oa, ob, and
oc.
Qo = : b/(c-a) : is the concurrence of aa, bb, cc, this point is
on the circumcircle.
ob and bb meet at Io + Ib, the midpoint of Io and Ib, also on the
circumcircle.

Now your answer -- these four pairs of lines: ac oc, ca oa, aa
ba, and bc cc meet on side b of ABC.

>
> Here are some of my results on the Schiffler
> configuration; I considered both the Euler lines
> and the Brocard axes. Most of the results above
> are from Hyacinthos message #6751:
>
> If the Euler line of triangle BIC intersects BC
> at Wa, and analogously define points Wb and Wc,
> then the lines AWa, BWb and CWc concur at the
> point X(86). This point is the isotomic conjugate
> of the Spieker point, and lies on the symmedian
> trail (the line joining the centroid with the
> symmedian point).
>
> The point Wa lies on the line joining the
> centroid and the a-excenter of triangle ABC.
>
> If the Brocard axis of triangle BIC intersects BC
> at Xa, and analogously define points Xb and Xc,
> then the lines AXa, BXb and CXc concur at the
> point X(81), with barycentrics
>
> / a b c \
> ( --- : --- : --- ).
> \ b+c c+a a+b /
>
> This point also lies on the symmedian trail.
> Moreover, it lies on the line joining the
> incenter, the "regular" Schiffler point and
> the isogonal of the Spieker point (i. e.,
> the intersection of the Brocard axes of
> triangles BIC, CIA, AIB).
>
> The point Xa lies on the line joining the
> symmedian point and the a-excenter of triangle
> ABC.
>
> The points X(86) and X(81), lying on the
> symmedian trail, must obviously have their
> extraversion lying on this trail too, so we
> get two new collinear quadruples of weak points!
>
> Perhaps this will be of interest for you.
>

oh yes, it was

Here is a summary of yours and my results side by side. I will
abbreviate the Euler line, the Brocard line, and circumcircle of ABC
by E, B, and CC. I will use the above notation for the individual
Euler and Brocard lines, and X--Y will denote the line between points
X, Y.

Euler schema
1. The traces Wa, Wb, Wc of lines oa, ob, oc on their respective sides
are perspective to ABC, with perspector tSo = :1/(c+a): , the
isotomic Spieker point on G--K. There are 4 such points.
2. tSo lies on the line joining the incenter and the isotomic incenter.
There are 4 such lines.
3. Wa is on Ia--G. There are 3 such lines.
4. oa, ob, and oc concur at Ho, the original Schiffler point, on E.
There are 4 such points.
5. aa, bb, cc concur at : b/(c-a) : on CC. There are 4 such points.
6. ob and bb meet at Io + Ib, the midpoint of Io and Ib, also on the
circumcircle. There are 6 such points.

Brocard Schema
1. The traces Xa, Xb, Xc of lines oa, ob, oc (now referring to Brocard
lines) on their respective sides are perspective to ABC, with
perspector : b/(c+a) : , also on G--K. There are 4 such points.
2. : b/(c+a) : lies on the line joining the incenter, Ho, and gSo
(the isogonal Spieker point). There are 4 such lines.
3. Xa is on Ia--K. There are 3 such lines.
4. oa, ob, and oc concur at gSo = : bb/(c+a):, the isogonal Spieker
pt, on B.. There are 4 such points.
5. aa, bb, cc concur at : bb/(c-a) : on CC. There are 4 such points.
6. ob and bb meet at Io + Ib, the midpoint of Io and Ib, also on the
circumcircle. There are 6 such points.

It seems to me likely that an analogous structure can be defined for
each point on the Neuberg cubic.

Steve
• Dear Steve Sigur, ... [...] ... Aha, thanks! ... A first experiment failed: I took the 1st Fermat point F+, and the Euler lines of triangles BF+C, CF+A, AF+B.
Message 3 of 11 , Sep 1, 2003
Dear Steve Sigur,

In Hyacinthos message #7727, you wrote:

>> >>> The 12 Euler lines also have a property not
>> >>> matched by the Brocard lines. They meet 2 at a
>> >>> time at 12 points on the edges of ABC (four on
>> >>> each edge), each line going between two such
>> >>> points.
>> >
>> > Sorry, could you please explain which ones concur
>> > at one point on BC ? There are many of them and I
>> > don't find the exact ones.
>>
>> Here are how the lines work:
>> We are looking at Euler lines of IxBC, AIxC, ABIx
>> where x has 4 possible o, a, b, c.
>> I denote each of the 12 Euler lines by two letters
>> xy, where x = o, a, b, c tells us which of the 4
>> incenters is being used and y tells which
>> triangle. Hence oa represents the Euler line of
>> IoBC and bc represents the Euler line of ABIb.
>>
[...]
>> ac oc, ca oa, aa ba, and bc cc meet on
>> side b of ABC.

Aha, thanks!

Now, I will comment on the last lines:

>> It seems to me likely that an analogous structure
>> can be defined for each point on the Neuberg cubic.

A first experiment failed: I took the 1st Fermat
point F+, and the Euler lines of triangles BF+C, CF+A,
AF+B. As you know, these Euler lines concur at the
centroid of triangle ABC, but they don't intercept
on the sides of triangle ABC the vertices of a
cevian triangle.

I am quite interested in the locus of all points P
so that the Euler lines of triangles BPC, CPA, APB
intercept the vertices of a cevian triangle. This
should pass through the incenters, the orthocenter,
the circumcenter, but any other well-known points??

Sincerely,
Darij Grinberg
• ... Hi to all members of Hyacinthos group from Sofia - Bulgaria! I joined the group in the beginning of August. I immediately checked that points Ia, Ib, Ic
Message 4 of 11 , Sep 2, 2003
--- In Hyacinthos@yahoogroups.com, "Darij Grinberg"
<darij_grinberg@w...> wrote:
> I am quite interested in the locus of all points P
> so that the Euler lines of triangles BPC, CPA, APB
> intercept the vertices of a cevian triangle. This
> should pass through the incenters, the orthocenter,
> the circumcenter, but any other well-known points??
>
> Sincerely,
> Darij Grinberg

Hi to all members of Hyacinthos group from Sofia - Bulgaria!

I joined the group in the beginning of August.
I immediately checked that points Ia, Ib, Ic and I are well
replacements of the above point P, so the Euler lines of triangles
APC, ABP, BCP and ABC concur at one point. This was also mentioned by
Steve Sigur and Darij Grinberg (and maybe others).
I also found that points X(13) TORRICELLI POINT makes the statement
true.
The statement is also true if the triangle has an angle larger than
120 degrees, and in this case points similar to X(13) are outside ABC.
If I have a mistake, the reason is because I am a new to both ETC and
our group, so sorry in this case.

My general questions arised while using the Geometer's Sketchpad (I
use version 1.0 from 1993)
Is there any library of sketches and scripts for Geometer's Sketchpad
that could be used for quick investigation of points in ETC or new
ones in our group.

ThanX

Bests

Andrey Antonov - Erol
• Dear Andrey Antonov, Welcome to Hyacinthos (as far as I, being none of the pioneers, but having myself entered Hyacinthos in December 2002, can say this); and
Message 5 of 11 , Sep 2, 2003
Dear Andrey Antonov,

Welcome to Hyacinthos (as far as I, being none of the
pioneers, but having myself entered Hyacinthos in
December 2002, can say this); and thanks for the mail.

In Hyacinthos message #7737, you wrote:

>> I immediately checked that points Ia, Ib, Ic and I
>> are well replacements of the above point P, so the
>> Euler lines of triangles APC, ABP, BCP and ABC
>> concur at one point. This was also mentioned by
>> Steve Sigur and Darij Grinberg (and maybe others).
>> I also found that points X(13) TORRICELLI POINT
>> makes the statement true.

Yes for the statement that the Euler lines concur.
But no for the other statement that they intercept
the vertices of a cevian triangle on the sides of
triangle ABC. In fact, if Wa, Wb, Wc are the
intersections of the Euler lines of triangles BPC,
CPA, APB (P = Fermat point) with the lines BC, CA,
AB, then the lines AWa, BWb, CWc generally do not
concur. Maybe, you have drawn a conclusion from an
imprecise dynamic sketch, but I rather guess you
simply meant the statement about the concurrent
Euler lines, which is true.

Sincerely,
Darij Grinberg
• ... Keep looking -- there are more! Steve
Message 6 of 11 , Sep 2, 2003
On Tuesday, September 2, 2003, at 07:53 AM, andreyantonov wrote:

> I joined the group in the beginning of August.
> I immediately checked that points Ia, Ib, Ic and I are well
> replacements of the above point P, so the Euler lines of triangles
> APC, ABP, BCP and ABC concur at one point. This was also mentioned by
> Steve Sigur and Darij Grinberg (and maybe others).
> I also found that points X(13) TORRICELLI POINT makes the statement
> true.

Keep looking -- there are more!

Steve
• Dear friends, ... the locus is a tricircular nonic (9th degree) having the perp. bisectors of ABC as real asymptotes. A, B, C are double. The tangents at A are
Message 7 of 11 , Sep 3, 2003
Dear friends,

--- In Hyacinthos@yahoogroups.com, "Darij Grinberg"
<darij_grinberg@w...> wrote:
> I am quite interested in the locus of all points P
> so that the Euler lines of triangles BPC, CPA, APB
> intercept the vertices of a cevian triangle. This
> should pass through the incenters, the orthocenter,
> the circumcenter, but any other well-known points??
>

the locus is a tricircular nonic (9th degree) having the perp.
bisectors of ABC as real asymptotes.

A, B, C are double. The tangents at A are the perp. at A to AB, AC.

it contains
- the antipodes of A, B, C on C(O,R),
- the reflections of A, B, C in the sidelines of ABC
- in/excenters
- O, H

but certainly not the Fermats !

I expect there are some other centers on the curve.

Best regards

Bernard

[Non-text portions of this message have been removed]
• ... The locus of these points is Neuberg Cubic ++ There was paper by a tetrad of hyacinthos listmembers on concurrent Euler lines (and other lines through O)
Message 8 of 11 , Sep 3, 2003
On Tuesday, September 2, 2003, at 07:53 AM, andreyantonov wrote:

>> I joined the group in the beginning of August.
>> I immediately checked that points Ia, Ib, Ic and I are well
>> replacements of the above point P, so the Euler lines of triangles
>> APC, ABP, BCP and ABC concur at one point. This was also mentioned by
>> Steve Sigur and Darij Grinberg (and maybe others).
>> I also found that points X(13) TORRICELLI POINT makes the statement
>> true.

The locus of these points is Neuberg Cubic ++

There was paper by a tetrad of hyacinthos listmembers
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!)

Antreas

--
• Dear Antreas ... You ve proved in your paper that the locus of P such as the Brocard axis of PBC, PCA, PAB concur is again the Neuberg cubic and that, if they
Message 9 of 11 , Sep 3, 2003
Dear Antreas

> There was paper by a tetrad of hyacinthos listmembers
> 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!)

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