## Re: I--incentral H parallel to Euler

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• ... incentral ... I think we met this before in #11564. This contained more information. Suppose we write Ix (x = o,a,b,c) for the in- and ex-centres, and
Message 1 of 7 , Oct 12, 2005
--- In Hyacinthos@yahoogroups.com, Steve Sigur <s.sigur@c...> wrote:
>
> The line from the incenter of ABC to the orthocenter of the
incentral
> triangle is parallel to the Euler line of ABC.
>
> I can see it on Sketchpad. I can verify this in coordinates, but
> there must be a reason why this is true, or at least a better way to
> understand this. Can anyone help?
>
> Thanks,
>
> Steve

I think we met this before in #11564.
Suppose we write Ix (x = o,a,b,c) for the in- and ex-centres,
and write Hx for the orthocentre of the cevian triangle of Ix.
Then ETC tells us that Ho = X500, and also that
Ho is on OK, and on IoX30, so X30 is on IoHo.
OK is a strong line, so Hx is on OK for all values of x, also
X30 is a strong point, so IoHo,IaHa,IbHb,IcHc concur at X30.

The former is the general result for the orthocentres of four
triangles obtained from four lines. Here, we have the sides of
the incentral triangle and the antiorthic axis. You mentioned
this in an earlier message.

I have two questions on this

Question 1
What is Ix relative to its cevian triangle,
i.e. is it the point Xn in ETC for some n?
Trivially, it is a point which is an in- or ex-centre
of its anticevian triangle. Does this help?

Question 2
Suppose we replace I above by a general centre P,
and write Hx for the orthocentre of the cevian triangle of Px.
What is the locus of P for which PxHx (x=o,a,b,c) concur?
What is the locus of the points of concurrence?

Cabri sketching suggests that the points are rare - indeed
the Ix may be the only non-degenerate ones.
Note that, if P is on the first locus, so are Pa,Pb,Pc,
so an equation of the locus ought to be quadratic in the
coordinates of P.

Wilson
• ... If I did not have ETC to look at, what principles in triangle geometry tell us that the line from the incenter of ABC to Orhocenter of incentral triangle
Message 2 of 7 , Oct 13, 2005
On Oct 12, 2005, at 8:20 AM, Wilson Stothers wrote:

> Then ETC tells us that Ho = X500, and also that
> Ho is on OK, and on IoX30, so X30 is on IoHo.

If I did not have ETC to look at, what principles in triangle
geometry tell us that the line from the incenter of ABC to Orhocenter
of incentral triangle is parallel to Euler?

[Non-text portions of this message have been removed]
• Dear Wilson, ... It seems that your locus is the sextic Q027. This is the locus of the fixed points of the isoconjugations with pole on the Thomson cubic.
Message 3 of 7 , Oct 14, 2005
Dear Wilson,

> [WS] Question 2
> Suppose we replace I above by a general centre P,
> and write Hx for the orthocentre of the cevian triangle of Px.
> What is the locus of P for which PxHx (x=o,a,b,c) concur?

It seems that your locus is the sextic Q027.
This is the locus of the fixed points of the isoconjugations with
pole on the Thomson cubic.
Hence, for any point M on Q027, the vertices of the anticevian
triangle of M lie on Q027.

In this case, these 4 lines are in fact parallel.

Best regards

Bernard

[Non-text portions of this message have been removed]
• ... Bernard, do not mean that the 4 lines are parallel for every line on this sextic, or just our particular points. The fact that the square roots of the
Message 4 of 7 , Oct 16, 2005
On Oct 14, 2005, at 7:41 AM, Bernard Gibert wrote:

> It seems that your locus is the sextic Q027.
> This is the locus of the fixed points of the isoconjugations with
> pole on the Thomson cubic.
> Hence, for any point M on Q027, the vertices of the anticevian
> triangle of M lie on Q027.
>
>
> In this case, these 4 lines are in fact parallel.

Bernard, do not mean that the 4 lines are parallel for every line on
this sextic, or just our particular points.

The fact that the square roots of the Steiner foci lie on this is
interesting. I wonder if sexic is the smallest degree that could hold
them?

Steve

[Non-text portions of this message have been removed]
• Dear Steve ... [SS] Bernard, do not mean that the 4 lines are parallel for every line [correction : point] on this sextic, or just our particular points. The 4
Message 5 of 7 , Oct 16, 2005
Dear Steve

> It seems that your locus is the sextic Q027.
> This is the locus of the fixed points of the isoconjugations with
> pole on the Thomson cubic.
> Hence, for any point M on Q027, the vertices of the anticevian
> triangle of M lie on Q027.
>
>
> In this case, these 4 lines are in fact parallel.

[SS] Bernard, do not mean that the 4 lines are parallel for every
line [correction : point] on
this sextic, or just our particular points.

The 4 lines are // for any point on Q027

[SS] The fact that the square roots of the Steiner foci lie on this is
interesting. I wonder if sexic is the smallest degree that could hold
them?

The diagonal hyperbola through G, the vertices of the antimedial
triangle which is tangent at G to the Steiner focal axis contains the
8 square roots of the Steiner foci.

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Bernard, Belated thanks for your work on this - it is a very nice result. ... A little background.. Previous messages have recalled the result of Steiner
Message 6 of 7 , Oct 17, 2005
Dear Bernard,

Belated thanks for your work on this - it is a very nice result.

>
> > [WS] Question 2
> > Suppose we take a general centre P(exversions Po,Pa,Pb,Pc)
> > Write Hx for the orthocentre of the cevian triangle of Px.
> > What is the locus of P for which PxHx (x=o,a,b,c) concur?
>
> It seems that your locus is the sextic Q027.
> This is the locus of the fixed points of the isoconjugations with
> pole on the Thomson cubic.
> Hence, for any point M on Q027, the vertices of the anticevian
> triangle of M lie on Q027.

> In this case, these 4 lines are in fact parallel.
>
A little background..

Previous messages have recalled the result of Steiner that
the points Hx lie on a line, and that this line contains O.

It appears to be less well known that the line also has the
point P^2, the barycentric square of P (new to me)

For P = I, the line is OK (K = I^2)

It is implicit in ETC (see X500) that the lines IxHx concur at X30
(i.e. are parallel to the Euler Line)
Your result shows that the lines concur (at infinity) if and only if
P^2 is also on the K002, Thomson Cubic.

I recently observed that, if Nx is the nine-point centre of Ix,
then the four lines IxNx concur at O.

I looked for the locus of P for which the PxNx concur.
It turns out to be precisely the same condition as for the Hx,
The lines PxNx concur if and only if P^2 is on K002.

I tried replacing H by O, but got a septic for P^2.

Then I tried replacing H by K, and got a surprise -
it appears that we cannot have all four concurent???
(except for a finite set of points)

I also looked at Px*, the isogonal of Px in Tx
(it was easy to compute, so I did)
For all P, the lines PxPx* concur at a point Q.
The point of concurrence is quite simple, if P = p:q:r
then Q has first barycentric p^2(b^2+c^2-a^2).

For P = I, O, Ix, Ix* and Nx are collinear,
which verifies that O is on each IxNx

Question - for what other points is the answer the
Thomson Cubic?

Regards,

Wilson.
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