## Re: [EMHL] Re: Locus (eguilateral triangles)

Expand Messages
• Dear Antreas, Angel, Francisco I think that If L1,L2,L3 are the peprendicular bisectors of the sides of ABC (BC, CA, AB) and the lines 1, 2, 3 pass through G
Message 1 of 7 , Jan 24, 2013
• 0 Attachment
Dear Antreas, Angel, Francisco

I think that

If L1,L2,L3 are the peprendicular
bisectors of the sides of ABC (BC, CA, AB)
and the lines 1, 2, 3 pass through G
then
La is the reflection of L1 in 1
Lb is the reflection of L2 in 2
Lc is the reflection of L3 in 3.
The lines are concurrent at Q and the
locus of Q is the circle with center G
that passes through the first Fermat point
and not the circumcenter of ABC.

Is this true or false?

Best regards

> Dear Francisco Javier,
>
> Indeed, through the circumcenter and therefore by their
> antipodal, X(381). These points correspond to the lines
> 1  parallel to the Euler line and its perpendicular
> direction, respectively.
>
> Best regards,
>
> Angel M.
>
>
> --- In Hyacinthos@yahoogroups.com,
> "Francisco Javier"  wrote:
> >
> > Dear Angel,
> >
> > this circle goes through the circumcenter!
> >
> > Best regards,
> >
> > Francisco Javier.
> >
> > --- In Hyacinthos@yahoogroups.com,
> "Angel"  wrote:
> > >
> > > Dear Antreas
> > >
> > > For every line 1, the lines La, Lb and Lc are
> concurrent. The intersection point is on the circle (with
> center the centroid) of the barycentric equation:
> > >
> > > CiclicSum [ a^2y*z + x(x+y+z)
> (a^4(b^2+c^2)-a^2(2b^4+b^2c^2+2c^4)+(b^2-c^2)^2(b^2+c^2))/
> (3(-a+b+c)(a+b-c)(a-b+c)(a+b+c)) ]=0.
> > >
> > >
> > > Angel M.
> > >
> > > --- In Hyacinthos@yahoogroups.com,
> "Antreas"  wrote:
> > > >
> > > > Let ABC be a triangle, 1 a line and 2,3 two
> lines such that
> > > > the triangle bounded by the lines (1,2,3) is
> equilateral.
> > > >
> > > > Denote:
> > > >
> > > > a1, a2, a3 = the lines through A parallel to
> 1,2,3,resp.
> > > > b1, b2, b3 = the lines through B parallel to
> 1,2,3,resp.
> > > > c1, c2, c3 = the lines through C parallel to
> 1,2,3,resp.
> > > >
> > > > La = the line joining the centers of the
> equilateral
> > > > triangles bounded by the lines (a1,b2,c3) and
> (a1,b3,c2)
> > > >
> > > > Lb = the line joining the centers of the
> equilateral
> > > > triangles bounded by the lines (b1,c2,a3) and
> (b1,c3,a2)
> > > >
> > > > Lc = the line joining the centers of the
> equilateral
> > > > triangles bounded by the lines (c1,a2,b3) and
> (c1,a3,b2)
> > > >
> > > > If 1 is the trilinear polar of point P, which
> is the
> > > > locus of P such tah La,Lb,Lc are concurrent?
> > > >
> > > > APH
> > > >
> > >
> >
>
>
>
>
> ------------------------------------
>
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
• I am wondering about a generalization. One possibly is this: Let 1,2,3 be three lines (concurrent with given angles or bounding a triangle). Let La, Lb, Lc be
Message 2 of 7 , Jan 24, 2013
• 0 Attachment
I am wondering about a generalization. One possibly is this:

Let 1,2,3 be three lines (concurrent with given angles or bounding a
triangle).

Let La, Lb, Lc be the lines joining the CENTROIDS of the respective
triangles.

Are always these lines concurrent??

Let's consider a special triad of such lines, namely:

1,2,3 = the reflections of the Euler line in the sidelines BC,CA,AB, resp.
(concurrent on the circumcircle).

Denote:

a1, a2, a3 = the lines through A parallel to 1,2,3,resp.
b1, b2, b3 = the lines through B parallel to 1,2,3,resp.
c1, c2, c3 = the lines through C parallel to 1,2,3,resp.

La = the line joining the centroids of the triangles bounded by the lines
(a1,b2,c3) and (a1,b3,c2)
[the triangle (a1,b2,c3) is a point, which we take as centroid]

Similarly Lb, Lc.

Are tle lines La,Lb,Lc concurrent?

APH

On Thu, Jan 24, 2013 at 1:18 PM, Antreas <anopolis72@...> wrote:

> **
>
>
> Let ABC be a triangle, 1 a line and 2,3 two lines such that
> the triangle bounded by the lines (1,2,3) is equilateral.
>
> Denote:
>
> a1, a2, a3 = the lines through A parallel to 1,2,3,resp.
> b1, b2, b3 = the lines through B parallel to 1,2,3,resp.
> c1, c2, c3 = the lines through C parallel to 1,2,3,resp.
>
> La = the line joining the centers of the equilateral
> triangles bounded by the lines (a1,b2,c3) and (a1,b3,c2)
>
> Lb = the line joining the centers of the equilateral
> triangles bounded by the lines (b1,c2,a3) and (b1,c3,a2)
>
> Lc = the line joining the centers of the equilateral
> triangles bounded by the lines (c1,a2,b3) and (c1,a3,b2)
>
> If 1 is the trilinear polar of point P, which is the
> locus of P such tah La,Lb,Lc are concurrent?
>
> APH
>

[Non-text portions of this message have been removed]
• Dear Antreas, Yes the lines are concurrent. If in barycentrics the point at infinity of line 1 is the point (1 : x :-1-x) of line 2 is the point (1 : y :-1-y)
Message 3 of 7 , Jan 24, 2013
• 0 Attachment
Dear Antreas,
Yes the lines are concurrent.
If in barycentrics the point at infinity
of line 1 is the point (1 : x :-1-x)
of line 2 is the point (1 : y :-1-y)
of line 3 is the point (1 : z :-1-z)
then the lines La, Lb, Lc are concurrent at the point

Q = {-x^4 y + x^3 y^2 - x^3 y^3 + x y^4 + 2 x^2 y^4 - x^4 z + 6 x^3 y z -
2 x^4 y z - 6 x^2 y^2 z + 5 x^3 y^2 z - 3 x y^3 z - 5 x^2 y^3 z -
2 x y^4 z + x^3 z^2 - 6 x^2 y z^2 + 5 x^3 y z^2 + 12 x y^2 z^2 -
6 x^2 y^2 z^2 - y^3 z^2 + 6 x y^3 z^2 + y^4 z^2 - x^3 z^3 -
3 x y z^3 - 5 x^2 y z^3 - y^2 z^3 + 6 x y^2 z^3 - 4 y^3 z^3 +
x z^4 + 2 x^2 z^4 - 2 x y z^4 + y^2 z^4,
x^4 y^2 - 2 x^3 y^3 + 2 x^2 y^4 + x^3 y^4 - 4 x^4 y z + 6 x^3 y^2 z -
x^4 y^2 z - 5 x^2 y^3 z - 3 x^3 y^3 z - x y^4 z + x^4 z^2 +
6 x^3 y z^2 - x^4 y z^2 - 6 x^2 y^2 z^2 + 12 x^3 y^2 z^2 +
5 x y^3 z^2 - 6 x^2 y^3 z^2 + x y^4 z^2 - 2 x^3 z^3 - 5 x^2 y z^3 -
3 x^3 y z^3 + 5 x y^2 z^3 - 6 x^2 y^2 z^3 - 2 y^3 z^3 +
6 x y^3 z^3 - y^4 z^3 + 2 x^2 z^4 + x^3 z^4 - x y z^4 + x y^2 z^4 -
y^3 z^4, x^4 y - x^3 y^2 + 2 x^4 y^2 - 3 x^3 y^3 - x y^4 - x^2 y^4 -
x^3 y^4 + x^4 z - 6 x^3 y z + 6 x^2 y^2 z - 5 x^3 y^2 z +
x^4 y^2 z + 3 x y^3 z + 16 x^2 y^3 z + 3 x^3 y^3 z - 3 x y^4 z -
x^3 z^2 + 2 x^4 z^2 + 6 x^2 y z^2 - 5 x^3 y z^2 + x^4 y z^2 -
12 x y^2 z^2 - 6 x^2 y^2 z^2 - 12 x^3 y^2 z^2 + y^3 z^2 -
5 x y^3 z^2 + 6 x^2 y^3 z^2 + 2 y^4 z^2 - x y^4 z^2 - 3 x^3 z^3 +
3 x y z^3 + 16 x^2 y z^3 + 3 x^3 y z^3 + y^2 z^3 - 5 x y^2 z^3 +
6 x^2 y^2 z^3 - 6 x y^3 z^3 + y^4 z^3 - x z^4 - x^2 z^4 - x^3 z^4 -
3 x y z^4 + 2 y^2 z^4 - x y^2 z^4 + y^3 z^4}

Nikos

> I am wondering about a
> generalization. One possibly is this:
>
> Let 1,2,3 be three lines (concurrent with given angles or
> bounding a
> triangle).
>
> Let La, Lb, Lc be the lines joining the CENTROIDS of the
> respective
> triangles.
>
> Are always these lines concurrent??
>
> Let's consider a special triad of such lines, namely:
>
> 1,2,3 = the reflections of the Euler line in the sidelines
> BC,CA,AB, resp.
> (concurrent on the circumcircle).
>
> Denote:
>
> a1, a2, a3 = the lines through A parallel to 1,2,3,resp.
> b1, b2, b3 = the lines through B parallel to 1,2,3,resp.
> c1, c2, c3 = the lines through C parallel to 1,2,3,resp.
>
> La = the line joining the centroids of the triangles bounded
> by the lines
> (a1,b2,c3) and (a1,b3,c2)
> [the triangle (a1,b2,c3) is a point, which we take as
> centroid]
>
> Similarly Lb, Lc.
>
> Are tle lines La,Lb,Lc concurrent?
>
> APH
>
>
>
> On Thu, Jan 24, 2013 at 1:18 PM, Antreas <anopolis72@...>
> wrote:
>
> > **
> >
> >
> > Let ABC be a triangle, 1 a line and 2,3 two lines such
> that
> > the triangle bounded by the lines (1,2,3) is
> equilateral.
> >
> > Denote:
> >
> > a1, a2, a3 = the lines through A parallel to
> 1,2,3,resp.
> > b1, b2, b3 = the lines through B parallel to
> 1,2,3,resp.
> > c1, c2, c3 = the lines through C parallel to
> 1,2,3,resp.
> >
> > La = the line joining the centers of the equilateral
> > triangles bounded by the lines (a1,b2,c3) and
> (a1,b3,c2)
> >
> > Lb = the line joining the centers of the equilateral
> > triangles bounded by the lines (b1,c2,a3) and
> (b1,c3,a2)
> >
> > Lc = the line joining the centers of the equilateral
> > triangles bounded by the lines (c1,a2,b3) and
> (c1,a3,b2)
> >
> > If 1 is the trilinear polar of point P, which is the
> > locus of P such tah La,Lb,Lc are concurrent?
> >
> > APH
> >
>
>
> [Non-text portions of this message have been removed]
>
>
>
> ------------------------------------
>