## Another perspector?

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• Dear friends: Let X be touch point at BC with the extangent circle to excircles (I_b) and (I_c) of ABC triangle. Y and Z are similarly the touch points on AC
Message 1 of 29 , Feb 2, 2006
Dear friends:
Let X be touch point at BC with the extangent circle to excircles
(I_b) and (I_c) of ABC triangle. Y and Z are similarly the touch
points on AC and AB. If we consider the triangle UVW performed by the
polars of vertex of ABC wrt the opposite excircles. Seems XYZ and UVW
are perspective at P. Is P in ETC?
Best regards
Juan Carlos
• Dear Juan P is the orthocenter of ABC. ABC and UVW are perspective and orthologic and so one... Friendly François ... [Non-text portions of this message have
Message 2 of 29 , Feb 2, 2006
Dear Juan
P is the orthocenter of ABC.
ABC and UVW are perspective and orthologic and so one...
Friendly
François

On 2/2/06, Juan Carlos Salazar <caiser@...> wrote:
>
> Dear friends:
> Let X be touch point at BC with the extangent circle to excircles
> (I_b) and (I_c) of ABC triangle. Y and Z are similarly the touch
> points on AC and AB. If we consider the triangle UVW performed by the
> polars of vertex of ABC wrt the opposite excircles. Seems XYZ and UVW
> are perspective at P. Is P in ETC?
> Best regards
> Juan Carlos
>
>
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>
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>

[Non-text portions of this message have been removed]
• Dear Juan Sorry, I mix ABC and XYZ for I just wake up! François ... [Non-text portions of this message have been removed]
Message 3 of 29 , Feb 3, 2006
Dear Juan
Sorry, I mix ABC and XYZ for I just wake up!
François

On 2/3/06, Francois Rideau <francois.rideau@...> wrote:
>
> Dear Juan
> P is the orthocenter of ABC.
> ABC and UVW are perspective and orthologic and so one...
> Friendly
> François
>
> On 2/2/06, Juan Carlos Salazar <caiser@...> wrote:
> >
> > Dear friends:
> > Let X be touch point at BC with the extangent circle to excircles
> > (I_b) and (I_c) of ABC triangle. Y and Z are similarly the touch
> > points on AC and AB. If we consider the triangle UVW performed by the
> > polars of vertex of ABC wrt the opposite excircles. Seems XYZ and UVW
> > are perspective at P. Is P in ETC?
> > Best regards
> > Juan Carlos
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
>

[Non-text portions of this message have been removed]
• Dear Juan ABC and XYZ have N the Nagel point as perspector . ABC and UVW have H the othocenter of ABC as perspector. So XYZ and UVW are perspective with
Message 4 of 29 , Feb 3, 2006
Dear Juan
ABC and XYZ have N the Nagel point as perspector .
ABC and UVW have H the othocenter of ABC as perspector. So XYZ and UVW are
perspective with perspector P on line HN.
I think P = X(n) is in ETC for some standard integer
Friendly
François

On 2/3/06, Francois Rideau <francois.rideau@...> wrote:
>
> Dear Juan
> Sorry, I mix ABC and XYZ for I just wake up!
> François
>
> On 2/3/06, Francois Rideau <francois.rideau@...> wrote:
> >
> > Dear Juan
> > P is the orthocenter of ABC.
> > ABC and UVW are perspective and orthologic and so one...
> > Friendly
> > François
> >
> > On 2/2/06, Juan Carlos Salazar < caiser@...> wrote:
> > >
> > > Dear friends:
> > > Let X be touch point at BC with the extangent circle to excircles
> > > (I_b) and (I_c) of ABC triangle. Y and Z are similarly the touch
> > > points on AC and AB. If we consider the triangle UVW performed by the
> > > polars of vertex of ABC wrt the opposite excircles. Seems XYZ and UVW
> > > are perspective at P. Is P in ETC?
> > > Thanks in advance
> > > Best regards
> > > Juan Carlos
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > > Yahoo! Groups Links
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> >
>

[Non-text portions of this message have been removed]
• Dear Francois: I am sorry by my statement, clarifying the circle that touches BC at X is a circle that is on the same side that excircles(I_b) and (I_c).
Message 5 of 29 , Feb 3, 2006
Dear Francois:
I am sorry by my statement, clarifying the circle that touches BC
at X is a circle that is on the same side that excircles(I_b) and
(I_c). Similarly to another two circles.Then P is not Nagel's point.
Sincerely
Juan Carlos

--- In Hyacinthos@yahoogroups.com, Francois Rideau
<francois.rideau@...> wrote:
>
> Dear Juan
> ABC and XYZ have N the Nagel point as perspector .
> ABC and UVW have H the othocenter of ABC as perspector. So XYZ and
UVW are
> perspective with perspector P on line HN.
> I think P = X(n) is in ETC for some standard integer
> Friendly
> François
>
> On 2/3/06, Francois Rideau <francois.rideau@...> wrote:
> >
> > Dear Juan
> > Sorry, I mix ABC and XYZ for I just wake up!
> > François
> >
> > On 2/3/06, Francois Rideau <francois.rideau@...> wrote:
> > >
> > > Dear Juan
> > > P is the orthocenter of ABC.
> > > ABC and UVW are perspective and orthologic and so one...
> > > Friendly
> > > François
> > >
> > > On 2/2/06, Juan Carlos Salazar < caiser@...> wrote:
> > > >
> > > > Dear friends:
> > > > Let X be touch point at BC with the extangent circle to
excircles
> > > > (I_b) and (I_c) of ABC triangle. Y and Z are similarly the
touch
> > > > points on AC and AB. If we consider the triangle UVW
performed by the
> > > > polars of vertex of ABC wrt the opposite excircles. Seems
XYZ and UVW
> > > > are perspective at P. Is P in ETC?
> > > > Thanks in advance
> > > > Best regards
> > > > Juan Carlos
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > Yahoo! Groups Links
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > >
> >
>
>
> [Non-text portions of this message have been removed]
>
• Dear Francois: I am sorry by my statement, clarifying the circle that touches BC at X is a circle that is on the same side that the excircles (I_b) and (I_c).
Message 6 of 29 , Feb 3, 2006
Dear Francois:
I am sorry by my statement, clarifying the circle that touches BC
at X is a circle that is on the same side that the excircles (I_b)
and (I_c). Similarly to another two circles.Then P is not Nagel's
point.
Sincerely
Juan carlos

--- In Hyacinthos@yahoogroups.com, Francois Rideau
<francois.rideau@...> wrote:
>
> Dear Juan
> ABC and XYZ have N the Nagel point as perspector .
> ABC and UVW have H the othocenter of ABC as perspector. So XYZ and
UVW are
> perspective with perspector P on line HN.
> I think P = X(n) is in ETC for some standard integer
> Friendly
> François
>
> On 2/3/06, Francois Rideau <francois.rideau@...> wrote:
> >
> > Dear Juan
> > Sorry, I mix ABC and XYZ for I just wake up!
> > François
> >
> > On 2/3/06, Francois Rideau <francois.rideau@...> wrote:
> > >
> > > Dear Juan
> > > P is the orthocenter of ABC.
> > > ABC and UVW are perspective and orthologic and so one...
> > > Friendly
> > > François
> > >
> > > On 2/2/06, Juan Carlos Salazar < caiser@...> wrote:
> > > >
> > > > Dear friends:
> > > > Let X be touch point at BC with the extangent circle to
excircles
> > > > (I_b) and (I_c) of ABC triangle. Y and Z are similarly the
touch
> > > > points on AC and AB. If we consider the triangle UVW
performed by the
> > > > polars of vertex of ABC wrt the opposite excircles. Seems
XYZ and UVW
> > > > are perspective at P. Is P in ETC?
> > > > Thanks in advance
> > > > Best regards
> > > > Juan Carlos
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > Yahoo! Groups Links
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > >
> >
>
>
> [Non-text portions of this message have been removed]
>
• Dear Francois: I posted in the files one figure for the first circle with the touch point X. Thanks Juan Carlos ... UVW are ... excircles ... touch ...
Message 7 of 29 , Feb 3, 2006
Dear Francois:
I posted in the files one figure for the first circle with the touch
point X.
Thanks
Juan Carlos

--- In Hyacinthos@yahoogroups.com, Francois Rideau
<francois.rideau@...> wrote:
>
> Dear Juan
> ABC and XYZ have N the Nagel point as perspector .
> ABC and UVW have H the othocenter of ABC as perspector. So XYZ and
UVW are
> perspective with perspector P on line HN.
> I think P = X(n) is in ETC for some standard integer
> Friendly
> François
>
> On 2/3/06, Francois Rideau <francois.rideau@...> wrote:
> >
> > Dear Juan
> > Sorry, I mix ABC and XYZ for I just wake up!
> > François
> >
> > On 2/3/06, Francois Rideau <francois.rideau@...> wrote:
> > >
> > > Dear Juan
> > > P is the orthocenter of ABC.
> > > ABC and UVW are perspective and orthologic and so one...
> > > Friendly
> > > François
> > >
> > > On 2/2/06, Juan Carlos Salazar < caiser@...> wrote:
> > > >
> > > > Dear friends:
> > > > Let X be touch point at BC with the extangent circle to
excircles
> > > > (I_b) and (I_c) of ABC triangle. Y and Z are similarly the
touch
> > > > points on AC and AB. If we consider the triangle UVW
performed by the
> > > > polars of vertex of ABC wrt the opposite excircles. Seems
XYZ and UVW
> > > > are perspective at P. Is P in ETC?
> > > > Thanks in advance
> > > > Best regards
> > > > Juan Carlos
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > Yahoo! Groups Links
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > >
> >
>
>
> [Non-text portions of this message have been removed]
>
• Dear all, this is the first time I gather courage to write. Even if I ve been working for a long time with synthetic geometry the reason is that I don t work
Message 8 of 29 , Feb 3, 2006
Dear all,

this is the first time I gather courage to write. Even if I've been working for a long time with synthetic geometry the reason is that I don't work with coordinates, so ETC and many of your messages are too much for me (Sorry, but I havn't havd the time!)

The thing is that I've been exploring Lemoine with Cabri II plus and have found things for which I do not have an explanation nor have I been able to find references:

One thing is that the locus of the intersecion of two symmedians with the Lemoine´s axis is an hyperbola that passes through the two other vertex as the one not involved moves in the circumcircle of the triangle.

The other one is the envelope of the Lemine´s axis as one vertex moves in the same way (which by the way can be an ellipse, parabola or hyperbola depending on the angles of the triangle at the vertex considered)

Then that the locus of the symmedian point as each vertx moves in a line to the paralel to the other side of the triangle is also an ellipse

Then I saw that the circumcenter of the tangential triangle is on the Euler line of the original triangle and that the locus of the intersections of the symmedian with the Euler line of one vertex as it moves in the same way, are three cirles tangent to the sides of the original triangle.

And still I can`t explain this!

So anyhelp will be very very welcome

Best regards,
Ma. de la Paz
Mèxico

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[Non-text portions of this message have been removed]
• you can look at some of the pictures in http://madelapaz.blogspot.com/ ... Do You Yahoo!? La mejor conexión a Internet y 2GB extra a tu correo por \$100 al
Message 9 of 29 , Feb 3, 2006
you can look at some of the pictures in

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[Non-text portions of this message have been removed]
• The circles I mentioned as locus of symmedian intersecion with Euler line are tangent to the tangential triangle NOT to the original triangle! ... Do You
Message 10 of 29 , Feb 3, 2006
The circles I mentioned as locus of symmedian intersecion with Euler line are tangent to the tangential triangle NOT to the original triangle!

---------------------------------
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[Non-text portions of this message have been removed]
• Dear Juan, ... There are two extangents circles. A small and a large. If we consider the case of the small circle then after tedious calculations I found that,
Message 11 of 29 , Feb 3, 2006
Dear Juan,

> Let X be touch point at BC with the extangent circle to excircles
> (I_b) and (I_c) of ABC triangle. Y and Z are similarly the touch
> points on AC and AB. If we consider the triangle UVW performed by the
> polars of vertex of ABC wrt the opposite excircles. Seems XYZ and UVW
> are perspective at P. Is P in ETC?
> Best regards
> Juan Carlos

There are two extangents circles. A small and a large.
If we consider the case of the small circle then after tedious
calculations I found that, yes the triangles XYZ and UVW
are perspective at
P ( S_B/sqrt(s-c)+S_C/sqrt(s-b) - a(b+c)/sqrt(s-a) : . . : . .)
in barycentrics with searsch number - 0.291573042804
but this point is not in ETC.
For the other case I didn't worked.
What construction for the small circle
do you have in mind?

Best regards
• Dear Nik: You wrote: ... Yes, you are right.In this case, I take the small circle. ... If we denote by P and Q the touch points between the excircles (I_c,r_c)
Message 12 of 29 , Feb 3, 2006
Dear Nik:

You wrote:
...
> There are two extangents circles. A small and a large.
Yes, you are right.In this case, I take the small circle.
> If we consider the case of the small circle then after tedious
> calculations I found that, yes the triangles XYZ and UVW
> are perspective at
> P ( S_B/sqrt(s-c)+S_C/sqrt(s-b) - a(b+c)/sqrt(s-a) : . . : . .)
> in barycentrics with searsch number - 0.291573042804
> but this point is not in ETC.
> For the other case I didn't worked.
> What construction for the small circle
> do you have in mind?
If we denote by P and Q the touch points between the excircles
(I_c,r_c) and (I_b,r_b) with BC, S and T the touch points of the
first small circle (O_1,r_1) with (I_c,r_c) and (I_b,r_b). Now:
PX=2sqrt(r_1.r_c) and QX =2sqrt(r_1.r_b),so:
PX/QX=sqrt(r_c)/sqrt(r_b)
PS and QT meets at D on the small circle (O_1,r_1).So, DX is
diameter of (O_1) and D is on the radical axis of the excircles
(I_c,r_c) and (I_b,r_b).
The construction for the small circle is the next:
1)We can to find the point X on PQ(BC) if we draw two segments
proportional to sqrt(r_b) and sqrt(r_c),next we using Thales or
similarity.
2)Drawing the perpendicular L_1 to PQ(BC) at X.
3)Drawing the radical axis L_2 of the excircles (I_c,r_c) and
(I_b,r_b).
4)We can to find D,the cut point of L_1 with L_2.
5)Finally we have the small circle (O_1) with diameter DX.
6)Similarly we can find the next two small circles.
Many thanks.
Best regards
Juan Carlos
• Dear Juan Carlos, your point X is the intersection of BC with the midcircle of the B- excircle and the C-excircle, lying between B and C For the large circle,
Message 13 of 29 , Feb 4, 2006
Dear Juan Carlos,

your point X is the intersection of BC with the midcircle of the B-
excircle and the C-excircle, lying between B and C

For the large circle, mentionned by Nikolaos, the point X is the other
intersection of BC with these excircles

Kind regards

Eric
• Dear Juan and Eric, the same construction I had in mind as you and Eric. If M is the intersection of PQ and IbIc and rc
Message 14 of 29 , Feb 4, 2006
Dear Juan and Eric,

the same construction I had in mind
as you and Eric.
If M is the intersection of PQ and IbIc and rc<rb
then the line PIc meets the circle with
diameter MQ at K. The circle (M, MK) is the
circle that gives on the line PQ the point X of
the small circle and the point X' of the large circle.

Best regards
• Dear Juan Carlos If Ab and Ac the points where the B and C-excircles touch BC, the circle with diameter AbAc (this circle is centered at the midpoint of BC
Message 15 of 29 , Feb 4, 2006
Dear Juan Carlos
If Ab and Ac the points where the B and C-excircles touch BC, the
circle with diameter AbAc (this circle is centered at the midpoint of
BC with diameter b+c) intersects the A-altitude at M1,M2. If IbIc
intersects BC at Za, the circle with center Za going through M1 and M2
intersects BC at the two required points X
Friendly. Jean-Pierre
• Dear Maria, you wrote ... In Cartesian coordinates the points B(-a, 0), C(a, 0) and the circumcenter Q(0, b) of triangle ABC are fixed points and the point
Message 16 of 29 , Feb 10, 2006
Dear Maria,
you wrote
> One thing is that the locus of the intersecion of two symmedians
> with the Lemoine´s axis is an hyperbola that passes through the
> two other vertex as the one not involved moves in the
> circumcircle of the triangle.
>
> The other one is the envelope of the Lemine´s axis as one vertex
> moves in the same way (which by the way can be an ellipse,
> parabola or hyperbola depending on the angles of the triangle at
> the vertex considered)
>
> Then that the locus of the symmedian point as each vertx moves in
> a line to the paralel to the other side of the triangle is also an ellipse
>
> Then I saw that the circumcenter of the tangential triangle is on
> the Euler line of the original triangle and that the locus of the
> intersections of the symmedian with the Euler line of one vertex
> as it moves in the same way, are three cirles tangent to the
> sides of the tangential triangle.

In Cartesian coordinates the points B(-a, 0), C(a, 0) and the circumcenter
Q(0, b) of triangle ABC are fixed points and the point A(p, q) is a
variable point on the circumcircle x^2 + (y - b)^2 = a^2 + b^2 of ABC
and hence we have p^2 + q^2 = 2bq + a^2 (1).
BC^2 = 4a^2
CA^2 = (p - a)^2 + q^2
AB^2 = (p + a)^2 + q^2
If the Cartesian coordinates of the Lemoine point K
are (x, y) then the signed areas (KBC) : (KCA) : (KAB)
are as the determinants
|x + a y| |x - a y | |x - p y - q |
|x - a y| : |x - p y - q | : |x + a y |
If A', B', C' are the points where AK, BK, CK meet the sides
BC, CA, AB of ABC then it is known that
BC' / C'A = BC^2 / CA^2 = (KBC) / (KCA) and from this
we get
y(p^2 + q^2) + 4ayp - 2axq - 2a^2q + 3a^2y = 0 (2)
and similarly
y(p^2 + q^2) - 4ayp + 2axq - 2a^2q + 3a^2y = 0 (3)
From (1), (2), (3) we find
p = x(2a^2 + bq)/(2a^2)
q = 2ya^2/(a^2 - by)
and by substitution in (1) we get the equation for K
a^2x^2 + (4a^2 + 3b^2)y^2 - 2ba^2 y - a^4 = 0
that is an ellipse.

The line AK has equation (a^2 + bq)x -bpy -pa^2 = 0
passes through the fixed point S( 0, a^2/b) that is the
antipode of Q in the circumcircle of triangle QBC.
Working with tedious known calculations you can find
the lines BK , CK are
qax - (2ap + bq + 2a^2)y + qa^2 = 0
qax + (-2ap + bq + 2a^2)y - qa^2 = 0
the Lemoine axis La is
2qa^2(a^2 + bq - px) + (p^2a^2 + q^2b^2 - a^4)y = 0
the intersection of AK with La moves on the conic
4a^2x^2 + (a^2 - 3b^2)y^2 - 8ba^2y - 4a^4 = 0

the intersections of La with BK or CK are lying on
the hyperbola
a^2x^2 - (2a^2 + 3b^2)y^2 - 2ba^2y - a^4 = 0

I think that' s enough.

Best regards
• Dear Nikos, dear all, First, thank you for your calculation for the hyperbola!!....I m still trying to work it out this case as the following: As I told you in
Message 17 of 29 , Feb 13, 2006
Dear Nikos, dear all,

First, thank you for your calculation for the hyperbola!!....I'm still trying to work it out this case as the following:

As I told you in my first mail, the envelope of the Lemoine's axis is, again, a conic. Now if we have a triangle ABC and let one of the vertex (say A) move along the circumcircle (with cener in O), we get Lemoine´s ellipse E. This can be seen as a point-conic. The ellipse part is due to Lemoine. This ellipse is tangent to the tangential triangle and to the circumcircle in B and C.

Now, the three polars of the symmedians (having the Lemoine's point in common) are collieneal (in fact they all are in Lemoine's axis l). If one takes the the intersection of the symmedian (by A) with the l (the point T) and takes the polar of this point, one gets the original ellipse E as a line-conic. By the way, this polar passes through the point of intersection of BC and l

Using the same arguments, we can get the envelope of the Lemoine's axis (i.e. a line-conic) as a point conic as the locus of T while A moves in the circumcircle, This line-conic is also tangent to the tengencial triangle and to the circumcircle in B and C.

This conic is an ellipse if the angle between OL and OT is acute, a parabola if Ol and OT are perpendicular, and an hyperbola in the obtuse case. (I still wonder why there is nothing literature about it!)

As for the hyperbola that results of taking the intersection of l with the symmedian by B (or by C), and letting A move. This hyperbola is....(I still have to work out the details about polars and quadrangles) as I said a couple of days ago, this hyerpbolas (and some more of all this conics) tend to grow in number with this approach... harmonics, croos-ratio etc are having fun with me.

Tomorrow at my office (where I have a better connection to intenert) I will try to post this figures
a) just as figures (paint brush) and
b) as cabri II plus files (which will have movement and you can replay the consructions). if you have the program.

Again thanks to Nikos, Francois and to all of you for your replies. As soon as the hyperbola is ready, I'll mail the result and post (if I understood how, of couse) the figures.

And I'm still curiuous about the fact that the circumcenter of the tangential triangle lies in the Euler line of the original triangle....I hope to have this worked out also soon.

Ma. de la Paz

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[Non-text portions of this message have been removed]
• Dear all: Happy new year. If we consider the three small extangent circles (O_1),(O_2),(O_3) and NPC of ABC, then the radicals axis to the three circles
Message 18 of 29 , Jan 12, 2007
Dear all:
Happy new year.
If we consider the three small extangent circles (O_1),(O_2),(O_3) and
NPC of ABC, then the radicals axis to the three circles (O_1),(O_2),
(O_3) with NPC performed a triangle QST, hence QST and ABC are
perspective at P'.
Is P'in ETC?
Best regards
Juan Carlos
• Dear Juan, you wrote ... What circles are these small (O_1),(O_2),(O_3)? Best regards Nikos Dergiades
Message 19 of 29 , Jan 13, 2007
Dear Juan,
you wrote

> If we consider the three small extangent circles
> (O_1),(O_2),(O_3) and
> NPC of ABC, then the radicals axis to the three
> circles (O_1),(O_2),
> (O_3) with NPC performed a triangle QST, hence QST
> and ABC are
> perspective at P'.
> Is P'in ETC?

What circles are these small (O_1),(O_2),(O_3)?

Best regards

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• Dear Juan Carlos ... and ... If I well understand, (O_1) is a circle touching BC and touching externally the B-excircle and the C-excircle. In this case, it is
Message 20 of 29 , Jan 13, 2007
Dear Juan Carlos
> If we consider the three small extangent circles (O_1),(O_2),(O_3)
and
> NPC of ABC, then the radicals axis to the three circles (O_1),(O_2),
> (O_3) with NPC performed a triangle QST, hence QST and ABC are
> perspective at P'.
> Is P'in ETC?

If I well understand, (O_1) is a circle touching BC and touching
externally the B-excircle and the C-excircle.
In this case, it is not necessary to choose the small one.
Let X, X' be the contact points with BC of the circles Cx and Cx'
touching BC and touching externally the B and C excircles.
Then the circle with diameter XX' is centered at Za = BC inter IbIc and
is orthogonal to the NPC.
It follows that Za is the radical center of (NPC,Cx,Cx')
As Za, Zb, Zc are on the same line, it follows from Desargues theorem
that the 8 triangles with side lines La,Lb,Lc are perspective with ABC
Of course, it could be interesting to look at the configuration of the
8 perspectors.
Friendly. Jean-Pierre
• Dear Nik and Jean Pierre: ... (O_2), ... Yes, there are small and large circles.Here the three radical axis are from the pairs of circles [NPC,(O_1)];
Message 21 of 29 , Jan 13, 2007
Dear Nik and Jean Pierre:

[JC]:
> > If we consider the three small extangent circles (O_1),(O_2),(O_3)
> and
> > NPC of ABC, then the radicals axis to the three circles (O_1),
(O_2),
> > (O_3) with NPC performed a triangle QST, hence QST and ABC are
> > perspective at P'.
> > Is P'in ETC?
[JP]:
> If I well understand, (O_1) is a circle touching BC and touching
> externally the B-excircle and the C-excircle.
Yes, there are small and large circles.Here the three radical axis are
from the pairs of circles [NPC,(O_1)]; [NPC,(O_2)]; [NPC,(O_3)].
Thanks.
Regards
Juan Carlos
• Dear Juan Carlos ... (O_3) ... [JC] ... are ... As I ve mailed before, you have two choices for each circle and you get 8 triangles perspective with ABC. The
Message 22 of 29 , Jan 13, 2007
Dear Juan Carlos
> [JC]:
> > > If we consider the three small extangent circles (O_1),(O_2),
(O_3)
> > and
> > > NPC of ABC, then the radicals axis to the three circles (O_1),
> (O_2),
> > > (O_3) with NPC performed a triangle QST, hence QST and ABC are
> > > perspective at P'.
> > > Is P'in ETC?
> [JP]:
> > If I well understand, (O_1) is a circle touching BC and touching
> > externally the B-excircle and the C-excircle.
[JC]
> Yes, there are small and large circles.Here the three radical axis
are
> from the pairs of circles [NPC,(O_1)]; [NPC,(O_2)]; [NPC,(O_3)].
As I've mailed before, you have two choices for each circle and you
get 8 triangles perspective with ABC.
The perspectrix is allways the trilinear polar of the incenter.
But what can be said about the configuration of the 8 perspectors ?
(two of them are triangle centers when we take the three "small"
circles of the three other ones)
Friendly. Jean-Pierre
• Dear friends, How can we construct the two lines given by the equation in barycentrics: CyclicSum (SA*SA*(SB-SC)xx + 2SB*SC*(SB-SC)yz) = 0 Best regards Nikos
Message 23 of 29 , Jan 14, 2007
Dear friends,
How can we construct the two lines
given by the equation in barycentrics:

CyclicSum (SA*SA*(SB-SC)xx + 2SB*SC*(SB-SC)yz) = 0

Best regards

___________________________________________________________
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• Dear Nikolaos, ... these are the parallels at H to the asymptotes of the Kiepert hyperbola. Best regards Bernard [Non-text portions of this message have been
Message 24 of 29 , Jan 14, 2007
Dear Nikolaos,

> [ND] How can we construct the two lines
> given by the equation in barycentrics:
>
> CyclicSum (SA*SA*(SB-SC)xx + 2SB*SC*(SB-SC)yz) = 0

these are the parallels at H to the asymptotes of the Kiepert hyperbola.

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Bernard, ... [BG] ... Thank you very much. These are the parallels from H to the Simson lines of the points where the Brocard axis meets the circumcircle.
Message 25 of 29 , Jan 14, 2007
Dear Bernard,

> > [ND] How can we construct the two lines
> > given by the equation in barycentrics:
> >
> > CyclicSum (SA*SA*(SB-SC)xx + 2SB*SC*(SB-SC)yz) = 0

[BG]
> these are the parallels at H to the asymptotes of
> the Kiepert hyperbola.

Thank you very much.
These are the parallels from H to the Simson
lines of the points where the Brocard axis meets
the circumcircle.

I tried to find for which points P the pedal triangle
of P A'B'C' has vertices equidistant from A,B,C
i.e. AA' = BB' = CC'.
These points must be symmetric wrt H.
I found that these points must be on these lines.
From a sketch it seems to exist only two such points
on one of these lines.
Can we construct these points?

Friendly

___________________________________________________________
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• Dear Nikolaos ... Your points are the homothetic of the focii of the Steiner circumellipse in (L,3/2) where L = de Longchamps point. For both points, AA =
Message 26 of 29 , Jan 14, 2007
Dear Nikolaos
> > > [ND] How can we construct the two lines
> > > given by the equation in barycentrics:
> > >
> > > CyclicSum (SA*SA*(SB-SC)xx + 2SB*SC*(SB-SC)yz) = 0
>
> [BG]
> > these are the parallels at H to the asymptotes of
> > the Kiepert hyperbola.
>
> Thank you very much.
> These are the parallels from H to the Simson
> lines of the points where the Brocard axis meets
> the circumcircle.
>
> I tried to find for which points P the pedal triangle
> of P A'B'C' has vertices equidistant from A,B,C
> i.e. AA' = BB' = CC'.
> These points must be symmetric wrt H.
> I found that these points must be on these lines.
> From a sketch it seems to exist only two such points
> on one of these lines.
> Can we construct these points?

Your points are the homothetic of the focii of the Steiner
circumellipse in (L,3/2) where L = de Longchamps point.
For both points, AA' = 3.T/2 where T = semimajor axis of the ellipse.

Friendly. Jean-Pierre
• Dear Nikolaos [ND] ... [JPE] ... ellipse. A little explanation : it is well known that the focii of the Steiner circumellipse have cevian of equal lengths. As
Message 27 of 29 , Jan 14, 2007
Dear Nikolaos
[ND]
> > I tried to find for which points P the pedal triangle
> > of P A'B'C' has vertices equidistant from A,B,C
> > i.e. AA' = BB' = CC'.
> > These points must be symmetric wrt H.
> > I found that these points must be on these lines.
> > From a sketch it seems to exist only two such points
> > on one of these lines.
> > Can we construct these points?

[JPE]
> Your points are the homothetic of the focii of the Steiner
> circumellipse in (L,3/2) where L = de Longchamps point.
> For both points, AA' = 3.T/2 where T = semimajor axis of the
ellipse.

A little explanation :
it is well known that the focii of the Steiner circumellipse have
cevian of equal lengths.
As they lie on the Lucas cubic, their cevian triangles are the pedal
triangles of the required points
Friendly. Jean-Pierre
• Dear Jean-Pierre, ... [JPE] ... Excellent! Many thanks Friendly Nikos ___________________________________________________________ ×ñçóéìïðïéåßôå
Message 28 of 29 , Jan 14, 2007
Dear Jean-Pierre,

> [ND]
> > > I tried to find for which points P the pedal
> triangle
> > > of P A'B'C' has vertices equidistant from A,B,C
> > > i.e. AA' = BB' = CC'.
> > > These points must be symmetric wrt H.
> > > I found that these points must be on these
> lines.
> > > From a sketch it seems to exist only two such
> points
> > > on one of these lines.
> > > Can we construct these points?
>
> [JPE]
> > Your points are the homothetic of the focii of the
> Steiner
> > circumellipse in (L,3/2) where L = de Longchamps
> point.
> > For both points, AA' = 3.T/2 where T = semimajor
> axis of the
> ellipse.
[JPE]
> A little explanation :
> it is well known that the focii of the Steiner
> circumellipse have
> cevian of equal lengths.
> As they lie on the Lucas cubic, their cevian
> triangles are the pedal
> triangles of the required points

Excellent!
Many thanks
Friendly
Nikos

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• ... and ... Dear Juan Carlos and all Related to this problem I can give a construction that uses inversion and the coordinates of your point P , not in ETC.
Message 29 of 29 , Jan 23, 2007
> If we consider the three small extangent circles (O_1),(O_2),(O_3)
and
> NPC of ABC, then the radicals axis to the three circles (O_1),(O_2),
> (O_3) with NPC performed a triangle QST, hence QST and ABC are
> perspective at P'.
> Is P'in ETC?

Dear Juan Carlos and all

Related to this problem I can give a construction that uses inversion
and the coordinates of your point P', not in ETC.

The construction can be found as a solution of problem 301 in Ricardo
Barroso's online magazine

http://www.aloj.us.es/rbarroso/trianguloscabri/sol/sol301garcap/

(server is slow, so you may need try it several times)

I gave there a formula for the radius of (O_1)

A rough translation of the Spanish text follows.

-------------------------------------------------------------------
Let K, L be the tangency points of (Ic),(Ib) with line BC.
We consider an inversion with center K and radius KL (or power KL^2).

The triangle KLIb is rectangle at L, so the circle (Ib) is orthogonal
to the inversion circle and thus invarible by the inversion. The circle
(Ic) pass trough the inversion center, so circle (Ic) inverts in a
line, namely, if KM is a diameter of (Ic), then circle (Ic) inverts in
the parallel to BC through M', the inverse of M.

The circle we want is the inverse of the circle (T) tangent to this
parallel, the line BC and the circle (Ib).

Then the center T must be in the median parallel to two lines. Let this
median parallel intersect KM at N. Then T is KN + rb away from Ib, and
this let us get the center T.
-------------------------------------------------------------------

To end this long post I give the coordinates of P', not in ETC.

Search Number is 1.367790355

We name 2 sqrt((s-b)(s-c)) as RA, and RB, RC the same ciclically.

{(a + b - c)*(a - b + c)*(b + c)*(2*a^4 + a^2*b^2 + b^4 + 2*a^2*b*c -
3*a^2*c^2 - 2*b^2*c^2 + c^4 - 2*a^3*RA - 2*a*b^2*RA +
2*a*c^2*RA)*(a^5*b + a^4*b^2 - a^3*b^3 - a^2*b^4 - 2*a^5*c -
3*a^4*b*c + 2*a^3*b^2*c + a^2*b^3*c - 2*a*b^4*c - 2*a^3*b*c^2 -
2*a^2*b^2*c^2 + a*b^3*c^2 - b^4*c^2 + 4*a^3*c^3 + 6*a^2*b*c^3 +
2*a*b^2*c^3 - b^3*c^3 + a*b*c^4 - 3*b^2*c^4 - 2*a*c^5 -
3*b*c^5 - a^5*RB - a^4*b*RB + a^3*b^2*RB + a^2*b^3*RB - a^4*c*RB -
4*a^3*b*c*RB - a^2*b^2*c*RB + 2*a*b^3*c*RB +
2*a^3*c^2*RB + 2*a^2*b*c^2*RB - a*b^2*c^2*RB + b^3*c^2*RB +
2*a^2*c^3*RB + 4*a*b*c^3*RB + b^2*c^3*RB - a*c^4*RB - b*c^4*RB -
c^5*RB)*(2*a^5*b - 4*a^3*b^3 + 2*a*b^5 + 3*a^5*c - a^4*b*c -
6*a^3*b^2*c + 2*a^2*b^3*c + 3*a*b^4*c - b^5*c + 3*a^4*c^2 -
2*a^3*b*c^2 + 2*a^2*b^2*c^2 - 2*a*b^3*c^2 - b^4*c^2 + a^3*c^3 -
a^2*b*c^3 - a*b^2*c^3 + b^3*c^3 + a^2*c^4 + 2*a*b*c^4 +
b^2*c^4 + a^5*RC + a^4*b*RC - 2*a^3*b^2*RC - 2*a^2*b^3*RC +
a*b^4*RC + b^5*RC + a^4*c*RC - 4*a^3*b*c*RC - 2*a^2*b^2*c*RC +
4*a*b^3*c*RC + b^4*c*RC - a^3*c^2*RC + a^2*b*c^2*RC + a*b^2*c^2*RC
- b^3*c^2*RC - a^2*c^3*RC - 2*a*b*c^3*RC - b^2*c^3*RC),
(a - b - c)*(a + b - c)*(a + c)*(a^4*b^2 + a^3*b^3 + 3*a^2*b^4 +
3*a*b^5 + 2*a^4*b*c - a^3*b^2*c - 2*a^2*b^3*c - a*b^4*c +
2*b^5*c + a^4*c^2 - a^3*b*c^2 + 2*a^2*b^2*c^2 - 6*a*b^3*c^2 +
a^3*c^3 - 2*a^2*b*c^3 + 2*a*b^2*c^3 - 4*b^3*c^3 - a^2*c^4 +
3*a*b*c^4 - a*c^5 + 2*b*c^5 - a^3*b^2*RA - a^2*b^3*RA + a*b^4*RA +
b^5*RA - 2*a^3*b*c*RA + a^2*b^2*c*RA - 4*a*b^3*c*RA +
b^4*c*RA - a^3*c^2*RA + a^2*b*c^2*RA - 2*a*b^2*c^2*RA -
2*b^3*c^2*RA - a^2*c^3*RA + 4*a*b*c^3*RA - 2*b^2*c^3*RA + a*c^4*RA +
b*c^4*RA + c^5*RA)*(a^4 - 3*a^2*b^2 + 2*b^4 + 2*a*b^2*c - 2*a^2*c^2
+ b^2*c^2 + c^4 + 2*a^2*b*RB - 2*b^3*RB - 2*b*c^2*RB)*
(2*a^5*b - 4*a^3*b^3 + 2*a*b^5 + 3*a^5*c - a^4*b*c - 6*a^3*b^2*c +
2*a^2*b^3*c + 3*a*b^4*c - b^5*c + 3*a^4*c^2 - 2*a^3*b*c^2 +
2*a^2*b^2*c^2 - 2*a*b^3*c^2 - b^4*c^2 + a^3*c^3 - a^2*b*c^3 -
a*b^2*c^3 + b^3*c^3 + a^2*c^4 + 2*a*b*c^4 + b^2*c^4 + a^5*RC +
a^4*b*RC - 2*a^3*b^2*RC - 2*a^2*b^3*RC + a*b^4*RC + b^5*RC +
a^4*c*RC - 4*a^3*b*c*RC - 2*a^2*b^2*c*RC + 4*a*b^3*c*RC +
b^4*c*RC - a^3*c^2*RC + a^2*b*c^2*RC + a*b^2*c^2*RC - b^3*c^2*RC -
a^2*c^3*RC - 2*a*b*c^3*RC - b^2*c^3*RC),
-((a + b)*(a - b - c)*(a - b + c)*(a^4*b^2 + a^3*b^3 + 3*a^2*b^4 +
3*a*b^5 + 2*a^4*b*c - a^3*b^2*c - 2*a^2*b^3*c - a*b^4*c +
2*b^5*c + a^4*c^2 - a^3*b*c^2 + 2*a^2*b^2*c^2 - 6*a*b^3*c^2 +
a^3*c^3 - 2*a^2*b*c^3 + 2*a*b^2*c^3 - 4*b^3*c^3 - a^2*c^4 +
3*a*b*c^4 - a*c^5 + 2*b*c^5 - a^3*b^2*RA - a^2*b^3*RA + a*b^4*RA +
b^5*RA - 2*a^3*b*c*RA + a^2*b^2*c*RA - 4*a*b^3*c*RA +
b^4*c*RA - a^3*c^2*RA + a^2*b*c^2*RA - 2*a*b^2*c^2*RA -
2*b^3*c^2*RA - a^2*c^3*RA + 4*a*b*c^3*RA - 2*b^2*c^3*RA + a*c^4*RA +
b*c^4*RA + c^5*RA)*(a^5*b + a^4*b^2 - a^3*b^3 - a^2*b^4 - 2*a^5*c
- 3*a^4*b*c + 2*a^3*b^2*c + a^2*b^3*c - 2*a*b^4*c -
2*a^3*b*c^2 - 2*a^2*b^2*c^2 + a*b^3*c^2 - b^4*c^2 + 4*a^3*c^3 +
6*a^2*b*c^3 + 2*a*b^2*c^3 - b^3*c^3 + a*b*c^4 - 3*b^2*c^4 -
2*a*c^5 - 3*b*c^5 - a^5*RB - a^4*b*RB + a^3*b^2*RB + a^2*b^3*RB -
a^4*c*RB - 4*a^3*b*c*RB - a^2*b^2*c*RB + 2*a*b^3*c*RB +
2*a^3*c^2*RB + 2*a^2*b*c^2*RB - a*b^2*c^2*RB + b^3*c^2*RB +
2*a^2*c^3*RB + 4*a*b*c^3*RB + b^2*c^3*RB - a*c^4*RB - b*c^4*RB -
c^5*RB)*(a^4 - 2*a^2*b^2 + b^4 + a^2*c^2 + 2*a*b*c^2 - 3*b^2*c^2 +
2*c^4 - 2*a^2*c*RC + 2*b^2*c*RC - 2*c^3*RC))}
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