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Perspector based on incenter and excircles
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 0 Attachment
Dear Hyacinthians,
here's a new point I've discovered by chance while playing with
extraversions.
Let I be the incenter.
Let A' be the point where the Aexcircle touches BC and let A* be
the second intersection of the line IA' with the Aexcircle.
Define B', C' and B*, C* cyclically.
==> Triangles ABC and A*B*C* are perspective
The barycentric coordinates of the perspector P are
P = ( a^5  a^4.(b + c)  2.a^3.(b + c)^2 + 2.a^2.(b + c)^3
+ a.(b + c)^4  (b + c)^5 :
b^5  b^4.(c + a)  2.b^3.(c + a)^2 + 2.b^2.(c + a)^3
+ b.(c + a)^4  (c + a)^5 :
c^5  c^4.(a + b)  2.c^3.(a + b)^2 + 2.c^2.(a + b)^3
+ c.(a + b)^4  (a + b)^5 )
It's amazing that:
 the most basic points still hide new points
 basic points and simple constructions are no guarantee for simple
coordinates
Greeting from Bruges
Eric Danneels 0 Attachment
Dear Eric,
I think that this property holds for every point P as follows:
If P is a point and P1, P2, P3 are its extraversions
If the line PX1 where X1 is the orthogonal projection of P1 on BC
meets the pedal circle of P1 at A* and similarly define the points
B*, C* then the triangles ABC, A*B*C* are perspective.
See the analogous results about the incircle and the Prasolov point
messages 6465 and before.
Best regards
Nikolaos Dergiades
[ED]
Let I be the incenter.
Let A' be the point where the Aexcircle touches BC and let A* be
the second intersection of the line IA' with the Aexcircle.
Define B', C' and B*, C* cyclically.
==> Triangles ABC and A*B*C* are perspective
The barycentric coordinates of the perspector P are
P = ( a^5  a^4.(b + c)  2.a^3.(b + c)^2 + 2.a^2.(b + c)^3
+ a.(b + c)^4  (b + c)^5 :
b^5  b^4.(c + a)  2.b^3.(c + a)^2 + 2.b^2.(c + a)^3
+ b.(c + a)^4  (c + a)^5 :
c^5  c^4.(a + b)  2.c^3.(a + b)^2 + 2.c^2.(a + b)^3
+ c.(a + b)^4  (a + b)^5 )
It's amazing that:
 the most basic points still hide new points
 basic points and simple constructions are no guarantee for simple
coordinates
Greeting from Bruges
Eric Danneels 0 Attachment
Dear Eric,
In Hyacinthos message #9595, you wrote:
>> Let I be the incenter.
This is indeed fascinating. It reminds me on the
>> Let A' be the point where the Aexcircle touches BC
>> and let A* be the second intersection of the line
>> IA' with the Aexcircle.
>> Define B', C' and B*, C* cyclically.
>>
>> ==> Triangles ABC and A*B*C* are perspective
NagelSchröder point X(1339) which is the point of
intersection of the circles AIA', BIB', CIC'
different from I. By the way, I have no synthetic
proof of the coaxality of these circles  can
anybody help?
Let me add some conjectures:
The poles of the lines IA', IB', IC' with respect
to the A, B, Cexcircles are collinear. In other
words, the tangents to the A, B, Cexcircles at
the points A*, B*, C* meet the lines BC, CA, AB at
three collinear points.
The polars of the incenter I with respect to the
excircles of triangle ABC enclose a triangle in
perspective to triangle ABC. The perspector is a
point not in the ETC.
Darij Grinberg 0 Attachment
Dear Nikolaos,
you wrote:
> I think that this property holds for every point P as follows:
What do you mean by the extraversions P1, P2 and P3 of P ?
> If P is a point and P1, P2, P3 are its extraversions
> If the line PX1 where X1 is the orthogonal projection of P1 on BC
> meets the pedal circle of P1 at A* and similarly define the points
> B*, C* then the triangles ABC, A*B*C* are perspective.
>
> See the analogous results about the incircle and the Prasolov point
> messages 6465 and before.
The recent messages about extraversions didn't make things clearer
to me, on the contrary...
I also don't see the link with messages 6465 and before.
However, about your generalization of the Prasolov point I would
like to add another generalization. (I thought it was known but I
couldn't find any reference).
Let A1B1BC1 be the pedal triangle of P and let PA1 intersect the
pedal circle again in A*. Define B* and C* cyclically.
==> AA*, BB* and CC* are also concurrent
Greetings from Bruges
Eric 0 Attachment
Dear Eric,
[ED]> What do you mean by the extraversions P1, P2 and P3 of P ?
Sorry. what I wrote about extraversions is nonsense.
> The recent messages about extraversions didn't make things clearer
> to me, on the contrary...
>
> I also don't see the link with messages 6465 and before.
>
> However, about your generalization of the Prasolov point I would
> like to add another generalization. (I thought it was known but I
> couldn't find any reference).
>
> Let A1B1BC1 be the pedal triangle of P and let PA1 intersect the
> pedal circle again in A*. Define B* and C* cyclically.
> ==> AA*, BB* and CC* are also concurrent
It doesn't hold. Take it as I didn't wrote it.
As for your Prasolov generalization you are riqht
because the point A* is antipode in the same pedal circle
of the projection on BC of the point P* isogonal conjugate of P.
Greetings from Thessaloniki
Nikolaos Dergiades 0 Attachment
Dear Eric and Darij
[ED]> >> Let I be the incenter.
Eric gave hugly barycentric coordinates for the perspector. In fact,
> >> Let A' be the point where the Aexcircle touches BC
> >> and let A* be the second intersection of the line
> >> IA' with the Aexcircle.
> >> Define B', C' and B*, C* cyclically.
> >>
> >> ==> Triangles ABC and A*B*C* are perspective
I think that they are (pa)^3 : (pb)^3 : (pc)^3.
> Let me add some conjectures:
They lie on the trilinear polar of X(346) = (pa)^2:(pb)^2:(pc)^2
>
> The poles of the lines IA', IB', IC' with respect
> to the A, B, Cexcircles are collinear. In other
> words, the tangents to the A, B, Cexcircles at
> the points A*, B*, C* meet the lines BC, CA, AB at
> three collinear points.
>
This point is 4X(9)3X(165), ie X(193) of the excentral triangle.
> The polars of the incenter I with respect to the
> excircles of triangle ABC enclose a triangle in
> perspective to triangle ABC. The perspector is a
> point not in the ETC.
He lies on the line X(9)X(165) (line GK of the excentral triangle)
and on the parallel through the Nagel point to the Soddy line.
He is barycentric a/((pa)^2(ba)(ca)):...
Friendly. JeanPierre 0 Attachment
Dear Darij> The polars of the incenter I with respect to the
If we condider the polars of a point M wrt the three incircles :
> excircles of triangle ABC enclose a triangle in
> perspective to triangle ABC. The perspector is a
> point not in the ETC.
1)they concur if and only if M lies on the radical circle of the
three circles, which is the circle with center the Spieker point,
radius 1/2*root(p^2+r^2)
2)they board a triangle perspective with ABC if and only if M lies
on the Spieker central cubic
http://perso.wanadoo.fr/bernard.gibert/Exemples/k033.html
Friendly. JeanPierre 0 Attachment
About Nikolaos' now withdrawn conjecture that
> If P is a point and P1, P2, P3 are its extraversions
Cabri suggests that for any point P there are a number of points P1 that
> If the line PX1 where X1 is the orthogonal projection of P1 on BC
> meets the pedal circle of P1 at A* and similarly define the points
> B*, C* then the triangles ABC, A*B*C* are perspective.
have the above property. I have not been able to determine a locus.
Dick Tahta 0 Attachment
Dear Eric and friends,
[ED]: Here's a new point I've discovered by chance while playing with
extraversions.
Let I be the incenter.
Let A' be the point where the Aexcircle touches BC and let A* be
the second intersection of the line IA' with the Aexcircle.
Define B', C' and B*, C* cyclically.
==> Triangles ABC and A*B*C* are perspective
The barycentric coordinates of the perspector P are
P = ( a^5  a^4.(b + c)  2.a^3.(b + c)^2 + 2.a^2.(b + c)^3
+ a.(b + c)^4  (b + c)^5 :
b^5  b^4.(c + a)  2.b^3.(c + a)^2 + 2.b^2.(c + a)^3
+ b.(c + a)^4  (c + a)^5 :
c^5  c^4.(a + b)  2.c^3.(a + b)^2 + 2.c^2.(a + b)^3
+ c.(a + b)^4  (a + b)^5 )
*** These coordinates simplify, as JeanPierre has pointed out, into
((b+ca)^3 : (c+ab)^3 : (a+bc)^3)). P is indeed the ``barycentric
cube'' of the Nagel point.
I have also investigated the question with the incenter I replaced by
a general point Q with barycentric coordinates (u:v:w).
With A* = second intersection of QA' with the Aexcircle etc., the
triangle A*B*C* is perspective with ABC if and only if Q lies on one
of the following curves:
(i) the line at infinity,
(ii) the inferior of the circumconic with center I, (which is the
conic through A', B', C' and the midpoints of the sides of ABC),
(iii) the Spieker central cubic, identified as K033 in Bernard's CTC).
The locus of the perspector is
(i) the isogonal conjugate of the incircle,
(ii) the inscribed conic with center X(1329), which is the
intersection of the lines joining the Nagel point to the Feuerbach
point, and the centroid to the outer Feuerbach point X(12),
(iii) to be determined, but here are the perspectors for various Q on
the Spieker central cubic:
Q perspector

I (b+ca)^3:...:...
H (a^3+a^2(b+c)a(b+c)^2(bc)^2(b+c))^2/(b+ca):...:...
X(8) X(8)
X(10) (b+ca)/a^2 : ... : ...
X(40) X(7)
X(65) X(1259)
X(72) X(55)
Best regards
Sincerely
Paul 0 Attachment
Dear JeanPierre,
my barycentrics were really ugly
I didn't notice they had a common divisor (a+b+c)^2
Greetings from Brussels
Eric 0 Attachment
Dear JeanPierre
> [JP] If we condider the polars of a point M wrt the three incircles :
If you replace the excircles with the three inconics centered at the
> 1)they concur if and only if M lies on the radical circle of the
> three circles, which is the circle with center the Spieker point,
> radius 1/2*root(p^2+r^2)
> 2)they board a triangle perspective with ABC if and only if M lies
> on the Spieker central cubic
vertices of the anticevian of a point Q, then :
1) the polars of M wrt to these conics concur iff M lies on a conic (C)
centered at the complement of Q.
The three inconics and (C) share the same directions of axis.
2) these polars bound a triangle perspective to ABC iff M lies on the
central pK
 with pole the center of the inconic with perspector Q,
 with pivot the anticomplement of Q,
 with center the complement of Q,
This is another way to approach our central pK in Special Isocubics §3.
Best regards
Bernard
[Nontext portions of this message have been removed] 0 Attachment
Dear Bernard and JeanPierre
[JP] If we condider the polars of a point M wrt the three incircles :
1)they concur if and only if M lies on the radical circle of the
three circles, which is the circle with center the Spieker point,
radius 1/2*root(p^2+r^2).
...
[BG]: If you replace the excircles with the three inconics centered
at the vertices of the anticevian of a point Q, then :
1) the polars of M wrt to these conics concur iff M lies on a conic (C)
centered at the complement of Q. The three inconics and (C) share the
same directions of axis.
...
*** Let Q' be the complement of Q. This conic (C) is the one through the
six intersections of the inconics with the polar of Q' with respect to
the three inconics. If M is a point on this conic, its polars with
respect to the inconic intersect at its antipode on the conic (C).
Best regards
Sincerely
Paul 0 Attachment
Dear Paul and Bernard> [JP] If we condider the polars of a point M wrt the three
incircles :
> 1)they concur if and only if M lies on the radical circle of the
centered
> three circles, which is the circle with center the Spieker point,
> radius 1/2*root(p^2+r^2).
> ...
>
> [BG]: If you replace the excircles with the three inconics
> at the vertices of the anticevian of a point Q, then :
conic (C)
>
> 1) the polars of M wrt to these conics concur iff M lies on a
> centered at the complement of Q. The three inconics and (C) share
the
> same directions of axis.
through the
> ...
>
> *** Let Q' be the complement of Q. This conic (C) is the one
> six intersections of the inconics with the polar of Q' with
respect to
> the three inconics. If M is a point on this conic, its polars with
We know that :
> respect to the inconic intersect at its antipode on the conic (C).
(1) The polar circle Cs of the three excircles is centered at the
complement of I and the polar line of S wrt an excircle is the
radical axis of Cs and the excircle.
Your results are now obvious if you change in (1) a,b,c to the
barycentric coordinates of Q.
Friendly. JeanPierre 0 Attachment
Dear Eric, JeanPierre and Paul,
I have finally found time to return to the problem
on the concurrent lines and the excircles. Thanks
for the very interesting data.
In Hyacinthos message #9607, Paul Yiu wrote:
>> These coordinates simplify, as JeanPierre has
In fact, now I see this coincides with another
>> pointed out, into
>> ((b+ca)^3 : (c+ab)^3 : (a+bc)^3)). P is
>> indeed the ``barycentric cube'' of the Nagel
>> point.
configuration.
The point A* is just the point on the Aexcircle
such that the circle BCA* is tangent to the
Aexcircle!!
And now the fact that the lines AA*, BB*, CC*
concur turns out to be equivalent to what
Antreas wrote in Hyacinthos message #5636
(our A*, B*, C* are his A2, B2, C2).
Now it wouldn't be surprising if there is a
synthetic proof of this all!
Sincerely,
Darij Grinberg 0 Attachment
Dear Darij
[DG]:>And now the fact that the lines AA*, BB*, CC*
That is,
>concur turns out to be equivalent to what
>Antreas wrote in Hyacinthos message #5636
>(our A*, B*, C* are his A2, B2, C2).
[APH]:>> Let A2, B2, C2 be the points of contact of the circles
We can consider other [than the three excircles] welldefined triads
>> passing through (B,C), (C,A), (A,B) and touching
>> the a,bcexcircles (resp.) [instead of the incircle].
>> Are the triangles ABC, A2B2C2 perspective?
of circles touching the sidelines, and ask similar questions.
Just an other one:
Let (Ka), (Kb), (Kc) be the three circles with diameters the three
altitudes AHa, BHb, CHc [==> they touch BC, CA, AB at Ha, Hb, Hc, resp.]
Consider now the circle passing through (B, C) and touching (Ka) at A*.
And similarly define the points B*, C*.
Are the triangles ABC, A*B*C* perspective?
If not :
Denote :
A# = BC /\ AA*, B# = CA /\ BB*, C# = AB /\ CC*
Are the points A#, B#, C# collinear?
Antreas
 0 Attachment
Dear Antreas> We can consider other [than the three excircles] welldefined
triads
> of circles touching the sidelines, and ask similar questions.
resp.]
>
> Just an other one:
>
> Let (Ka), (Kb), (Kc) be the three circles with diameters the three
> altitudes AHa, BHb, CHc [==> they touch BC, CA, AB at Ha, Hb, Hc,
>
at A*.
> Consider now the circle passing through (B, C) and touching (Ka)
> And similarly define the points B*, C*.
A# = BC inter HbHc = harmonic conjugate of Ha wrt B,C; hence A#, B#,
>
> Are the triangles ABC, A*B*C* perspective?
>
> If not :
>
> Denote :
>
> A# = BC /\ AA*, B# = CA /\ BB*, C# = AB /\ CC*
>
> Are the points A#, B#, C# collinear?
C# lie on the trilinear polar of H (orthic axis)
Of course, A* = projection of Ha upon AA#.
Friendly. JeanPierre 0 Attachment
Dear Antreas> [APH]:
triads
> >> Let A2, B2, C2 be the points of contact of the circles
> >> passing through (B,C), (C,A), (A,B) and touching
> >> the a,bcexcircles (resp.) [instead of the incircle].
> >> Are the triangles ABC, A2B2C2 perspective?
>
> We can consider other [than the three excircles] welldefined
> of circles touching the sidelines, and ask similar questions.
If the circle Ca touches BC at U, let U' = harmonic conjugate of U
wrt (B,C) and U'' = antipode of U on Ca. Then A2 is the projection
of U upon U'U''.
I've tried two particular cases : the three circles Ca, Cb, Cc touch
respectively BC, CA, AB at the vertices of the pedal  or cevian 
triangle of P and go respectively through A, B, C. Unfortunately,
I'm absolutely unable to find even one point P such as A2B2C2 and
ABC are perspective (their locus is of degree 12 in the case of the
pedal triangle and of degree 9 in the case of the cevian triangle).
Friendly. JeanPierre 0 Attachment
Dear JeanPierre
[APH]:
>> We can consider other [than the three excircles] welldefined triads
[JPE]:
>> of circles touching the sidelines, and ask similar questions.
>I've tried two particular cases : the three circles Ca, Cb, Cc touch
Here is a variation:
>respectively BC, CA, AB at the vertices of the pedal  or cevian 
>triangle of P and go respectively through A, B, C. Unfortunately,
>I'm absolutely unable to find even one point P such as A2B2C2 and
>ABC are perspective (their locus is of degree 12 in the case of the
>pedal triangle and of degree 9 in the case of the cevian triangle).
Each one of three circles Ka,Kb,Kc passes through TWO vertices
of the pedal (cevian) triangle of P, and touches ONE sideline of ABC.
Let A2, B2, C2 be the points of contact of the circles
passing through (B,C), (C,A), (A,B) and touching
the Ka,Kb,Kc circles (resp.)
Which is now the locus of P such that
ABC, A2B2C2 are perspective?
Greetings
Antreas
 0 Attachment
Dear Antreas and JeanPierre
[APH]:
>> We can consider other [than the three excircles] welldefined triads
[JPE]:
>> of circles touching the sidelines, and ask similar questions.
>I've tried two particular cases : the three circles Ca, Cb, Cc touch
Another variation:
>respectively BC, CA, AB at the vertices of the pedal  or cevian 
>triangle of P and go respectively through A, B, C. Unfortunately,
>I'm absolutely unable to find even one point P such as A2B2C2 and
>ABC are perspective (their locus is of degree 12 in the case of the
>pedal triangle and of degree 9 in the case of the cevian triangle).
Let P be a point and A1B1C1 its pedal triangle. Define Ka the circle
with center P and tangent to BC at A1. Similarly define the circles Kb,Kc.
Let A2, B2, C2 be the points of contact of the circles
passing through (B,C), (C,A), (A,B) and touching
the Ka,Kb,Kc circles (resp.)
Which is the locus of P such that
ABC, A2B2C2 are perspective?
An obvious point on this locus is the circumcenter O of ABC.
It is easy to give a synthetic proof that the triangles
ABC, A2B2C2 are homothetic and that the perspector is the
De Longchamps point.
Another point on this locus (I think) is the incenter I of ABC
and that the perspector is not in ETC.
Best regards
Nikos 0 Attachment
Dear Nikolaos and Antreas> Let P be a point and A1B1C1 its pedal triangle. Define Ka the
circle
> with center P and tangent to BC at A1. Similarly define the
circles Kb,Kc.
>
I think that this point is barycentric 1/(pa)^3:.., which is X(479)
> Let A2, B2, C2 be the points of contact of the circles
> passing through (B,C), (C,A), (A,B) and touching
> the Ka,Kb,Kc circles (resp.)
>
> Which is the locus of P such that
> ABC, A2B2C2 are perspective?
>
> An obvious point on this locus is the circumcenter O of ABC.
> It is easy to give a synthetic proof that the triangles
> ABC, A2B2C2 are homothetic and that the perspector is the
> De Longchamps point.
>
> Another point on this locus (I think) is the incenter I of ABC
> and that the perspector is not in ETC.
in ETC.
It seems that your locus is a ugly curve of degree 15.
Friendly. JeanPierre 0 Attachment
Dear JeanPierre,
[JPE]
I think that this point is barycentric 1/(pa)^3:.., which is X(479)
in ETC.
It seems that your locus is a ugly curve of degree 15.
*************
Thanks for the locus.
You are right for X(479) I was looking for a wrong search number.
It seems to me that on the locus are also the three excenters of ABC.
Best regards
Nikolaos Dergiades 0 Attachment
Another triad of circles touching the sidelines of ABC.
Let Abc be a triangle and A1B1C1 a triangle inscribed in ABC.
Construct three CONGRUENT and CONCURRENT circles Ka, Kb, Kc such that
Ka touches BC at A1, Kb touches CA at B1, and Kc touches AB at C1.
If A1B1C1 is the pedal (or cevian) triangle of P, and Ca, Cb, Cc
are the three circles passing through (B,C), (C,A), (A,B)
and touching Ka, Kb, Kc at A2,B2,C2 resp., then which is the locus
of P such that ABC, A2B2C2 are perspective?
APH 0 Attachment
[APH]:
>Let ABC be a triangle and A1B1C1 a triangle inscribed in ABC.
And which is the locus of P such that the triangles
>Construct three CONGRUENT and CONCURRENT circles (Ka), (Kb), (Kc) such that
>Ka touches BC at A1, Kb touches CA at B1, and Kc touches AB at C1.
>
>If A1B1C1 is the pedal (or cevian) triangle of P, and (Ca), (Cb), (Cc)
>are the three circles passing through (B,C), (C,A), (A,B)
>and touching (Ka), (Kb), (Kc) at A2,B2,C2 resp., then which is the locus
>of P such that ABC, A2B2C2 are perspective?
ABC, KaKbKc are perspective? (if A1B1C1 = pedal or cevian tr. of P)
APH

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