Browse Groups

• Let ABC be a triangle and Pa,Pb,Pc points on the altitudes AHa, BHb, CHc, such that: APa = t*AHa, BPb = t*BHb, CPc = t*CHc A /| / | / | / Pa / |
Message 1 of 30 , Apr 1, 2002
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Let ABC be a triangle and Pa,Pb,Pc points on the
altitudes AHa, BHb, CHc, such that:

APa = t*AHa, BPb = t*BHb, CPc = t*CHc

A
/|\
/ | \
/ | \
/ Pa \
/ | \
/ | \
B------Ha-----C

1. A1 := (reflection of BC in BPa) /\ (reflection of BC in CPa)
B1, C1 similarly.

The triangles ABC, A1B1C1 are perspective.

Perspector in Normals:

cosA
(--------------- ::), where T(t) = t^2 / (1 - 2t)
1 + T(t)sin^2A

[The locus of the perspectors is Jerabek c/h.]

2. A2 := (reflection of BA in BPa) /\ (reflection of CA in CPa)
B2, C2 similarly.

The triangles ABC, A2B2C2 are perspective.

Perspector in Normals:

1 1 - T(t)sin^2A
(------ * ------------------ ::)
cosA 1 - T'(t)sin^2A

1 - 2t + (1 - t)^2
where T(t) = -------------------
1 - 2t

t^2 - 2
T'(t) = ----------
t - 2

3. A3 := (reflection of BPa in BA) /\ (reflection of CPa in CA)
B3, C3 similarly.

The triangles ABC, A3B3C3 are perspective.

Perspector in Normals:

1 1 - T(t)sin^2A
(------ * ------------------ ::)
cosA 1 - T'(t)sin^2A

Where T(t) = 2t / (1 + t), T'(t) = 2 / t

Antreas

A mantinada (= Cretan distich) in modern style I learned
in the days (last Friday to Monday) I was in Crete:

MODEM qa balw sto xwrio, COMPUTER sto mhtato,
sto INTERNET qa to poulw to gala twn probatw.

aph
• Dear Antreas, you must add in your mantinada ... kai me EMAIL s olh th gh qa steilw to mantato, me mpalwqies ton tolmhth HACKER qa balw katw. Nikos
Message 1 of 30 , Apr 2, 2002
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Dear Antreas,

> A mantinada (= Cretan distich) in modern style I learned
> in the days (last Friday to Monday) I was in Crete:
>
> MODEM qa balw sto xwrio, COMPUTER sto mhtato,
> sto INTERNET qa to poulw to gala twn probatw.

kai me EMAIL s'olh th gh qa steilw to mantato,
me mpalwqies ton tolmhth HACKER qa balw katw.

Nikos

--- In Hyacinthos@y..., "Antreas P. Hatzipolakis" <xpolakis@m...>
wrote:
> Let ABC be a triangle and Pa,Pb,Pc points on the
> altitudes AHa, BHb, CHc, such that:
>
> APa = t*AHa, BPb = t*BHb, CPc = t*CHc
>
> A
> /|\
> / | \
> / | \
> / Pa \
> / | \
> / | \
> B------Ha-----C
>
>
> 1. A1 := (reflection of BC in BPa) /\ (reflection of BC in CPa)
> B1, C1 similarly.
>
> The triangles ABC, A1B1C1 are perspective.
>
> Perspector in Normals:
>
> cosA
> (--------------- ::), where T(t) = t^2 / (1 - 2t)
> 1 + T(t)sin^2A
>
> [The locus of the perspectors is Jerabek c/h.]
>
> 2. A2 := (reflection of BA in BPa) /\ (reflection of CA in CPa)
> B2, C2 similarly.
>
> The triangles ABC, A2B2C2 are perspective.
>
> Perspector in Normals:
>
> 1 1 - T(t)sin^2A
> (------ * ------------------ ::)
> cosA 1 - T'(t)sin^2A
>
> 1 - 2t + (1 - t)^2
> where T(t) = -------------------
> 1 - 2t
>
> t^2 - 2
> T'(t) = ----------
> t - 2
>
>
> 3. A3 := (reflection of BPa in BA) /\ (reflection of CPa in CA)
> B3, C3 similarly.
>
> The triangles ABC, A3B3C3 are perspective.
>
> Perspector in Normals:
>
> 1 1 - T(t)sin^2A
> (------ * ------------------ ::)
> cosA 1 - T'(t)sin^2A
>
>
> Where T(t) = 2t / (1 + t), T'(t) = 2 / t
>
>
> Antreas
>
> A mantinada (= Cretan distich) in modern style I learned
> in the days (last Friday to Monday) I was in Crete:
>
> MODEM qa balw sto xwrio, COMPUTER sto mhtato,
> sto INTERNET qa to poulw to gala twn probatw.
>
> aph
• Dear Antreas and Nikos, ... (free translation) In my village I ll set a MODEM, in my ranch a COMPUTER through INTERNET I ll sell, my cow s milk and butter
Message 1 of 30 , Apr 2, 2002
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Dear Antreas and Nikos,

>-----Original Message-----
>Sent: Tuesday, April 02, 2002 11:51 AM
>To: Hyacinthos@yahoogroups.com
>Subject: [EMHL] Re: Reflections

>
>> A mantinada (= Cretan distich) in modern style I learned
>> in the days (last Friday to Monday) I was in Crete:
>>
>> MODEM qa balw sto xwrio, COMPUTER sto mhtato,
>> sto INTERNET qa to poulw to gala twn probatw.

(free translation)

In my village I' ll set a MODEM, in my ranch a COMPUTER
through INTERNET I'll sell, my cow's milk and butter

>
> kai me EMAIL s'olh th gh qa steilw to mantato,
> me mpalwqies ton tolmhth HACKER qa balw katw.
>

(free translation)

via E-MAIL to all I will let it be known
if I see a HACKER, I will gun him down

Michael
• Dear Michael, ... It is sheeps milk. Cretans are not cow-boys ! (Although they have much in common with Texans ! :-) A nice book about Cretan shepherds: David
Message 1 of 30 , Apr 2, 2002
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Dear Michael,

[APH]:

>>> A mantinada (= Cretan distich) in modern style I learned
>>> in the days (last Friday to Monday) I was in Crete:
>>>
>>> MODEM qa balw sto xwrio, COMPUTER sto mhtato,
>>> sto INTERNET qa to poulw to gala twn probatw.
>
>

[ML]:

> (free translation)
>
> In my village I' ll set a MODEM, in my ranch a COMPUTER
> through INTERNET I'll sell, my cow's milk and butter
>

It is sheeps' milk. Cretans are not cow-boys !
(Although they have much in common with Texans ! :-)

A nice book about Cretan shepherds:

David Outerbridge - Julie Thayer: The Last Shepherds.
New York, N.Y.: The Viking Press, 1979
(In pp. 108 - 130: Miltiades Xsilouris "voskos" Anoyia)

[ND]:

>
>>
>> kai me EMAIL s'olh th gh qa steilw to mantato,
>> me mpalwqies ton tolmhth HACKER qa balw katw.
>>
>

[ML]:

> (free translation)
>
> via E-MAIL to all I will let it be known
> if I see a HACKER, I will gun him down
>
>

And two very good books about modern Cretan social life:

Michael Herzfeld: Poetics of Manhood.
Contest and Identity in a Cretan Mountain Village.
Princeton, N.J.: Princeton University Press, 1985.

Michael Herzfeld: A Place in History. Social and Monumental
Time in a Cretan Town.
Princeton, N.J.: Princeton University Press, 1991.

Antreas
• Let ABC be a triangle, P a point and p a line passing through P. Let A , B , C be the reflections of A,B,C in p. Which is the locus of p such that ABC, A B C
Message 1 of 30 , Jun 17, 2002
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Let ABC be a triangle, P a point and p a line passing through P.

Let A', B', C' be the reflections of A,B,C in p.

Which is the locus of p such that ABC, A'B'C' are rthologic?
(the whole plane?)

APH
• Dear Antreas, yes the locus is the whole plane. If S is the orthopole of p and H is the reflection of H = Orthocenter of ABC in S then the perpendicular
Message 1 of 30 , Jun 17, 2002
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Dear Antreas, yes the locus is the whole plane.
If S is the orthopole of p and H' is the reflection of
H = Orthocenter of ABC in S
then the perpendicular from A' to BC passes through H'
and the same for the perpendiculars from B', C' to CA, AB.
Hence ABC, A'B'C' are orthologic.

Best regards

--- In Hyacinthos@y..., Antreas P. Hatzipolakis <xpolakis@m...> wrote:
> Let ABC be a triangle, P a point and p a line passing through P.
>
> Let A', B', C' be the reflections of A,B,C in p.
>
> Which is the locus of p such that ABC, A'B'C' are rthologic?
> (the whole plane?)
>
> APH
• Dear Antreas, ... Two indirectly similar triangles are allways orthologic. Friendly. Jean-Pierre
Message 1 of 30 , Jun 17, 2002
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Dear Antreas,

> Let ABC be a triangle, P a point and p a line passing through P.
>
> Let A', B', C' be the reflections of A,B,C in p.
>
> Which is the locus of p such that ABC, A'B'C' are rthologic?
> (the whole plane?)

Two indirectly similar triangles are allways orthologic.
Friendly. Jean-Pierre
• Let ABC be a triangle P a point and PaPbPc the pedal triangle of P. Which is the locus of P such that the orthocenter of PaPbPc is lying on the Euler line of
Message 1 of 30 , Aug 12, 2003
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Let ABC be a triangle P a point and PaPbPc the pedal triangle
of P.

Which is the locus of P such that the orthocenter of PaPbPc
is lying on the Euler line of ABC?

I think that it is equivalent to:

Which is the locus of P such that the reflections
of the Euler line of ABC in the sides of PaPbPc
are concurrent.

Antreas

__________________________________________________________________
McAfee VirusScan Online from the Netscape Network.
http://channels.netscape.com/ns/computing/mcafee/index.jsp?promo=393397

• Dear Antreas, ... Yes, it is equivalent. Thank you for the interesting problem. The locus is a cubic with equation a² SA (b²-c²) (b²z+c²y)
Message 1 of 30 , Aug 12, 2003
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Dear Antreas,

In Hyacinthos message #7495, you wrote:

>> Let ABC be a triangle P a point and PaPbPc the pedal
>> triangle of P.
>>
>> Which is the locus of P such that the orthocenter of
>> PaPbPc is lying on the Euler line of ABC?
>>
>> I think that it is equivalent to:
>>
>> Which is the locus of P such that the reflections
>> of the Euler line of ABC in the sides of PaPbPc
>> are concurrent.

Yes, it is equivalent. Thank you for the interesting
problem.

The locus is a cubic with equation

a² SA (b²-c²) (b²z+c²y) (b²SBzx+c²SCxy-a²SAyz+b²c²x²)
+ ... cyclic
= 0.

I don't know any remarkable points on this cubic except
of the circumcenter X(3).

In fact, the orthocenter of the pedal triangle of a
point with barycentrics ( x : y : z ) has barycentrics

( a² (b²z+c²y) (b²SBzx+c²SCxy-a²SAyz+b²c²x²)
: ...
: ... ).

We can take some other lines instead of the Euler line.
The Brocard axis gives a cubic through the
circumcenter and the orthocenter, for instance.

The same question with cevian instead of pedal triangles
would be interesting too, but I have not tried yet.

Sincerely,
Darij Grinberg
• Dear Antreas and Darij, [APH] ... [DG] ... Let L be the trilinear polar of point P. The locus of P such that the orthocenter of PaPbPc lies on L is a circular
Message 1 of 30 , Aug 12, 2003
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Dear Antreas and Darij,

[APH]
> >> Let ABC be a triangle P a point and PaPbPc the pedal
> >> triangle of P.
> >>
> >> Which is the locus of P such that the orthocenter of
> >> PaPbPc is lying on the Euler line of ABC?

[DG]
> We can take some other lines instead of the Euler line.
> The Brocard axis gives a cubic through the
> circumcenter and the orthocenter, for instance.

Let L be the trilinear polar of point P.
The locus of P such that the orthocenter of PaPbPc lies on L is a
circular circum-cubic.
All those cubics form a mesh generated by three decomposed members Ca,
Cb, Cc.

If KaKbKc is the tangential triangle, Ca is the union of the line KbKc
and the circle centered at Ka passing through B and C.

Those cubics are not very exciting. The "best choice" would be P=H and
L=orthic axis where we obtain a nK0 with pole X571, root X6.
The real point at infinity is X924 and the cubic contains also X110 and
the centers of the Apollonian circles.

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Antreas, ... Now I have. The orthocenter of the cevian triangle of a point P with barycentrics ( 1/x : 1/y : 1/z ) has barycentrics ( (a²x + SCy + SBz)
Message 1 of 30 , Aug 12, 2003
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Dear Antreas,

In Hyacinthos message #7498, I wrote:

>> The same question with cevian instead of pedal triangles
>> would be interesting too, but I have not tried yet.

Now I have. The orthocenter of the cevian triangle of a
point P with barycentrics ( 1/x : 1/y : 1/z ) has
barycentrics

( (a²x + SCy + SBz) (2SAyz + b²y² + c²z² - a²x²)
: ...
: ... ).

The locus of the point P so that this orthocenter lies on
any special line must be a sextic. If the special line
passes through the circumcenter of triangle ABC, this
sextic decomposes into the union of the circumcircle [see
next message] and a quartic which is actually the
isogonal conjugate of a conic through the circumcenter.
If our special line is the Brocard axis, this quartic has
equation

a²(b²-c²)y²z² + b²(c²-a²)z²x² + c²(a²-b²)x²y² = 0.

This quartic passes through X(1), X(2) and X(4) and is the
isogonal conjugate of the conic

a²(b²-c²)x² + b²(c²-a²)y² + c²(a²-b²)z² = 0,

which known as the Stammler hyperbola.

Sincerely,
Darij Grinberg
• Dear Antreas, ... ... where ( x : y : z ) are the barycentrics of the point; not ( 1/x : 1/y : 1/z ). Sorry for the ambiguous use of variables. Sincerely,
Message 1 of 30 , Aug 12, 2003
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Dear Antreas,

In Hyacinthos message #7505, I wrote:

>> If our special line is the Brocard axis, this quartic has
>> equation
>>
>> a²(b²-c²)y²z² + b²(c²-a²)z²x² + c²(a²-b²)x²y² = 0.

... where ( x : y : z ) are the barycentrics of the point;
not ( 1/x : 1/y : 1/z ).

Sorry for the ambiguous use of variables.

Sincerely,
Darij Grinberg
• Let ABC be a triangle, HaHbHc its orthic triangle and P a point. Which is the locus of P such that the reflections of PHa, PHb, PHc in HHa, HHb, HHc, resp are
Message 1 of 30 , Sep 12, 2003
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Let ABC be a triangle, HaHbHc its orthic triangle
and P a point.

Which is the locus of P such that the reflections
of PHa, PHb, PHc in HHa, HHb, HHc, resp
are concurrent?

If the locus is the whole plane,
then which is the locus of the point of
concurrence, if P moves on a notable line (OH, OK,...)?

Now, let P, P* be two isogonal points
and PaPbPc the pedal triangle of P.

Which is the locus of P such that the reflections of
P*Pa, P*Pb, P*Pc in PPa, PPb, PPc, resp.
are concurrent?

Greetings from Athens

Antreas

--
• Dear Antreas, ... The whole plane: the reflections concur at the isogonal conjugate of P with respect to the orthic triangle HaHbHc. In fact, HHa, HHb, HHc are
Message 1 of 30 , Sep 12, 2003
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Dear Antreas,

In Hyacinthos message #7868, you wrote:

>> Let ABC be a triangle, HaHbHc its orthic triangle
>> and P a point.
>>
>> Which is the locus of P such that the reflections
>> of PHa, PHb, PHc in HHa, HHb, HHc, resp
>> are concurrent?

The whole plane: the reflections concur at the isogonal
conjugate of P with respect to the orthic triangle
HaHbHc. In fact, HHa, HHb, HHc are the angle bisectors
of the orthic triangle.

Sincerely,
Darij Grinberg
• Let ABC be a triangle, P a point, and Oa;Ha, Ob;Hb, Oc;Hc, the circumcenters, Orthocenters of PBC, PCA, PAB. The locus of P such that the lines OaHa, ObHb,
Message 1 of 30 , Jan 30, 2004
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Let ABC be a triangle, P a point, and Oa;Ha, Ob;Hb, Oc;Hc,
the circumcenters, Orthocenters of PBC, PCA, PAB.

The locus of P such that the lines
OaHa, ObHb, OcHc [Euler Lines] are concurrent is well-known.

Now, let O'a, H'a be the reflections of Oa, Ha in BC, resp.
Similarly O'b, H'b; O'c,H'c.

Which is the locus of P such that the lines

1. O'aHa, O'bHb, O'cHc

2. OaH'a, ObH'b, OcH'c

are concurrent?

Greetings from Athens

Antreas
• Dear Antreas, ... the McCay cubic + Linf + C(O,R) ... its inverse in the circumcircle, a nice bicircular quintic with a triple point at O where the tangents
Message 1 of 30 , Jan 30, 2004
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Dear Antreas,

> [APH] Let ABC be a triangle, P a point, and Oa;Ha, Ob;Hb, Oc;Hc,
> the circumcenters, Orthocenters of PBC, PCA, PAB.
>
> The locus of P such that the lines
> OaHa, ObHb, OcHc [Euler Lines] are concurrent is well-known.
>
> Now, let O'a, H'a be the reflections of Oa, Ha in BC, resp.
> Similarly O'b, H'b; O'c,H'c.
>
> Which is the locus of P such that the lines
>
> 1. O'aHa, O'bHb, O'cHc

the McCay cubic + Linf + C(O,R)

> 2. OaH'a, ObH'b, OcH'c

its inverse in the circumcircle, a nice bicircular quintic with a
triple point at O where the tangents make 60° angles with one another.

> are concurrent?

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Bernard ... Naturally we can replace the points O, H with another ones, but probably the case: O, K [instead of H] is most interesting (since the locus of
Message 1 of 30 , Jan 30, 2004
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Dear Bernard

[APH]:
>> [APH] Let ABC be a triangle, P a point, and Oa;Ha, Ob;Hb, Oc;Hc,
>> the circumcenters, Orthocenters of PBC, PCA, PAB.
>>
>> The locus of P such that the lines
>> OaHa, ObHb, OcHc [Euler Lines] are concurrent is well-known.
>>
>> Now, let O'a, H'a be the reflections of Oa, Ha in BC, resp.
>> Similarly O'b, H'b; O'c,H'c.
>>
>> Which is the locus of P such that the lines
>>
>> 1. O'aHa, O'bHb, O'cHc

[BG]:
>the McCay cubic + Linf + C(O,R)

[APH]:
>> 2. OaH'a, ObH'b, OcH'c

[BG]:
>its inverse in the circumcircle, a nice bicircular quintic with a
>triple point at O where the tangents make 60 deg. angles with one another.

[APH]:
>> are concurrent?

Naturally we can replace the points O, H with another ones,
but probably the case: O, K [instead of H] is most interesting
(since the locus of the Euler lines mentioned above, is the same with
the locus of the Brocard axes)

So, let O'a, K'a be the reflections of Oa, Ka in BC, resp.
Similarly O'b, K'b; O'c,K'c, where K's are the Lemoines
of PBC, PCA, PAB.

Which is the locus of P such that the lines

1. O'aKa, O'bKb, O'cKc

2. OaK'a, ObK'b, OcK'c

are concurrent?

Greetings from Athens

Antreas

--
• Let ABC be a triangle, P a point, A ,B ,C the reflections of P in BC,CA,AB, resp., and A ,B ,C the reflections of P in AA ,BB ,CC resp. Which is the locus
Message 1 of 30 , Mar 12, 2004
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Let ABC be a triangle, P a point, A',B',C' the reflections
of P in BC,CA,AB, resp., and A",B",C" the reflections of P
in AA',BB',CC' resp.

Which is the locus of p such that ABC, A"B"C" are
(1) Perspective (2) Orthologic ?

Also, if PA" /\ BC = A*, PB" /\ CA = B*, PC" /\ AB = C*,
then which is the locus of P such that A*,B*,C*, are collinear?

Greetings from Athens

Antreas

--
• A well-known theorem is: Let L be a line passing through H. The reflections La,Lb,Lc of L in the sidelines BC,CA,AB, resp. of ABC, are concurrent. Now, denote
Message 1 of 30 , Jul 13, 2006
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A well-known theorem is:

Let L be a line passing through H.
The reflections La,Lb,Lc of L in the sidelines BC,CA,AB, resp.
of ABC, are concurrent.

Now, denote Ma,Mb,Mc the reflections of BC,CA,AB in La,Lb,Lc, resp.

For which line(s) L [passing through H] these reflections
Ma,Mb,Mc are concurrent?

In general:

Let L be a line, La,Lb,Lc its reflections in BC,CA,AB,resp.,
and Ma,Mb,Mc the reflections of BC,CA,AB in La,Lb,Lc, resp.

Which is the envelope of L such that Ma,Mb,Mc be concurrent?

Antreas
--
• Let ABC be a triangle P = (x:y:z) a point and L a line through P. The Reflections of AP, BP, CP in L intersect BC, CA, AB at A ,B ,C , resp. The points
Message 1 of 30 , Feb 11, 2009
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Let ABC be a triangle P = (x:y:z) a point and L a line through P.

The Reflections of AP, BP, CP in L intersect BC, CA, AB
at A',B',C', resp.

The points A',B',C' are collinear.
Denote this line as L(P).

If L = PP* (ie the line passing through two isogonal points
P,P*) which is the intersection of L(P) and L(P*) ?
(Special case: P = H, P* = O)

APH
• For the moment, I see that L(P) and L(P*) are tangent to the in-conic with foci P and P* but that s all I can do! Friendly Francois [Non-text portions of this
Message 1 of 30 , Feb 11, 2009
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For the moment, I see that L(P) and L(P*) are tangent to the in-conic with
foci P and P* but that's all I can do!
Friendly
Francois

[Non-text portions of this message have been removed]
• Dear Antreas and François, ... The intersection Q of L(P) and L(P*) is a 15th degree point with respect to the coordinates p:q:r of P ! Very ugly ! When P =
Message 1 of 30 , Feb 11, 2009
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Dear Antreas and François,

> [APH] Let ABC be a triangle P = (x:y:z) a point and L a line through
> P.
>
> The Reflections of AP, BP, CP in L intersect BC, CA, AB
> at A',B',C', resp.
>
> The points A',B',C' are collinear.
> Denote this line as L(P).
>
> If L = PP* (ie the line passing through two isogonal points
> P,P*) which is the intersection of L(P) and L(P*) ?
> (Special case: P = H, P* = O)

The intersection Q of L(P) and L(P*) is a 15th degree point with
respect to the coordinates p:q:r of P ! Very ugly !

When P = O, it is X3258, the complement of X476.

Note that L(P) and L(P*) are parallel when P and P* are two points on
Kjp = K024.

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Bernard Given that X3258 lies on the NPC of ABC (pedal circle of O and H), I make this generalization (conjecture): Let ABC be a triangle, and P,P* two
Message 1 of 30 , Feb 14, 2009
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Dear Bernard

Given that X3258 lies on the NPC of ABC (pedal circle of O and H),
I make this generalization (conjecture):

Let ABC be a triangle, and P,P* two isogonal
conjugate points.

The reflections of AP,BP,CP in the line PP* intersect
BC,CA,AB, at three collinear points A',B',C'.

The reflections of AP*,BP*,CP* in the line PP* intersect
BC,CA,AB, at three collinear points A*,B*,C*.

Conjecture:

The Lines A'B'C' and A*B*C* intersect at a point
lying on the common pedal circle of P and P*.

Is it true for all P's, or only for some ones? Locus??

Antreas

> > [APH] Let ABC be a triangle P = (x:y:z) a point and L a line through
> > P.
> >
> > The Reflections of AP, BP, CP in L intersect BC, CA, AB
> > at A',B',C', resp.
> >
> > The points A',B',C' are collinear.
> > Denote this line as L(P).
> >
> > If L = PP* (ie the line passing through two isogonal points
> > P,P*) which is the intersection of L(P) and L(P*) ?
> > (Special case: P = H, P* = O)

[BG]
> The intersection Q of L(P) and L(P*) is a 15th degree point with
> respect to the coordinates p:q:r of P ! Very ugly !
>
> When P = O, it is X3258, the complement of X476.
>
> Note that L(P) and L(P*) are parallel when P and P* are two points on
> Kjp = K024.
• Dear Antreas, ... This is true for any point P on the McCay cubic and on a tricircular isogonal 12th degree curve with six real asymptotes parallel to those of
Message 1 of 30 , Feb 15, 2009
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Dear Antreas,

> [APH] Given that X3258 lies on the NPC of ABC (pedal circle of O and
> H),
> I make this generalization (conjecture):
>
> Let ABC be a triangle, and P,P* two isogonal
> conjugate points.
>
> The reflections of AP,BP,CP in the line PP* intersect
> BC,CA,AB, at three collinear points A',B',C'.
>
> The reflections of AP*,BP*,CP* in the line PP* intersect
> BC,CA,AB, at three collinear points A*,B*,C*.
>
> Conjecture:
>
> The Lines A'B'C' and A*B*C* intersect at a point
> lying on the common pedal circle of P and P*.
>
> Is it true for all P's, or only for some ones? Locus??

This is true for any point P on the McCay cubic and on a tricircular
isogonal 12th degree curve with six real asymptotes parallel to those
of the Thomson cubic and the Kjp cubic.

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Francois [APH] ... [FR] ... Probably is interesting the envelope of the lines L(P) as P moves on the line L. Special Case: L = Euler Line. By the way,
Message 1 of 30 , Feb 17, 2009
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Dear Francois

[APH]
>>Let ABC be a triangle P = (x:y:z) a point and L a line through P.
>>
>> The Reflections of AP, BP, CP in L intersect BC, CA, AB
>> at A',B',C', resp.
>>
>> The points A',B',C' are collinear.
>> Denote this line as L(P).
>>
>> If L = PP* (ie the line passing through two isogonal points
>> P,P*) which is the intersection of L(P) and L(P*) ?
>> (Special case: P = H, P* = O)

[FR]
> For the moment, I see that L(P) and L(P*) are tangent to
> the in-conic with
> foci P and P* but that's all I can do!

Probably is interesting the envelope of the lines L(P)
as P moves on the line L.
Special Case: L = Euler Line.

By the way, which point is the intersection of the trilinear
polars of two isogonal conjugate points P,P*?

Greetings from Greece

APH
• Dear Andreas, in case of the Euler line the envelope is the inconic with foci H, O and perspector the isotomic conjugate of O. If the line L passes through the
Message 1 of 30 , Feb 17, 2009
View Source
Dear Andreas,
in case of the Euler line the envelope is
the inconic with foci H, O and perspector the
isotomic conjugate of O.
If the line L passes through the incenter I
then the envelope of L(P) is always the incircle.

Best regards
Nikos

> [APH]
> >>Let ABC be a triangle P = (x:y:z) a point and L a
> line through P.
> >>
> >> The Reflections of AP, BP, CP in L intersect BC,
> CA, AB
> >> at A',B',C', resp.
> >>
> >> The points A',B',C' are collinear.
> >> Denote this line as L(P).
>
> [FR]
> > For the moment, I see that L(P) and L(P*) are tangent
> to
> > the in-conic with
> > foci P and P* but that's all I can do!
>
> Probably is interesting the envelope of the lines L(P)
> as P moves on the line L.
> Special Case: L = Euler Line.
>
> By the way, which point is the intersection of the
> trilinear
> polars of two isogonal conjugate points P,P*?
>
> Greetings from Greece
>
> APH
>

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• Let ABC be a triangle, P a point, A*B*C* the cevian triangle of P, and L a line through P. The reflections of AP,BP,CP in L intersect the BC,CA,AB at the
Message 1 of 30 , May 31, 2009
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Let ABC be a triangle, P a point, A*B*C*
the cevian triangle of P, and L a line through P.

The reflections of AP,BP,CP in L intersect
the BC,CA,AB at the collinear points A',B',C', resp.
and the B*C*,C*A*,A*B* at the collinear points A",B",C", resp.

Which is the locus of the intersection of the lines
A'B'C' and A"B"C" as P moves on the line L?

Special Case: L = Euler Line of ABC.

Antreas

[APH]
> Let ABC be a triangle P = (x:y:z) a point and L a line through P.
>
> The Reflections of AP, BP, CP in L intersect BC, CA, AB
> at A',B',C', resp.
>
> The points A',B',C' are collinear.
> Denote this line as L(P).
>
> If L = PP* (ie the line passing through two isogonal points
> P,P*) which is the intersection of L(P) and L(P*) ?
> (Special case: P = H, P* = O)
>
> APH
>
• Let ABC be a triangle, P a point and L a line passing through P. The reflections of AP, BP, CP in L intersect BC, CA, AB in three collinear points. Call this
Message 1 of 30 , Jun 20, 2009
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Let ABC be a triangle, P a point and L a line
passing through P.

The reflections of AP, BP, CP in L
intersect BC, CA, AB in three collinear points.
Call this new line as f(L).

Reversely:

For a pointP and a line L, there are two lines
L1,L2, perpendicular at P, such that:
f(L1) = f(L2) = L.

Now, let L be the line tangent to incircle
at the Feuerbach Point.

Which are the perpendicular at I lines L1, L2,
so that f(L1) = f(L2) = L ?

Antreas

[APH]
> > >>Let ABC be a triangle P = (x:y:z) a point and L a
> > line through P.
> > >>
> > >> The Reflections of AP, BP, CP in L intersect BC,
> > CA, AB
> > >> at A',B',C', resp.
> > >>
> > >> The points A',B',C' are collinear.
> > >> Denote this line as L(P).
> >
> > Probably is interesting the envelope of the lines L(P)
> > as P moves on the line L.
> > Special Case: L = Euler Line.
> >
> > By the way, which point is the intersection of the
> > trilinear
> > polars of two isogonal conjugate points P,P*?
> >

[FR]
> > > For the moment, I see that L(P) and L(P*) are tangent
> > > to the in-conic with
> > > foci P and P* but that's all I can do!

[ND]
> in case of the Euler line the envelope is
> the inconic with foci H, O and perspector the
> isotomic conjugate of O.
> If the line L passes through the incenter I
> then the envelope of L(P) is always the incircle.
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