 Hyacinthos

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Reflections
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Dear all,
Let us start with our friend ABC.
Let us consider the lines a_c through C and a_b through B such that the
reflection of a_c through AB and the reflection of a_b through AC are
the same line a_a. So a_b is: a_c reflected in AB and the result
reflected in AC.
In the same way we can make triples b_i and c_i for i = a,b,c.
The triangle formed by a_a, b_b and c_c is of course the reflection
triangle (A reflected in BC, B in AC and C in AB).
"Problem" for you:
The lines a_b and a_c intersect in a point A_1. Idem B_1 and C_1.
Triangle A_1B_1C_1 is perspective to ABC. Identify the perspector.
The point A_1 can sometimes be seen as the point where a laser beam
meets itself when sent from C to B via reflection in AB and AC.
The easiest way to solve this problem is of course to use
Pperpendicularity.
Kind regards,
Floor van Lamoen <f.v.lamoen@...>.
PS. Barry, does adding the address after my name help you to read the
address?. 0 Attachment
Dear all,
I posted this problem:
> Let us start with our friend ABC.
I will have to add my excuses on bad English to the ones that our French
>
> Let us consider the lines a_c through C and a_b through B such that the
> reflection of a_c through AB and the reflection of a_b through AC are
> the same line a_a. So a_b is: a_c reflected in AB and the result
> reflected in AC.
>
> In the same way we can make triples b_i and c_i for i = a,b,c.
>
> The triangle formed by a_a, b_b and c_c is of course the reflection
> triangle (A reflected in BC, B in AC and C in AB).
>
> "Problem" for you:
> The lines a_b and a_c intersect in a point A_1. Idem B_1 and C_1.
> Triangle A_1B_1C_1 is perspective to ABC. Identify the perspector.
>
> The point A_1 can sometimes be seen as the point where a laser beam
> meets itself when sent from C to B via reflection in AB and AC.
>
> The easiest way to solve this problem is of course to use
> Pperpendicularity.
and Greek friends gave. For example, "Reflection in ..." is a Dutchism
for "Reflection through ...".
The solution:
Consider P=(f:g:h) as base for perpendicularity.
The point CP/\AB = (f:g:0) so the reflection of C through AB is the
reflection of C through (f:g:0) is (2f:2g:fg). This point I call C',
the Cvertex of the Preflection triangle.
Similarly A' is (gh:2g:2h).
The point X=A'C'/\AB is found as ( 3fhg : 2g(1+h) : 0 ).
The point Y=A'C'/\BC is found as ( 0 : 2g(1+f) : 3fhg ).
Line CX and BY are of course [2g(1+h) : g3fh : 0] and [0 : g3fh :
2g(1+f)] respectively.
2g(1+f)(1+h)
These lines intersect in B_1 = ( 1+f :  : 1+h ).
3fhg
Finding similar results for A_1 and C_1 we see that the perspector of
ABC and A_1B_1C_1 must be (1+f:1+g:1+h), which is the point dividing
segment GP in ratio 1:3. It is the Pnine point center.
Best regards,
Floor van Lamoen <f.v.lamoen AT wxs.nl> 0 Attachment
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 \
/  \
/  \
BHaC
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 0 Attachment
Dear Antreas,
you must add in your mantinada
> A mantinada (= Cretan distich) in modern style I learned
kai me EMAIL s'olh th gh qa steilw to mantato,
> 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.
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 \
> /  \
> /  \
> BHaC
>
>
> 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 0 Attachment
Dear Antreas and Nikos,
>Original Message
(free translation)
>From: ndergiades [mailto:ndergiades@...]
>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.
In my village I' ll set a MODEM, in my ranch a COMPUTER
through INTERNET I'll sell, my cow's milk and butter
>
(free translation)
> kai me EMAIL s'olh th gh qa steilw to mantato,
> me mpalwqies ton tolmhth HACKER qa balw katw.
>
via EMAIL to all I will let it be known
if I see a HACKER, I will gun him down
Michael 0 Attachment
Dear Michael,
[APH]:
>>> A mantinada (= Cretan distich) in modern style I learned
[ML]:
>>> 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)
It is sheeps' milk. Cretans are not cowboys !
>
> In my village I' ll set a MODEM, in my ranch a COMPUTER
> through INTERNET I'll sell, my cow's milk and butter
>
(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]:
>
[ML]:
>>
>> kai me EMAIL s'olh th gh qa steilw to mantato,
>> me mpalwqies ton tolmhth HACKER qa balw katw.
>>
>
> (free translation)
And two very good books about modern Cretan social life:
>
> via EMAIL to all I will let it be known
> if I see a HACKER, I will gun him down
>
>
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 0 Attachment
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 0 Attachment
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
Nikos Dergiades
 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 0 Attachment
Dear Antreas,
> Let ABC be a triangle, P a point and p a line passing through P.
Two indirectly similar triangles are allways orthologic.
>
> 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?)
Friendly. JeanPierre 0 Attachment
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
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Dear Antreas,
In Hyacinthos message #7495, you wrote:
>> Let ABC be a triangle P a point and PaPbPc the pedal
Yes, it is equivalent. Thank you for the interesting
>> 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.
problem.
The locus is a cubic with equation
a² SA (b²c²) (b²z+c²y) (b²SBzx+c²SCxya²SAyz+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+c²SCxya²SAyz+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 0 Attachment
Dear Antreas and Darij,
[APH]> >> Let ABC be a triangle P a point and PaPbPc the pedal
[DG]
> >> triangle of P.
> >>
> >> Which is the locus of P such that the orthocenter of
> >> PaPbPc is lying on the Euler line of ABC?
> We can take some other lines instead of the Euler line.
Let L be the trilinear polar of point P.
> The Brocard axis gives a cubic through the
> circumcenter and the orthocenter, for instance.
The locus of P such that the orthocenter of PaPbPc lies on L is a
circular circumcubic.
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
[Nontext portions of this message have been removed] 0 Attachment
Dear Antreas,
In Hyacinthos message #7498, I wrote:
>> The same question with cevian instead of pedal triangles
Now I have. The orthocenter of the cevian triangle of a
>> would be interesting too, but I have not tried yet.
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(b²c²)y²z² + b²(c²a²)z²x² + c²(a²b²)x²y² = 0.
This quartic passes through X(1), X(2) and X(4) and is the
isogonal conjugate of the conic
a²(b²c²)x² + b²(c²a²)y² + c²(a²b²)z² = 0,
which known as the Stammler hyperbola.
Sincerely,
Darij Grinberg 0 Attachment
Dear Antreas,
In Hyacinthos message #7505, I wrote:
>> If our special line is the Brocard axis, this quartic has
... where ( x : y : z ) are the barycentrics of the point;
>> equation
>>
>> a²(b²c²)y²z² + b²(c²a²)z²x² + c²(a²b²)x²y² = 0.
not ( 1/x : 1/y : 1/z ).
Sorry for the ambiguous use of variables.
Sincerely,
Darij Grinberg 0 Attachment
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?
(Variations: isotomic instead of isogonal;
cevian instead of pedal)
Greetings from Athens
Antreas
 0 Attachment
Dear Antreas,
In Hyacinthos message #7868, you wrote:
>> Let ABC be a triangle, HaHbHc its orthic triangle
The whole plane: the reflections concur at the isogonal
>> 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?
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 0 Attachment
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 wellknown.
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 0 Attachment
Dear Antreas,
> [APH] Let ABC be a triangle, P a point, and Oa;Ha, Ob;Hb, Oc;Hc,
the McCay cubic + Linf + C(O,R)
> the circumcenters, Orthocenters of PBC, PCA, PAB.
>
> The locus of P such that the lines
> OaHa, ObHb, OcHc [Euler Lines] are concurrent is wellknown.
>
> 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
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
[Nontext portions of this message have been removed] 0 Attachment
Dear Bernard
[APH]:>> [APH] Let ABC be a triangle, P a point, and Oa;Ha, Ob;Hb, Oc;Hc,
[BG]:
>> the circumcenters, Orthocenters of PBC, PCA, PAB.
>>
>> The locus of P such that the lines
>> OaHa, ObHb, OcHc [Euler Lines] are concurrent is wellknown.
>>
>> 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)
[APH]:
>> 2. OaH'a, ObH'b, OcH'c
[BG]:
>its inverse in the circumcircle, a nice bicircular quintic with a
[APH]:
>triple point at O where the tangents make 60 deg. angles with one another.
>> 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
 0 Attachment
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
 0 Attachment
A wellknown 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
 0 Attachment
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 0 Attachment
For the moment, I see that L(P) and L(P*) are tangent to the inconic with
foci P and P* but that's all I can do!
Friendly
Francois
[Nontext portions of this message have been removed] 0 Attachment
Dear Antreas and François,
> [APH] Let ABC be a triangle P = (x:y:z) a point and L a line through
The intersection Q of L(P) and L(P*) is a 15th degree point with
> 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)
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
[Nontext portions of this message have been removed] 0 Attachment
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
[BG]
> > 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. 0 Attachment
Dear Antreas,
> [APH] Given that X3258 lies on the NPC of ABC (pedal circle of O and
This is true for any point P on the McCay cubic and on a tricircular
> 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??
isogonal 12th degree curve with six real asymptotes parallel to those
of the Thomson cubic and the Kjp cubic.
Best regards
Bernard
[Nontext portions of this message have been removed] 0 Attachment
Dear Francois
[APH]>>Let ABC be a triangle P = (x:y:z) a point and L a line through P.
[FR]
>>
>> 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)
> For the moment, I see that L(P) and L(P*) are tangent to
Probably is interesting the envelope of the lines L(P)
> the inconic with
> foci P and P* but that's all I can do!
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 0 Attachment
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 inconic 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 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
> 0 Attachment
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
[FR]
> > 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*?
> >
> > > For the moment, I see that L(P) and L(P*) are tangent
[ND]
> > > to the inconic with
> > > foci P and P* but that's all I can do!
> 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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