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A concyclic problem
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 Dear friends of Hyacinthos,
I found an interesting concyclic problem. Can anyone solve it synthetically?
Given triangle ABC with its circumcircle (O) and its orthocenter H. Let H_aH_bH_c be the orthic triangle of triangle ABC. P is an arbitrary point on Euler line. AP, BP, CP intersect (O) again at A_1, B_1, C_1. Let A_2, B_2, C_2 be the reflections of A_1, B_1, C_1 wrt H_a, H_b, H_c, respectively. Prove that H, A_2,B_2,C_2 are concyclic.
Best regard,
Linh Nguyen Van  Dear Linh Nguyen Van,
We have that the result is also true when P is on the circumcircle.
In this case A1=B1=C1=P and the circumcircle of A2B2C2 is image of the nine point circle by a homothety with P as center and ratio 2. Is easy that H is on this circumcircle.
Finally, we have a degenerate case when P is on an altitude, say AH. In this case we have A2=H, thus H,A_2,B_2,C_2 are trivially concyclic.
 In Hyacinthos@yahoogroups.com, "lovemathforever" <lovemathforever@...> wrote:
>
> Dear friends of Hyacinthos,
> I found an interesting concyclic problem. Can anyone solve it synthetically?
> Given triangle ABC with its circumcircle (O) and its orthocenter H. Let H_aH_bH_c be the orthic triangle of triangle ABC. P is an arbitrary point on Euler line. AP, BP, CP intersect (O) again at A_1, B_1, C_1. Let A_2, B_2, C_2 be the reflections of A_1, B_1, C_1 wrt H_a, H_b, H_c, respectively. Prove that H, A_2,B_2,C_2 are concyclic.
> Best regard,
> Linh Nguyen Van
>  Dear Linh Nguyen Van
See Hagge circle construction in FG 2007 pp. 231247
M.T.
 In Hyacinthos@yahoogroups.com, "lovemathforever" <lovemathforever@...> wrote:
>
> Dear friends of Hyacinthos,
> I found an interesting concyclic problem. Can anyone solve it synthetically?
> Given triangle ABC with its circumcircle (O) and its orthocenter H. Let H_aH_bH_c be the orthic triangle of triangle ABC. P is an arbitrary point on Euler line. AP, BP, CP intersect (O) again at A_1, B_1, C_1. Let A_2, B_2, C_2 be the reflections of A_1, B_1, C_1 wrt H_a, H_b, H_c, respectively. Prove that H, A_2,B_2,C_2 are concyclic.
> Best regard,
> Linh Nguyen Van
>  Dear Francisco,
Yes, two cases you mentioned are trivival. But my problem is stated that P lies on Euler line.
Dear Armpist,
Please read the message above carefully! My problem is only similar to Hagge circle, but it is not Hagge.
Best regard,
Linh Nguyen Van  Dear Linh Nguyen Van,
I haven't a synthetic solution at the moment, but perhaps is useful to know that there are another (although trivial) solutions of the problem.
 In Hyacinthos@yahoogroups.com, "lovemathforever" <lovemathforever@...> wrote:
>
> Dear friends of Hyacinthos,
> I found an interesting concyclic problem. Can anyone solve it synthetically?
> Given triangle ABC with its circumcircle (O) and its orthocenter H. Let H_aH_bH_c be the orthic triangle of triangle ABC. P is an arbitrary point on Euler line. AP, BP, CP intersect (O) again at A_1, B_1, C_1. Let A_2, B_2, C_2 be the reflections of A_1, B_1, C_1 wrt H_a, H_b, H_c, respectively. Prove that H, A_2,B_2,C_2 are concyclic.
> Best regard,
> Linh Nguyen Van
>  [Linh Nguyen Van]
> Given triangle ABC with its circumcircle (O) and its orthocenter
As locus problem:
> H. Let H_aH_bH_c be the orthic triangle of triangle ABC. P is
> an arbitrary point on Euler line. AP, BP, CP intersect (O) again
> at A_1, B_1, C_1. Let A_2, B_2, C_2 be the reflections of A_1,
> B_1, C_1 wrt H_a, H_b, H_c, respectively. Prove that H, A_2,B_2,C_2
> are concyclic.
Let ABC be a triangle, HaHbHc the pedal triangle of H (orthic
triangle), P a point, A1B1C1 the circumcevian triangle of P
and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc, resp.
Which is the locus of P such that the circumcircle of
A2B2C2 passes through H?
Questions:
1. Is the answer: "Circumcircle of ABC + Three Altitudes + Euler
Line of ABC" complete?
2. As P moves on the Euler line, where is moving the circumcenter
of A2B2C2?
Variation:
Let MaMbMc be the pedal triangle of O (medial triangle) and A2,
B2,C2 the reflections of A1,B1,C1 in Ma,Mb,Mc, resp.
Which is the locus of P such that the circumcircle of
A2B2C2 passes through H?
Antreas  Dear Antreas
> As locus problem:
It seems to be part of the Euler line.
>
> Let ABC be a triangle, HaHbHc the pedal triangle of H
> (orthic
> triangle), P a point, A1B1C1 the circumcevian triangle of P
>
> and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
> resp.
>
> Which is the locus of P such that the circumcircle of
> A2B2C2 passes through H?
>
> Questions:
>
>
> 2. As P moves on the Euler line, where is moving the
> circumcenter
> of A2B2C2?
Best regards
Nikos
__________________________________________________
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http://mail.yahoo.gr  We may want to know what is the locus of points P such that the orthocenter lies on one of the sides of A2B2C2.
Answer: For example, B2, C2 and H are collinear if and only if P is on the inverse with respect to the circumcircle of ABC of the circumcircle of BCH.
 In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
> [Linh Nguyen Van]
> > Given triangle ABC with its circumcircle (O) and its orthocenter
> > H. Let H_aH_bH_c be the orthic triangle of triangle ABC. P is
> > an arbitrary point on Euler line. AP, BP, CP intersect (O) again
> > at A_1, B_1, C_1. Let A_2, B_2, C_2 be the reflections of A_1,
> > B_1, C_1 wrt H_a, H_b, H_c, respectively. Prove that H, A_2,B_2,C_2
> > are concyclic.
>
> As locus problem:
>
> Let ABC be a triangle, HaHbHc the pedal triangle of H (orthic
> triangle), P a point, A1B1C1 the circumcevian triangle of P
> and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc, resp.
>
> Which is the locus of P such that the circumcircle of
> A2B2C2 passes through H?
>
> Questions:
>
> 1. Is the answer: "Circumcircle of ABC + Three Altitudes + Euler
> Line of ABC" complete?
>
> 2. As P moves on the Euler line, where is moving the circumcenter
> of A2B2C2?
>
> Variation:
>
> Let MaMbMc be the pedal triangle of O (medial triangle) and A2,
> B2,C2 the reflections of A1,B1,C1 in Ma,Mb,Mc, resp.
> Which is the locus of P such that the circumcircle of
> A2B2C2 passes through H?
>
> Antreas
>  Dear Linh Nguyen Van,
you wrote> Given triangle ABC with its circumcircle (O) and
I tried to give a synthetic proof but did nothing.
> its orthocenter H. Let HaHbHc be the orthic triangle of
> triangle ABC. P is an arbitrary point on Euler line.
> AP, BP, CP intersect (O) again at A1, B1, C1.
> Let A2, B2, C2 be the reflections of A1, B1, C1 wrt Ha, Hb, Hc,
> respectively. Prove that H, A2, B2, C2 are concyclic.
I tried to give a proof with barycentrics but also did nothing
because the computations were very complicated.
Here is a proof with complex numbers.
Let the circumcircle of ABC be the unit circle and at the
points A, B, C, O be the complex numbers a, b, c, 0.
Since the conjugate(a) = 1/a ... we can do all we need.
Then it is known and easy to prove that at H is the number h=a+b+c.
At the point P on the Euler line is the number t*(a+b+c).
At the point A1 is the number a1=bc[at(a+b+c)]/[t(ab+bc+ca)bc].
At the point Ha is the number ha=[a(a+b+c)bc]/(2a).
The reflection of O in Ha is the point O1 at the number o1=2ha.
The point A2 is at the number a2 = 2ha  a1.
The triangle O1A2H is isosceles because
O1A2 = OA1 = R = OH1 = O1H where H1 (on the circumcircle)
is the reflection of H in Ha.
The mid point M1 of A2H is at the point m1 = (a2+h)/2.
Finally the line O1M1 meets the Euler line at the point Q
that is equidistant from A2 and H.
This point Q is at the number q = m*(a+b+c)
where m = [(s+2abc)t2abc]/[(s+abc)tabc]
where s = (a+b)(b+c)(c+a).
Since this point Q is symmetric relative to a,b,c
is equidistant from B2, H and from C2, H.
Hence the points A2, B2, C2, H are concyclic
with center Q on the Euler line.
If P = H then t = 1 and Q = H
If P = O then t = 0 and Q = X382 the reflection of O in H.
Best regards
Nikos Dergiades
__________________________________________________
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http://mail.yahoo.gr  Dear friends,
Francisco mentioned me that the points
A2, B2, C2, H are collinear if P is the inverse
of H in (O).
Yes this is true if Q lies at infinity
or the denominator of q is zero
that is (s+abc)tabc = 0 or
t = abc / [(a+b)(b+c)(c+a)+abc]
and for this t the point P is the inverse
of H in (O).
Thank you Francisco.
Best regards
Nikos Dergiades
> Dear Linh Nguyen Van,
__________________________________________________
>
> you wrote
> > Given triangle ABC with its circumcircle (O) and
> > its orthocenter H. Let HaHbHc be the orthic triangle
> of
> > triangle ABC. P is an arbitrary point on Euler line.
> > AP, BP, CP intersect (O) again at A1, B1, C1.
> > Let A2, B2, C2 be the reflections of A1, B1, C1 wrt
> Ha, Hb, Hc,
> > respectively. Prove that H, A2, B2, C2 are concyclic.
>
> I tried to give a synthetic proof but did nothing.
> I tried to give a proof with barycentrics but also did
> nothing
> because the computations were very complicated.
>
> Here is a proof with complex numbers.
> Let the circumcircle of ABC be the unit circle and at the
> points A, B, C, O be the complex numbers a, b, c, 0.
> Since the conjugate(a) = 1/a ... we can do all we need.
> Then it is known and easy to prove that at H is the number
> h=a+b+c.
> At the point P on the Euler line is the number t*(a+b+c).
> At the point A1 is the number
> a1=bc[at(a+b+c)]/[t(ab+bc+ca)bc].
> At the point Ha is the number ha=[a(a+b+c)bc]/(2a).
> The reflection of O in Ha is the point O1 at the number
> o1=2ha.
> The point A2 is at the number a2 = 2ha  a1.
> The triangle O1A2H is isosceles because
> O1A2 = OA1 = R = OH1 = O1H where H1 (on the
> circumcircle)
> is the reflection of H in Ha.
> The mid point M1 of A2H is at the point m1 = (a2+h)/2.
> Finally the line O1M1 meets the Euler line at the point Q
> that is equidistant from A2 and H.
> This point Q is at the number q = m*(a+b+c)
> where m = [(s+2abc)t2abc]/[(s+abc)tabc]
> where s = (a+b)(b+c)(c+a).
> Since this point Q is symmetric relative to a,b,c
> is equidistant from B2, H and from C2, H.
> Hence the points A2, B2, C2, H are concyclic
> with center Q on the Euler line.
>
> If P = H then t = 1 and Q = H
> If P = O then t = 0 and Q = X382 the reflection of O in H.
>
> Best regards
> Nikos Dergiades
>
>
>
>
> __________________________________________________
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http://mail.yahoo.gr  Thanks you dear Nikolaos for your solution. I am also finding the synthetic solution but false.
>
Dear Francisco,
>[Francisco Javier]
> Variation:
>
> Let MaMbMc be the pedal triangle of O (medial triangle) and A2,
> B2,C2 the reflections of A1,B1,C1 in Ma,Mb,Mc, resp.
> Which is the locus of P such that the circumcircle of
> A2B2C2 passes through H?
>
For this variation you can see 'Hagge circle'. Every points P on the plane satisfy this condition.  Dear Linh Nguyen Van!
I haven't a synthetic proof, but using complex numbers it is easy to prove your assertion and also obtain that the circumcenter of A_2B_2C_2 lie on the Euelr line.
Sincerely Alexey
Dear friends of Hyacinthos,
I found an interesting concyclic problem. Can anyone solve it synthetically?
Given triangle ABC with its circumcircle (O) and its orthocenter H. Let H_aH_bH_c be the orthic triangle of triangle ABC. P is an arbitrary point on Euler line. AP, BP, CP intersect (O) again at A_1, B_1, C_1. Let A_2, B_2, C_2 be the reflections of A_1, B_1, C_1 wrt H_a, H_b, H_c, respectively. Prove that H, A_2,B_2,C_2 are concyclic.
Best regard,
Linh Nguyen Van
.
[Nontext portions of this message have been removed]  [APH]
> > As locus problem:
[ND]
> >
> > Let ABC be a triangle, HaHbHc the pedal triangle of H
> > (orthic
> > triangle), P a point, A1B1C1 the circumcevian triangle of P
> >
> > and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
> > resp.
> >
> > Which is the locus of P such that the circumcircle of
> > A2B2C2 passes through H?
> >
> > Questions:
> >
> >
> > 2. As P moves on the Euler line, where is moving the
> > circumcenter
> > of A2B2C2?
> It seems to be part of the Euler line.
Dear Nikos
So if P lies on the Euler line, then the circumcenter
of A2B2C2 lies also on the Euler line.
Reversely: If the circumcenter of A2B2C2 is on the
Euler line, then where is lying P? That is:
Let ABC be a triangle, HaHbHc the pedal triangle of H
(orthic triangle), P a point, A1B1C1 the circumcevian triangle
of P and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
resp.
Which is the locus of P such that the circumcenter of
A2B2C2 is lying on the Euler Line of ABC?
It is the Euler Line + ?????
APH  [APH]
> Let ABC be a triangle, HaHbHc the pedal triangle of H
Variation (easier?):
> (orthic triangle), P a point, A1B1C1 the circumcevian triangle
> of P and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
> resp.
> Which is the locus of P such that the circumcenter of
> A2B2C2 is lying on the Euler Line of ABC?
>
> It is the Euler Line + ?????
Let ABC be a triangle, HaHbHc the pedal triangle of H
(orthic triangle), P a point, and A',B', and C' the reflections
of P in Ha, Hb, Hc, resp.
Which is the locus of P such that the circumcenter of
A'B'C' is lying on the Euler Line of ABC?
Is it Euler Line +???
APH  Thear Antreas,
> [APH]
This is the Euler line + A quintic though A, B, C, H, Ha, Hb, Hc
> > Let ABC be a triangle, HaHbHc the pedal triangle of H
> > (orthic triangle), P a point, A1B1C1 the circumcevian triangle
> > of P and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
> > resp.
> > Which is the locus of P such that the circumcenter of
> > A2B2C2 is lying on the Euler Line of ABC?
> >
> > It is the Euler Line + ?????
Equation:
a^4 c^2 x^3 y^2 + 2 a^2 b^2 c^2 x^3 y^2  b^4 c^2 x^3 y^2 +
a^2 c^4 x^3 y^2  b^2 c^4 x^3 y^2  a^4 c^2 x^2 y^3 +
2 a^2 b^2 c^2 x^2 y^3  b^4 c^2 x^2 y^3  a^2 c^4 x^2 y^3 +
b^2 c^4 x^2 y^3 + a^6 x^3 y z  2 a^4 b^2 x^3 y z +
a^2 b^4 x^3 y z  2 a^4 c^2 x^3 y z + 4 a^2 b^2 c^2 x^3 y z 
2 b^4 c^2 x^3 y z + a^2 c^4 x^3 y z  2 b^2 c^4 x^3 y z +
a^6 x^2 y^2 z  a^4 b^2 x^2 y^2 z  a^2 b^4 x^2 y^2 z +
b^6 x^2 y^2 z  2 a^4 c^2 x^2 y^2 z + 4 a^2 b^2 c^2 x^2 y^2 z 
2 b^4 c^2 x^2 y^2 z + c^6 x^2 y^2 z + a^4 b^2 x y^3 z 
2 a^2 b^4 x y^3 z + b^6 x y^3 z  2 a^4 c^2 x y^3 z +
4 a^2 b^2 c^2 x y^3 z  2 b^4 c^2 x y^3 z  2 a^2 c^4 x y^3 z +
b^2 c^4 x y^3 z  a^4 b^2 x^3 z^2 + a^2 b^4 x^3 z^2 +
2 a^2 b^2 c^2 x^3 z^2  b^4 c^2 x^3 z^2  b^2 c^4 x^3 z^2 +
a^6 x^2 y z^2  2 a^4 b^2 x^2 y z^2 + b^6 x^2 y z^2 
a^4 c^2 x^2 y z^2 + 4 a^2 b^2 c^2 x^2 y z^2  a^2 c^4 x^2 y z^2 
2 b^2 c^4 x^2 y z^2 + c^6 x^2 y z^2 + a^6 x y^2 z^2 
2 a^2 b^4 x y^2 z^2 + b^6 x y^2 z^2 + 4 a^2 b^2 c^2 x y^2 z^2 
b^4 c^2 x y^2 z^2  2 a^2 c^4 x y^2 z^2  b^2 c^4 x y^2 z^2 +
c^6 x y^2 z^2 + a^4 b^2 y^3 z^2  a^2 b^4 y^3 z^2 
a^4 c^2 y^3 z^2 + 2 a^2 b^2 c^2 y^3 z^2  a^2 c^4 y^3 z^2 
a^4 b^2 x^2 z^3  a^2 b^4 x^2 z^3 + 2 a^2 b^2 c^2 x^2 z^3 +
b^4 c^2 x^2 z^3  b^2 c^4 x^2 z^3  2 a^4 b^2 x y z^3 
2 a^2 b^4 x y z^3 + a^4 c^2 x y z^3 + 4 a^2 b^2 c^2 x y z^3 +
b^4 c^2 x y z^3  2 a^2 c^4 x y z^3  2 b^2 c^4 x y z^3 +
c^6 x y z^3  a^4 b^2 y^2 z^3  a^2 b^4 y^2 z^3 + a^4 c^2 y^2 z^3 +
2 a^2 b^2 c^2 y^2 z^3  a^2 c^4 y^2 z^3
>
This is Euler line + nothing
> Variation (easier?):
>
> Let ABC be a triangle, HaHbHc the pedal triangle of H
> (orthic triangle), P a point, and A',B', and C' the reflections
> of P in Ha, Hb, Hc, resp.
> Which is the locus of P such that the circumcenter of
> A'B'C' is lying on the Euler Line of ABC?
>
> Is it Euler Line +???
>
> APH
>  Dear Linh Nguyen Van and Nikos!
If triangle ABC is regular we obtain next interesting result.
Given regular triangle ABC and point P. A_1B_1C_1 is the circucmevian
triangle of P, points A_2, B_2, C_2 are the reflections of A_1, B_1, C_1 in
the midpoints of BC, CA, AB
Then the circumcircle of A_2B_2C_2 pass through the center O of ABC and its
center is the reflection of P in O.
Proof. Points O and A_2 lie on the circle with center in p[oint A_0 opposite
to A. Thus the bisector of segment OA_2 is also the bisectrix of angle
OA_0A_2 which is parallel to AP.
It is clear that this line passes through point Q symmetric to P wrt O.
Similarly the bisectors of OB_2 and OC_2 pass through Q.
Sincerely
Alexey  Another problem with circumcenters:
Let ABC be a triangle and X,Y,Z three collinear points and
XaXbXc, YaYbYc the pedal triangles of X,Y resp.
Let A1,B1,C1 be the reflections of Z in Xa,Xb,Xc, resp.
and A2,B2,C2 the reflections of A1,B1,C1 in Ya,Yb,Yc, resp.
The circumcircle of A2B2C2 passes through Z and has center
on the line XYZ.
True??
Antreas  Yes, it is true,
If XZ:ZY = t and W is the circumcenter of A2B2C2 then W is on line XY and XW:WY = 2  1/t.
Francisco Javier.
 In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
>
> Another problem with circumcenters:
>
> Let ABC be a triangle and X,Y,Z three collinear points and
> XaXbXc, YaYbYc the pedal triangles of X,Y resp.
>
> Let A1,B1,C1 be the reflections of Z in Xa,Xb,Xc, resp.
> and A2,B2,C2 the reflections of A1,B1,C1 in Ya,Yb,Yc, resp.
>
> The circumcircle of A2B2C2 passes through Z and has center
> on the line XYZ.
>
> True??
>
> Antreas
> >
Is it listed in Bernard's list of 5ics?
>
> > [APH]
> > > Let ABC be a triangle, HaHbHc the pedal triangle of H
> > > (orthic triangle), P a point, A1B1C1 the circumcevian triangle
> > > of P and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
> > > resp.
> > > Which is the locus of P such that the circumcenter of
> > > A2B2C2 is lying on the Euler Line of ABC?
> > >
> > > It is the Euler Line + ?????
>
> [Francisco]
> This is the Euler line + A quintic though A, B, C, H, Ha, Hb, Hc
>
> Equation:
>
APH
[Nontext portions of this message have been removed] [APH]
> > Let ABC be a triangle, HaHbHc the pedal triangle of H
[Francisco]:
> > (orthic triangle), P a point, and A',B', and C' the reflections
> > of P in Ha, Hb, Hc, resp.
> > Which is the locus of P such that the circumcenter of
> > A'B'C' is lying on the Euler Line of ABC?
> >
> > Is it Euler Line +???
> This is Euler line + nothing
Another variation:
Let ABC be a triangle, HaHbHc the pedal triangle of H
(orthic triangle), Q a fixed point on the Euler line, and
P a point.
Let A1,B1,C1 be the reflections of Q in Ha,Hb,Hc, resp.
and A2,B2,C2 the reflections of A1,B1,C1 in P, resp.
Which is the locus of P such that the circumcenter
of A2B2C2 is on the Euler line of ABC?
Is it the Euler Line ??
Perspectivity:
Are the triangles HaHbHc, A2B2C2 perspective?
In general:
Let X,Y,Z be three collinear points, XaXbXc the
pedal triangle of X, and A1,B1,C1 the reflections
of Y in Xa,Xb,Xc resp. and A2,B2,C2 the reflections
of A1,B1,C1 in Z, resp.
Are the triangles A2B2C2, XaXbXc perspective
(with perspector on the line XYZ)?
APH  Dear friends of Hyacinthos:
The synthetics solutions of this problem can be found at here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&p=1863882&sid=ce1cf3b9df63a1477cc4699f020559dc#p1863882
Best regard,
Linh Nguyen Van
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