## 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
Message 1 of 21 , Apr 10, 2010
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
Message 2 of 21 , Apr 10, 2010
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. 231-247 M.T.
Message 3 of 21 , Apr 11, 2010
Dear Linh Nguyen Van

See Hagge circle construction in FG 2007 pp. 231-247

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
Message 4 of 21 , Apr 11, 2010
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
Message 5 of 21 , Apr 11, 2010
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] ... As locus problem: Let ABC be a triangle, HaHbHc the pedal triangle of H (orthic triangle), P a point, A1B1C1 the circumcevian triangle of
Message 6 of 21 , Apr 12, 2010
[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 Antreas ... It seems to be part of the Euler line. Best regards Nikos __________________________________________________ Χρησιμοποιείτε
Message 7 of 21 , Apr 12, 2010
Dear Antreas

> 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:
>
>
> 2. As P moves on the Euler line, where is moving the
> circumcenter
> of A2B2C2?

It seems to be part of the Euler line.

Best regards
Nikos

__________________________________________________
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• 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
Message 8 of 21 , Apr 12, 2010
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 ... I tried to give a synthetic proof but did nothing. I tried to give a proof with barycentrics but also did nothing because
Message 9 of 21 , Apr 12, 2010
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[a-t(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)t-2abc]/[(s+abc)t-abc]
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

__________________________________________________
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• 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
Message 10 of 21 , Apr 12, 2010
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)t-abc = 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

> 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[a-t(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)t-2abc]/[(s+abc)t-abc]
> 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
>
>
>
>
> __________________________________________________
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>
>
> ------------------------------------
>
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
>

__________________________________________________
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• Thanks you dear Nikolaos for your solution. I am also finding the synthetic solution but false. ... Dear Francisco, For this variation you can see Hagge
Message 11 of 21 , Apr 12, 2010
Thanks you dear Nikolaos for your solution. I am also finding the synthetic solution but false.

>
>[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?
>
Dear Francisco,
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
Message 12 of 21 , Apr 12, 2010
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

.

[Non-text portions of this message have been removed]
• [APH] ... [ND] ... 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
Message 13 of 21 , Apr 13, 2010
[APH]
> > 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:
> >
> >
> > 2. As P moves on the Euler line, where is moving the
> > circumcenter
> > of A2B2C2?

[ND]
> 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] ... 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
Message 14 of 21 , Apr 13, 2010
[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 + ?????

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
• Thear Antreas, ... This is the Euler line + A quintic though A, B, C, H, Ha, Hb, Hc 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
Message 15 of 21 , Apr 13, 2010
Thear Antreas,

> [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 + ?????

This is the Euler line + A quintic though A, B, C, H, Ha, Hb, Hc

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

>
> 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 +???
>

This is Euler line + nothing

> 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
Message 16 of 21 , Apr 13, 2010
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
Message 17 of 21 , Apr 13, 2010
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.
Message 18 of 21 , Apr 13, 2010
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 [Non-text portions of this message have been removed]
Message 19 of 21 , Apr 14, 2010
>
>
> > [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:
>

Is it listed in Bernard's list of 5ics?

APH

[Non-text portions of this message have been removed]
• [APH] ... 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
Message 20 of 21 , Apr 14, 2010
[APH]
> > 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 +???

[Francisco]:
> 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:
Message 21 of 21 , May 2, 2010
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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