## Re: [EMHL] Concurrent Circles Problem

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• ... Let P = (x:y:z) be a point. As L moves around P, which is the locus of the point of concurrence? (Special case P = O) APH [Non-text portions of this
Message 1 of 8 , Nov 30, 2010
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On Wed, Dec 1, 2010 at 12:47 AM, Antreas <anopolis72@...> wrote:

>
>
> Let ABC be a trianle, L a line intersecting
> BC,CA,AB at A',B',C' resp.
> Let Ab,Ac be the orthogonal projections of
> A' on AB, AC, resp., Bc,Ba of B' on BC,BA, resp.
> and Ca,Cb of C' on CA,CB, resp.
> The circles (A'AbAc),(B'BcBa) and (C'CaCb)
> are concurrent.
>
> For which lines L the point of concurrence
> is on the circumcircle of ABC ?
>
> APH
>
>
>
Let P = (x:y:z) be a point. As L moves around P, which is the locus
of the point of concurrence? (Special case P = O)

APH

[Non-text portions of this message have been removed]
• Dear Antreas If L: px + qy + rz = 0, the circle (A AbAc), whose equation a^2yz+b^2zx+c^2xy - (x+y+z) ((q*SB)/(q-r) y - (r*SC)/(q-r) z)=0, passing through A.
Message 2 of 8 , Dec 2, 2010
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Dear Antreas

If L: px + qy + rz = 0, the circle (A'AbAc), whose equation a^2yz+b^2zx+c^2xy - (x+y+z) ((q*SB)/(q-r) y - (r*SC)/(q-r) z)=0, passing through A. The circles (A'AbAc), (B'BcBa) and (C'CaCb) are coaxial, radical axis d: p(q-r)SAx + q(r-p)SBy + r(p-q)zSC= 0, which passes through the orthocenter, for any line L

Angel

--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...> wrote:
>
> On Wed, Dec 1, 2010 at 12:47 AM, Antreas <anopolis72@...> wrote:
>
> >
> >
> > Let ABC be a trianle, L a line intersecting
> > BC,CA,AB at A',B',C' resp.
> > Let Ab,Ac be the orthogonal projections of
> > A' on AB, AC, resp., Bc,Ba of B' on BC,BA, resp.
> > and Ca,Cb of C' on CA,CB, resp.
> > The circles (A'AbAc),(B'BcBa) and (C'CaCb)
> > are concurrent.
> >
> > For which lines L the point of concurrence
> > is on the circumcircle of ABC ?
> >
> > APH
> >
> >
> >
> Let P = (x:y:z) be a point. As L moves around P, which is the locus
> of the point of concurrence? (Special case P = O)
>
> APH
>
>
> [Non-text portions of this message have been removed]
>
• Dear Angel I think that the general problem is this: Let P be a point and A B C the pedal triangle of P. Let Ab,Ac be the orthogonal projections of A on AB,
Message 3 of 8 , Dec 2, 2010
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Dear Angel

I think that the general problem is this:

Let P be a point and A'B'C' the pedal triangle of P.
Let Ab,Ac be the orthogonal projections of
A' on AB, AC, resp., Bc,Ba of B' on BC,BA, resp.
and Ca,Cb of C' on CA,CB, resp.
Which is the locus of P such that the circles (A'AbAc),(B'BcBa)
and (C'CaCb) are concurrent ?

Part of the locus is the circumcircle of ABC.

APH

On Thu, Dec 2, 2010 at 3:05 PM, Angel <amontes1949@...> wrote:

>
>
> Dear Antreas
>
> If L: px + qy + rz = 0, the circle (A'AbAc), whose equation
> a^2yz+b^2zx+c^2xy - (x+y+z) ((q*SB)/(q-r) y - (r*SC)/(q-r) z)=0, passing
> through A. The circles (A'AbAc), (B'BcBa) and (C'CaCb) are coaxial, radical
> axis d: p(q-r)SAx + q(r-p)SBy + r(p-q)zSC= 0, which passes through the
> orthocenter, for any line L
>
> Angel
>
> --- In Hyacinthos@yahoogroups.com <Hyacinthos%40yahoogroups.com>, Antreas
> Hatzipolakis <anopolis72@...> wrote:
>
> >
> > On Wed, Dec 1, 2010 at 12:47 AM, Antreas <anopolis72@...> wrote:
> >
> > >
> > >
> > > Let ABC be a trianle, L a line intersecting
> > > BC,CA,AB at A',B',C' resp.
> > > Let Ab,Ac be the orthogonal projections of
> > > A' on AB, AC, resp., Bc,Ba of B' on BC,BA, resp.
> > > and Ca,Cb of C' on CA,CB, resp.
> > > The circles (A'AbAc),(B'BcBa) and (C'CaCb)
> > > are concurrent.
> > >
> > > For which lines L the point of concurrence
> > > is on the circumcircle of ABC ?
> > >
> > > APH
> > >
> > >
> > >
> > Let P = (x:y:z) be a point. As L moves around P, which is the locus
> > of the point of concurrence? (Special case P = O)
> >
> > APH
> >
> >
> > [Non-text portions of this message have been removed]
> >
>
>
>

--
http://anopolis72000.blogspot.com/

[Non-text portions of this message have been removed]
• Dear Antreas If A B C is the pedal triangle of P(u:v:w), the circle (A AbAc) is c^2*x*y + b^2*x*z + a^2*y*z -(x+y+z)*(2*SB*(2*SB*u + 2*a^2*w)*y + 2*SC*(2*SC*u
Message 4 of 8 , Dec 3, 2010
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Dear Antreas

If A'B'C'is the pedal triangle of P(u:v:w), the circle (A'AbAc) is

c^2*x*y + b^2*x*z + a^2*y*z -(x+y+z)*(2*SB*(2*SB*u + 2*a^2*w)*y +
2*SC*(2*SC*u + 2*a^2*v)*z)/(4*a^2*(u + v + w))=0, (through A).

The radical axis of circles (B'BcBa) and (C'CaCb) is

SA*(b^2*c^2*v-c^2*SA*v-b^2*c^2*w+b^2*SA*w)x + b^2*SB*(c^2*u + SB*w)y - c^2*SC*(b^2*u + SC*v)z=0.

The radical center of circles (A'AbAc), (B'BcBa) and (C'CaCb) is the orthocenter of ABC. Moreover, the orthocenter never belongs to the circles (A'AbAc), (B'BcBa) and (C'CACB).

Angel

--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...> wrote:
>
> Dear Angel
>
> I think that the general problem is this:
>
> Let P be a point and A'B'C' the pedal triangle of P.
> Let Ab,Ac be the orthogonal projections of
> A' on AB, AC, resp., Bc,Ba of B' on BC,BA, resp.
> and Ca,Cb of C' on CA,CB, resp.
> Which is the locus of P such that the circles (A'AbAc),(B'BcBa)
> and (C'CaCb) are concurrent ?
>
> Part of the locus is the circumcircle of ABC.
>
> APH
>
>
> On Thu, Dec 2, 2010 at 3:05 PM, Angel <amontes1949@...> wrote:
>
> >
> >
> > Dear Antreas
> >
> > If L: px + qy + rz = 0, the circle (A'AbAc), whose equation
> > a^2yz+b^2zx+c^2xy - (x+y+z) ((q*SB)/(q-r) y - (r*SC)/(q-r) z)=0, passing
> > through A. The circles (A'AbAc), (B'BcBa) and (C'CaCb) are coaxial, radical
> > axis d: p(q-r)SAx + q(r-p)SBy + r(p-q)zSC= 0, which passes through the
> > orthocenter, for any line L
> >
> > Angel
> >
> > --- In Hyacinthos@yahoogroups.com <Hyacinthos%40yahoogroups.com>, Antreas
> > Hatzipolakis <anopolis72@> wrote:
> >
> > >
> > > On Wed, Dec 1, 2010 at 12:47 AM, Antreas <anopolis72@> wrote:
> > >
> > > >
> > > >
> > > > Let ABC be a trianle, L a line intersecting
> > > > BC,CA,AB at A',B',C' resp.
> > > > Let Ab,Ac be the orthogonal projections of
> > > > A' on AB, AC, resp., Bc,Ba of B' on BC,BA, resp.
> > > > and Ca,Cb of C' on CA,CB, resp.
> > > > The circles (A'AbAc),(B'BcBa) and (C'CaCb)
> > > > are concurrent.
> > > >
> > > > For which lines L the point of concurrence
> > > > is on the circumcircle of ABC ?
> > > >
> > > > APH
> > > >
> > > >
> > > >
> > > Let P = (x:y:z) be a point. As L moves around P, which is the locus
> > > of the point of concurrence? (Special case P = O)
> > >
> > > APH
> > >
> > >
> > > [Non-text portions of this message have been removed]
> > >
> >
> >
> >
>
>
>
> --
> http://anopolis72000.blogspot.com/
>
>
> [Non-text portions of this message have been removed]
>
• ... Note that the circle (A AbAc) is the pedal circle of A . More generally, if P1 and P2 are any 2 points on the sidelines of ABC (except the vertices), then
Message 5 of 8 , Dec 3, 2010
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--- On Fri, 12/3/10, Angel <amontes1949@...> wrote:
> Dear Antreas
>
> If A'B'C'is the pedal triangle of P(u:v:w), the circle
> (A'AbAc) is
>
> c^2*x*y + b^2*x*z + a^2*y*z -(x+y+z)*(2*SB*(2*SB*u +
> 2*a^2*w)*y +
>     2*SC*(2*SC*u + 2*a^2*v)*z)/(4*a^2*(u + v +
> w))=0, (through A).
>
> The radical axis of circles (B'BcBa) and (C'CaCb) is
>
> SA*(b^2*c^2*v-c^2*SA*v-b^2*c^2*w+b^2*SA*w)x + b^2*SB*(c^2*u
> + SB*w)y -  c^2*SC*(b^2*u + SC*v)z=0.
>
> The radical center of circles (A'AbAc), (B'BcBa) and
> (C'CaCb) is the  orthocenter of ABC. Moreover, the
> orthocenter never belongs to the circles (A'AbAc), (B'BcBa)
> and (C'CACB).
>
> Angel

Note that the circle (A'AbAc) is the pedal circle of A'.

More generally, if P1 and P2 are any 2 points on the sidelines of ABC (except the vertices), then the orthocenter H lies on the radical axis of the pedal circles of P1 and of P2.
--
Barry Wolk
• Dear Barry Wolk, This means that if a point P moves on the line BC then the orthocenter H of ABC has constant power relative to the pedal circle of the point
Message 6 of 8 , Dec 5, 2010
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Dear Barry Wolk,

This means that if a point P moves on the line BC
then the orthocenter H of ABC has constant power
relative to the pedal circle of the point P.

This can be generalized to the following:
If a point P moves on a line L then the orthopole Q
of L has constant power relative to the pedal circle
of P, or if two points P1, P2, are on L then the
radical axis of the pedal circles of P1, P2 passes
through Q.

If you had in mind this property do you know if
there is a more extension to other curves and not lines?
Best regards

> More generally, if P1 and P2 are any 2 points on the
> sidelines of ABC (except the vertices), then the orthocenter
> H lies on the radical axis of the pedal circles of P1 and of
> P2.
> --
> Barry Wolk
>
>
>
>
>
> ------------------------------------
>
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
>
• ... If P is on line BC, then the pedal circle of P has diameter PA, so the case of P1 and P2 both on the same sideline is easy. And it s not hard to get a
Message 7 of 8 , Dec 6, 2010
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> Dear Barry Wolk,
>
> This means that if a point P moves on the line BC
> then the orthocenter H of ABC has constant power
> relative to the pedal circle of the point P.
>
> This can be generalized to the following:
> If a point P moves on a line L then the orthopole Q
> of L has constant power relative to the pedal circle
> of P, or if two points P1, P2, are on L then the
> radical axis of the pedal circles of P1, P2 passes
> through Q.
>
> If you had in mind this property do you know if
> there is a more extension to other curves and not lines?
> Best regards
>
>
> > More generally, if P1 and P2 are any 2 points on the
> > sidelines of ABC (except the vertices), then the orthocenter
> > H lies on the radical axis of the pedal circles of P1 and of
> > P2.
> > --
> > Barry Wolk

If P is on line BC, then the pedal circle of P has diameter PA, so the case of P1 and P2 both on the same sideline is easy. And it's not hard to get a synthetic proof when P1 and P2 are on different sidelines.

I haven't considered curves other than lines.
--
Barry Wolk
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