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• Russian olympiad 2011 proposed the problem http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2272708 It can be viewed as Let ABC be a triangle,
Message 1 of 7 , Aug 1, 2011
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Russian olympiad 2011 proposed the problem http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2272708

It can be viewed as

Let ABC be a triangle, circumcenter O. P is a point on BC. Circumcircle (OBP) cuts AB again at M, circumcircle (OCP) cuts AC again at N. Then O is orthocenter of PMN.

I found some other

Let ABC be a triangle, orthocenter H. P is a point on BC. Circumcircle (HBP) cuts AB again at M, circumcircle (HCP) cuts again AC at N. Then H is incenter of PMN.

Let ABC be a triangle, Lemoine point L. P is a point on BC. Circumcircle (LBP) cuts again AB at M, circumcircle (LCP) cuts again AC at N. Then L is centroid of PMN.

Is there law of these problems ?
• Why don t you post this nice result on mathlinks? :)
Message 2 of 7 , Aug 2, 2011
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Why don't you post this nice result on mathlinks? :)
• You should try to solve for the general barycentric relation of two points. Thank you for the nice result. Sung Hyun [Non-text portions of this message have
Message 3 of 7 , Aug 2, 2011
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You should try to solve for the general barycentric relation of two points.
Thank you for the nice result.

Sung Hyun

[Non-text portions of this message have been removed]
• Yes there is a law. If O is an arbitrary fixed point A is a moving point on BC and the circles OA C, OBA meet AC and AB at B , C respectively then the
Message 4 of 7 , Aug 2, 2011
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Yes there is a law.
If O is an arbitrary fixed point
A' is a moving point on BC
and the circles OA'C, OBA' meet
AC and AB at B', C' respectively then
the triangle A'B'C' is similar to the
pedal triangle A*B*C* of O.
It is easy to show that A'B'C' is a rotation
of A*B*C* around O.
then the point O is in triangle A'B'C'
as is in triangle A*B*C*

If O is circumcenter in ABC then
O is in A*B*C* and A'B'C' orthocenter.

If O is orthocenter of an acute triangle ABC
then since O is incenter in A*B*C* it is incenter in A'B'C'.

if ABC is not acute then the orthocenter is an excenter of A'B'C'

If O is the Lemoine point of ABC then O it is known
that the isogonal conjugate of O (centroid of ABC)
has the same barycentrics in ABC (1:1:1)
as O in A*B*C*. Hence O is the centroid of A*B*C* and
hence is the cetroid of A'B'C'.

Best regards

> Russian olympiad 2011 proposed the
> problem http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2272708
>
> It can be viewed as
>
> Let ABC be a triangle, circumcenter O. P is a point on BC.
> Circumcircle (OBP) cuts AB again at M, circumcircle (OCP)
> cuts AC again at N. Then O is orthocenter of PMN.
>
> I found some other
>
>
> Let ABC be a triangle, orthocenter H. P is a point on BC.
> Circumcircle (HBP) cuts AB again at M, circumcircle (HCP)
> cuts again AC at N. Then H is incenter of PMN.
>
> Let ABC be a triangle, Lemoine point L. P is a point on BC.
> Circumcircle (LBP) cuts again AB at M, circumcircle (LCP)
> cuts again AC at N. Then L is centroid of PMN.
>
> Is there law of these problems ?
• Thank you much dear Nikolaos Dergiades for your clear explanation. That s I mean but I can t see that. I try to find comments from experts of Hyacinthos group.
Message 5 of 7 , Aug 5, 2011
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Thank you much dear Nikolaos Dergiades for your clear explanation. That's I mean but I can't see that. I try to find comments from experts of Hyacinthos group.

So we can see if O is X(i) center of triangle A*B*C* then O is X(i) center of triangle A'B'C', with X(i) is Kimberling center.

Best regards.

Subject: Re: [EMHL] From Russian olympiad
To: Hyacinthos@yahoogroups.com
Date: Tuesday, August 2, 2011, 10:44 PM

Yes there is a law.

If O is an arbitrary fixed point

A' is a moving point on BC

and the circles OA'C, OBA' meet

AC and AB at B', C' respectively then

the triangle A'B'C' is similar to the

pedal triangle A*B*C* of O.

It is easy to show that A'B'C' is a rotation

of A*B*C* around O.

then the point O is in triangle A'B'C'

as is in triangle A*B*C*

If O is circumcenter in ABC then

O is in A*B*C* and A'B'C' orthocenter.

If O is orthocenter of an acute triangle ABC

then since O is incenter in A*B*C* it is incenter in A'B'C'.

if ABC is not acute then the orthocenter is an excenter of A'B'C'

If O is the Lemoine point of ABC then O it is known

that the isogonal conjugate of O (centroid of ABC)

has the same barycentrics in ABC (1:1:1)

as O in A*B*C*. Hence O is the centroid of A*B*C* and

hence is the cetroid of A'B'C'.

Best regards

> Russian olympiad 2011 proposed the

> problem http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2272708

>

> It can be viewed as

>

> Let ABC be a triangle, circumcenter O. P is a point on BC.

> Circumcircle (OBP) cuts AB again at M, circumcircle (OCP)

> cuts AC again at N. Then O is orthocenter of PMN.

>

> I found some other

>

>

> Let ABC be a triangle, orthocenter H. P is a point on BC.

> Circumcircle (HBP) cuts AB again at M, circumcircle (HCP)

> cuts again AC at N. Then H is incenter of PMN.

>

> Let ABC be a triangle, Lemoine point L. P is a point on BC.

> Circumcircle (LBP) cuts again AB at M, circumcircle (LCP)

> cuts again AC at N. Then L is centroid of PMN.

>

> Is there law of these problems ?

[Non-text portions of this message have been removed]
• Dear Tran Quang Hung If A , B , C are arbitrary points on lines BC, CA, AB of a triangle ABC and the circumcircles of triangles BA C , CB A meet again at O
Message 6 of 7 , Aug 5, 2011
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Dear Tran Quang Hung
If A', B', C' are arbitrary points on lines
BC, CA, AB of a triangle ABC and the circumcircles
of triangles BA'C', CB'A' meet again at O
and the acute angle of OA' and BC is f = ang OA'C
then f = ang OC'B = ang OB'A and the qudrilateral
AC'OB' is cyclic.
If A*B*C* is the pedal triangle of O then
ang B*OC* = ang B'OC'
ang C*OA* = ang C'OA'
ang A*OB* = ang A'OB' and
OA* = OA'.sinf
OB* = OB'.sinf
OC* = OC'.sinf
Hence A'B'C' is a rotation and enlargement of A*B*C*
by angle f and ratio 1/sinf.
Hence A'B'C' and A*B*C* are similar triangles and
if O is triangle center X(i) in A*B*C* it is also
triangle center X(i) in A'B'C'.
You are right that in order to produce similar problems
to the problems you proposed it is needed to knwo
for a triangle center X(i) in ABC what triangle center
is relative to its pedal triangle.
This problem is not easy to be solved, and I think
that there are not many known triangle centers in ABC
that are known triangle centers in their pedal triangles.
Best regards

> Thank you much dear Nikolaos
> can't see that. I try to find comments from experts of
> Hyacinthos group.
>
> So we can see if O is X(i) center of triangle A*B*C* then O
> is X(i) center of triangle A'B'C', with X(i) is Kimberling
> center.
>
> Best regards.
>
> wrote:
>
> Subject: Re: [EMHL] From Russian olympiad
> To: Hyacinthos@yahoogroups.com
> Date: Tuesday, August 2, 2011, 10:44 PM
>
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>       Yes there is a law.
>
> If O is an arbitrary fixed point
>
> A' is a moving point on BC
>
> and the circles OA'C, OBA' meet
>
> AC and AB at B', C' respectively then
>
> the triangle A'B'C' is similar to the
>
> pedal triangle A*B*C* of O.
>
> It is easy to show that A'B'C' is a rotation
>
> of A*B*C* around O.
>
> then the point O is in triangle A'B'C'
>
> as is in triangle A*B*C*
>
>
>
> If O is circumcenter in ABC then
>
> O is in A*B*C* and A'B'C' orthocenter.
>
>
>
> If O is orthocenter of an acute triangle ABC
>
> then since O is incenter in A*B*C* it is incenter in
> A'B'C'.
>
>
>
> if ABC is not acute then the orthocenter is an excenter of
> A'B'C'
>
>
>
> If O is the Lemoine point of ABC then O it is known
>
> that the isogonal conjugate of O (centroid of ABC)
>
> has the same barycentrics in ABC (1:1:1)
>
> as O in A*B*C*. Hence O is the centroid of A*B*C* and
>
> hence is the cetroid of A'B'C'.
>
>
>
> Best regards
>
>
>
>
> > Russian olympiad 2011 proposed the
>
> > problem http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2272708
>
> >
>
> > It can be viewed as
>
> >
>
> > Let ABC be a triangle, circumcenter O. P is a point on
> BC.
>
> > Circumcircle (OBP) cuts AB again at M, circumcircle
> (OCP)
>
> > cuts AC again at N. Then O is orthocenter of PMN.
>
> >
>
> > I found some other
>
> >
>
> >
>
> > Let ABC be a triangle, orthocenter H. P is a point on
> BC.
>
> > Circumcircle (HBP) cuts AB again at M, circumcircle
> (HCP)
>
> > cuts again AC at N. Then H is incenter of PMN.
>
> >
>
> > Let ABC be a triangle, Lemoine point L. P is a point
> on BC.
>
> > Circumcircle (LBP) cuts again AB at M, circumcircle
> (LCP)
>
> > cuts again AC at N. Then L is centroid of PMN.
>
> >
>
> > Is there law of these problems ?
>
>
>
>
>
>
>
>
>
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> [Non-text portions of this message have been removed]
>
>
>
> ------------------------------------
>
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
>
• Dear Mr. Nikos Dergiades Thank you so much for your helping. You are right, it is hard to know triangle centers in a triangle that are known triangle
Message 7 of 7 , Aug 6, 2011
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Thank you so much for your helping. You are right, it is hard to know triangle centers in a triangle that are known triangle centers in their pedal triangles. "Name" of point is hard to distinguish in many cases. It is same when we choose different system of coordiante.
Thank you so much again.

Subject: Re: [EMHL] From Russian olympiad
To: Hyacinthos@yahoogroups.com
Date: Saturday, August 6, 2011, 2:45 AM

Dear Tran Quang Hung
If A', B', C' are arbitrary points on lines
BC, CA, AB of a triangle ABC and the circumcircles
of triangles BA'C', CB'A' meet again at O
and the acute angle of OA' and BC is f = ang OA'C
then f = ang OC'B = ang OB'A and the qudrilateral
AC'OB' is cyclic.
If A*B*C* is the pedal triangle of O then
ang B*OC* = ang B'OC'
ang C*OA* = ang C'OA'
ang A*OB* = ang A'OB' and
OA* = OA'.sinf
OB* = OB'.sinf
OC* = OC'.sinf
Hence A'B'C' is a rotation and enlargement of A*B*C*
by angle f and ratio 1/sinf.
Hence A'B'C' and A*B*C* are similar triangles and
if O is triangle center X(i) in A*B*C* it is also
triangle center X(i) in A'B'C'.
You are right that in order to produce similar problems
to the problems you proposed it is needed to knwo
for a triangle center X(i) in ABC what triangle center
is relative to its pedal triangle.
This problem is not easy to be solved, and I think
that there are not many known triangle centers in ABC
that are known triangle centers in their pedal triangles.
Best regards

> Thank you much dear Nikolaos
> can't see that. I try to find comments from experts of
> Hyacinthos group.
>
> So we can see if O is X(i) center of triangle A*B*C* then O
> is X(i) center of triangle A'B'C', with X(i) is Kimberling
> center.
>
> Best regards.
>
> wrote:
>
> Subject: Re: [EMHL] From Russian olympiad
> To: Hyacinthos@yahoogroups.com
> Date: Tuesday, August 2, 2011, 10:44 PM
>
>
>
>
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>
>
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>

>
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>       Yes there is a law.
>
> If O is an arbitrary fixed point
>
> A' is a moving point on BC
>
> and the circles OA'C, OBA' meet
>
> AC and AB at B', C' respectively then
>
> the triangle A'B'C' is similar to the
>
> pedal triangle A*B*C* of O.
>
> It is easy to show that A'B'C' is a rotation
>
> of A*B*C* around O.
>
> then the point O is in triangle A'B'C'
>
> as is in triangle A*B*C*
>
>
>
> If O is circumcenter in ABC then
>
> O is in A*B*C* and A'B'C' orthocenter.
>
>
>
> If O is orthocenter of an acute triangle ABC
>
> then since O is incenter in A*B*C* it is incenter in
> A'B'C'.
>
>
>
> if ABC is not acute then the orthocenter is an excenter of
> A'B'C'
>
>
>
> If O is the Lemoine point of ABC then O it is known
>
> that the isogonal conjugate of O (centroid of ABC)
>
> has the same barycentrics in ABC (1:1:1)
>
> as O in A*B*C*. Hence O is the centroid of A*B*C* and
>
> hence is the cetroid of A'B'C'.
>
>
>
> Best regards
>
>
>
>
> > Russian olympiad 2011 proposed the
>
> > problem http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2272708
>
> >
>
> > It can be viewed as
>
> >
>
> > Let ABC be a triangle, circumcenter O. P is a point on
> BC.
>
> > Circumcircle (OBP) cuts AB again at M, circumcircle
> (OCP)
>
> > cuts AC again at N. Then O is orthocenter of PMN.
>
> >
>
> > I found some other
>
> >
>
> >
>
> > Let ABC be a triangle, orthocenter H. P is a point on
> BC.
>
> > Circumcircle (HBP) cuts AB again at M, circumcircle
> (HCP)
>
> > cuts again AC at N. Then H is incenter of PMN.
>
> >
>
> > Let ABC be a triangle, Lemoine point L. P is a point
> on BC.
>
> > Circumcircle (LBP) cuts again AB at M, circumcircle
> (LCP)
>
> > cuts again AC at N. Then L is centroid of PMN.
>
> >
>
> > Is there law of these problems ?
>
>
>
>
>
>
>
>
>
>
>
>

>
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> [Non-text portions of this message have been removed]
>
>
>
> ------------------------------------
>
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
>

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