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From Russian olympiad

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  • libebook
    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 ?
    • AHP
      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? :)
      • Sung Hyun Lim
        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]
        • Nikolaos Dergiades
          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
            Nikos Dergiades



            > 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 ?
          • Tran Quang Hung
            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.

              --- On Tue, 8/2/11, Nikolaos Dergiades <ndergiades@...> wrote:

              From: Nikolaos Dergiades <ndergiades@...>
              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

              Nikos Dergiades



              > 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]
            • Nikolaos Dergiades
              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
                Nikos Dergiades


                > 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.
                >
                > --- On Tue, 8/2/11, Nikolaos Dergiades <ndergiades@...>
                > wrote:
                >
                > From: Nikolaos Dergiades <ndergiades@...>
                > 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
                >
                > Nikos Dergiades
                >
                >
                >
                > > 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]
                >
                >
                >
                > ------------------------------------
                >
                > Yahoo! Groups Links
                >
                >
                >     Hyacinthos-fullfeatured@yahoogroups.com
                >
                >
                >
              • Tran Quang Hung
                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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                  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 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.
                  Best reagrads.Tran Quang Hung.

                  --- On Sat, 8/6/11, Nikolaos Dergiades <ndergiades@...> wrote:

                  From: Nikolaos Dergiades <ndergiades@...>
                  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
                  Nikos Dergiades


                  > 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.
                  >
                  > --- On Tue, 8/2/11, Nikolaos Dergiades <ndergiades@...>
                  > wrote:
                  >
                  > From: Nikolaos Dergiades <ndergiades@...>
                  > 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
                  >
                  > Nikos Dergiades
                  >
                  >
                  >
                  > > 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]
                  >
                  >
                  >
                  > ------------------------------------
                  >
                  > Yahoo! Groups Links
                  >
                  >
                  >     Hyacinthos-fullfeatured@yahoogroups.com
                  >
                  >
                  >


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