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property of points of Soddy

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  • Сталислав Тахаев
    Respected Hyacinthists! Respected Jean-Pierre! Article s Jean-Pierre in Forum Geometricorum(Volume6,2006) and messages 14503,14505,14506,14507,14520 in
    Message 1 of 1 , Jan 6, 2007
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      Respected Hyacinthists!
      Respected Jean-Pierre!
      Article's Jean-Pierre in Forum Geometricorum(Volume6,2006) and messages 14503,14505,14506,14507,14520 in Hyacinthos induce me to be divided similar result.I received their quite other way.
      Let P is point inside triangle ABC and BB',CC'-cevians.The quadrileral AC'PB' have an incircle if and only if:
      tanCBP/2*tanC/2= tanBCP/2*tanB/2. Vertexs P of family similar quadrilerals lie on the hyperbola with foci B,C.Analogous quadrilerals may be constructed relatively every vertex ABC. Solving
      system of three equations we receive formulies for such point P,that three quadrilerals AC'PB',BA'PC',CA'PB' will have incircles simultaneity,where AA',BB',CC'-cevians.Point P is intersection of three hyperbolies going via vertexs ABC.
      tan(CBP/2)= tanB/2:(1+tanB/2+tanC/2)
      tan(BCP/2)= tanC/2:(1+tanB/2+tanC/2)
      It is proved that these formulies determine the centers of the inner (sign"+")
      and outer(sign"-") Soddy's circles.Received formulies corresponding to Trilinears Kimberling -X(176),X(175).
      Let S1=tanA/2+tanB/2 S1,S2,S3 determine position of the center
      S2=tanA/2+tanC/2 outer Soddy's circle-inside or outside plane
      S3=tanB/2+tanC/2 triangle ABC.If S1 or S2 or S3 equal 1,then center of outer Soddy's circle lie on side joined indicated angles.
      If either (tanA/2)^2+tanA/2*tanB/2+(tanB/2)^2=1
      or (tanA/2)^2+tanA/2*tanC/2+(tanC/2)^2=1
      or (tanB/2)^2+tanB/2*tanC/2+(tanC/2)^2=1
      These formulies determine line Soddy parallel to side triangle ABC
      joined indicated angles.
      Particular case is degeneration of outer Soddy's circle in straight
      line,which touches with of three circles (A,p-a),(B,p-b),(C,p-c),
      where "p" is semiperimeter ABC.In this case line Soddy is perpendicu-
      late to touch line always,moreover tanA/2+tanB/2+tanC/2=2.
      Besides It is proved that incircle ABC is outer Soddy's circle for
      three incircles triangles BPC,CPA,APB and were calculated radiuses of these incircles.Also were calculated radiuses of incircles the
      three quadrilerals AC'PB',BA'PC',CA'PB'.AS a result-sums curves ap-
      propriate to pair incircles AC'PB'and BPC,BA'PC'and CPA,CA'PB'and
      APB are constant and equal curve of inner Soddy's circle ABC,i.e.:
      r1,r2,r3-radiuses of incircles quadrilerals-AC'PB',BA'PC',CA'PB';
      r4,r5,r6-radiuses of incircles triangles-APB,BPC,CPA;
      rs-radius of inner Soddy's circle ABC.
      If either tanA/2+tanB/2=tanC/2 -condition,when r3=r4=2rs;
      or tanA/2+tanC/2=tanB/2 -condition,when r2=r6=2rs;
      or tanB/2+tanC/2=tanA/2 -condition,when r1=r5=2rs.
      These corelations determine conditions,when radiuses appropriate to pair incircles(quadrilerals and triangles) are equal.
      In conclusion-points X(176),X(175) are foci of ellipce,on which lie vertexs triangle ABC.However if one of angles triangle ABC more,than
      2arctan4/3,then X(176),X(175) are foci of hyperbola.
      Note,that in case ellipce perimeters triangles ASS',BSS',CSS' are
      equal,i.e.: AS'+AS=BS'+BS=CS'+CS=rs'+rs,where
      S,S'-centers of inner and of outer Soddy's circles;
      rs,rs'-radiuses inner and outer Soddy's circles.
      In case hyperbola-- AS'-AS=BS'-BS=CS'-CS=rs'-rs.

      Exuse me my bad English. Sincerely,Stalislav

      (sign"+") and outer(sign"-") Soddy's circles.Received formulies cor-
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