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25581Re: Points on the Incircle

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  • Antreas Hatzipolakis
    Mar 7, 2017
      [APH]:

      Excentral Triangle version of a problem (orthic triangle version) by Hình Học [ = Dao Thanh Oai]  (*)

      Let ABC be a triangle and A'B'C' the antipedal triangle of I (excentral triangle).

      Denote:

      (Ja), (Jb), (Jc) = the incircles of A'BC, B'CA, C'AB, resp.

      The circle tangent externally the circles (Ja), (Jb), (Jc) touches internally (I)

      Which is its center and which is its point of contact with (I) ?

      Variation:

      Denote:

      (Jaa), (Jbb), (Jcc) = the excircles of A'BC, B'CA, C'AB corresponding to A', B', C', resp.

      I think the circle tangent internally (Jaa), (Jbb), (Jcc) touches internally (I).

      Which is its center and which is its point of contact with (I) ?

      (*) Posted to FB group BÀI TOÁN HAY - LỜI GIẢI ĐẸP - ĐAM MÊ TOÁN HỌC


      [Peter Moses];

      Hi Antreas,
       
      >(Jaa), (Jbb), (Jcc) = the excircles of A'BC, B'CA, C'AB corresponding
      to A', B', C', resp.
      >I think the circle tangent internally (Jaa), (Jbb), (Jcc) touches
      internally (I).

      Yes! On the incircle at a somewhat unwieldly point:

      a (a+b-c) (a-b+c) (a^6 b^3-2 a^4 b^5+a^2 b^7+3 a^6 b^2 c-5 a^5 b^3 c-2 a^4 b^4 c+6 a^3 b^5 c-a^2 b^6 c-a b^7 c+3 a^6 b c^2-6 a^5 b^2 c^2+8 a^4 b^3 c^2-a^3 b^4 c^2-7 a^2 b^5 c^2+a b^6 c^2+2 b^7 c^2+a^6 c^3-5 a^5 b c^3+8 a^4 b^2 c^3-10 a^3 b^3 c^3+7 a^2 b^4 c^3+9 a b^5 c^3-6 b^6 c^3-2 a^4 b c^4-a^3 b^2 c^4+7 a^2 b^3 c^4-18 a b^4 c^4+4 b^5 c^4-2 a^4 c^5+6 a^3 b c^5-7 a^2 b^2 c^5+9 a b^3 c^5+4 b^4 c^5-a^2 b c^6+a b^2 c^6-6 b^3 c^6+a^2 c^7-a b c^7+2 b^2 c^7+(-6 a^5 b^3 c+3 a^4 b^4 c+7 a^3 b^5 c-3 a^2 b^6 c-a b^7 c-4 a^5 b^2 c^2+13 a^4 b^3 c^2-6 a^3 b^4 c^2-6 a^2 b^5 c^2+2 a b^6 c^2+b^7 c^2-6 a^5 b c^3+13 a^4 b^2 c^3-18 a^3 b^3 c^3+9 a^2 b^4 c^3+9 a b^5 c^3-7 b^6 c^3+3 a^4 b c^4-6 a^3 b^2 c^4+9 a^2 b^3 c^4-20 a b^4 c^4+6 b^5 c^4+7 a^3 b c^5-6 a^2 b^2 c^5+9 a b^3 c^5+6 b^4 c^5-3 a^2 b c^6+2 a b^2 c^6-7 b^3 c^6-a b c^7+b^2 c^7) Sin[A/2]+(-3 a^6 b^2 c+5 a^5 b^3 c+3 a^4 b^4 c-5 a^3 b^5 c+2 a^6 b c^2+2 a^5 b^2 c^2-16 a^4 b^3 c^2+10 a^3 b^4 c^2+6 a^2 b^5 c^2-4 a b^6 c^2+a^6 c^3-7 a^5 b c^3+17 a^4 b^2 c^3+7 a^3 b^3 c^3-29 a^2 b^4 c^3+11 a b^5 c^3-2 a^4 b c^4-20 a^3 b^2 c^4+34 a^2 b^3 c^4-4 a b^4 c^4-2 a^4 c^5+8 a^3 b c^5-12 a^2 b^2 c^5-10 a b^3 c^5+8 a b^2 c^6+a^2 c^7-a b c^7) Sin[B/2]+(a^6 b^3-2 a^4 b^5+a^2 b^7+2 a^6 b^2 c-7 a^5 b^3 c-2 a^4 b^4 c+8 a^3 b^5 c-a b^7 c-3 a^6 b c^2+2 a^5 b^2 c^2+17 a^4 b^3 c^2-20 a^3 b^4 c^2-12 a^2 b^5 c^2+8 a b^6 c^2+5 a^5 b c^3-16 a^4 b^2 c^3+7 a^3 b^3 c^3+34 a^2 b^4 c^3-10 a b^5 c^3+3 a^4 b c^4+10 a^3 b^2 c^4-29 a^2 b^3 c^4-4 a b^4 c^4-5 a^3 b c^5+6 a^2 b^2 c^5+11 a b^3 c^5-4 a b^2 c^6) Sin[C/2])::
      on lines {{65,2089},{177,10505},...}

      Best regards,
      Peter.

      [Peter Moses]:

      A remarkable simplification from my earlier message!

      Touch point = a (b-c) (a+b-c) (a-b+c) ((b-c) (-a+b+c) Sin[A/2]+(a-c) (a-b+c) Sin[B/2]+(-a+b) (a+b-c) Sin[C/2])::

      on lines {{65,2089},{177,10505},{1122,7 371},{6018,10508}}.
      X(7371)-Ceva conjugate of X(3669).

      Search {0.619008183563320582741855096 805,2.320431033558261101243544 60292,1.7485161431840503794535 0725580}.
       
      I wonder about simplification.  How do we know it has been reduced to the simplest form?  I only have some clue by replacing the barycentrics with integers.  If they simplify greatly, then chances are that the barys could too.  However it is often far from obvious when Sin[A/2] or square roots are involved.  I tried this particular problem a while ago, but decided it was too time intensive when we have to deal with intersections with circles, orthic projections on lines of similitude, inverses in circles ..., when they all start with square roots!  Happily, one of the radii could be written as a square, so the similitude only had terms in Sin[A/2] rather than Sqrt[ ... Sin[A/2] ..].  Even so, took quite a few hours of coaxing to arrive at the result above.
       
      Best regards
      Peter.
       


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