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[EMHL] Re: Distance between incenter and orthocenter

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  • Ercole Suppa
    between incenter and orthocenter Dear Barry and Walther , I post again my last message because the formulas are badly formatted. Sorry ! ... thank you for your
    Message 1 of 5 , Sep 30, 2001
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      between incenter and orthocenter


      Dear Barry and Walther ,

      I post again my last message because the formulas are badly formatted. Sorry
      !

      >I wrote:
      >> Dear friends,
      >>
      >> Can you help me to prove the formula :
      >>
      >> IH^2 = 4R^2 + 4Rr + 3r^2 -p^2

      thank you for your nice answers.
      I am not an expert with barycentrics (can you recommend some references ?)


      I have find a vectorial proof (similar to barycentric solution). The use of
      vectors in geometry problems is explained , for example, in the book :

      (*) R. Rusczyk, S. Lehoczky -The art of problem solving, vol. 2
      Greater Testing Concepts, P.O. 5014
      New York, NY 10185-5014


      Choosing for origin the circumcenter of the triangle, we have [see (*)] :

      --> --> --> -->
      OH = OA + OB + OC

      --> 1 --> --> -->
      OI = ------- [a OA + b OB + c OC ]
      a+b+c


      Conseguently:

      --> --> --> 1 --> --> -->
      IH = OH - OI = ------- [(b+c)OA + (a+c)OB + (a+b)OC ]
      a+b+c


      and


      -->
      IH^2 = (Norm( IH ))^2 =


      1
      = ----[(b+c)^2R^2 +(a+c)^2R^2 +(a+b)^2R^2 +2R^2(b+c)(a+c)cos(2C) + ... ]=
      4p^2


      R^2
      = ----[2(a^2 +b^2 +c^2) +2(ab +ac +bc) +2(b+c)(a+c)(1 -2(sinC)^2) +...] =
      4p^2


      R^2 c^2
      = ----[2(a^2 +b^2 +c^2) +2(ab +ac +bc) +2(b+c)(a+c)(1 - 2------ ) +...] =
      4p^2 4R^2


      R^2 a^4+b^4+c^4
      =----[2(a^2+b^2+c^2)+2(ab+ac+bc)+2(a^+b^2+c^2+3ab+3ac+3bc)- -----------]=
      4p^2 R^2


      R^2 a^4+b^4+c^4
      =----[2(a^2+b^2+c^2)+2(ab+ac+bc)+2(a^+b^2+c^2+3ab+3ac+3bc)- -----------]=
      4p^2 R^2


      R^2 (a^2+b^2+c^2)^2-2(a^2b^2+a^2c^2+b^2c^2)
      =----[4(a^2+b^2+c^2)+5(ab+ac+bc)- --------------------------------------]
      4p^2 R^2


      By substituing the well-known formulas


      (ab + ac + bc) = p^2 + 4Rr + r^2

      (a^2 + b^2 + c^2) = 2p^2 - 8Rr -2r^2

      (a^2b^2 + a^2c^2+ b^2c^2) = (ab + ac + bc)^2 - 2abc(a + b + c) =

      = (p^2 + 4Rr + r^2)^2 - 16Rrp^2


      we obtain the result:

      IH^2 = 4R^2 + 4Rr + 3r^2 - p^2


      Best regards

      Ercole Suppa







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