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Inversion of excircles in incircle, and vice versa

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  • Paul Yiu
    Dear friends, (1) The A-excircle is inverted in the incircle. The center of the inversive image is the point (a(a^2+3(b-c)^2) : b(3a^2+(b-c)^2) :
    Message 1 of 1 , Feb 21, 2013
      Dear friends,

      (1) The A-excircle is inverted in the incircle. The center of the inversive image is the point

      (a(a^2+3(b-c)^2) : b(3a^2+(b-c)^2) : c(3a^2+(b-c)^2)).

      The triangle of centers of the inversive images of the excircles in the incircle is perspective with

      (i) the intouch triangle at X_{4907} = (a(b+c-a)^2(a^2+3(b-c)^2) : ... : ...),

      (ii) the medial triangle at
      (a(a^2+3(b-c)^2) : b(b^2+3(c-a)^2) : c(c^2+3(a-b)^2))
      with (6-9-13)-search number 0.778919595267....

      (2) The incircle is inverted in the A-excircle. The center of the inversive image is the point

      (-a(a^2+3(b-c)^2) : b(3a^2+(b-c)^2) : c(3a^2+(b-c)^2)).

      The triangle of centers of the inversive images of the incircle in the excircles is perspective with

      (i) the intouch triangle at
      ( a(a^2+3(b-c)^2)/(b+c-a) : b(b^2+3(c-a)^2)/(c+a-b) : c(c^2+3(a-b)^2)/(a+b-c))

      with (6-9-13)-search number 0.16447696491...,

      (ii) the medial triangle at

      (a(b+c-a)^3(a^2+3(b-c)^2) : ... : ...)

      with (6-9-13)-search number 3.93642587545...

      Best regards
      Sincerely
      Paul


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