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2697Re: [EMHL] A simple construction

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  • xpolakis@otenet.gr
    Apr 2 7:56 AM
      >>Given triangle ABC, it is easy to construct the two circles,
      >>each passing through A, and tangent to BC at one of the vertices B
      >>and C.
      >>How does one construct the circle tangent to BC, and to each of these
      >>two circles externally?
      >>There is of course a construction by inversion. I have found a very
      >>simple construction of this circle, and verified the correctness by
      >>calculations. I wonder if the following construction is known and if
      >>one can find a synthetic proof:
      >>Let the bisector of angle A intersect BC at A'. Construct the
      >>perpendicular to BC at A' to intersect the median AA" at Pa.
      >>The cirlce with A'Pa as diameter is tangent BC and to each of the two
      >>circles above.
      [I changed the notation of the points. Similarly we define the points
      B', C'; B", C"; Pb, Pc]

      1. Let Ma, Mb, Mc be the centers of Paul's circles (ie midpoints
      of A'Pa, B'Pb, C'Pc).
      Are the lines AMa, BMb, CMc concurrent?

      2. Let P'a, P'b, P'c be the reflections of Pa, Pb, Pc in BC, CA, AB, resp.
      Are the lines AP'a, BP'b, CP'c concurrent?

      3. Let M'a, M'b, M'c be the reflections of Ma, Mb, Mc on BC, CA, AB, resp.
      Are the lines AM'a, BM'b, CM'c concurrent?


      PS: A typo in my proof:

      > KN^2 + KC^2 + NP^2 + 2KC*NP (1)

      KN^2 = KC^2 + NP^2 + 2KC*NP (1)
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