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Re: Cevians and wqual circles

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  • peter_mows
    Hello Vladimir, ... two ... concurrence, but ... I have done a similar experiment and find that although close, I do not think they are perspective. Call A
    Message 1 of 3 , Feb 18, 2005
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      Hello Vladimir,

      > I've been asked the following question:
      >
      > do the three cevians each of which divides a given triangle into
      two
      > triangles with equal inradii concur?
      >
      > A quick GSP experiment shows that they are very close to
      concurrence, but
      > there's a lot of room for doubt.
      >
      > Does anybody know the answer?

      I have done a similar experiment and find that although close, I do
      not think they are perspective.

      Call A' the meet of the A cevian with BC. Then

      A' = {0, c (b + Sqrt[s^2 - a s]), b (c + Sqrt[s^2 - a s])}
      B' = {c (a + Sqrt[s^2 - b s]), 0, a (c + Sqrt[s^2 - b s])}
      C' = {b (a + Sqrt[s^2 - c s]), a (b + Sqrt[s^2 - c s]), 0}

      from which A'B'C' is not perspective.

      The line I of BAA' to I of BCC'
      The line I of CBB' to I of CAA'
      The line I of ACC' to I of ABB'
      are of course concurrent in I of ABC.

      The radii of the incircles are rA' = r (s - Sqrt[s^2 - a s]) / a

      Best regards,
      Peter.
    • jpehrmfr
      Dear Vladimir ... two ... concurrence, but ... If A is the foot of the A-cevian, we have cos
      Message 2 of 3 , Feb 19, 2005
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        Dear Vladimir
        > I've been asked the following question:
        >
        > do the three cevians each of which divides a given triangle into
        two
        > triangles with equal inradii concur?
        >
        > A quick GSP experiment shows that they are very close to
        concurrence, but
        > there's a lot of room for doubt.
        >
        > Does anybody know the answer?

        If A' is the foot of the A-cevian, we have cos <BA'A = (b-c)/a and
        it follows that the three cevian lines don't concur.
        Friendly. Jean-Pierre
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