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Subject: "Crux Converse"     Previous Topic | Next Topic
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Member since Jun-9-03
Dec-04-05, 07:11 PM (EST)
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"Crux Converse"
   This problem is a converse to one that I found in the journal "Crux Mathematicorum with Mathematical Mayhem" published by the Canadian Mathematical Society".

Let I and J be the incircle and excircle of triangle ABC in the interior of angle ABC. I and J touch side AC at points D and E respectively. Line BE intersects I at points F and G with F between B and G. Prove that DF is a diameter of I.

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1701 posts
Dec-04-05, 07:41 PM (EST)
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1. "RE: Crux Converse"
In response to message #0
   Yes. This is because

  1. the two circles are homothetic in B,
  2. F corresponds to E, OI corresponds to OJ,
  3. if EK is the diameter of J, naturally through OJ, then K corresponds to D, (this is because EOJ and DOI are perpendicular to AC)
  4. FD is the diameter of I,

where OI and OJ are the centers of I and J, repsectively.

Here's a relevant problem:


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