Thursday, November 11, 2010

Fox 310

4 comments:


  1. http://img831.imageshack.us/img831/4227/fox310.jpg

    Let a, b, c, d ratios of circles having points A,B,C,D, 2L the length of the segment passing through centers and E and F the centers of circles a and d. Ratios are related and it's easy to get:
    b=L-c; a=(L+c)/2; d=L-c/2

    Apply Pythagora's theorem in triangles QKE and PHF to prove AB=CD.

    If we complete lower semi-circles and trace CD tangent to the upper ones, we can also prove that AC=BD.

    César Lozada
    Venezuela

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  2. Sorry. I forgot to mention QK perpendicular to AE and PH perpendicular to FD

    César Lozada
    Venezuela

    ReplyDelete
  3. Prove: AB=CD=sqrt(2bc)

    César Lozada
    Venezuela

    ReplyDelete
  4. You can also use a=c+b/2 and d=b+c/2

    ReplyDelete