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Geometry Puzzle

  1. Jul 11, 2005 #1
  2. jcsd
  3. Jul 11, 2005 #2


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    [itex]\sqrt{2.3925} - 1.35 \approx 0.197[/itex]
  4. Jul 11, 2005 #3
    Yes, that appears to be the correct answer. I am interested how you got it. The expression I got was
    sin((60-2*(90-arccos(.45)))/2) * 3^.5 * 2
  5. Jul 11, 2005 #4


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    2(sin(120)/sin(30))sin((60 - Y)/2)
    = 2(sin(120)/sin(30))(sin(30)cos(Y/2) - cos(30)sin(Y/2))
    = 2(30.5)(0.5(1 - 0.45²)0.5 - 0.5(30.5)(0.45))
    = (3(1 - 0.45²))0.5 - 1.35
    = (2.3925)0.5 - 1.35

    Y is the small angle in the triangle the contains the blue line (the bottom angle). I didn't want to use any inverse trigonometric functions, so I used Pythogras' Theorem, Sine Law, and the angle addition formula for sine.
  6. Jul 11, 2005 #5


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    Hmm, if you bisect the figure, then you have an isoscoles trianlge at the top so the interior angle of the top triangle is
    Moreover, it's not hard to see that the lenght from the center intersection to the edge of the red segment has length

    That means that you can go with:
    [tex]2\left(\sqrt{3} \sin\left(30-\sin^{-1}(.45)\right)\right)[/tex]

    Which can easily be simplified using angle addition forumlas.
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