Geometry Puzzle

Answers and Replies

  • #2
AKG
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[itex]\sqrt{2.3925} - 1.35 \approx 0.197[/itex]
 
  • #3
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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
 
  • #4
AKG
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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.
 
  • #5
NateTG
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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
[tex]2\sin^{-1}(.45)[/tex]
Moreover, it's not hard to see that the lenght from the center intersection to the edge of the red segment has length
[tex]\sqrt{3}[/tex]

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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