Thursday, March 4, 2010

Fox 259

An extension of what Shelby submitted before. Enjoy...
We have not checked if the triangle was constructable. Can it exist with the given lengths?
Posted by 8foxes at 8:19 AM
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1 comment:

  1. AjitMarch 4, 2010 at 9:15 PM

    By Stewart's Theorem: AD^2+ 4=2(AE^2+x^2)
    and AE^2+1=2(AD^2+x^2) if BD=DE=EC=x. Eliminate x from the two equationa and use AD + AE = 2 to obtain AD = 3/4 and AE = 5/4 and this gives x^2 = 23/32. Now Tr. ADE gives: 23/32 = 9/16 + 25/16 - 2(3*5*/16)cos(Φ) or cos(Φ) = 3/4
    Ajit

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