Sunday, February 28, 2010

Fox 255

http://www.8foxes.com/
Shelby suggested this problem. Thanks!

4 comments:

  1. So we also know that side BC must be less than 3 but more than 1. That's proven by the triangle leg addition postulate.

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  2. Answer is D.

    1^2 + 2^2 = 2m^2 + m^2/2

    10 = 5m^2

    m = sqrt(2).

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  3. There is at least one geometric solution, using similarity

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  4. AD=BC=2x
    BD=DC=x

    also AB=1,AC=2

    using Stewart's Theorem

    AB^2 . CD + AC^2 . BD = BC . (AD^2 + BD.CD)

    1.x + 4.x = 2x(4x^2 + x.x)
    x=1/sqrt2

    AD=2x=2/sqrt2=sqrt2

    so option D is answer.

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