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Thread: geometry olympics 2014

  1. #1
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    geometry olympics 2014

    https://postimg.org/image/epnzx6clr/

    I couldn't solve that and need help.
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    Re: geometry olympics 2014



    I get $x = 6 \pm 2\sqrt{2}$
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    Re: geometry olympics 2014

    Answer is 12 according to the book.How did you find it?
    I triedsimilar triangles but no result.
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    Re: geometry olympics 2014

    Can you show your attempt using similar triangles?

    Using instead triangle ACD: x^2 = AD^2 + 2^2
    Last edited by DenisB; Jan 18th 2017 at 11:45 AM.
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    Re: geometry olympics 2014

    Quote Originally Posted by kastamonu View Post
    Answer is 12 according to the book.How did you find it?
    I triedsimilar triangles but no result.
    yes, the sum of x's is $(6+2\sqrt{2})+(6-2\sqrt{2}) = 12$

    First, I drew a picture to draw in an auxiliary line segment & label some variables ...



    equations I came up with using Pythagoras & similar triangles ...

    $x^2+2^2=(b\sqrt{2})^2$

    $a^2+b^2=12^2$

    $x^2+14^2 = (a+b)^2$

    $\dfrac{x}{a+b}=\dfrac{b}{12}$

    Found an equation quadratic in $x$ and solved ...
    Thanks from topsquark
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    Re: geometry olympics 2014

    Please can you show the quadratic equation? I worked a lot on that one.
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    Re: geometry olympics 2014

    Quote Originally Posted by kastamonu View Post
    Please can you show the quadratic equation? I worked a lot on that one.
    using the first Pythagoras equation, $b^2 = \dfrac{x^2+4}{2}$

    using the second & third Pythagoras equations, $x^2+14^2 = 12^2 + 2ab \implies ab = \dfrac{x^2+52}{2}$

    using the proportion from similar triangles, $12x = ab+b^2$

    substitute for $ab$ and $b^2$ ... you get an equation quadratic in $x$
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    Re: geometry olympics 2014

    many thanks.
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    Re: geometry olympics 2014

    I will disturb you one more time.
    https://postimg.org/image/rzkw43x1l/
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    Re: geometry olympics 2014

    Quote Originally Posted by kastamonu View Post
    I will disturb you one more time.
    https://postimg.org/image/rzkw43x1l/
    Draw an auxiliary line between B and E ... pretty much the same equations as your last problem. You'll get a quadratic in $x$. Choose the larger of the two solutions. Do it.
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  11. #11
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    Re: geometry olympics 2014

    Many Thanks.x is 3.
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