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Thread: Olympics1

  1. #1
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    Olympics1

    ABCD is a quadrilateral.
    AB|=2
    BC=16
    CD=13
    DA=5

    Find |DBC|/|ABD|
    Attached Thumbnails Attached Thumbnails Olympics1-yesso.png  
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  2. #2
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    Re: Olympics1

    Quote Originally Posted by kastamonu View Post
    ABCD is a quadrilateral.
    AB|=2
    BC=16
    CD=13
    DA=5

    Find |DBC|/|ABD|
    Again, please show some attempt so we may help you from there.
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  3. #3
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    Re: Olympics1

    ıF I could do anything I sjouldn't have asked.
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  4. #4
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    Re: Olympics1

    Use Heron's Formula:

    $\dfrac{\sqrt{ \dfrac{|BD|+|DC|+|CB|}{2} \left(\dfrac{|BD|+|DC|+|CB|}{2}-|BD|\right)\left(\dfrac{|BD|+|DC|+|CB|}{2}-|DC|\right)\left(\dfrac{|BD|+|DC|+|CB|}{2}-|CB|\right)}}{\sqrt{\dfrac{|AB|+|BD|+|DA|}{2}\left (\dfrac{|AB|+|BD|+|DA|}{2}-|AB|\right)\left(\dfrac{|AB|+|BD|+|DA|}{2}-|BD|\right)\left(\dfrac{|AB|+|BD|+|DA|}{2}-|DA|\right)}}$

    Just plug in the lengths you know and see if anything cancels out.

    I get:
    $\sqrt{\dfrac{29^2-|BD|^2}{7^2-|BD|^2}}$
    Last edited by SlipEternal; May 22nd 2017 at 12:16 PM.
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  5. #5
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    Re: Olympics1

    Answer is 6 according to the book.
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  6. #6
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    Re: Olympics1

    Quote Originally Posted by kastamonu View Post
    Answer is 6 according to the book.
    If $|BD|^2 = \dfrac{223}{35}$, my answer is 6, as well. I have not tried to figure out how to get $|BD|^2 = \dfrac{223}{35}$, but if you can prove that, then my solution gives the book's answer.
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  7. #7
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    Re: Olympics1

    Quote Originally Posted by kastamonu View Post
    ıF I could do anything I sjouldn't have asked.
    No, you're lazy and an excuse-maker. Your posting history speaks for itself.
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  8. #8
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    Re: Olympics1

    Many Thanks. I tried to solve by Brahmahupta but I couldn't. I couldn't find BD also. Maybe there is something wrong.
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  9. #9
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    Re: Olympics1

    A combination of Heron's formula and Brahmagupta's formula together give $|BD| = \sqrt{\dfrac{5141}{109}}$ and the ratio of areas is $\dfrac{104}{5}$.
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  10. #10
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    Re: Olympics1

    Problem is BD. Question maybe wrong or there may be a drawing that makes it very easy.
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  11. #11
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    Re: Olympics1

    I sent an email to the author. I hope he will help.
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  12. #12
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    Re: Olympics1

    Quote Originally Posted by kastamonu View Post
    Problem is BD. Question maybe wrong or there may be a drawing that makes it very easy.
    Why is there a problem with BD?
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  13. #13
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    Re: Olympics1

    I couldn't find BD. Answer is 10 according to the book.
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  14. #14
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    Re: Olympics1

    I was able to find BD. Use Heron's formula for each triangle. The sum of the areas is equal to the area of the quadrilateral. Use Brahmagupta's formula for the area of the quadrilateral. BD cannot equal 10. That would mean ABD is not a triangle because $2+5 <10$. The answer that I found involves a simple application of algebra to solve.

    $\sqrt{ \dfrac{(29+x)(29-x)(x+3)(x-3)}{16} }+\sqrt{\dfrac{(7+x)(7-x)(x+3)(x-3)}{16} }= \sqrt{2080}$

    Note: if the quadrilateral is not cyclic, then the area formula would involve trig and be based of the sum of two opposite interior angles (meaning an infinite number of solutions).
    Last edited by SlipEternal; May 23rd 2017 at 03:28 PM.
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  15. #15
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    Re: Olympics1

    DBC/ABD is 10 according to the book. It is not 6. I mistyped the answer.
    To apply Brahmagupta we have to know 2 opposite angles. The formula above looks like Brahmagupta except the part with cosinus.
    Last edited by kastamonu; May 23rd 2017 at 03:41 PM.
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