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Thread: Olympics1

  1. #16
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    Re: Olympics1

    Like I said, I had to assume the quadrilateral was cyclic or there would be an infinite number of answers.
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  2. #17
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    Re: Olympics1

    I calculated a ratio of 10 by assuming angle ABC is 90 degrees (cause it looked like it) ... see diagram

    area of triangle ABD = $x$, area of triangle BDC = $8h$ ... ratio is $\dfrac{8h}{x}$

    $x^2 + (h-2)^2 = 5^2 \implies x^2+h^2 = 4h+21$

    $(16-x)^2 + h^2 = 13^2 \implies h = 8x-27$

    $x^2+(8x-27)^2 = 4(8x-27) + 21$

    $65x^2 - 464x + 816 = 0$

    $(x-4)(65x-204) = 0$

    $x = 4 \implies h = 5 \implies \dfrac{8h}{x} = 10$

    $x = \dfrac{204}{65}$ makes $h < 0$
    Attached Thumbnails Attached Thumbnails Olympics1-olympic1.jpg  
    Thanks from SlipEternal
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  3. #18
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    Re: Olympics1

    Many Thanks. I am not showing my work because I am working on the book. Sometimes I am erasing something and trying an other thing. Also Ican't type some math symbols.
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  4. #19
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    Re: Olympics1

    If I don't solve these questions my self who will solve them for me in the exam?
    I am not lazy, I trying my best to solve. But geometry is not just knowledge you have to see something.
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