Like I said, I had to assume the quadrilateral was cyclic or there would be an infinite number of answers.

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- May 23rd 2017, 04:01 PM #16

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- May 23rd 2017, 04:56 PM #17
## 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 $h < 0$

- May 24th 2017, 08:05 AM #18

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- May 24th 2017, 08:10 AM #19

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