2003 AIME I Problems/Problem 12
Problem
In convex quadrilateral 
and 
The perimeter of 
is 640. Find 
(The notation 
means the greatest integer that is less than or equal to 
)
Solution
Let $AD = x$ so $BC = 640 - 360 - x = 280 - x$. Let $BD = d$ so by the Law of Cosines in $\triangle ABD$ at angle $A$ and in $\triangle BCD$ at angle $C$, $180^2 + x^2 - 2\cdot180 \cdot x \cdot \cos A = d^2 = 180^2 + (280 - x)^2 - 2\cdot180\cdot(280 - x) \cdot \cos A$. Then $x^2 - 360x\cos A = (280 -x)^2 -360(280 - x)\cos A$ and grouping the $\cos A$ terms gives 360(280 - 2x)\cos A = 280(280 - 2x)$.
Since $x \neq 280 - x$, $280 - 2x \neq 0$ and thus $360\cos A = 280$ so $\cos A = \frac{7}{9} = 0.7777\ldots$ and so $\lfloor 1000\cos A\rfloor = 777$.
