2003 AIME I Problems/Problem 12

Problem

In convex quadrilateral $ABCD, \angle A \cong \angle C, AB = CD = 180,$ and $AD \neq BC.$ The perimeter of $ABCD$ is 640. Find $\lfloor 1000 \cos A \rfloor.$ (The notation $\lfloor x \rfloor$ means the greatest integer that is less than or equal to $x.$ )

Solution

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Let $AD = x$ so $BC = 640 - 360 - x = 280 - x$ . By the Law of Cosines in $\triangle ABD$ at angle $A$ and in $\triangle BCD$ at angle $C$ , $180^2 + x^2 - 2\cdot180 \cdot x \cdot \cos A = BD^2 = 180^2 + (280 - x)^2 - 2\cdot180\cdot(280 - x) \cdot \cos A.$ 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 $\lfloor 1000\cos A\rfloor = \boxed{777}$ .

2003 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2003 AIME I Problems/Problem 12

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