Difference between revisions of "2003 AIME I Problems/Problem 12"
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== Problem == | == Problem == | ||
| − | In convex quadrilateral <math> ABCD, \angle A \cong \angle C, AB = CD = 180, </math> and <math> AD \neq BC. </math> The perimeter of <math> ABCD </math> is 640. Find <math> \lfloor 1000 \cos A \rfloor. </math> (The notation <math> \lfloor x \rfloor </math> means the greatest integer that is less than or equal to <math> x. </math>) | + | In [[convex]] [[quadrilateral]] <math> ABCD, \angle A \cong \angle C, AB = CD = 180, </math> and <math> AD \neq BC. </math> The [[perimeter]] of <math> ABCD </math> is 640. Find <math> \lfloor 1000 \cos A \rfloor. </math> (The notation <math> \lfloor x \rfloor </math> means the greatest [[integer]] that is less than or equal to <math> x. </math>) |
== Solution == | == 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 $\lfloor 1000\cos A\rfloor = 777$. | ||
== See also == | == See also == | ||
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* [[2003 AIME I Problems/Problem 13 | Next problem]] | * [[2003 AIME I Problems/Problem 13 | Next problem]] | ||
* [[2003 AIME I Problems]] | * [[2003 AIME I Problems]] | ||
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| + | [[Category:Intermediate Geometry Problems]] | ||
Revision as of 17:46, 19 January 2007
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 $\lfloor 1000\cos A\rfloor = 777$.
