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Difference between revisions of "2003 AIME I Problems/Problem 12"

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== Solution ==
 
== Solution ==
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Let <math>AD = x</math> so <math>BC = 640 - 360 - x = 280 - x</math>. Let <math>BD = d</math> so by the [[Law of Cosines]] in <math>\triangle ABD</math> at [[angle]] <math>A</math> and in <math>\triangle BCD</math> at angle <math>C</math>,  
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<math>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</math>. Then
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Let <math>AD = x</math> so <math>BC = 640 - 360 - x = 280 - x</math>. By the [[Law of Cosines]] in <math>\triangle ABD</math> at angle <math>A</math> and in <math>\triangle BCD</math> at angle <math>C</math>,  
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<cmath>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.</cmath>  Then
 
<math>x^2 - 360x\cos A = (280 -x)^2 -360(280 - x)\cos A</math> and grouping the <math>\cos A</math> terms gives
 
<math>x^2 - 360x\cos A = (280 -x)^2 -360(280 - x)\cos A</math> and grouping the <math>\cos A</math> terms gives
 
<math>360(280 - 2x)\cos A = 280(280 - 2x)</math>.
 
<math>360(280 - 2x)\cos A = 280(280 - 2x)</math>.
  
 
Since <math>x \neq 280 - x</math>, <math>280 - 2x \neq 0</math> and thus
 
Since <math>x \neq 280 - x</math>, <math>280 - 2x \neq 0</math> and thus
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<math>360\cos A = 280</math> so <math>\cos A = \frac{7}{9} = 0.7777\ldots</math> and so <math>\lfloor 1000\cos A\rfloor = 777</math>.
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<math>360\cos A = 280</math> so <math>\cos A = \frac{7}{9} = 0.7777\ldots</math> and so <math>\lfloor 1000\cos A\rfloor = \boxed{777}</math>.
  
 
== See also ==
 
== See also ==
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* [[2003 AIME I Problems/Problem 11 | Previous problem]]
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{{AIME box|year=2003|n=I|num-b=11|num-a=13}}
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* [[2003 AIME I Problems/Problem 13 | Next problem]]
 
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* [[2003 AIME I Problems]]
 
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 15:46, 10 June 2008

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

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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 $\lfloor 1000\cos A\rfloor = \boxed{777}$.

See also

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
All AIME Problems and Solutions
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