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

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<cmath>(AD - BC)(AD + BC) = 360(AD - BC) \cos A</cmath>
 
<cmath>(AD - BC)(AD + BC) = 360(AD - BC) \cos A</cmath>
 
<cmath>(AD + BC) = 360 \cos A</cmath>
 
<cmath>(AD + BC) = 360 \cos A</cmath>
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We know that <math>AD + BC = 640 - 360 = 280</math>. <math>\cos A = \dfrac{240}{360} =  \dfrac{7}{9} = 0.777 \ldots</math>
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We know that <math>AD + BC = 640 - 360 = 280</math>. <math>\cos A = \dfrac{280}{360} =  \dfrac{7}{9} = 0.777 \ldots</math>
  
 
<math>\lfloor 1000 \cos  A \rfloor = \boxed{777}</math>.
 
<math>\lfloor 1000 \cos  A \rfloor = \boxed{777}</math>.

Revision as of 17:25, 3 July 2015

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

Solution 1

[asy] real x = 1.60; /* arbitrary */ pointpen = black; pathpen = black+linewidth(0.7); size(180); real BD = x*x + 1.80*1.80 - 2 * 1.80 * x * 7 / 9; pair A=(0,0),B=(1.8,0),D=IP(CR(A,x),CR(B,BD)),C=OP(CR(D,1.8),CR(B,2.80 - x)); D(MP("A",A)--MP("B",B)--MP("C",C)--MP("D",D,N)--B--A--D); MP("180",(A+B)/2); MP("180",(C+D)/2,NE); D(anglemark(B,A,D)); D(anglemark(D,C,B)); [/asy]

By the Law of Cosines on $\triangle ABD$ at angle $A$ and on $\triangle BCD$ at angle $C$ (note $\angle C = \angle A$),

\[180^2 + AD^2 - 360 \cdot AD \cos A = 180^2 + BC^2 - 360 \cdot BC \cos A\] \[(AD^2 - BC^2) = 360(AD - BC) \cos A\] \[(AD - BC)(AD + BC) = 360(AD - BC) \cos A\] \[(AD + BC) = 360 \cos A\] We know that $AD + BC = 640 - 360 = 280$. $\cos A = \dfrac{280}{360} = \dfrac{7}{9} = 0.777 \ldots$

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

Solution 2

Notice that $AB = CD$, and $BD = DB$, and $\angle{DAB} \cong \angle{BCD}$, so we have side-side-angle matching on triangles $ABD$ and $CDB$. Since the problem does not allow $\triangle{ABD} \cong \triangle{CDB}$, we know that $\angle{ADB}$ is not a right angle, and there is a unique other triangle with the matching side-side-angle.

Extend $AD$ to $C'$ so that $\triangle{ABC'}$ is isosceles with $AB = C'B$. Then notice that $\triangle{DC'B}$ has matching side-side-angle, and yet $\triangle{ADB} \not\cong \triangle{C'DB}$ because $\angle{ADB}$ is not right. Therefore $\triangle{C'DB}$ is the unique triangle mentioned above, so $\triangle{CDB}$ is congruent, in some order of vertices, to $\triangle{C'DB}$. Since $\triangle{CDB} \cong \triangle{C'DB}$ would imply $\triangle{CDB} = \triangle{C'DB}$, making quadrilateral $ABCD$ degenerate, we must have $\triangle{CDB} \cong \triangle{C'BD}$.

Since the perimeter of $ABCD$ is $640$, $AD + BC = 640 - 180 - 180 = 280$. Hence $280 = AD + BC = AD + DC'$. Drop the altitude of $\triangle{ABC'}$ from $B$ and call the foot $P$. Then right triangle trigonometry on $\triangle{APB}$ shows that $\cos{A} = AP/AB = 140/180 = 7/9$, 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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