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

(→‎Solution: construct asy using x = 160)
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== Solution ==
 
== Solution ==
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===Solution 1===
 
<center><asy>
 
<center><asy>
 
real x = 1.60; /* arbitrary */
 
real x = 1.60; /* arbitrary */
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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
 
<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>.
 
<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>.
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===Solution 2===
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Notice that <math>AB = CD</math>, and <math>BD = DB</math>, and <math>\angle{DAB} \cong \angle{BCD}</math>, so we have side-side-angle matching on triangles <math>ABD</math> and <math>CDB</math>.  Since the problem does not allow <math>\triangle{ABD} \cong \triangle{CDB}</math>, we know that <math>\angle{ADB}</math> is not a right angle, and there is a unique other triangle with the matching side-side-angle.
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Extend <math>AD</math> to <math>C'</math> so that <math>\triangle{ABC'}</math> is isosceles with <math>AB = C'B</math>.  Then notice that <math>\triangle{DC'B}</math> has matching side-side-angle, and yet <math>\triangle{ADB} \not\cong \triangle{C'DB}</math> because <math>\angle{ADB}</math> is not right.  Therefore <math>\triangle{C'DB}</math> is the unique triangle mentioned above, so <math>\triangle{CDB}</math> is congruent, in some order of vertices, to <math>\triangle{C'DB}</math>.  Since <math>\triangle{CDB} \cong \triangle{C'DB}</math> would imply <math>\triangle{CDB} = \triangle{C'DB}</math>, making quadrilateral <math>ABCD</math> degenerate, we must have <math>\triangle{CDB} \cong \triangle{C'BD}</math>.
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Since the perimeter of <math>ABCD</math> is <math>640</math>, <math>AD + BC = 640 - 180 - 180 = 280</math>.  Hence <math>280 = AD + BC = AD + DC'</math>.  Drop the altitude of <math>\triangle{ABC'}</math> from <math>B</math> and call the foot <math>P</math>.  Then right triangle trigonometry on <math>\triangle{APB}</math> shows that <math>\cos{A} = AP/AB = 140/180 = 7/9</math>, so <math>\lfloor 1000 \cos A \rfloor = \boxed{777}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 23:27, 27 February 2009

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]

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


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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