Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

({{img}})
(→‎Solution: construct asy using x = 160)
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
−
{{image}}
 
−
<!--
 
 
<center><asy>
 
<center><asy>
−
pair A=(0,0),B=(1.8,0);
+
real x = 1.60; /* arbitrary */
−
</asy></center>-->
+
 
 +
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></center>
  
 
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>,  
 
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>,  

Revision as of 15:59, 11 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

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

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_I_Problems/Problem_12&oldid=26462"