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 "2007 AMC 12B Problems/Problem 20"
m (→‎Solution)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
−
Plotting the parallelogram on the coordinate plane, the 4 corners are at <math>(0,c),(0,d),(\frac{d-c}{a-b},\frac{ad-bc}{a-b}),(\frac{c-d}{a-b},\frac{bc-ad}{a-b})</math>. Because <math>72= 4\cdot 18</math>, we have that <math>4(c-d)(\frac{c-d}{a-b}) = (c+d)(\frac{c+d}{a-b})</math> or that <math>2(c-d)=c+d</math>, which gives <math>c=3d</math>. From the 4 corners of the parallelogram, we have that <math>(c-d)(\frac{bc-ad}{a-b})=18</math>. Substituting <math>c=3d</math>,
+
Plotting the parallelogram on the coordinate plane, the 4 corners are at <math>(0,c),(0,d),\left(\frac{d-c}{a-b},\frac{ad-bc}{a-b}\right),\left(\frac{c-d}{a-b},\frac{bc-ad}{a-b}\right)</math>. Because <math>72= 4\cdot 18</math>, we have that <math>4(c-d)\left(\frac{c-d}{a-b}\right) = (c+d)\left(\frac{c+d}{a-b}\right)</math> or that <math>2(c-d)=c+d</math>, which gives <math>c=3d</math>. From the 4 corners of the parallelogram, we have that <math>(c-d)\left(\frac{bc-ad}{a-b}\right)=18</math>. Substituting <math>c=3d</math>,
  
 
<math>\begin{eqnarray*}
 
<math>\begin{eqnarray*}

Revision as of 09:44, 18 January 2009

Problem

The parallelogram bounded by the lines $y=ax+c$, $y=ax+d$, $y=bx+c$, and $y=bx+d$ has area $18$. The parallelogram bounded by the lines $y=ax+c$, $y=ax-d$, $y=bx+c$, and $y=bx-d$ has area $72$. Given that $a$, $b$, $c$, and $d$ are positive integers, what is the smallest possible value of $a+b+c+d$?

$\mathrm {(A)} 13\qquad \mathrm {(B)} 14\qquad \mathrm {(C)} 15\qquad \mathrm {(D)} 16\qquad \mathrm {(E)} 17$

Solution

Plotting the parallelogram on the coordinate plane, the 4 corners are at $(0,c),(0,d),\left(\frac{d-c}{a-b},\frac{ad-bc}{a-b}\right),\left(\frac{c-d}{a-b},\frac{bc-ad}{a-b}\right)$. Because $72= 4\cdot 18$, we have that $4(c-d)\left(\frac{c-d}{a-b}\right) = (c+d)\left(\frac{c+d}{a-b}\right)$ or that $2(c-d)=c+d$, which gives $c=3d$. From the 4 corners of the parallelogram, we have that $(c-d)\left(\frac{bc-ad}{a-b}\right)=18$. Substituting $c=3d$,

$\begin{eqnarray*} 2d\left(\frac{3bd-ad}{a-b}\right) &=& 18 \\ d\left(\frac{bd-ad}{a-b}+\frac{2bd}{a-b}\right) &=& 9 \\ d\left(-d+\frac{2bd}{a-b}\right) &=& 9 \\ -d^2 + \frac{2b}{a-b}d^2 &=& 9 \\ \frac{2b}{a-b}d^2 &=& 9 + d^2 \end{eqnarray*}$ (Error compiling LaTeX. Unknown error_msg)

Taking $d=1$, we have that $\frac{2b}{a-b} = 10$. This gives us that $6b=5a$. Taking the minimum values of $a=6,b=5$ we have that $a+b+c+d=6+5+3+1=15 \Rightarrow \mathrm {(C)}$

See also

2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_20&oldid=29604"