Difference between revisions of "2007 AMC 12B Problems/Problem 20"
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==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>, |
+ | |||
+ | <math>\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*}</math> | ||
+ | |||
+ | Taking <math>d=1</math>, we have that <math>\frac{2b}{a-b} = 10</math>. This gives us that <math>6b=5a</math>. Taking the minimum values of <math>a=6,b=5</math> we have that <math>a+b+c+d=6+5+3+1=15 \Rightarrow \mathrm {(C)}</math> | ||
==See also== | ==See also== |
Revision as of 03:25, 18 January 2009
Problem
The parallelogram bounded by the lines , , , and has area . The parallelogram bounded by the lines , , , and has area . Given that , , , and are positive integers, what is the smallest possible value of ?
Solution
Plotting the parallelogram on the coordinate plane, the 4 corners are at . Because , we have that or that , which gives . From the 4 corners of the parallelogram, we have that . Substituting ,
$\begin{eqnarray*} 2d\left(\frac{3bd-ad}{a-b}\right) &=& 18 \\ d\left(\frac{bd-ad}{a-b}+\frac{2bd}{a-b}\right) &=& 9 \\ d\left(-d+\frac{2bd}{a-b}\right) &=& 9 \\ -d^2 + \frac{2b}{a-b}d^2 &=& 9 \\ \frac{2b}{a-b}d^2 &=& 9 + d^2 \end{eqnarray*}$ (Error compiling LaTeX. Unknown error_msg)
Taking , we have that . This gives us that . Taking the minimum values of we have that
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 |