Difference between revisions of "2007 AMC 12B Problems/Problem 20"

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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> (consider a [[homothety]], or dilation, that carries the first parallelogram to the second parallelogram; because the area increases by <math>4\times</math>, it follows that the stretch along the diagonal is <math>2\times</math>). The area of triangular half of the parallelogram on the right side of the y-axis is given by <math>9 = \frac{1}{2} (c-d)\left(\frac{d-c}{a-b}\right)</math>, so 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> (consider a [[homothety]], or dilation, that carries the first parallelogram to the second parallelogram; because the area increases by <math>4\times</math>, it follows that the stretch along the diagonal is <math>2\times</math>). The area of triangular half of the parallelogram on the right side of the y-axis is given by <math>9 = \frac{1}{2} (c-d)\left(\frac{d-c}{a-b}\right)</math>, so substituting <math>c = 3d</math>:
 
<center><cmath>
 
<center><cmath>
\frac{1}{2} (c-d)\left(\frac{c-d}{a-b}\right) &= 9 \quad \Longrightarrow \quad 2d^2 &= 9(a-b)
+
\frac{1}{2} (c-d)\left(\frac{c-d}{a-b}\right) &= 9 \quad \Longrightarrow \quad 2d^2 &= 9(a-b)</cmath></center>
\end{align*}</cmath></center>
 
 
Thus <math>3|d</math>, and we verify that <math>d = 3</math>, <math>a-b = 2 \Longrightarrow a = 3, b = 1</math> will give us a minimum value for <math>a+b+c+d</math>. Then <math>a+b+c+d = 3 + 1 + 9 + 3 = 16\ \mathbf{(D)}</math>.
 
Thus <math>3|d</math>, and we verify that <math>d = 3</math>, <math>a-b = 2 \Longrightarrow a = 3, b = 1</math> will give us a minimum value for <math>a+b+c+d</math>. Then <math>a+b+c+d = 3 + 1 + 9 + 3 = 16\ \mathbf{(D)}</math>.
  

Revision as of 00:14, 25 February 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

Template:Incomplete 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 $c=3d$ (consider a homothety, or dilation, that carries the first parallelogram to the second parallelogram; because the area increases by $4\times$, it follows that the stretch along the diagonal is $2\times$). The area of triangular half of the parallelogram on the right side of the y-axis is given by $9 = \frac{1}{2} (c-d)\left(\frac{d-c}{a-b}\right)$, so substituting $c = 3d$:

\[\frac{1}{2} (c-d)\left(\frac{c-d}{a-b}\right) &= 9 \quad \Longrightarrow \quad 2d^2 &= 9(a-b)\] (Error compiling LaTeX. Unknown error_msg)

Thus $3|d$, and we verify that $d = 3$, $a-b = 2 \Longrightarrow a = 3, b = 1$ will give us a minimum value for $a+b+c+d$. Then $a+b+c+d = 3 + 1 + 9 + 3 = 16\ \mathbf{(D)}$.

See also

2007 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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All AMC 12 Problems and Solutions