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2007 AMC 12B Problems/Problem 20: Difference between revisions

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==Solution==
==Solution==
{{incomplete|solution}}
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real a = 3, b = 1, c = 9, d = 3;
D((0,c) -- ((d-c)/(a-b),(a*d-b*c)/(a-b)) -- (0,d) -- ((c-d)/(a-b),(b*c-a*d)/(a-b)) -- cycle);
D((0,c) -- ((-d-c)/(a-b),(-a*d-b*c)/(a-b)) -- (0,-d) -- ((c+d)/(a-b),-(-a*d-b*c)/(a-b)) -- cycle);
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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>\begin{align*}
<center><cmath>
\frac{1}{2} (c-d)\left(\frac{c-d}{a-b}\right) &= 9 \\
\frac{1}{2} (c-d)\left(\frac{c-d}{a-b}\right) &= 9 \quad \Longrightarrow \quad 2d^2 &= 9(a-b)
2d^2 &= 9(a-b) \\
\end{align*}</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 23:14, 24 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 $c = 3d$: 

\[\frac{1}{2} (c-d)\left(\frac{c-d}{a-b}\right) &= 9 \quad \Longrightarrow \quad 2d^2 &= 9(a-b)
\end{align*}\] (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 (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
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