Difference between revisions of "2007 AMC 12B Problems/Problem 20"
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\frac{1}{2} (c-d)\left(\frac{c-d}{a-b}\right) &= 9 \quad \Longrightarrow \quad 2d^2 &= 9(a-b)</cmath></center> | \frac{1}{2} (c-d)\left(\frac{c-d}{a-b}\right) &= 9 \quad \Longrightarrow \quad 2d^2 &= 9(a-b)</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>. | ||
− | + | ==Solution 2== | |
+ | The key to this solution is that area is invariant under translation. By suitably shifting the plane, the problem is mapped to the lines <math>c,d,(b-a)x+c,(b-a)x+d</math> and <math>c,-d,(b-a)x+c,(b-a)x-d</math>. Now, the area of the parallelogram contained by is the former is equal to the area of a rectangle with sides <math>d-c</math> and <math>\frac{d-c}{b-a}</math>, <math>\frac{(d-c)^2}{b-a}=18</math>, and the area contained by the latter is <math>\frac{(c+d)^2}{b-a}=72</math>. Thus, <math>d=3c</math> and <math>b-a</math> must be even if the former quantity is to equal <math>18</math>. <math>c^2=18(b-a)</math> so <math>c</math> is a multiple of <math>3</math>. Putting this all together, the minimal solution is <math>(a,b,c,d)=(3,1,3,9)</math> | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2007|ab=B|num-b=21|num-a=23}} | {{AMC12 box|year=2007|ab=B|num-b=21|num-a=23}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] |
Revision as of 00:41, 25 February 2009
Contents
[hide]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
Template:Incomplete Plotting the parallelogram on the coordinate plane, the 4 corners are at . Because , we have that or that , which gives (consider a homothety, or dilation, that carries the first parallelogram to the second parallelogram; because the area increases by , it follows that the stretch along the diagonal is ). The area of triangular half of the parallelogram on the right side of the y-axis is given by , so substituting :
\[\frac{1}{2} (c-d)\left(\frac{c-d}{a-b}\right) &= 9 \quad \Longrightarrow \quad 2d^2 &= 9(a-b)\] (Error compiling LaTeX. Unknown error_msg)
Thus , and we verify that , will give us a minimum value for . Then .
Solution 2
The key to this solution is that area is invariant under translation. By suitably shifting the plane, the problem is mapped to the lines and . Now, the area of the parallelogram contained by is the former is equal to the area of a rectangle with sides and , , and the area contained by the latter is . Thus, and must be even if the former quantity is to equal . so is a multiple of . Putting this all together, the minimal solution is
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 |