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, or the ratio of side lengths, 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, or the ratio of side lengths, 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) | + | \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== | ==Solution 2== |
Revision as of 16:22, 2 May 2015
Contents
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, or the ratio of side lengths, is ). The area of triangular half of the parallelogram on the right side of the y-axis is given by , so substituting :
Thus , and we verify that , will give us a minimum value for . Then .
Solution 2
Template:Incomplete 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 for , so the sum is .
See also
2007 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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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