Difference between revisions of "2007 AMC 12B Problems/Problem 20"
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<math>\mathrm {(A)} 13\qquad \mathrm {(B)} 14\qquad \mathrm {(C)} 15\qquad \mathrm {(D)} 16\qquad \mathrm {(E)} 17</math> | <math>\mathrm {(A)} 13\qquad \mathrm {(B)} 14\qquad \mathrm {(C)} 15\qquad \mathrm {(D)} 16\qquad \mathrm {(E)} 17</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | |||
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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); | 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, 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 the 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)</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 = | + | 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 = \boxed{\mathbf{(D)} 16}</math>. |
+ | |||
==Solution 2== | ==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 for <math>(a,b,c,d)=(3,1,3,9)</math>, so the sum is <math> \boxed{\textbf{(D)} 16} </math>. | 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 for <math>(a,b,c,d)=(3,1,3,9)</math>, so the sum is <math> \boxed{\textbf{(D)} 16} </math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | Let <math>a</math> and <math>b</math> be the slopes of the lines such that <math>b > a</math> (i.e. the line <math>bx+c</math> is steeper than <math>ax+c</math>) and <math>c > d</math> (i.e. the point <math>(0, c)</math> is higher than the point <math>(0, d)</math>. Upon drawing a diagram, we see that both the smaller and the larger parallelogram can be split along the x-axis, such that both of their areas are the combinations of two corresponding triangles. The area of a triangle is <math>\frac{bh}{2}</math>, but since a parallelogram is two such triangles, the area becomes <math>bh</math>. | ||
+ | |||
+ | Let <math>b_1</math> and <math>h_1</math> denote the base length and height, respectively, of the triangles pertaining to the smaller parallelogram and <math>b_2</math> and <math>h_2</math> denote those of the larger parallelogram. Notice that <math>b_1</math> is simple the distance from <math>(0, d)</math> to <math>(0, c)</math>, or <math>(c-d)</math>. Also notice that <math>h_1</math> is the distance from the <math>x</math>-axis to the intersection of lines <math>ax+c</math> and <math>bx+d</math>. This is equivalent to the value of the <math>x</math>-coordinate of intersection, so we solve for <math>x</math>: | ||
+ | |||
+ | |||
+ | <math>ax+c=bx+d</math> | ||
+ | |||
+ | <math>\Rightarrow bx-ax = c-d</math> | ||
+ | |||
+ | <math>\Rightarrow (b-a)x = c-d</math> | ||
+ | |||
+ | <math>\Rightarrow x = \frac{c-d}{b-a}</math>. | ||
+ | |||
+ | |||
+ | The area of the smaller parallelogram is <math>b_1*h_1</math>, or | ||
+ | |||
+ | |||
+ | <math>(c-d) * \frac{c-d}{b-a}</math> | ||
+ | |||
+ | |||
+ | <math>\Rightarrow \frac{(c-d)^2}{b-a}</math>. | ||
+ | |||
+ | |||
+ | <math>b_2</math> is the distance from <math>(0, -d)</math> to <math>(0, c)</math>, or <math>(c+d)</math>. <math>h_2</math> is the <math>x</math>-coordinate of the intersection of the lines <math>ax+c</math> and <math>bx-d</math>. Again, we solve for x: | ||
+ | |||
+ | |||
+ | <math>ax+c=bx-d</math> | ||
+ | |||
+ | <math>\Rightarrow bx-ax = c+d</math> | ||
+ | |||
+ | <math>\Rightarrow (b-a)x = c+d</math> | ||
+ | |||
+ | <math>\Rightarrow x = \frac{c+d}{b-a}</math>. | ||
+ | |||
+ | |||
+ | The area of the larger parallelogram is <math>b-1*h_1</math>, or | ||
+ | |||
+ | |||
+ | <math>(c+d) * \frac{c+d}{b-a}</math> | ||
+ | |||
+ | |||
+ | <math>\Rightarrow \frac{(c+d)^2}{b-a}</math>. | ||
+ | |||
+ | |||
+ | The areas of the parallelograms are given to us: <math>18</math> and <math>72</math>. Therefore we can set up a ratio: | ||
+ | |||
+ | |||
+ | <math>\frac{18}{72} = \frac{\frac{(c-d)^2}{b-a}}{\frac{(c+d)^2}{b-a}}</math> | ||
+ | |||
+ | <math>\Rightarrow 18(c+d)^2 = 72(c-d)^2</math> | ||
+ | |||
+ | <math>\Rightarrow (c+d)^2 = 4(c-d)^2</math> | ||
+ | |||
+ | <math>\Rightarrow c^2 + 2cd + d^2 = 4c^2 - 8cd + 4d^2</math> | ||
+ | |||
+ | <math>\Rightarrow 3c^2 - 10cd +3d^2 = 0</math> | ||
+ | |||
+ | <math>\Rightarrow (3c-d)(c-3d)=0</math> | ||
+ | |||
+ | <math>\Rightarrow c=3d, c=\frac{d}{3}</math> | ||
+ | |||
+ | |||
+ | We established earlier that <math>c>d</math>, so <math>c=3d</math>. Plugging this into the intial equations yields | ||
+ | |||
+ | |||
+ | <math>\frac{16d^2}{b-a} = 72</math> | ||
+ | |||
+ | |||
+ | and | ||
+ | |||
+ | |||
+ | <math>\frac{4d^2}{b-a} = 18</math> | ||
+ | |||
+ | Solving for <math>d</math>, we get | ||
+ | |||
+ | |||
+ | <math>d = 3\sqrt{\frac{b-a}{2}}</math> | ||
+ | |||
+ | |||
+ | We want the sum of <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math>. We can now rewrite this | ||
+ | |||
+ | |||
+ | <math>a + b + 12\sqrt{\frac{b-a}{2}}</math> | ||
+ | |||
+ | |||
+ | We are told that <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are all positive integers. Therefore the value under the radical must be a perfect square greater than 0. We can rewrite this | ||
+ | |||
+ | |||
+ | <math>\frac{b-a}{2} = k^2</math> | ||
+ | |||
+ | |||
+ | where <math>k</math> is some positive integer. | ||
+ | |||
+ | Rearranging, we get | ||
+ | |||
+ | |||
+ | <math>b = a + 2k^2</math> | ||
+ | |||
+ | Now we can rewrite the sum as | ||
+ | |||
+ | |||
+ | <math>a + a + 2k^2 +12k</math>. | ||
+ | |||
+ | |||
+ | Since both <math>a</math> and <math>k</math> must be at least <math>1</math>, the minimum value is | ||
+ | |||
+ | |||
+ | <math>1 + 1 + 2(1)^2 + 12(1) = 16 \Rightarrow \boxed{\text{D}}</math>. | ||
==See also== | ==See also== |
Latest revision as of 21:25, 20 December 2020
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 1
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 the 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
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 .
Solution 3
Let and be the slopes of the lines such that (i.e. the line is steeper than ) and (i.e. the point is higher than the point . Upon drawing a diagram, we see that both the smaller and the larger parallelogram can be split along the x-axis, such that both of their areas are the combinations of two corresponding triangles. The area of a triangle is , but since a parallelogram is two such triangles, the area becomes .
Let and denote the base length and height, respectively, of the triangles pertaining to the smaller parallelogram and and denote those of the larger parallelogram. Notice that is simple the distance from to , or . Also notice that is the distance from the -axis to the intersection of lines and . This is equivalent to the value of the -coordinate of intersection, so we solve for :
.
The area of the smaller parallelogram is , or
.
is the distance from to , or . is the -coordinate of the intersection of the lines and . Again, we solve for x:
.
The area of the larger parallelogram is , or
.
The areas of the parallelograms are given to us: and . Therefore we can set up a ratio:
We established earlier that , so . Plugging this into the intial equations yields
and
Solving for , we get
We want the sum of , , , and . We can now rewrite this
We are told that , , , and are all positive integers. Therefore the value under the radical must be a perfect square greater than 0. We can rewrite this
where is some positive integer.
Rearranging, we get
Now we can rewrite the sum as
.
Since both and must be at least , the minimum value 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.