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==
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==Solution 1==
{{incomplete|solution}}
 
 
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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 = 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 = \boxed{\mathbf{(D)} 16}</math>.
 +
 
 
==Solution 2==
 
==Solution 2==
{{incomplete|solution}}
 
 
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>.
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 +
==Solution 3==
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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>:
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<math>ax+c=bx+d</math>
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<math>\Rightarrow bx-ax = c-d</math>
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<math>\Rightarrow (b-a)x = c-d</math>
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<math>\Rightarrow x = \frac{c-d}{b-a}</math>.
 +
 +
 +
The area of the smaller parallelogram is <math>b_1*h_1</math>, or
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<math>(c-d) * \frac{c-d}{b-a}</math>
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<math>\Rightarrow \frac{(c-d)^2}{b-a}</math>.
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<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:
 +
 +
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<math>ax+c=bx-d</math>
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 +
<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>
 +
 +
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<math>\Rightarrow \frac{(c+d)^2}{b-a}</math>.
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 +
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The areas of the parallelograms are given to us: <math>18</math> and <math>72</math>. Therefore we can set up a ratio:
 +
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<math>\frac{18}{72} = \frac{\frac{(c-d)^2}{b-a}}{\frac{(c+d)^2}{b-a}}</math>
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<math>\Rightarrow 18(c+d)^2 = 72(c-d)^2</math>
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<math>\Rightarrow (c+d)^2 = 4(c-d)^2</math>
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<math>\Rightarrow c^2 + 2cd + d^2 = 4c^2 - 8cd + 4d^2</math>
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<math>\Rightarrow 3c^2 - 10cd +3d^2 = 0</math>
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<math>\Rightarrow (3c-d)(c-3d)=0</math>
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<math>\Rightarrow c=3d, c=\frac{d}{3}</math>
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We established earlier that <math>c>d</math>, so <math>c=3d</math>. Plugging this into the intial equations yields
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<math>\frac{16d^2}{b-a} = 72</math>
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and
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<math>\frac{4d^2}{b-a} = 18</math>
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Solving for <math>d</math>, we get
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<math>d = 3\sqrt{\frac{b-a}{2}}</math>
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We want the sum of <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math>. We can now rewrite this
 +
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<math>a + b + 12\sqrt{\frac{b-a}{2}}</math>
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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
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<math>\frac{b-a}{2} = k^2</math>
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 +
 +
where <math>k</math> is some positive integer.
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Rearranging, we get
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<math>b = a + 2k^2</math>
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Now we can rewrite the sum as
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<math>a + a + 2k^2 +12k</math>.
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Since both <math>a</math> and <math>k</math> must be at least <math>1</math>, the minimum value is
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<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 $y=ax+c$, $y=ax+d$, $y=bx+c$, 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 1

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, or the ratio of side lengths, is $2\times$). The area of the 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 $c = 3d$:

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

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 $a+b+c+d = 3 + 1 + 9 + 3 = \boxed{\mathbf{(D)} 16}$.

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 $c,d,(b-a)x+c,(b-a)x+d$ and $c,-d,(b-a)x+c,(b-a)x-d$. Now, the area of the parallelogram contained by is the former is equal to the area of a rectangle with sides $d-c$ and $\frac{d-c}{b-a}$, $\frac{(d-c)^2}{b-a}=18$, and the area contained by the latter is $\frac{(c+d)^2}{b-a}=72$. Thus, $d=3c$ and $b-a$ must be even if the former quantity is to equal $18$. $c^2=18(b-a)$ so $c$ is a multiple of $3$. Putting this all together, the minimal solution for $(a,b,c,d)=(3,1,3,9)$, so the sum is $\boxed{\textbf{(D)} 16}$.

Solution 3

Let $a$ and $b$ be the slopes of the lines such that $b > a$ (i.e. the line $bx+c$ is steeper than $ax+c$) and $c > d$ (i.e. the point $(0, c)$ is higher than the point $(0, d)$. 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 $\frac{bh}{2}$, but since a parallelogram is two such triangles, the area becomes $bh$.

Let $b_1$ and $h_1$ denote the base length and height, respectively, of the triangles pertaining to the smaller parallelogram and $b_2$ and $h_2$ denote those of the larger parallelogram. Notice that $b_1$ is simple the distance from $(0, d)$ to $(0, c)$, or $(c-d)$. Also notice that $h_1$ is the distance from the $x$-axis to the intersection of lines $ax+c$ and $bx+d$. This is equivalent to the value of the $x$-coordinate of intersection, so we solve for $x$:


$ax+c=bx+d$

$\Rightarrow bx-ax = c-d$

$\Rightarrow (b-a)x = c-d$

$\Rightarrow x = \frac{c-d}{b-a}$.


The area of the smaller parallelogram is $b_1*h_1$, or


$(c-d) * \frac{c-d}{b-a}$


$\Rightarrow \frac{(c-d)^2}{b-a}$.


$b_2$ is the distance from $(0, -d)$ to $(0, c)$, or $(c+d)$. $h_2$ is the $x$-coordinate of the intersection of the lines $ax+c$ and $bx-d$. Again, we solve for x:


$ax+c=bx-d$

$\Rightarrow bx-ax = c+d$

$\Rightarrow (b-a)x = c+d$

$\Rightarrow x = \frac{c+d}{b-a}$.


The area of the larger parallelogram is $b-1*h_1$, or


$(c+d) * \frac{c+d}{b-a}$


$\Rightarrow \frac{(c+d)^2}{b-a}$.


The areas of the parallelograms are given to us: $18$ and $72$. Therefore we can set up a ratio:


$\frac{18}{72} = \frac{\frac{(c-d)^2}{b-a}}{\frac{(c+d)^2}{b-a}}$

$\Rightarrow 18(c+d)^2 = 72(c-d)^2$

$\Rightarrow (c+d)^2 = 4(c-d)^2$

$\Rightarrow c^2 + 2cd + d^2 = 4c^2 - 8cd + 4d^2$

$\Rightarrow 3c^2 - 10cd +3d^2 = 0$

$\Rightarrow (3c-d)(c-3d)=0$

$\Rightarrow c=3d, c=\frac{d}{3}$


We established earlier that $c>d$, so $c=3d$. Plugging this into the intial equations yields


$\frac{16d^2}{b-a} = 72$


and


$\frac{4d^2}{b-a} = 18$

Solving for $d$, we get


$d = 3\sqrt{\frac{b-a}{2}}$


We want the sum of $a$, $b$, $c$, and $d$. We can now rewrite this


$a + b + 12\sqrt{\frac{b-a}{2}}$


We are told that $a$, $b$, $c$, and $d$ are all positive integers. Therefore the value under the radical must be a perfect square greater than 0. We can rewrite this


$\frac{b-a}{2} = k^2$


where $k$ is some positive integer.

Rearranging, we get


$b = a + 2k^2$

Now we can rewrite the sum as


$a + a + 2k^2 +12k$.


Since both $a$ and $k$ must be at least $1$, the minimum value is


$1 + 1 + 2(1)^2 + 12(1) = 16 \Rightarrow \boxed{\text{D}}$.

See also

2007 AMC 12B (ProblemsAnswer KeyResources)
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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