Difference between revisions of "2007 AMC 12B Problems/Problem 20"

Latest revision as of 20:25, 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

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

@@ Line 4: / Line 4: @@
 <math>\mathrm {(A)} 13\qquad \mathrm {(B)} 14\qquad \mathrm {(C)} 15\qquad \mathrm {(D)} 16\qquad \mathrm {(E)} 17</math>
-==Solution==
+==Solution 1==
-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>. From the 4 corners of the parallelogram, we have that <math>(c-d)\left(\frac{bc-ad}{a-b}\right)=18</math>. Substituting <math>c=3d</math>,
+<!-- <center><asy>
+pathpen = linewidth(0.7);
+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);
+</asy></center> -->
+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>
+\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 = \boxed{\mathbf{(D)} 16}</math>.
-<math>\begin{eqnarray*}
+==Solution 2==
-d\left(\frac{3bd-ad}{a-b}\right) &=& 18 \\
+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>.
-d\left(\frac{bd-ad}{a-b}+\frac{2bd}{a-b}\right) &=& 9 \\
-d\left(-d+\frac{2bd}{a-b}\right) &=& 9 \\
--d^2 + \frac{2b}{a-b}d^2 &=& 9 \\
-\frac{2b}{a-b}d^2 &=& 9 + d^2
-\end{eqnarray*}</math>
-Taking <math>d=1</math>, we have that <math>\frac{2b}{a-b} = 10</math>. This gives us that <math>6b=5a</math>. Taking the minimum values of <math>a=6,b=5</math> we have that <math>a+b+c+d=6+5+3+1=15 \Rightarrow \mathrm {(C)}</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==
-{{AMC12 box|year=2007|ab=B|num-b=21|num-a=23}}
+{{AMC12 box|year=2007|ab=B|num-b=19|num-a=21}}
 [[Category:Introductory Geometry Problems]]
+{{MAA Notice}}

2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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