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> | ||
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==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>. |
+ | |||
+ | ==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 | + | 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
:
![\[\frac{1}{2} (c-d)\left(\frac{c-d}{a-b}\right) = 9 \quad \Longrightarrow \quad 2d^2 = 9(a-b)\]](http://latex.artofproblemsolving.com/5/1/e/51edc5dfccf7d46f4df539eb3d687c0b4d8f4c04.png)
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 |
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