Difference between revisions of "2007 AMC 12A Problems/Problem 19"
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<math>\mathrm{(A)}\ 282 \qquad \mathrm{(B)}\ 300 \qquad \mathrm{(C)}\ 600 \qquad \mathrm{(D)}\ 900 \qquad \mathrm{(E)}\ 1200</math> | <math>\mathrm{(A)}\ 282 \qquad \mathrm{(B)}\ 300 \qquad \mathrm{(C)}\ 600 \qquad \mathrm{(D)}\ 900 \qquad \mathrm{(E)}\ 1200</math> | ||
− | == | + | ==Solution== |
[[Image:2007_12A_AMC-19.png]] | [[Image:2007_12A_AMC-19.png]] | ||
− | + | ==Solution 1== | |
From <math>k = [ABC] = \frac 12bh</math>, we have that the height of <math>\triangle ABC</math> is <math>h = \frac{2k}{b} = \frac{2007 \cdot 2}{223} = 18</math>. Thus <math>A</math> lies on the lines <math>y = \pm 18 \quad \mathrm{(1)}</math>. | From <math>k = [ABC] = \frac 12bh</math>, we have that the height of <math>\triangle ABC</math> is <math>h = \frac{2k}{b} = \frac{2007 \cdot 2}{223} = 18</math>. Thus <math>A</math> lies on the lines <math>y = \pm 18 \quad \mathrm{(1)}</math>. | ||
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We just need to find the intersections of these two lines and sum up the values of the x-coordinates. Substituting the <math>\mathrm{(1)}</math> into <math>\mathrm{(2)}</math>, we get <math>x = \pm 18 + 300 \pm 1556 = 4(300) = 1200 \Longrightarrow \mathrm{(E)}</math>. | We just need to find the intersections of these two lines and sum up the values of the x-coordinates. Substituting the <math>\mathrm{(1)}</math> into <math>\mathrm{(2)}</math>, we get <math>x = \pm 18 + 300 \pm 1556 = 4(300) = 1200 \Longrightarrow \mathrm{(E)}</math>. | ||
− | + | ==Solution 2== | |
We are finding the intersection of two pairs of [[parallel]] lines, which will form a [[parallelogram]]. The [[centroid]] of this parallelogram is just the intersection of <math>\overline{BC}</math> and <math>\overline{DE}</math>, which can easily be calculated to be <math>(300,0)</math>. Now the sum of the x-coordinates is just <math>4(300) = 1200</math>. | We are finding the intersection of two pairs of [[parallel]] lines, which will form a [[parallelogram]]. The [[centroid]] of this parallelogram is just the intersection of <math>\overline{BC}</math> and <math>\overline{DE}</math>, which can easily be calculated to be <math>(300,0)</math>. Now the sum of the x-coordinates is just <math>4(300) = 1200</math>. | ||
− | + | ==Solution 3 (Bashing but very straightforward)== | |
After we compute that the y-value can be either <math>y = \pm 18</math> and realize there are four total values (each pair being equally spaced on their respective y-lines of <math>\pm 18</math>), we can use an easy application of the [[Shoelace Theorem]] to figure out the values of X. Since we already know the two distances (positive y and negative y) will be the same, then we can simply plug in y=18, compute the sum of the two corresponding x-values and multiply it by two to get our answer which is <math>2(600) = 1200 \Longrightarrow \mathrm{(E)}</math> | After we compute that the y-value can be either <math>y = \pm 18</math> and realize there are four total values (each pair being equally spaced on their respective y-lines of <math>\pm 18</math>), we can use an easy application of the [[Shoelace Theorem]] to figure out the values of X. Since we already know the two distances (positive y and negative y) will be the same, then we can simply plug in y=18, compute the sum of the two corresponding x-values and multiply it by two to get our answer which is <math>2(600) = 1200 \Longrightarrow \mathrm{(E)}</math> | ||
Revision as of 01:18, 31 January 2023
Contents
Problem
Triangles and
have areas
and
respectively, with
and
What is the sum of all possible x coordinates of
?
Solution
Solution 1
From , we have that the height of
is
. Thus
lies on the lines
.
using 45-45-90 triangles, so in
we have that
. The slope of
is
, so the equation of the line is
. The point
lies on one of two parallel lines that are
units away from
. Now take an arbitrary point on the line
and draw the perpendicular to one of the parallel lines; then draw a line straight down from the same arbitrary point. These form a 45-45-90
, so the straight line down has a length of
. Now we note that the y-intercept of the parallel lines is either
units above or below the y-intercept of line
; hence the equation of the parallel lines is
.
We just need to find the intersections of these two lines and sum up the values of the x-coordinates. Substituting the into
, we get
.
Solution 2
We are finding the intersection of two pairs of parallel lines, which will form a parallelogram. The centroid of this parallelogram is just the intersection of and
, which can easily be calculated to be
. Now the sum of the x-coordinates is just
.
Solution 3 (Bashing but very straightforward)
After we compute that the y-value can be either and realize there are four total values (each pair being equally spaced on their respective y-lines of
), we can use an easy application of the Shoelace Theorem to figure out the values of X. Since we already know the two distances (positive y and negative y) will be the same, then we can simply plug in y=18, compute the sum of the two corresponding x-values and multiply it by two to get our answer which is
- Zephyrica
See Also
2007 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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.