Difference between revisions of "2007 AMC 12A Problems/Problem 19"
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=== Solution 2 === | === 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>. | ||
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+ | === 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> | ||
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+ | - Zephyrica | ||
== See Also == | == See Also == |
Latest revision as of 10:10, 8 July 2021
Contents
Problem
Triangles and have areas and respectively, with and What is the sum of all possible x coordinates of ?
Solutions
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 |
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