Difference between revisions of "2007 AMC 12A Problems/Problem 19"
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We can use the shoelace theorem to first find that the y-coordinate of <math>A</math> can be <math>-18</math> or <math>18</math>. Then we can apply shoelace again to find the <math>4</math> possible x-coordinates, namely <math>-1274</math>, <math>-1238</math>, <math>1838</math>, and <math>1874</math>. Adding these up, we get <math>1200 \Longrightarrow \mathrm{(E)}</math>. | We can use the shoelace theorem to first find that the y-coordinate of <math>A</math> can be <math>-18</math> or <math>18</math>. Then we can apply shoelace again to find the <math>4</math> possible x-coordinates, namely <math>-1274</math>, <math>-1238</math>, <math>1838</math>, and <math>1874</math>. Adding these up, we get <math>1200 \Longrightarrow \mathrm{(E)}</math>. | ||
Latest revision as of 11:09, 3 November 2024
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
Solution 4 (intense bashing, similar to Solution 3)
We can use the shoelace theorem to first find that the y-coordinate of can be or . Then we can apply shoelace again to find the possible x-coordinates, namely , , , and . Adding these up, we get .
~ erinb28lms
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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