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> | ||
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[[Image:2007_12A_AMC-19.png]] | [[Image:2007_12A_AMC-19.png]] | ||
Revision as of 16:39, 19 December 2020
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 .
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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