2007 AMC 12A Problems/Problem 19
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
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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