2007 AMC 12A Problems/Problem 19

Problem

$DE = 9\sqrt{2}$ using 45-45-90 triangles, so in $\triangle ADE$ we have that $h = \frac{2 \cdot 7002}{9\sqrt{2}} = 778\sqrt{2}$ . The slope of $DE$ is $1$ , so the equation of the line is $y = x + b \Longrightarrow b = (380) - (680) = -300 \Longrightarrow y = x - 300$ . The point $A$ lies on one of two parallel lines that are $778\sqrt{2}$ units away from $\overline{DE}$ . Now take an arbitrary point on the line $\overline{DE}$ 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 $\triangle$ , so the straight line down has a length of $778\sqrt{2} \cdot \sqrt{2} = 1556$ . Now we note that the y-intercept of the parallel lines is either $1556$ units above or below the y-intercept of line $\overline{DE}$ ; hence the equation of the parallel lines is $y = x - 300 \pm 1556 \Longrightarrow x = y + 300 \pm 1556 \quad \mathrm{(2)}$ .

We just need to find the intersections of these two lines and sum up the values of the x-coordinates. Substituting the $\mathrm{(1)}$ into $\mathrm{(2)}$ , we get $x = \pm 18 + 300 \pm 1556 = 4(300) = 1200 \Longrightarrow \mathrm{(E)}$ .

We are finding the intersection of two parallel lines, which will form a parallelogram. The centroid of this parallelogram is just the intersection of $\overline{BC}$ amd $\overline{DE}$ , which can easily be calculated to be $(300,0)$ . Now the sum of the x-coordinates is just $4(300) = 1200$ .