Difference between revisions of "2007 AMC 12A Problems/Problem 19"

Revision as of 00:03, 19 October 2020

Problem

Triangles $ABC$ and $ADE$ have areas $2007$ and $7002,$ respectively, with $B = (0,0),$ $C = (223,0),$ $D = (680,380),$ and $E = (689,389).$ What is the sum of all possible x coordinates of $A$ ?

$\mathrm{(A)}\ 282 \qquad \mathrm{(B)}\ 300 \qquad \mathrm{(C)}\ 600 \qquad \mathrm{(D)}\ 900 \qquad \mathrm{(E)}\ 1200$

Solution

Solution 1

From $k = [ABC] = \frac 12bh$ , we have that the height of $\triangle ABC$ is $h = \frac{2k}{b} = \frac{2007 \cdot 2}{223} = 18$ . Thus $A$ lies on the lines $y = \pm 18 \quad \mathrm{(1)}$ .

$DE = 9\sqrt{2}$ 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 $x = \pm 18 + 300 \pm 1556 = 4(300) = 1200 \Longrightarrow \mathrm{(E)}$ .

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 $\overline{BC}$ and $\overline{DE}$ , which can easily be calculated to be $(300,0)$ . Now the sum of the x-coordinates is just $4(300) = 1200$ .

@@ Line 12: / Line 12: @@
 <math>DE = 9\sqrt{2}</math> using 45-45-90 triangles, so in <math>\triangle ADE</math> we have that <math>h = \frac{2 \cdot 7002}{9\sqrt{2}} = 778\sqrt{2}</math>. The slope of <math>DE</math> is <math>1</math>, so the equation of the line is <math>y = x + b \Longrightarrow b = (380) - (680) = -300 \Longrightarrow y = x - 300</math>. The point <math>A</math> lies on one of two [[parallel]] lines that are <math>778\sqrt{2}</math> units away from <math>\overline{DE}</math>. Now take an arbitrary point on the line <math>\overline{DE}</math> 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 <math>\triangle</math>, so the straight line down has a length of <math>778\sqrt{2} \cdot \sqrt{2} = 1556</math>. Now we note that the [[y-intercept]] of the parallel lines is either <math>1556</math> units above or below the y-intercept of line <math>\overline{DE}</math>; hence the equation of the parallel lines is <math>y = x - 300 \pm 1556 \Longrightarrow x = y + 300 \pm 1556 \quad \mathrm{(2)}</math>.
-We just need to find the intersections of these two lines and sum up the values of the [[x-coordinates]]. Substituting the <math>\mathrm{(1)}</math> into <math>\mathrm{(2)}</math>, we get <math>x = \pm 18 + 300 \pm 1556 = 4(300) = 1200 \Longrightarrow \mathrm{(E)}</math>.
+We just need to find the intersections of these two lines and sum up the values of the x-coordinates. Substituting the <math>\mathrm{(1)}</math> into <math>\mathrm{(2)}</math>, we get <math>x = \pm 18 + 300 \pm 1556 = 4(300) = 1200 \Longrightarrow \mathrm{(E)}</math>.
 === Solution 2 ===

2007 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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