Difference between revisions of "2007 AMC 12A Problems/Problem 19"
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=== Solution 1 === | === Solution 1 === | ||
− | From <math> | + | From <math>k = [ABC] = \frac 12bh</math>, we have that the height of <math>\triangle ABC</math> is <math>h = \frac{2k}{b} = \frac{2007 \cdot 2}{223} = 18</math>. Thus <math>A</math> lies on the lines <math>y = \pm 18 \quad \mathrm{(1)}</math>. |
<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>. | <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>. |
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=== Solution 2 === | === Solution 2 === | ||
We are finding the intersection of two [[parallel]] lines, which will form a [[parallelogram]]. The [[centroid]] of this parallelogram is just the intersection of <math>\overline{BC}</math> amd <math>\overline{DE}</math>, which can easily be calculated to be <math>(300,0)</math>. Now the sum of the x-coordinates is just <math>4(300) = 1200</math>. | We are finding the intersection of two [[parallel]] lines, which will form a [[parallelogram]]. The [[centroid]] of this parallelogram is just the intersection of <math>\overline{BC}</math> amd <math>\overline{DE}</math>, which can easily be calculated to be <math>(300,0)</math>. Now the sum of the x-coordinates is just <math>4(300) = 1200</math>. |
Revision as of 17:41, 20 September 2007
Problem
Triangles and have areas and respectively, with and What is the sum of all possible x-coordinates of ?
$\matrhm{(A)}\ 282 \qquad \mathrm{(B)}\ 300 \qquad \mathrm{(C)}\ 600 \qquad \mathrm{(D)}\ 900 \qquad \mathrm{(E)}\ 1200$ (Error compiling LaTeX. ! Undefined control sequence.)
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 parallel lines, which will form a parallelogram. The centroid of this parallelogram is just the intersection of amd , 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 |