Difference between revisions of "2013 AMC 10A Problems/Problem 18"
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==Solution== | ==Solution== | ||
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− | + | First, various area formulas (shoelace, splitting, etc) allow us to find that <math>[ABCD] = \frac{15}{2}</math>. Therefore, each equal piece that the line separates <math>ABCD</math> into must have an area of <math>\frac{15}{4}</math>. | |
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− | + | Call the point where the line through <math>A</math> intersects <math>\overline{CD}</math> <math>E</math>. We know that <math>[ADE] = \frac{15}{4} = \frac{bh}{2}</math>. Furthermore, we know that <math>b = 4</math>, as <math>AD = 4</math>. Thus, solving for <math>h</math>, we find that <math>2h = \frac{15}{4}</math>, so <math>h = \frac{15}{8}</math>. This gives that the y coordinate of E is <math>\frac{15}{8}</math>. | |
− | 27+8 | + | |
+ | Line CD can be expressed as <math>y = -3x+12</math>, so the <math>x</math> coordinate of E satisfies <math>\frac{15}{8} = -3x + 12</math>. Solving for <math>x</math>, we find that <math>x = \frac{27}{8}</math>. | ||
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+ | From this, we know that <math>E = (\frac{27}{8}, \frac{15}{8})</math>. <math>27 + 15 + 8 + 8 = 58</math>, <math>\textbf{(A)}</math>. | ||
==See Also== | ==See Also== |
Revision as of 12:02, 8 February 2013
Problem
Let points , , , and . Quadrilateral is cut into equal area pieces by a line passing through . This line intersects at point , where these fractions are in lowest terms. What is ?
Solution
First, various area formulas (shoelace, splitting, etc) allow us to find that . Therefore, each equal piece that the line separates into must have an area of .
Call the point where the line through intersects . We know that . Furthermore, we know that , as . Thus, solving for , we find that , so . This gives that the y coordinate of E is .
Line CD can be expressed as , so the coordinate of E satisfies . Solving for , we find that .
From this, we know that . , .
See Also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 10 Problems and Solutions |
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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 |