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Difference between revisions of "2013 AMC 10A Problems/Problem 18"

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==Solution==
 
==Solution==
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The area of the quadrilateral ABCD is 7.5, therefore the area of each of the equal area pieces is 3.75.
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Since the dividing line passes through point A, the lower piece must be a triangle with base AD.
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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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Since the formula for the area of a triangle is b*h/2 and b=4, h must equal 15/8.
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The point on line CD with y-coordinate 15/8 is (27/8, 15/8).
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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>.
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27+8+15+8=58, therefore the answer is (B).
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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 13:02, 8 February 2013

Problem

Let points $A = (0, 0)$, $B = (1, 2)$, $C=(3, 3)$, and $D = (4, 0)$. Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$. This line intersects $\overline{CD}$ at point $(\frac{p}{q}, \frac{r}{s})$, where these fractions are in lowest terms. What is $p+q+r+s$?


$\textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75$

Solution

First, various area formulas (shoelace, splitting, etc) allow us to find that $[ABCD] = \frac{15}{2}$. Therefore, each equal piece that the line separates $ABCD$ into must have an area of $\frac{15}{4}$.

Call the point where the line through $A$ intersects $\overline{CD}$ $E$. We know that $[ADE] = \frac{15}{4} = \frac{bh}{2}$. Furthermore, we know that $b = 4$, as $AD = 4$. Thus, solving for $h$, we find that $2h = \frac{15}{4}$, so $h = \frac{15}{8}$. This gives that the y coordinate of E is $\frac{15}{8}$.

Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\frac{15}{8} = -3x + 12$. Solving for $x$, we find that $x = \frac{27}{8}$.

From this, we know that $E = (\frac{27}{8}, \frac{15}{8})$. $27 + 15 + 8 + 8 = 58$, $\textbf{(A)}$.

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
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