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

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Splitting: Drop perpendiculars from <math>B</math> and <math>C</math> to the x-axis to divide the quadrilateral into triangles and trapezoids, and so the area is <math>1 + 5 + \dfrac{3}{2} = \dfrac{15}{2}.</math>
 
Splitting: Drop perpendiculars from <math>B</math> and <math>C</math> to the x-axis to divide the quadrilateral into triangles and trapezoids, and so the area is <math>1 + 5 + \dfrac{3}{2} = \dfrac{15}{2}.</math>
  
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Shoelace Theorem: The area is half of <math>|1 \cdot 3 - 2 \cdot 3 - 3 \cdot 4| = 15</math>, or <math>\dfrac{15}{2}</math>.
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Shoelace Method: The area is half of <math>|1 \cdot 3 - 2 \cdot 3 - 3 \cdot 4| = 15</math>, or <math>\dfrac{15}{2}</math>.
  
 
<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>.
 
<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>.

Revision as of 07:08, 7 February 2017

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, we shall find the area of quadrilateral $ABCD$. This can be done in any of three ways:

Pick's Theorem: $[ABCD] = I + \dfrac{B}{2} - 1 = 5 + \dfrac{7}{2} - 1 = \dfrac{15}{2}.$

Splitting: Drop perpendiculars from $B$ and $C$ to the x-axis to divide the quadrilateral into triangles and trapezoids, and so the area is $1 + 5 + \dfrac{3}{2} = \dfrac{15}{2}.$

Shoelace Method: The area is half of $|1 \cdot 3 - 2 \cdot 3 - 3 \cdot 4| = 15$, or $\dfrac{15}{2}$.

$[ABCD] = \frac{15}{2}$. 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 $E = (\frac{27}{8}, \frac{15}{8})$. $27 + 15 + 8 + 8 = \boxed{\textbf{(B) }58}$

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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