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

## 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 $\left(\frac{p}{q}, \frac{r}{s}\right)$, 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 1

$[asy] size(8cm); pair A, B, C, D, E, EE; A = (0,0); B = (1,2); C = (3,3); D = (4,0); E = (27/8,15/8); EE = (27/8,0); draw(A--B--C--D--A--E); draw(E--EE,linetype("8 8")); dot(A); dot(B); dot(C); dot(D); dot(E); draw(rightanglemark(E,EE,D,4)); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,NE); label("4",(A+D)/2,S); label("\frac{27}{8}",(A+EE)/2,S); label("\frac{15}{8}",(E+EE)/2,W); [/asy]$

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 $E = \left(\frac{27}{8}, \frac{15}{8}\right)$. $27 + 15 + 8 + 8 = \boxed{\textbf{(B) }58}$

## Solution 2

$[asy] size(8cm); pair A, B, C, D, E, F; A = (0,0); B = (1,2); C = (3,3); D = (4,0); E = (27/8,15/8); F = (27/8,0); draw(A--B--C--D--A--E); draw(E--F,linetype("8 8")); dot(A); dot(B); dot(C); dot(D); dot(E); draw(rightanglemark(E,F,D,4)); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,NE); label("F",F,S); label("4",(A+D)/2,S); label("x",(A+F)/2,S); label("\frac{15}{8}",(E+F)/2,W); [/asy]$

Let the point where the altitude from $E$ to $\overline{AD}$ be labeled $F$. Following the steps above, you can find that the height of $\triangle ADE$ is $\frac{15}{8}$, and from there split the base into two parts, $x$, and $4-x$, such that $x$ is the segment from the origin to the point $F$, and $4-x$ is the segment from point $F$ to point $D$. Then, by the Pythagorean Theorem, $x=\frac{27}{8}$, and the answer is $\boxed{\textbf{(B) }58}$

## See Also

 2013 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 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 byProblem 12 Followed byProblem 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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