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2013 AMC 10A Problems/Problem 18: Difference between revisions

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==Solution==
==Solution==
The area of the quadrilateral ABCD is 7.5, therefore the area of each of the equal area pieces is 3.75.
 
Since the dividing line passes through point A, the lower piece must be a triangle with base AD.
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>.
Since the formula for the area of a triangle is b*h/2 and b=4, h must equal 15/8.
 
The point on line CD with y-coordinate 15/8 is (27/8, 15/8).
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+15+8=58, therefore the answer is (B).
 
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>.
 
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 $A = (0, 0)$, $B = (1, 2)$, $C=(3, 3)$, 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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