Difference between revisions of "2013 AMC 12A Problems/Problem 13"

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[[Category:Introductory Geometry Problems]]

Revision as of 09:56, 7 April 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 $\left (\frac{p}{q}, \frac{r}{s} \right )$, 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

Solution 1

If you have graph paper, use Pick's Theorem to quickly and efficiently find the area of the quadrilateral. If not, just find the area by other methods.

Pick's Theorem states that

$A$ = $I$ $+$ $\frac{B}{2}$ - $1$, where $I$ is the number of lattice points in the interior of the polygon, and $B$ is the number of lattice points on the boundary of the polygon.

In this case,

$A$ = $5$ $+$ $\frac{7}{2}$ - $1$ = $7.5$

so

$\frac{A}{2}$ = $3.75$

The bottom half of the quadrilateral makes a triangle with base $4$ and half the total area, so we can deduce that the height of the triangle must be $\frac{15}{8}$ in order for its area to be $3.75$. This height is the y coordinate of our desired intersection point.


Note that segment CD lies on the line $y = -3x + 12$. Substituting in $\frac{15}{8}$ for y, we can find that the x coordinate of our intersection point is $\frac{27}{8}$.

Therefore the point of intersection is ($\frac{27}{8}$, $\frac{15}{8}$), and our desired result is $27+8+15+8=58$, which is $B$.

See also

2013 AMC 12A (ProblemsAnswer KeyResources)
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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