2013 AMC 12A Problems/Problem 13

Revision as of 01:41, 7 February 2013 by Epicwisdom (talk | contribs) (moved 2013 AMC 12A Problems/Problems 13 to 2013 AMC 12A Problems/Problem 13: Wrong title ("Problems 13"))

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 $B$.

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