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2013 AMC 12A Problems/Problem 13

Revision as of 22:43, 6 February 2013 by Indianninja707 (talk | contribs)

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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