2013 AMC 12A Problems/Problem 13

If you have graph paper, use Pick's Theorem to quickly and efficiently find the area. If not, just find the area of the quadrilateral 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 part of the quadrilateral makes a triangle with base $4$ , so we can deduce that the height of the triangle must be $\frac{15}{8}$ in order for the 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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2013 AMC 12A Problems/Problem 13