imc

First, we shall find the area of quadrilateral $ABCD$. This can be done in any of three ways: Pick's Theorem: $[ABCD]=I+\frac{B}{2}-1=5+\frac{7}{2}-1=\frac{15}{2}.$ Splitting: Drop perpendiculars from $B$ and $C$ to the x-axis to divide the quadrilateral into triangles and trapezoids, and so the area is $1+5+\frac{3}{2}=\frac{15}{2}.$ Shoelace Theorem: The area is half of $|1\cdot 3-2\cdot 3-3\cdot 4|=15$, or $\frac{15}{2}$. $[ABCD]=\frac{15}{2}$. 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 $E=\left(\frac{27}{8},\frac{15}{8}\right)$. $27+15+8+8=\boxed{58}$