2007 AMC 10A Problem 18

The vertices $A, B, G, H$ form a $4 \times 12$ rectangle: $AB = 4$ is horizontal, $AH = 12$ and $BG = 12$ are vertical, and $C$ lies on segment $BG$ with $BC = 4$, $CG = 8$.

Locate $M$. Segments $AH$ and $CG$ are both vertical, hence parallel, so triangles $AMH$ and $GMC$ are similar (vertical angles at $M$, alternate interior angles elsewhere). Their ratio is $AH : GC = 12 : 8 = 3 : 2$, so $M$'s distances to the parallel lines $AH$ and $BG$ are in ratio $3 : 2$. Those distances total $4$, so $M$ is $\frac{8}{5}$ from line $BG$.

Now decompose. Diagonal $AG$ cuts the rectangle into two triangles; $ABG$ has area $\frac12 \cdot 4 \cdot 12 = 24$. Quadrilateral $ABCM$ is triangle $ABG$ with triangle $CMG$ removed. Triangle $CMG$ has base $CG = 8$ and height $\frac85$ (the distance from $M$ to line $BG$), so its area is $\frac12 \cdot 8 \cdot \frac85 = \frac{32}{5}$.

$$[ABCM] = 24 - \frac{32}{5} = \frac{88}{5}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{88}{5}}$.

Two crossing segments between parallel lines create a pair of similar triangles whose ratio is read off from the parallel segments' lengths, which pins down the crossing point. Irregular quadrilaterals are then handled by subtraction from a triangle or rectangle whose area is trivial. Recognize the "hourglass" (two triangles meeting at a vertex between parallels) and its ratio is the key.

Coordinates with $H = (0,0)$: $A = (0,12)$, $B = (4,12)$, $C = (4,8)$, $G = (4,0)$. Line $AG$ is $y = 12 - 3x$ and line $HC$ is $y = 2x$; they meet at $M = \left(\frac{12}{5}, \frac{24}{5}\right)$. Shoelace on $A, B, C, M$ gives $\frac12 \left| -48 - 16 + 0 + \frac{144}{5} \right| = \frac{88}{5}$.

• Assuming $M$ is the midpoint of $AG$ or of $CH$, or computing the area of triangle $ACM$ alone and forgetting the rest of the quadrilateral.
• Using the ratio $AH : CG$ as $12 : 4$ (confusing $CG$ with $BC$), which places $M$ at distance $1$ from $BG$ and gives area $20$.