# 2007 AMC 10A Problems/Problem 18

## Problem

Consider the $12$-sided polygon $ABCDEFGHIJKL$, as shown. Each of its sides has length $4$, and each two consecutive sides form a right angle. Suppose that $\overline{AG}$ and $\overline{CH}$ meet at $M$. What is the area of quadrilateral $ABCM$?

$[asy] unitsize(13mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2); pair M=intersectionpoints(A--G,H--C)[0]; draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle); draw(A--G); draw(H--C); dot(M); label("A",A,NW); label("B",B,NE); label("C",C,NE); label("D",D,NE); label("E",Ep,SE); label("F",F,SE); label("G",G,SE); label("H",H,SW); label("I",I,SW); label("J",J,SW); label("K",K,NW); label("L",L,NW); label("M",M,W); [/asy]$

$\text{(A)}\ \frac {44}{3}\qquad \text{(B)}\ 16 \qquad \text{(C)}\ \frac {88}{5}\qquad \text{(D)}\ 20 \qquad \text{(E)}\ \frac {62}{3}$

## Solution

### Solution 1

We can obtain the solution by calculating the area of rectangle $ABGH$ minus the combined area of triangles $\triangle AHG$ and $\triangle CGM$.

We know that triangles $\triangle AMH$ and $\triangle GMC$ are similar because $\overline{AH} \parallel \overline{CG}$. Also, since $\frac{AH}{CG} = \frac{3}{2}$, the ratio of the distance from $M$ to $\overline{AH}$ to the distance from $M$ to $\overline{CG}$ is also $\frac{3}{2}$. Solving with the fact that the distance from $\overline{AH}$ to $\overline{CG}$ is 4, we see that the distance from $M$ to $\overline{CG}$ is $\frac{8}{5}$.

The area of $\triangle AHG$ is simply $\frac{1}{2} \cdot 4 \cdot 12 = 24$, the area of $\triangle CGM$ is $\frac{1}{2} \cdot \frac{8}{5} \cdot 8 = \frac{32}{5}$, and the area of rectangle $ABGH$ is $4 \cdot 12 = 48$.

Taking the area of rectangle $ABGH$ and subtracting the combined area of $\triangle AHG$ and $\triangle CGM$ yields $48 - (24 + \frac{32}{5}) = \boxed{\frac{88}{5}}\ \text{(C)}$.

### Solution 2

$[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2); pair M=intersectionpoints(A--G,H--C)[0]; pair Z=(2.5,3); draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle); draw(A--G); draw(H--C); draw(B--Z--C); draw(C--F); dot(M); label("A",A,NW); label("B",B,N); label("C",C,SE); label("D",D,NE); label("E",Ep,SE); label("F",F,SE); label("G",G,SE); label("H",H,SW); label("I",I,SW); label("J",J,SW); label("K",K,NW); label("L",L,NW); label("M",M,W); label("N",Z,NE); [/asy]$

Extend $AB$ and $CH$ and call their intersection $N$.

The triangles $CBN$ and $CGH$ are clearly similar with ratio $1:2$, hence $BN=2$ and thus $AN=6$. The area of the triangle $BCN$ is $\frac{2\cdot 4}2 = 4$.

The triangles $MAN$ and $MGH$ are similar as well, and we now know that the ratio of their dimensions is $AN:GH = 6:4 = 3:2$.

Draw altitudes from $M$ onto $AN$ and $GH$, let their feet be $M_1$ and $M_2$. We get that $MM_1 : MM_2 = 3:2$. Hence $MM_1 = \frac 35 \cdot 12 = \frac {36}5$. (An alternate way is by seeing that the set-up AHGCM is similar to the 2 pole problem (http://www.artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_30). Therefore, $MM_2$ must be $\frac{1}{\frac{1}{8}+\frac{1}{12}} = \frac{24}{5}$, by the harmonic mean. Thus, $MM_1$ must be $\frac{36}{5}$.)

Then the area of $AMN$ is $\frac 12 \cdot AN \cdot MM_1 = \frac{108}5$, and the area of $ABCM$ can be obtained by subtracting the area of $BCN$, which is $4$. Hence the answer is $\frac{108}5 - 4 = \boxed{\frac{88}5}$.

## Solution 3

We can use coordinates to solve this. Let $H=(0,0).$ Thus, we have $A=(0,12), C=(4,8), G=(4,0).$ Therefore, $AG$ has equation $-3x+12=y$ and $HC$ has equation $2x=y.$ Solving, we have $M=(12/5,24/5).$ Using Shoelace Theorem (or you could connect $LC$ and solve for the resulting triangle + trapezoid areas), we find $[ABCM]=\boxed{\mathrm{(C) \ }\dfrac{88}{5}}.$