Difference between revisions of "2007 AMC 10A Problems/Problem 18"

Latest revision as of 23:17, 19 May 2023

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 $48 - \left(24 + \frac{32}{5}\right) = \boxed{\frac{88}{5}}\ \text{(C)}$ .

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

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

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 $M=(\frac {12}{5},\frac {24}{5}).$ Using the Shoelace Theorem (or you could connect $LC$ and solve for the resulting triangle + trapezoid areas), we find $[ABCM]=\boxed{\mathrm{(C) \ }\dfrac{88}{5}}.$

~(minor edits by Arcticturn)

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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