2007 AMC 10A Problem 18

Below is the professionally curated solution for Problem 18 of the 2007 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 10A solutions, or check the answer key.

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Concepts:coordinate geometry shoelace formula

Difficulty rating: 1790

18.

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 $ABCM?$

$\dfrac{44}{3}$

$16$

$\dfrac{88}{5}$

$20$

$\dfrac{62}{3}$

Solution:

Put the figure on coordinates with $A = (-2, 6),$ $B = (2, 6),$ $C = (2, 2),$ $G = (2, -6),$ and $H = (-2, -6).$

Line $AG$ is $y = -3x,$ and line $CH$ is $y = 2x - 2.$

Their intersection is $M = \left(\tfrac25, -\tfrac65\right).$

Applying the shoelace formula to $A, B, C, M$ gives area $\dfrac{|{-35.2}|}{2} = \dfrac{88}{5}.$

Thus, the correct answer is C.

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