2003 AMC 10A Problem 22

Below is the professionally curated solution for Problem 22 of the 2003 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 10A solutions, or check the answer key.

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Concepts:coordinate geometry linear equation

Difficulty rating: 1800

22.

In rectangle $ABCD,$ we have $AB = 8,$ $BC = 9,$ $H$ is on $\overline{BC}$ with $BH = 6,$ $E$ is on $AD$ with $DE = 4,$ line $EC$ intersects line $AH$ at $G,$ and $F$ is on line $AD$ with $\overline{GF} \perp \overline{AF}.$ Find the length $\overline{GF}.$

$16$

$20$

$24$

$28$

$30$

Solution:

Place $D = (0, 0),$ $A = (9, 0),$ $B = (9, 8),$ $C = (0, 8),$ $H = (3, 8),$ and $E = (4, 0).$

Line $AH$ has equation $y = -\dfrac{4}{3}x + 12,$ and line $EC$ has equation $y = -2x + 8.$

Setting them equal gives $x = -6$ and $y = 20,$ so $G = (-6, 20).$ Since $\overline{GF}$ is perpendicular to line $AD$ (the $x$ -axis), its length is the height $20.$

Thus, the correct answer is B.

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