2003 AMC 10A Problem 22

Attempt Problem 22 of the 2003 AMC 10A below, then check your answer against the professionally curated solution 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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22.

In rectangle ABCD,ABCD, we have AB=8,AB = 8, BC=9,BC = 9, HH is on BC\overline{BC} with BH=6,BH = 6, EE is on ADAD with DE=4,DE = 4, line ECEC intersects line AHAH at G,G, and FF is on line ADAD with GFAF.\overline{GF} \perp \overline{AF}. Find the length GF.\overline{GF}.

1616

2020

2424

2828

3030

Answer: B
Concepts:coordinate geometrylinear equation
Difficulty rating: 1800
Solution:

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

Line AHAH has equation y=43x+12,y = -\dfrac{4}{3}x + 12, and line ECEC has equation y=2x+8.y = -2x + 8.

Setting them equal gives x=6x = -6 and y=20,y = 20, so G=(6,20).G = (-6, 20). Since GF\overline{GF} is perpendicular to line ADAD (the xx-axis), its length is the height 20.20.

Thus, the correct answer is B.

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