2002 AMC 10A Problem 20

Attempt Problem 20 of the 2002 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 2002 AMC 10A solutions, or check the answer key.

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20.

Points A,A, B,B, C,C, D,D, E,E, and FF lie, in that order, on AF,\overline{AF}, dividing it into five segments, each of length 1.1. Point GG is not on line AF.AF. Point HH lies on GD,\overline{GD}, and point JJ lies on GF.\overline{GF}. The line segments HC,\overline{HC}, JE,\overline{JE}, and AG\overline{AG} are parallel. Find HC/JE.HC/JE.

54\dfrac{5}{4}

43\dfrac{4}{3}

32\dfrac{3}{2}

53\dfrac{5}{3}

22

Answer: D
Concepts:similarityparallel lines
Difficulty rating: 1460
Solution:

Since HCAG,HC\parallel AG, DHCDGA,\triangle DHC\sim\triangle DGA, so HCAG=DCDA=13,\dfrac{HC}{AG}=\dfrac{DC}{DA}=\dfrac{1}{3}, giving HC=AG3.HC=\dfrac{AG}{3}.

Since JEAG,JE\parallel AG, FJEFGA,\triangle FJE\sim\triangle FGA, so JEAG=FEFA=15,\dfrac{JE}{AG}=\dfrac{FE}{FA}=\dfrac{1}{5}, giving JE=AG5.JE=\dfrac{AG}{5}.

Therefore HCJE=AG/3AG/5=53.\dfrac{HC}{JE}=\dfrac{AG/3}{AG/5}=\dfrac{5}{3}.

Thus, the correct answer is D.

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