2002 AMC 10A Problem 20

Below is the professionally curated solution for Problem 20 of the 2002 AMC 10A, 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.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:similarityparallel lines

Difficulty rating: 1460

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

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.

Problem 20 in Other Years