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,

B,

C,

D,

E,

and

F

lie, in that order, on

\overline{AF},

dividing it into five segments, each of length

1.

Point

G

is not on line

AF.

Point

H

lies on

\overline{GD},

and point

J

lies on

\overline{GF}.

The line segments

\overline{HC},

\overline{JE},

and

\overline{AG}

are parallel. Find

\frac{HC}{JE}.

\dfrac{5}{4}

\dfrac{4}{3}

\dfrac{3}{2}

\dfrac{5}{3}

2

Answer: D

Concepts:similarity parallel lines

Difficulty rating: 1460

Small hint:

\triangle DHC\sim\triangle DGA

gives

HC

in terms of

AG

Big hint:

\triangle FJE\sim\triangle FGA

gives

JE

in terms of

AG

Solution:

Since

HC\parallel AG,

\triangle DHC\sim\triangle DGA,

so

\dfrac{HC}{AG}=\dfrac{DC}{DA}=\dfrac{1}{3},

giving

HC=\dfrac{AG}{3}.

Since

JE\parallel AG,

\triangle FJE\sim\triangle FGA,

so

\dfrac{JE}{AG}=\dfrac{FE}{FA}=\dfrac{1}{5},

giving

JE=\dfrac{AG}{5}.

Therefore

\dfrac{HC}{JE}=\dfrac{\frac{AG}{3}}{\frac{AG}{5}}=\dfrac{5}{3}.

Thus, the correct answer is D.