2002 AMC 10A Problems/Problem 20
Problem
Points 
and 
lie, in that order, on 
, dividing it into five segments, each of length 1. Point 
is not on line 
. Point 
lies on 
, and point 
lies on 
. The line segments 
and 
are parallel. Find 
.
![$[asy] pair A,B,C,D,EE,F,G,H,J; A = (0,0); B = (0.2,0); C = 2*B; D = 3*B; EE = 4*B; F = 5*B; G = (-0.2,0.8); H = intersectionpoint(G--D,C -- (C + G)); J = intersectionpoint(G--F,EE--(EE+G)); draw(G--F--A--G--B); draw(H--C--G--D); draw(J--EE--G); label("$](http://latex.artofproblemsolving.com/8/b/f/8bf4490635cbde0a1eb4ed74efda1f527edd516d.png)
A
B
C
D
E
F
J
G
H![$",H,NE,UnFill(0.1mm)); [/asy]$](http://latex.artofproblemsolving.com/f/8/3/f839d1c25eea5bd0bc87453919a1d385e26b66b0.png)
Solution
Solution $\text{#1}:
Since$ (Error compiling LaTeX. Unknown error_msg)AG
CH
GADHCD
CH/AG = CD/AD = 1/3$.
Since$ (Error compiling LaTeX. Unknown error_msg)AGJEGAFJEFEJ/AG = EF/AF = 1/5
CH/EJ = (CH/AG)/(EJ/AG) = (1/3)/(1/5) = \boxed{5/3}$. The answer is (D).
Solution \text{#2}:
As$ (Error compiling LaTeX. Unknown error_msg)\overline{JE}
\overline{AG}
\triangle AGF \sim \triangle EJF
\frac {AG}{JE} =5
\frac {AG}{HC} = 3
\frac {HC}{JE} = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}$.
See Also
| 2002 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
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