2002 AMC 10A Problems/Problem 20: Difference between revisions
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==Solution 2== | ==Solution 2== | ||
As | As angle F is clearly congruent to itself, we get from AA similarity, <math>\triangle AGF \sim \triangle EJF</math>; hence <math>\frac {AG}{JE} =5</math>. Similarly, <math>\frac {AG}{HC} = 3</math>. Thus, <math>\frac {HC}{JE}=\left(\frac{AG}{JE}\right)\left(\frac{HC}{AG}\right) = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}</math>. | ||
==See Also== | ==See Also== | ||
Revision as of 12:03, 4 January 2016
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 
.
Solution 1
Since and 
are parallel, triangles 
and 
are similar. Hence, 
.
Since and 
are parallel, triangles 
and 
are similar. Hence, 
. Therefore, 
. The answer is 
.
Solution 2
As angle F is clearly congruent to itself, we get from AA similarity, 
; hence 
. Similarly, 
. Thus, 
.
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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