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("
",A,SW);
label("
",B,S);
label("
",C,S);
label("
",D,S);
label("
",EE,S);
label("",F,SE);
label("",J,NE);
label("",G,N);
label(scale(0.9)*"",H,NE,UnFill(0.1mm));
[/asy]
Solution
Solution #1:
Since 
and 
are parallel, triangles 
and 
are similar. Hence, 
.
Since and 
are parallel, triangles 
and 
are similar. Hence, 
. Therefore, 
. The answer is (D).
Solution #2:
As 
is parallel to , angles FJE and FGA are congruent. Also, angle F is clearly congruent to itself. 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
