Difference between revisions of "2002 AMC 10A Problems/Problem 20"
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<math>\text{(A)}\ 5/4 \qquad \text{(B)}\ 4/3 \qquad \text{(C)}\ 3/2 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 2</math> | <math>\text{(A)}\ 5/4 \qquad \text{(B)}\ 4/3 \qquad \text{(C)}\ 3/2 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 2</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | + | First we can draw an image. | |
| + | <asy> | ||
| + | unitsize(0.8 cm); | ||
| + | |||
| + | pair A, B, C, D, E, F, G, H, J; | ||
| + | |||
| + | A = (0,0); | ||
| + | B = (1,0); | ||
| + | C = (2,0); | ||
| + | D = (3,0); | ||
| + | E = (4,0); | ||
| + | F = (5,0); | ||
| + | G = (-1.5,4); | ||
| + | H = extension(D, G, C, C + G - A); | ||
| + | J = extension(F, G, E, E + G - A); | ||
| + | |||
| + | draw(A--F--G--cycle); | ||
| + | draw(B--G); | ||
| + | draw(C--G); | ||
| + | draw(D--G); | ||
| + | draw(E--G); | ||
| + | draw(C--H); | ||
| + | draw(E--J); | ||
| + | |||
| + | label("$A$", A, SW); | ||
| + | label("$B$", B, S); | ||
| + | label("$C$", C, S); | ||
| + | label("$D$", D, S); | ||
| + | label("$E$", E, S); | ||
| + | label("$F$", F, SE); | ||
| + | label("$G$", G, NW); | ||
| + | label("$H$", H, W); | ||
| + | label("$J$", J, NE); | ||
| + | </asy> | ||
| + | |||
| + | Since <math>\overline{AG}</math> and <math>\overline{CH}</math> are parallel, triangles <math>\triangle GAD</math> and <math>\triangle HCD</math> are similar. Hence, <math>\frac{CH}{AG} = \frac{CD}{AD} = \frac{1}{3}</math>. | ||
| + | |||
| + | Since <math>\overline{AG}</math> and <math>\overline{JE}</math> are parallel, triangles <math>\triangle GAF</math> and <math>\triangle JEF</math> are similar. Hence, <math>\frac{EJ}{AG} = \frac{EF}{AF} = \frac{1}{5}</math>. Therefore, <math>\frac{CH}{EJ} = \left(\frac{CH}{AG}\right)\div\left(\frac{EJ}{AG}\right) = \left(\frac{1}{3}\right)\div\left(\frac{1}{5}\right) = \boxed{\frac{5}{3}}</math>. The answer is <math>\boxed{(D) 5/3}</math>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | 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>. | ||
| + | |||
| + | ==Solution 3== | ||
| + | Assume an arbitrary value of <math>AG=15</math> WLOG. <math>\overline{AG}</math> and <math>\overline{CH}</math> are parallel, so <math>\triangle GAD</math> and <math>\triangle HCD</math> are similar. So, <math>\frac{AD}{CD}=3=\frac{GA}{HC}</math> which means <math>HC=5</math>. By the same logic, <math>JE=3</math>, so <math>\frac{HC}{JE}=\frac{5}{3}</math>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | |||
| + | https://www.youtube.com/watch?v=AU2PJeMZ7R0 ~David | ||
==See Also== | ==See Also== | ||
Latest revision as of 21:15, 9 June 2023
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
First we can draw an image.
![[asy] unitsize(0.8 cm); pair A, B, C, D, E, F, G, H, J; A = (0,0); B = (1,0); C = (2,0); D = (3,0); E = (4,0); F = (5,0); G = (-1.5,4); H = extension(D, G, C, C + G - A); J = extension(F, G, E, E + G - A); draw(A--F--G--cycle); draw(B--G); draw(C--G); draw(D--G); draw(E--G); draw(C--H); draw(E--J); label("$A$", A, SW); label("$B$", B, S); label("$C$", C, S); label("$D$", D, S); label("$E$", E, S); label("$F$", F, SE); label("$G$", G, NW); label("$H$", H, W); label("$J$", J, NE); [/asy]](http://latex.artofproblemsolving.com/f/3/6/f36a37fb3fee1ec7e6c9eb3326ff0df1b01aedbe.png)
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, 
.
Solution 3
Assume an arbitrary value of 
WLOG. and are parallel, so and are similar. So, 
which means 
. By the same logic, 
, so 
.
Video Solution
https://www.youtube.com/watch?v=AU2PJeMZ7R0 ~David
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. 
