Difference between revisions of "2002 AMC 10A Problems/Problem 20"

(Solution 1)
(Solution 1)
Line 5: Line 5:
  
 
==Solution 1==
 
==Solution 1==
Since <math>\overline{AG}</math> and <math>\overline{CH}</math> are parallel, from AA similarity, triangles <math>GAD</math> and <math>HCD</math> are similar. Hence, <math>\frac{CH}{AG} = \frac{CD}{AD} = \frac{1}{3}</math>.
+
Since <math>\overline{AG}</math> and <math>\overline{CH}</math> are parallel, triangles <math>GAD</math> and <math>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, from AA similarity, triangles <math>GAF</math> and <math>JEF</math> are similar. Hence, <math>\frac{EJ}{AG} = \frac{EF}{AF} = \frac{1}{5}</math>. Therefore, <math>\frac{CH}{EJ} = (\frac{CH}{AG})\div(\frac{EJ}{AG}) = (\frac{1}{3})\div(\frac{1}{5}) = \boxed{\frac{5}{3}}</math>. The answer is <math>\boxed{(D)}</math>.
+
Since <math>\overline{AG}</math> and <math>\overline{JE}</math> are parallel, triangles <math>GAF</math> and <math>JEF</math> are similar. Hence, <math>\frac{EJ}{AG} = \frac{EF}{AF} = \frac{1}{5}</math>. Therefore, <math>\frac{CH}{EJ} = (\frac{CH}{AG})\div(\frac{EJ}{AG}) = (\frac{1}{3})\div(\frac{1}{5}) = \boxed{\frac{5}{3}}</math>. The answer is <math>\boxed{(D)}</math>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 21:36, 28 December 2015

Problem

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 $HC/JE$.

$\text{(A)}\ 5/4 \qquad \text{(B)}\ 4/3 \qquad \text{(C)}\ 3/2 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 2$

Solution 1

Since $\overline{AG}$ and $\overline{CH}$ are parallel, triangles $GAD$ and $HCD$ are similar. Hence, $\frac{CH}{AG} = \frac{CD}{AD} = \frac{1}{3}$.

Since $\overline{AG}$ and $\overline{JE}$ are parallel, triangles $GAF$ and $JEF$ are similar. Hence, $\frac{EJ}{AG} = \frac{EF}{AF} = \frac{1}{5}$. Therefore, $\frac{CH}{EJ} = (\frac{CH}{AG})\div(\frac{EJ}{AG}) = (\frac{1}{3})\div(\frac{1}{5}) = \boxed{\frac{5}{3}}$. The answer is $\boxed{(D)}$.

Solution 2

As $\overline{JE}$ is parallel to $\overline{AG}$, angles FJE and FGA are congruent. Also, angle F is clearly congruent to itself. From AA similarity, $\triangle AGF \sim \triangle EJF$; hence $\frac {AG}{JE} =5$. Similarly, $\frac {AG}{HC} = 3$. Thus, $\frac {HC}{JE} = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}$.

See Also

2002 AMC 10A (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS