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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>
  
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==Solution==
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==Solution 1==
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Solution <math>\text{1}</math>:
 
 
Since <math>AG</math> and <math>CH</math> are parallel, triangles <math>GAD</math> and <math>HCD</math> are similar. Hence, <math>CH/AG = CD/AD = 1/3</math>.
 
Since <math>AG</math> and <math>CH</math> are parallel, triangles <math>GAD</math> and <math>HCD</math> are similar. Hence, <math>CH/AG = CD/AD = 1/3</math>.
  
 
Since <math>AG</math> and <math>JE</math> are parallel, triangles <math>GAF</math> and <math>JEF</math> are similar. Hence, <math>EJ/AG = EF/AF = 1/5</math>. Therefore, <math>CH/EJ = (CH/AG)/(EJ/AG) = (1/3)/(1/5) = \boxed{5/3}</math>. The answer is (D).
 
Since <math>AG</math> and <math>JE</math> are parallel, triangles <math>GAF</math> and <math>JEF</math> are similar. Hence, <math>EJ/AG = EF/AF = 1/5</math>. Therefore, <math>CH/EJ = (CH/AG)/(EJ/AG) = (1/3)/(1/5) = \boxed{5/3}</math>. The answer is (D).
  
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Solution <math>\text{2}</math>:
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==Solution 2==
 
As <math>\overline{JE}</math> is parallel to <math>\overline{AG}</math>, angles FJE and FGA are congruent. Also, angle F is clearly congruent to itself. 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} = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}</math>.
 
As <math>\overline{JE}</math> is parallel to <math>\overline{AG}</math>, angles FJE and FGA are congruent. Also, angle F is clearly congruent to itself. 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} = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}</math>.
  

Revision as of 18:15, 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 $AG$ and $CH$ are parallel, triangles $GAD$ and $HCD$ are similar. Hence, $CH/AG = CD/AD = 1/3$.

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