2002 AMC 10A Problems/Problem 20

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

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

Solution 2

As angle F is clearly congruent to itself, we get from AA similarity, $\triangle AGF \sim \triangle EJF$ ; hence $\frac {AG}{JE} =5$ . Similarly, $\frac {AG}{HC} = 3$ . Thus, $\frac {HC}{JE}=\left(\frac{AG}{JE}\right)\left(\frac{HC}{AG}\right) = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}$ .

2002 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2002 AMC 10A Problems/Problem 20

Contents

Problem

Solution 1

Solution 2

See Also