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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>
 +
 +
[asy]
 +
pair A,B,C,D,EE,F,G,H,J;
 +
A = (0,0);
 +
B = (0.2,0);
 +
C = 2*B;
 +
D = 3*B;
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EE = 4*B;
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F = 5*B;
 +
G = (-0.2,0.8);
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H = intersectionpoint(G--D,C -- (C + G));
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J = intersectionpoint(G--F,EE--(EE+G));
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draw(G--F--A--G--B);
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draw(H--C--G--D);
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draw(J--EE--G);
 +
label("<math>A</math>",A,SW);
 +
label("<math>B</math>",B,S);
 +
label("<math>C</math>",C,S);
 +
label("<math>D</math>",D,S);
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label("<math>E</math>",EE,S);
 +
label("<math>F</math>",F,SE);
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label("<math>J</math>",J,NE);
 +
label("<math>G</math>",G,N);
 +
label(scale(0.9)*"<math>H</math>",H,NE,UnFill(0.1mm));
 +
[/asy]
  
 
==Solution==
 
==Solution==

Revision as of 19:30, 5 November 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 $HC/JE$.

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

[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("$A$",A,SW); label("$B$",B,S); label("$C$",C,S); label("$D$",D,S); label("$E$",EE,S); label("$F$",F,SE); label("$J$",J,NE); label("$G$",G,N); label(scale(0.9)*"$H$",H,NE,UnFill(0.1mm)); [/asy]

Solution

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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