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

(Problem)
(Problem)
Line 3: Line 3:
  
 
<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>
 
<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;
 
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("</math>A<math>",A,SW);
 
label("</math>B<math>",B,S);
 
label("</math>C<math>",C,S);
 
label("</math>D<math>",D,S);
 
label("</math>E<math>",EE,S);
 
label("</math>F<math>",F,SE);
 
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]</math>
 
Please fix code!
 
  
 
==Solution==
 
==Solution==

Revision as of 20:10, 5 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

Solution $\text{1}$: 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 $\text{2}$: 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

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