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

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(Solution 1)
 
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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>
  
<math>[asy]
+
==Solution 1==
pair A,B,C,D,EE,F,G,H,J;
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First we can draw an image.
 +
<asy>
 +
unitsize(0.8 cm);
 +
 
 +
pair A, B, C, D, E, F, G, H, J;
 +
 
 
A = (0,0);
 
A = (0,0);
B = (0.2,0);
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B = (1,0);
C = 2*B;
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C = (2,0);
D = 3*B;
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D = (3,0);
EE = 4*B;
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E = (4,0);
F = 5*B;
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F = (5,0);
G = (-0.2,0.8);
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G = (-1.5,4);
H = intersectionpoint(G--D,C -- (C + G));
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H = extension(D, G, C, C + G - A);
J = intersectionpoint(G--F,EE--(EE+G));
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J = extension(F, G, E, E + G - A);
draw(G--F--A--G--B);
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draw(H--C--G--D);
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draw(A--F--G--cycle);
draw(J--EE--G);
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draw(B--G);
label("</math>A<math>",A,SW);
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draw(C--G);
label("</math>B<math>",B,S);
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draw(D--G);
label("</math>C<math>",C,S);
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draw(E--G);
label("</math>D<math>",D,S);
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draw(C--H);
label("</math>E<math>",EE,S);
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draw(E--J);
label("</math>F<math>",F,SE);
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label("</math>J<math>",J,NE);
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label("$A$", A, SW);
label("</math>G<math>",G,N);
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label("$B$", B, S);
label(scale(0.9)*"</math>H<math>",H,NE,UnFill(0.1mm));
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label("$C$", C, S);
[/asy]</math>
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label("$D$", D, S);
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label("$E$", E, S);
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label("$F$", F, SE);
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label("$G$", G, NW);
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label("$H$", H, W);
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label("$J$", J, NE);
 +
</asy>
  
==Solution==
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Since <math>\overline{AG}</math> and <math>\overline{CH}</math> are parallel, triangles <math>\triangle GAD</math> and <math>\triangle HCD</math> are similar. Hence, <math>\frac{CH}{AG} = \frac{CD}{AD} = \frac{1}{3}</math>.
Solution <math>\text{#1}:
 
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).
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Since <math>\overline{AG}</math> and <math>\overline{JE}</math> are parallel, triangles <math>\triangle GAF</math> and <math>\triangle JEF</math> are similar. Hence, <math>\frac{EJ}{AG} = \frac{EF}{AF} = \frac{1}{5}</math>. Therefore, <math>\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}}</math>. The answer is <math>\boxed{(D) 5/3}</math>.
  
Solution \text{#2}:
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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)}}$.
+
As angle F is clearly congruent to itself, we get 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}=\left(\frac{AG}{JE}\right)\left(\frac{HC}{AG}\right) = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 16:14, 20 April 2016

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

First we can draw an image. [asy] unitsize(0.8 cm);  pair A, B, C, D, E, F, G, H, J;  A = (0,0); B = (1,0); C = (2,0); D = (3,0); E = (4,0); F = (5,0); G = (-1.5,4); H = extension(D, G, C, C + G - A); J = extension(F, G, E, E + G - A);  draw(A--F--G--cycle); draw(B--G); draw(C--G); draw(D--G); draw(E--G); draw(C--H); draw(E--J);  label("$A$", A, SW); label("$B$", B, S); label("$C$", C, S); label("$D$", D, S); label("$E$", E, S); label("$F$", F, SE); label("$G$", G, NW); label("$H$", H, W); label("$J$", J, NE); [/asy]

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, $\frac {HC}{JE}=\left(\frac{AG}{JE}\right)\left(\frac{HC}{AG}\right) = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}$.

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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