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

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==Solution 2==
 
==Solution 2==
 
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>.
 
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>.
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==Solution 3==
 +
Assume an arbitrary value of <math>AG=15</math> WLOG. <math>\overline{AG}</math> and <math>\overline{CH}</math> are parallel, so <math>\triangle GAD</math> and <math>\triangle HCD</math> are similar. So, <math>\frac{AD}{CD}=3=\frac{GA}{HC}</math> which means <math>HC=5</math>. By the same logic, <math>JE=3</math>, so <math>\frac{HC}{JE}=\frac{5}{3}</math>.
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==Video Solution==
 +
 +
https://www.youtube.com/watch?v=AU2PJeMZ7R0    ~David
  
 
==See Also==
 
==See Also==

Latest revision as of 21:15, 9 June 2023

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

Solution 3

Assume an arbitrary value of $AG=15$ WLOG. $\overline{AG}$ and $\overline{CH}$ are parallel, so $\triangle GAD$ and $\triangle HCD$ are similar. So, $\frac{AD}{CD}=3=\frac{GA}{HC}$ which means $HC=5$. By the same logic, $JE=3$, so $\frac{HC}{JE}=\frac{5}{3}$.

Video Solution

https://www.youtube.com/watch?v=AU2PJeMZ7R0 ~David

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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