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> | ||
− | ==Solution== | + | ==Solution 1== |
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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>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). | 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). | ||
− | Solution | + | ==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)}}</math>. | 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)}}</math>. | ||
Revision as of 19:15, 28 December 2015
Contents
Problem
Points and lie, in that order, on , dividing it into five segments, each of length 1. Point is not on line . Point lies on , and point lies on . The line segments and are parallel. Find .
Solution 1
Since and are parallel, triangles and are similar. Hence, .
Since and are parallel, triangles and are similar. Hence, . Therefore, . The answer is (D).
Solution 2
As is parallel to , angles FJE and FGA are congruent. Also, angle F is clearly congruent to itself. From AA similarity, ; hence . Similarly, . Thus, .
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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