Difference between revisions of "2002 AMC 10A Problems/Problem 20"
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==Solution 1== | ==Solution 1== | ||
− | + | First we can draw an image. | |
+ | <asy> | ||
+ | unitsize(0.8 cm); | ||
− | Since <math>\overline{AG}</math> and <math>\overline{JE}</math> are parallel, triangles <math>GAF</math> and <math>JEF</math> are similar. Hence, <math>\frac{EJ}{AG} = \frac{EF}{AF} = \frac{1}{5}</math>. Therefore, <math>\frac{CH}{EJ} = (\frac{CH}{AG}) | + | 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 <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>. | ||
+ | |||
+ | 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 2== | ==Solution 2== | ||
− | As | + | 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
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
First we can draw an image.
Since and are parallel, triangles and are similar. Hence, .
Since and are parallel, triangles and are similar. Hence, . Therefore, . The answer is .
Solution 2
As angle F is clearly congruent to itself, we get 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.