Difference between revisions of "2017 AMC 12A Problems/Problem 4"
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Let <math>j</math> represent how far Jerry walked, and <math>s</math> represent how far Sylvia walked. Since the field is a square, and Jerry walked two sides of it, while Silvia walked the diagonal, we can simply define the side of the square field to be one, and find the distances they walked. Since Jerry walked two sides, | Let <math>j</math> represent how far Jerry walked, and <math>s</math> represent how far Sylvia walked. Since the field is a square, and Jerry walked two sides of it, while Silvia walked the diagonal, we can simply define the side of the square field to be one, and find the distances they walked. Since Jerry walked two sides, | ||
<math>j = 2</math> | <math>j = 2</math> | ||
− | Since Silvia walked the diagonal, she walked the hypotenuse of a 45, 45, 90 triangle with leg length 1. Thus, | + | Since Silvia walked the diagonal, she walked the hypotenuse of a <math>45</math>, <math>45</math>, <math>90</math> triangle with leg length <math>1</math>. Thus, |
<math>s = \sqrt{2} = 1.414...</math> | <math>s = \sqrt{2} = 1.414...</math> | ||
We can then take | We can then take |
Revision as of 13:51, 19 December 2020
Problem
Jerry and Silvia wanted to go from the southwest corner of a square field to the northeast corner. Jerry walked due east and then due north to reach the goal, but Silvia headed northeast and reached the goal walking in a straight line. Which of the following is closest to how much shorter Silvia's trip was, compared to Jerry's trip?
Solution
Let represent how far Jerry walked, and
represent how far Sylvia walked. Since the field is a square, and Jerry walked two sides of it, while Silvia walked the diagonal, we can simply define the side of the square field to be one, and find the distances they walked. Since Jerry walked two sides,
Since Silvia walked the diagonal, she walked the hypotenuse of a
,
,
triangle with leg length
. Thus,
We can then take
.
See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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 |
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 3 |
Followed by Problem 5 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.