2009 AMC 12A Problems/Problem 13
Contents
Problem
A ship sails miles in a straight line from to , turns through an angle between and , and then sails another miles to . Let be measured in miles. Which of the following intervals contains ?
Solution
Answering the question
To answer the question we are asked, it is enough to compute for two different angles, preferably for both extremes ( and degrees). You can use the law of cosines to do so.
Alternately, it is enough to compute for one of the extreme angles. In case it falls inside one of the given intervals, we are done. In case it falls on the boundary between two options, we also have to argue whether our is the minimal or the maximal possible value of .
Below we show a complete solution in which we also show that all possible values of do indeed lie in the given interval.
Complete solution
Let be the point the ship would reach if it turned , and the point it would reach if it turned . Obviously, is the furthest possible point from , and is the closest possible point to .
Hence the interval of possible values for is .
We can find and as follows:
Let and be the feet of the heights from and onto . The angles in the triangle are , , and , hence . Similarly, the angles in the triangle are , , and , hence and .
Hence we get:
Therefore for any valid the value is surely in the interval .
Alternate Solution
From the law of cosines, . This is essentially the same solution as above. The answer is .
See Also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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 |
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