2003 AIME I Problems/Problem 11
Problem
An angle is chosen at random from the interval Let be the probability that the numbers and are not the lengths of the sides of a triangle. Given that where is the number of degrees in and and are positive integers with find
Solution
Note that the three expressions are symmetric with respect to interchanging and , and so the probability is symmetric around . Thus, take so that . Then is the largest of the three given expressions and those three lengths not forming a triangle is equivalent to a violation of the triangle inequality
This is equivalent to
and, using some of our trigonometric identities, we can re-write this as . Since we've chosen , so
The probability that lies in this range is so that , and our answer is .
Solution 2 (Complementary Counting)
We seek a complementary counting argument, where we look for the probability that , and form the side lengths of a triangle.
By the triangle inequality, we must have the following three inequalities to be true: The first inequality will always hold since we have , and for all (The maximum value of is when ).
Now, we examine the second inequality . If we subtract from both sides, we have . Aha! This resembles our sine and cosine double angle identities. Therefore, our inequality is now . We can divide both sides by and we have . The solutions to this occur when .
(To understand why it must be , we can draw the unit circle, and notice as x moves from to , approaches . We must cap at , since if , , and will be negative.)
Next, we examine the third inequality, . Once again, we can get our double angle identities for sine and cosine by subtracting from both sides. We have, .
Next, we again, divide by to produce a (we do this because one trig function is easier to deal with than 2). However, if , we do not need to flip the sign since , and so if , all values for which that is true satisfy the inequality. So we only consider if , and when we divide by a negative, we must flip the sign. Thus we have .
We can take the of both sides, and we have . Once again, to better understand this, we can draw the angle for which , and we notice as moves to , approaches . We must cap at since if , we have .
Notice that if we draw the terminal points for and , they have the same smaller angle with the x and y axis respectively. This means the range of degree measures for which our inequalities hold is which has an area of . However, we want the complement of this, which has an area of . Therefore, the desired probability is , and so .
-BossLu99
See also
2003 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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