2019 AIME II Problems/Problem 10
Problem
There is a unique angle between and such that for nonnegative integers , the value of is positive when is a multiple of , and negative otherwise. The degree measure of is , where and are relatively prime integers. Find .
Solution 1
Note that if is positive, then is in the first or third quadrant, so .
Furthermore, the only way can be positive for all that are multiples of is when: (This is because if it isn't the same value, the terminal angle will gradually shift from the first quadrant into different quadrants, making the condition for positive tan untrue. This must also be true in order for to be unique.)
This is the case if , so . Therefore, recalling that the possible are:
does not work since is positive.
does not work because is positive.
Thus, , and a quick check verifies that it does work. Our desired answer is .
Solution 2
As in the previous solution, we note that is positive when is in the first or third quadrant. In order for to be positive for all divisible by , we must have , , , etc to lie in the first or third quadrants. We already know that . We can keep track of the range of for each by considering the portion in the desired quadrants, which gives at which point we realize a pattern emerging. Specifically, the intervals repeat every after . We can use these repeating intervals to determine the desired value of since the upper and lower bounds will converge to such a value (since it is unique, as indicated in the problem). Let's keep track of the lower bound.
Initially, the lower bound is (at ), then increases to at . This then becomes at , at , at , at , etc. Due to the observed pattern of the intervals, the lower bound follows a partial geometric series. Hence, as approaches infinity, the lower bound converges to -ktong
Extra note: If you are still unsure, you can check the upper bound to see if the top converges. This value comes out to be:
.
Thus, the limit does in fact converge to the same value on both sides. ~eevee9406
Solution 3
Since , . Since , has to be in the second half of the interval (0, 90) ie (45, 90). Since , has to be in the second half of that interval ie (67.5, 90). And since , has to be in the first half of (67.5, 90). Inductively, the pattern repeats: is in the second half of the second half of the first half of the second half of the second half... of the interval (0, 90). Consider the binary representation of numbers in the interval (0, 1). Numbers in the first half of the interval start with 0.0... and numbers in the second half start with 0.1... . Similarly, numbers in the second half of the second half start with 0.11... etc. So if we want a number in the first half of the second half of the second half... of the interval, we want its binary representation to be . So we want the number which is 6/7 of the way through the interval (0, 90) so and
~minor edit by Mathkiddie
Solution 4
With some simple arithmetic and guess and check, we can set the lower bound and upper bounds for the "first round of powers of two", which are and . Going on to the "second round of powers of two, we set the new lower and upper bounds as and using some guess and check and bashing. Now, it is obvious that the bounds for the "zeroth round of powers of two" are and , and notice that and and . This is obviously a geometric series, so setting as , we obtain = = = which simplifies to . We can now finally subtract from and then we get as the unique angle, so is our answer. -fidgetboss_4000
Solution 5
Since is the only number n such that f(x) = has a period of 3, we find that is a multiple of . Note that the tangents of , , are positive while those of , , and are negative. With a bit of trial and error, we find is and is .
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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