2019 AIME II Problems/Problem 10
Problem 10
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 . Also notice that the only way can be positive for all that are multiples of is when , etc. are all the same value . This happens if , so . Therefore, the only possible values of theta between and are , , and . However does not work since is positive, and does not work because is positive. Thus, . .
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 second 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 . 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
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 first half of 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 $\theat = \frac{6}{7}\cdot 90 = \frac{540}{7}$ (Error compiling LaTeX. ! Undefined control sequence.) and
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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