# Difference between revisions of "2019 AIME II Problems/Problem 10"

## Problem 10

There is a unique angle $\theta$ between $0^{\circ}$ and $90^{\circ}$ such that for nonnegative integers $n$, the value of $\tan{\left(2^{n}\theta\right)}$ is positive when $n$ is a multiple of $3$, and negative otherwise. The degree measure of $\theta$ is $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime integers. Find $p+q$.

## Solution 1

Note that if $\tan \theta$ is positive, then $\theta$ is in the first or third quadrant, so $0^{\circ} < \theta < 90^{\circ} \pmod{180^{\circ}}$. Also notice that the only way $\tan{\left(2^{n}\theta\right)}$ can be positive for all $n$ that are multiples of $3$ is when $2^0\theta, 2^3\theta, 2^6\theta$, etc. are all the same value $\pmod{180^{\circ}}$. This happens if $8\theta = \theta \pmod{180^{\circ}}$, so $7\theta = 0^{\circ} \pmod{180^{\circ}}$. Therefore, the only possible values of theta between $0^{\circ}$ and $90^{\circ}$ are $\frac{180}{7}^{\circ}$, $\frac{360}{7}^{\circ}$, and $\frac{540}{7}^{\circ}$. However $\frac{180}{7}^{\circ}$ does not work since $\tan{2 \cdot \frac{180}{7}^{\circ}}$ is positive, and $\frac{360}{7}^{\circ}$ does not work because $\tan{4 \cdot \frac{360}{7}^{\circ}}$ is positive. Thus, $\theta = \frac{540}{7}^{\circ}$. $540 + 7 = \boxed{547}$.

## Solution 2

As in the previous solution, we note that $\tan \theta$ is positive when $\theta$ is in the first or third quadrant. In order for $\tan\left(2^n\theta\right)$ to be positive for all $n$ divisible by $3$, we must have $\theta$, $2^3\theta$, $2^6\theta$, etc to lie in the first or second quadrants. We already know that $\theta\in(0,90)$. We can keep track of the range of $2^n\theta$ for each $n$ by considering the portion in the desired quadrants, which gives $$n=1 \implies (90,180)$$ $$n=2\implies (270,360)$$ $$n=3 \implies (180,270)$$ $$n=4 \implies (90,180)$$ $$n=5\implies(270,360)$$ $$n=6 \implies (180,270)$$ $$\cdots$$ at which point we realize a pattern emerging. Specifically, the intervals repeat every $3$ after $n=1$. We can use these repeating intervals to determine the desired value of $\theta$ 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 $0$ (at $n=0$), then increases to $\frac{90}{2}=45$ at $n=1$. This then becomes $45+\frac{45}{2}$ at $n=2$, $45+\frac{45}{2}$ at $n=3$, $45+\frac{45}{2}+\frac{45}{2^3}$ at $n=4$, $45+\frac{45}{2}+\frac{45}{2^3}+\frac{45}{2^4}$ at $n=5$. Due to the observed pattern of the intervals, the lower bound follows a partial geometric series. Hence, as $n$ approaches infinity, the lower bound converges to $$\sum_{k=0}^{\infty}\left(45+\frac{45}{2}\right)\cdot \left(\frac{1}{8}\right)^k=\frac{45+\frac{45}{2}}{1-\frac{1}{8}}=\frac{\frac{135}{2}}{\frac{7}{8}}=\frac{540}{7}\implies p+q=540+7=\boxed{547}$$-ktong

## Solution 3

Since $tan\left(\theta\right) > 0$, $0 < \theta < 90$. Since $tan\left(2\theta\right) < 0$, $\theta$ has to be in the second half of the interval (0, 90) ie (45, 90). Since $tan\left(4\theta\right) < 0$, $\theta$ has to be in the second half of that interval ie (67.5, 90). And since $tan\left(8\theta\right) > 0$, $\theta$ has to be in the first half of (67.5, 90). Inductively, the pattern repeats: $\theta$ 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 $0.11011011011..._2 = \frac{6}{7}_{10}$. So we want the number which is 6/7 of the way through the interval (0, 90) so $\theta = \frac{6}{7}\cdot 90 = \frac{540}{7}$ and $p+q = 540 + 7 = \boxed{547}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 