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

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, $\theta = \frac{540}{7}^{\circ}$ . $540 + 7 = \boxed{547}$ .

2019 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2019 AIME II Problems/Problem 10

Problem 10

Solution

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