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

m (Problem 10)
(Solution)
Line 3: Line 3:
  
 
==Solution==
 
==Solution==
 +
Note that if <math>\tan \theta</math> is positive, then <math>\theta</math> is in the first or third quadrant, so <math>0^{\circ} < \theta < 90^{\circ} \pmod{180^{\circ}}</math>. Also notice that the only way <math>\tan{\left(2^{n}\theta\right)}</math> can be positive for all <math>n</math> that are multiples of <math>3</math> is when <math>2^0\theta, 2^3\theta, 2^6\theta</math>, etc. are all the same value <math>\pmod{180^{\circ}}</math>. This happens if <math>8\theta = \theta \pmod{180^{\circ}}</math>, so <math>7\theta = 0^{\circ} \pmod{180^{\circ}}</math>. Therefore, the only possible values of theta between <math>0^{\circ}</math> and <math>90^{\circ}</math>  are <math>\frac{180}{7}^{\circ}</math>, <math>\frac{360}{7}^{\circ}</math>, and <math>\frac{540}{7}^{\circ}</math>. However <math>\frac{180}{7}^{\circ}</math> does not work since <math>\tan{2 \cdot \frac{180}{7}^{\circ}}</math> is positive, and <math>\frac{360}{7}^{\circ}</math> does not work because <math>\tan{4 \cdot \frac{360}{7}^{\circ}}</math> is positive. Thus, <math>\theta = \frac{540}{7}^{\circ}</math>. <math>540 + 7 = \boxed{547}</math>.
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2019|n=II|num-b=9|num-a=11}}
 
{{AIME box|year=2019|n=II|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:41, 22 March 2019

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

See Also

2019 AIME II (ProblemsAnswer KeyResources)
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

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