Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 2: Line 2:
 
There is a unique angle <math>\theta</math> between <math>0^{\circ}</math> and <math>90^{\circ}</math> such that for nonnegative integers <math>n</math>, the value of <math>\tan{\left(2^{n}\theta\right)}</math> is positive when <math>n</math> is a multiple of <math>3</math>, and negative otherwise. The degree measure of <math>\theta</math> is <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime integers. Find <math>p+q</math>.
 
There is a unique angle <math>\theta</math> between <math>0^{\circ}</math> and <math>90^{\circ}</math> such that for nonnegative integers <math>n</math>, the value of <math>\tan{\left(2^{n}\theta\right)}</math> is positive when <math>n</math> is a multiple of <math>3</math>, and negative otherwise. The degree measure of <math>\theta</math> is <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime integers. Find <math>p+q</math>.
  
−
==Solution==
+
==Solution 1==
 
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>.
 
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>.
 +
 +
==Solution 2==
 +
As in the previous solution, we note that <math>\tan \theta</math> is positive when <math>\theta</math> is in the first or third quadrant. In order for <math>\tan\left(2^n\theta\right)</math> to be positive for all <math>n</math> divisible by <math>3</math>, we must have <math>\theta</math>, <math>2^3\theta</math>, <math>2^6\theta</math>, etc to lie in the first or second quadrants. We already know that <math>\theta\in(0,90)</math>. We can keep track of the range of <math>2^n\theta</math> for each <math>n</math> by considering the portion in the desired quadrants, which gives
 +
<cmath>n=1 \implies (90,180)</cmath>
 +
<cmath>n=2\implies (270,360)</cmath>
 +
<cmath>n=3 \implies (180,270)</cmath>
 +
<cmath>n=4 \implies (90,180)</cmath>
 +
<cmath>n=5\implies(270,360)</cmath>
 +
<cmath>n=6 \implies (180,270)</cmath>
 +
<cmath>\cdots</cmath>
 +
at which point we realize a pattern emerging. Specifically, the intervals repeat every <math>3</math> after <math>n=1</math>. We can use these repeating intervals to determine the desired value of <math>\theta</math> 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 <math>0</math> (at <math>n=0</math>), then increases to <math>\frac{90}{2}=45</math> at <math>n=1</math>. This then becomes <math>45+\frac{45}{2}</math> at <math>n=2</math>, <math>45+\frac{45}{2}</math> at <math>n=3</math>, <math>45+\frac{45}{2}+\frac{45}{2^3}</math> at <math>n=4</math>,<math>45+\frac{45}{2}+\frac{45}{2^3}+\frac{45}{2^4}</math> at <math>n=5</math>. Due to the observed pattern of the intervals, the lower bound follows a partial geometric series. Hence, as <math>n</math> approaches infinity, the lower bound converges to
 +
<cmath>\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}</cmath>-ktong
  
 
==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 23:27, 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 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

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_10&oldid=104846"