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