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Difference between revisions of "1991 AIME Problems/Problem 4"

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== Solution ==
 
== Solution ==
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{{image}}
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<center>[[Image:AIME_1991_Solution_04.png]]</center>
 
The [[range]] of the [[sine]] function is <math>-1 \le y \le 1</math>. It is [[periodic function|periodic]] (in this problem) with a period of <math>\frac{2}{5}</math>.  
 
The [[range]] of the [[sine]] function is <math>-1 \le y \le 1</math>. It is [[periodic function|periodic]] (in this problem) with a period of <math>\frac{2}{5}</math>.  
  
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Thus, <math>-1 \le \frac{1}{5} \log_2 x \le 1</math>, and <math>-5 \le \log_2 x \le 5</math>. The solutions for <math>x</math> occur in the domain of <math>\frac{1}{32} \le x \le 32</math>. When <math>x > 1</math> the [[logarithm]] function returns a [[positive]] value; up to <math>x = 32</math> it will pass through the sine curve. There are exactly 10 intersections of five periods (every two integral values of <math>x</math>) of the sine curve and another curve that is <math>< 1</math>, so there are <math>\frac{32}{2} \cdot 10 - 6 = 160 - 6 = 154</math> values (the subtraction of 6 since all the “intersections” when <math>x < 1</math> must be disregarded). When <math>y = 0</math>, there is exactly <math>1</math> touching point between the two functions: <math>(\frac{1}{5},0)</math>. When <math>y < 0</math> or <math>x < 1</math>, we can count <math>4</math> more solutions. The solution is <math>154 + 1 + 4 = 159</math>.
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Thus, <math>-1 \le \frac{1}{5} \log_2 x \le 1</math>, and <math>-5 \le \log_2 x \le 5</math>. The solutions for <math>x</math> occur in the domain of <math>\frac{1}{32} \le x \le 32</math>. When <math>x > 1</math> the [[logarithm]] function returns a [[positive]] value; up to <math>x = 32</math> it will pass through the sine curve. There are exactly 10 intersections of five periods (every two integral values of <math>x</math>) of the sine curve and another curve that is <math>< 1</math>, so there are <math>\frac{32}{2} \cdot 10 - 6 = 160 - 6 = 154</math> values (the subtraction of 6 since all the “intersections” when <math>x < 1</math> must be disregarded). When <math>y = 0</math>, there is exactly <math>1</math> touching point between the two functions: <math>\left(\frac{1}{5},0\right)</math>. When <math>y < 0</math> or <math>x < 1</math>, we can count <math>4</math> more solutions. The solution is <math>154 + 1 + 4 = \boxed{159}</math>.
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==Solution 2==
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Notice that the equation is satisfied twice for every sine period (which is <math>\frac{2}{5}</math>), except in the sole case when the two equations equate to <math>0</math>. In that case, the equation is satisfied twice but only at the one instance when <math>y=0</math>. Hence, it is double-counted in our final solution, so we have to subtract it out. We then compute:
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<math>32 \cdot \frac{5}{2} \cdot 2 - 1 = \boxed {159}</math>
  
 
== See also ==
 
== See also ==
 
*[[Trigonometry]]
 
*[[Trigonometry]]
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*[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70180 Aops Topic]
 
{{AIME box|year=1991|num-b=3|num-a=5}}
 
{{AIME box|year=1991|num-b=3|num-a=5}}
  
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[[Category:Intermediate Algebra Problem]]
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[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 15:01, 26 November 2019

Problem

How many real numbers $x^{}_{}$ satisfy the equation $\frac{1}{5}\log_2 x = \sin (5\pi x)$?

Solution

AIME 1991 Solution 04.png

The range of the sine function is $-1 \le y \le 1$. It is periodic (in this problem) with a period of $\frac{2}{5}$.

Thus, $-1 \le \frac{1}{5} \log_2 x \le 1$, and $-5 \le \log_2 x \le 5$. The solutions for $x$ occur in the domain of $\frac{1}{32} \le x \le 32$. When $x > 1$ the logarithm function returns a positive value; up to $x = 32$ it will pass through the sine curve. There are exactly 10 intersections of five periods (every two integral values of $x$) of the sine curve and another curve that is $< 1$, so there are $\frac{32}{2} \cdot 10 - 6 = 160 - 6 = 154$ values (the subtraction of 6 since all the “intersections” when $x < 1$ must be disregarded). When $y = 0$, there is exactly $1$ touching point between the two functions: $\left(\frac{1}{5},0\right)$. When $y < 0$ or $x < 1$, we can count $4$ more solutions. The solution is $154 + 1 + 4 = \boxed{159}$.

Solution 2

Notice that the equation is satisfied twice for every sine period (which is $\frac{2}{5}$), except in the sole case when the two equations equate to $0$. In that case, the equation is satisfied twice but only at the one instance when $y=0$. Hence, it is double-counted in our final solution, so we have to subtract it out. We then compute: $32 \cdot \frac{5}{2} \cdot 2 - 1 = \boxed {159}$

See also

1991 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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