Difference between revisions of "2021 AIME I Problems/Problem 7"

(Solution 2)
m (Use \ldots instead of \cdots for lists.)
(28 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
Find the number of pairs <math>(m,n)</math> of positive integers with <math>1\le m<n\le 30</math> such that there exists a real number <math>x</math> satisfying<cmath>\sin(mx)+\sin(nx)=2.</cmath>
+
Find the number of pairs <math>(m,n)</math> of positive integers with <math>1\le m<n\le 30</math> such that there exists a real number <math>x</math> satisfying <cmath>\sin(mx)+\sin(nx)=2.</cmath>
  
 
==Solution 1==
 
==Solution 1==
Line 40: Line 40:
 
Now it is easy to tell that <math>a \equiv 1 (</math>mod <math>4)</math> and <math>b \equiv 1 (</math>mod <math>4)</math>. However, there is another case: that  
 
Now it is easy to tell that <math>a \equiv 1 (</math>mod <math>4)</math> and <math>b \equiv 1 (</math>mod <math>4)</math>. However, there is another case: that  
  
<math>a \equiv 3 (</math>mod <math>4)</math> and <math>b \equiv 3 (</math>mod <math>4)</math>. This is because multiplying both <math>a</math> and <math>b</math> by <math>-1</math> will not change the fraction, but each congruence will be changed to <math>-1 (</math>mod <math>4) \equiv 3 (</math>mod <math>4)</math>.
+
<math>a \equiv 3 (</math>mod <math>4)</math> and <math>b \equiv 3 (</math>mod <math>4)</math>. This is because multiplying both <math>4\alpha+1</math> and <math>4\beta+1</math> by <math>-1</math> will not change the fraction, but each congruence will be changed to <math>-1 (</math>mod <math>4) \equiv 3 (</math>mod <math>4)</math>.
  
 
From the first set of congruences, we find that <math>a</math> and <math>b</math> can be two of  
 
From the first set of congruences, we find that <math>a</math> and <math>b</math> can be two of  
<math>\{1, 5, 9, \cdots, 29\}</math>.
+
<math>\{1, 5, 9, \ldots, 29\}</math>.
  
 
From the second set of congruences, we find that <math>a</math> and <math>b</math> can be two of  
 
From the second set of congruences, we find that <math>a</math> and <math>b</math> can be two of  
<math>\{3, 7, 11, \cdots, 27\}</math>.
+
<math>\{3, 7, 11, \ldots, 27\}</math>.
  
Now all we have to do is multiply by <math>2^p</math> to get back to <math>m</math> and <math>n</math>. Let’s organize the solutions in a table with increasing values of <math>p</math>, keeping in mind that <math>m</math> and <math>n</math> are bounded between 1 and 30.
+
Now all we have to do is multiply by <math>2^p</math> to get back to <math>m</math> and <math>n</math>.  
 +
Let’s organize the solutions in order of increasing values of <math>p</math>, keeping in mind that <math>m</math> and <math>n</math> are bounded between 1 and 30.
  
 +
For <math>p = 0</math> we get <math>\{1, 5, 9, \ldots, 29\}, \{3, 7, 11, \ldots, 27\}</math>.
  
I WILL FINISH THE SOLUTION SOON, PLEASE DO NOT EDIT THIS BEFORE THEN, THANK YOU!
+
For <math>p = 1</math> we get <math>\{2, 10, 18, 26\}, \{6, 14, 22, 30\}</math>
 +
 
 +
For <math>p = 2</math> we get <math>\{4, 20\}, \{12, 28\}</math>
 +
 
 +
If we increase the value of <math>p</math> more, there will be less than two integers in our sets, so we are done there.
 +
 
 +
There are 8 numbers in the first set, 7 in the second, 4 in the third, 4 in the fourth, 2 in the fifth, and 2 in the sixth.
 +
 +
In each of these sets we can choose 2 numbers to be <math>m</math> and <math>n</math> and then assign them in increasing order. Thus there are:
 +
 
 +
<cmath>\dbinom{8}{2}+\dbinom{7}{2}+\dbinom{4}{2}+\dbinom{4}{2}+\dbinom{2}{2}+\dbinom{2}{2} = 28+21+6+6+1+1 = \boxed{63}</cmath> possible pairs <math>(m,n)</math> that satisfy the conditions.
  
 
-KingRavi
 
-KingRavi
  
==See also==
+
==Solution 3==
 +
We know that the range of <math>sin</math> is between <math>-1</math> and <math>1</math>.
 +
 
 +
Thus, the only way for the sum to be <math>2</math> is for <math>sin</math> of <math>mx</math> and <math>nx</math> to both be <math>1</math>.
 +
 
 +
The <math>sin</math> of <math>(90+360k)</math> is equal to 1.
 +
 
 +
Assuming <math>mx</math> and <math>nm</math> are both positive, m and n could be <math>1,5,9,13,17,21,25,29</math>. There are <math>8</math> ways, so <math>\dbinom{8}{2}</math>.
 +
 
 +
If both are negative, m and n could be <math>3,7,11,15,19,23,27</math>. There are <math>7</math> ways, so <math>\dbinom{7}{2}</math>.
 +
 
 +
However, the pair <math>(1,5)</math> could also be <math>(2, 10)</math> and so on. The same goes for some other pairs.
 +
 
 +
In total there are <math>14</math> of these extra pairs.
 +
 
 +
The answer is <math>28+21+14 = \boxed{063}</math>
 +
 
 +
==Remark==
 +
The graphs of <math>r\leq\sin(m\theta)+\sin(n\theta)</math> and <math>r=2</math> are shown here in Desmos: https://www.desmos.com/calculator/busxadywja
 +
 
 +
Move the sliders around for <math>1\leq m \leq 29</math> and <math>2\leq m+1\leq n\leq30</math> to observe the geometric representation generated by each pair <math>(m,n).</math>
 +
 
 +
~MRENTHUSIASM (inspired by TheAMCHub)
 +
 
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=LUkQ7R1DqKo
 +
~Mathematical Dexterity
 +
 
 +
==See Also==
 
{{AIME box|year=2021|n=I|num-b=6|num-a=8}}
 
{{AIME box|year=2021|n=I|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:37, 8 September 2021

Problem

Find the number of pairs $(m,n)$ of positive integers with $1\le m<n\le 30$ such that there exists a real number $x$ satisfying \[\sin(mx)+\sin(nx)=2.\]

Solution 1

The maximum value of $\sin \theta$ is $1$, which is achieved at $\theta = \frac{\pi}{2}+2k\pi$ for some integer $k$. This is left as an exercise to the reader.

This implies that $\sin(mx) = \sin(nx) = 1$, and that $mx = \frac{\pi}{2}+2a\pi$ and $nx = \frac{\pi}{2}+2b\pi$, for integers $a, b$.

Taking their ratio, we have \[\frac{mx}{nx} = \frac{\frac{\pi}{2}+2a\pi}{\frac{\pi}{2}+2b\pi} \implies \frac{m}{n} = \frac{4a + 1}{4b + 1} \implies \frac{m}{4a + 1} = \frac{n}{4b + 1} = k.\] It remains to find all $m, n$ that satisfy this equation.

If $k = 1$, then $m \equiv n \equiv 1 \pmod 4$. This corresponds to choosing two elements from the set $\{1, 5, 9, 13, 17, 21, 25, 29\}$. There are $\binom 82$ ways to do so.

If $k < 1$, by multiplying $m$ and $n$ by the same constant $c = \frac{1}{k}$, we have that $mc \equiv nc \equiv 1 \pmod 4$. Then either $m \equiv n \equiv 1 \pmod 4$, or $m \equiv n \equiv 3 \pmod 4$. But the first case was already counted, so we don't need to consider that case. The other case corresponds to choosing two numbers from the set $\{3, 7, 11, 15, 19, 23, 27\}$. There are $\binom 72$ ways here.

Finally, if $k > 1$, note that $k$ must be an integer. This means that $m, n$ belong to the set $\{k, 5k, 9k, \dots\}$, or $\{3k, 7k, 11k, \dots\}$. Taking casework on $k$, we get the sets $\{2, 10, 18, 26\}, \{6, 14, 22, 30\}, \{4, 20\}, \{12, 28\}$. Some sets have been omitted; this is because they were counted in the other cases already. This sums to $\binom 42 + \binom 42 + \binom 22 + \binom 22$.

In total, there are $\binom 82 + \binom 72 + \binom 42 + \binom 42 + \binom 22 + \binom 22 = \boxed{63}$ pairs of $(m, n)$.

This solution was brought to you by ~Leonard_my_dude~

Solution 2

In order for $\sin(mx) + \sin(nx) = 2$, $\sin(mx) = \sin(nx) = 1$.

This happens when $mx \equiv nx \equiv \frac{\pi}{2} ($mod $2\pi).$

This means that $mx = \frac{\pi}{2} + 2\pi\alpha$ and $nx = \frac{\pi}{2} + 2\pi\beta$ for any integers $\alpha$ and $\beta$.

As in Solution 1, take the ratio of the two equations: \[\frac{mx}{nx} = \frac{\frac{\pi}{2}+2\pi\alpha}{\frac{\pi}{2}+2\pi\beta} \implies \frac{m}{n} = \frac{\frac{1}{2}+2\alpha}{\frac{1}{2}+2\beta} \implies \frac{m}{n} = \frac{4\alpha+1}{4\beta+1}\]

Now notice that the numerator and denominator of $\frac{4\alpha+1}{4\beta+1}$ are both odd, which means that $m$ and $n$ have the same power of two (the powers of 2 cancel out).

Let the common power be $p$: then $m = 2^p\cdot a$, and $n = 2^p\cdot b$ where $a$ and $b$ are integers between 1 and 30.

We can now rewrite the equation: \[\frac{2^p\cdot a}{2^p\cdot b} = \frac{4\alpha+1}{4\beta+1} \implies \frac{a}{b} = \frac{4\alpha+1}{4\beta+1}\]

Now it is easy to tell that $a \equiv 1 ($mod $4)$ and $b \equiv 1 ($mod $4)$. However, there is another case: that

$a \equiv 3 ($mod $4)$ and $b \equiv 3 ($mod $4)$. This is because multiplying both $4\alpha+1$ and $4\beta+1$ by $-1$ will not change the fraction, but each congruence will be changed to $-1 ($mod $4) \equiv 3 ($mod $4)$.

From the first set of congruences, we find that $a$ and $b$ can be two of $\{1, 5, 9, \ldots, 29\}$.

From the second set of congruences, we find that $a$ and $b$ can be two of $\{3, 7, 11, \ldots, 27\}$.

Now all we have to do is multiply by $2^p$ to get back to $m$ and $n$. Let’s organize the solutions in order of increasing values of $p$, keeping in mind that $m$ and $n$ are bounded between 1 and 30.

For $p = 0$ we get $\{1, 5, 9, \ldots, 29\}, \{3, 7, 11, \ldots, 27\}$.

For $p = 1$ we get $\{2, 10, 18, 26\}, \{6, 14, 22, 30\}$

For $p = 2$ we get $\{4, 20\}, \{12, 28\}$

If we increase the value of $p$ more, there will be less than two integers in our sets, so we are done there.

There are 8 numbers in the first set, 7 in the second, 4 in the third, 4 in the fourth, 2 in the fifth, and 2 in the sixth.

In each of these sets we can choose 2 numbers to be $m$ and $n$ and then assign them in increasing order. Thus there are:

\[\dbinom{8}{2}+\dbinom{7}{2}+\dbinom{4}{2}+\dbinom{4}{2}+\dbinom{2}{2}+\dbinom{2}{2} = 28+21+6+6+1+1 = \boxed{63}\] possible pairs $(m,n)$ that satisfy the conditions.

-KingRavi

Solution 3

We know that the range of $sin$ is between $-1$ and $1$.

Thus, the only way for the sum to be $2$ is for $sin$ of $mx$ and $nx$ to both be $1$.

The $sin$ of $(90+360k)$ is equal to 1.

Assuming $mx$ and $nm$ are both positive, m and n could be $1,5,9,13,17,21,25,29$. There are $8$ ways, so $\dbinom{8}{2}$.

If both are negative, m and n could be $3,7,11,15,19,23,27$. There are $7$ ways, so $\dbinom{7}{2}$.

However, the pair $(1,5)$ could also be $(2, 10)$ and so on. The same goes for some other pairs.

In total there are $14$ of these extra pairs.

The answer is $28+21+14 = \boxed{063}$

Remark

The graphs of $r\leq\sin(m\theta)+\sin(n\theta)$ and $r=2$ are shown here in Desmos: https://www.desmos.com/calculator/busxadywja

Move the sliders around for $1\leq m \leq 29$ and $2\leq m+1\leq n\leq30$ to observe the geometric representation generated by each pair $(m,n).$

~MRENTHUSIASM (inspired by TheAMCHub)

Video Solution

https://www.youtube.com/watch?v=LUkQ7R1DqKo ~Mathematical Dexterity

See Also

2021 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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