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

(Solution 3)
(Solution 3)
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The <math>sin</math> of <math>(90+360k)</math> is equal to 1.
 
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>8C2</math>
+
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 bother are negative, m and n could be <math>3,7,11,15,19,23,27</math>. There are <math>7</math> ways, so <math>7C2</math>
+
If bother 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>
  
 
==See also==
 
==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 22:08, 13 March 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 $a$ and $b$ 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 $\{1, 5, 9, \cdots, 29\}$.

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

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 $\{1, 5, 9, \cdots, 29\}, \{3, 7, 11, \cdots, 27\}$.

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 pairings of $m, n$ that satisfy the conditions.


-KingRavi

Solution 3

We know that the range of $sin$ is between $-1$ and $0$.

Thus, the only way for the sum to be $2$ is the $sin$ 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 bother 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 $28+21+14 = \boxed{063}$

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

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