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 "2021 AIME I Problems/Problem 7"

m
(→‎Solution 1)
 
(33 intermediate revisions by 10 users not shown)
Line 3: Line 3:
  
 
==Solution 1==
 
==Solution 1==
−
The maximum value of <math>\sin \theta</math> is <math>1</math>, which is achieved at <math>\theta = \frac{\pi}{2}+2k\pi</math> for some integer <math>k</math>. This is left as an exercise to the reader.
+
It is trivial that the maximum value of <math>\sin \theta</math> is <math>1</math>, is achieved at <math>\theta = \frac{\pi}{2}+2k\pi</math> for some integer <math>k</math>.
  
 
This implies that <math>\sin(mx) = \sin(nx) = 1</math>, and that <math>mx = \frac{\pi}{2}+2a\pi</math> and <math>nx = \frac{\pi}{2}+2b\pi</math>, for integers <math>a, b</math>.
 
This implies that <math>\sin(mx) = \sin(nx) = 1</math>, and that <math>mx = \frac{\pi}{2}+2a\pi</math> and <math>nx = \frac{\pi}{2}+2b\pi</math>, for integers <math>a, b</math>.
Line 12: Line 12:
 
If <math>k = 1</math>, then <math>m \equiv n \equiv 1 \pmod 4</math>. This corresponds to choosing two elements from the set <math>\{1, 5, 9, 13, 17, 21, 25, 29\}</math>. There are <math>\binom 82</math> ways to do so.
 
If <math>k = 1</math>, then <math>m \equiv n \equiv 1 \pmod 4</math>. This corresponds to choosing two elements from the set <math>\{1, 5, 9, 13, 17, 21, 25, 29\}</math>. There are <math>\binom 82</math> ways to do so.
  
−
If <math>k < 1</math>, by multiplying <math>m</math> and <math>n</math> by the same constant <math>c = \frac{1}{k}</math>, we have that <math>mc \equiv nc \equiv 1 \pmod 4</math>. Then either <math>m \equiv n \equiv 1 \pmod 4</math>, or <math>m \equiv n \equiv 3 \pmod 4</math>. 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 <math>\{3, 7, 11, 15, 19, 23, 27\}</math>. There are <math>\binom 72</math> ways here.
+
If <math>k < 1</math>, by multiplying <math>m</math> and <math>n</math> by the same constant <math>c = \frac{1}{k}</math>, we have that <math>mc \equiv nc \equiv 1 \pmod 4</math>. Then either <math>m \equiv n \equiv 1 \pmod 4</math>, or <math>m \equiv n \equiv 3 \pmod 4</math>. 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 <math>\{3, 7, 11, 15, 19, 23, 27\}</math>. There are <math>\binom 72</math> ways here. (This argument seems to have a logical flaw)
  
 
Finally, if <math>k > 1</math>, note that <math>k</math> must be an integer. This means that <math>m, n</math> belong to the set <math>\{k, 5k, 9k, \dots\}</math>, or <math>\{3k, 7k, 11k, \dots\}</math>. Taking casework on <math>k</math>, we get the sets <math>\{2, 10, 18, 26\}, \{6, 14, 22, 30\}, \{4, 20\}, \{12, 28\}</math>. Some sets have been omitted; this is because they were counted in the other cases already. This sums to <math>\binom 42 + \binom 42 + \binom 22 + \binom 22</math>.
 
Finally, if <math>k > 1</math>, note that <math>k</math> must be an integer. This means that <math>m, n</math> belong to the set <math>\{k, 5k, 9k, \dots\}</math>, or <math>\{3k, 7k, 11k, \dots\}</math>. Taking casework on <math>k</math>, we get the sets <math>\{2, 10, 18, 26\}, \{6, 14, 22, 30\}, \{4, 20\}, \{12, 28\}</math>. Some sets have been omitted; this is because they were counted in the other cases already. This sums to <math>\binom 42 + \binom 42 + \binom 22 + \binom 22</math>.
  
−
In total, there are <math>\binom 82 + \binom 72 + \binom 42 + \binom 42 + \binom 22 + \binom 22 = \boxed{63}</math> pairs of <math>(m, n)</math>.
+
In total, there are <math>\binom 82 + \binom 72 + \binom 42 + \binom 42 + \binom 22 + \binom 22 = \boxed{063}</math> pairs of <math>(m, n)</math>.
  
 
This solution was brought to you by ~Leonard_my_dude~
 
This solution was brought to you by ~Leonard_my_dude~
Line 56: Line 56:
  
 
For <math>p = 2</math> we get <math>\{4, 20\}, \{12, 28\}</math>
 
For <math>p = 2</math> we get <math>\{4, 20\}, \{12, 28\}</math>
 +
 +
Note that <math>16\mid{a}</math> since <math>m</math> will cancel out a factor of 4 from <math>a</math>, and <math>\frac{a}{m}</math> must contain a factor of 4. Again, <math>1-4X</math> will never contribute a factor of 2. Simply inspecting, we see two feasible values for <math>a</math> and <math>m</math> such that <math>a+m\leq30</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.
 
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.
Line 63: Line 65:
 
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:
 
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.
+
<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{063}</cmath> possible pairs <math>(m,n)</math> that satisfy the conditions.
  
 
-KingRavi
 
-KingRavi
  
 
==Solution 3==
 
==Solution 3==
−
We know that the range of <math>sin</math> is between <math>-1</math> and <math>1</math>.
+
We know that the range of sine is between <math>-1</math> and <math>1</math>, inclusive.
  
−
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>.
+
Thus, the only way for the sum to be <math>2</math> is for <math>\sin(mx)=\sin(nx)=1</math>.
  
−
The <math>sin</math> of <math>(90+360k)</math> is equal to 1.
+
Note that <math>\sin(90+360k)=1</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>.
+
Assuming <math>mx</math> and <math>nx</math> are both positive, <math>m</math> and <math>n</math> 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>.
+
If both are negative, <math>m</math> and <math>n</math> 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.
 
However, the pair <math>(1,5)</math> could also be <math>(2, 10)</math> and so on. The same goes for some other pairs.
Line 82: Line 84:
 
In total there are <math>14</math> of these extra pairs.
 
In total there are <math>14</math> of these extra pairs.
  
−
The answer is <math>28+21+14 = \boxed{063}</math>
+
The answer is <math>28+21+14 = \boxed{063}</math>.
 +
 
 +
==Solution 4==
 +
 
 +
The equation implies that <math>\sin(mx)=\sin(nx)=1</math>. Therefore, we can write <math>mx</math> as <math>2{\pi}k_1+\frac{\pi}{2}</math> and <math>nx</math> as <math>2{\pi}k_2+\frac{\pi}{2}</math> for integers <math>k_1</math> and <math>k_2</math>. Then, <math>\frac{mx}{nx}=\frac{m}{n}=\frac{2k_1+\frac{1}{2}}{2k_2+\frac{1}{2}}</math>. Cross multiplying, we get <math>m\cdot{(2k_2+\frac{1}{2})}=n\cdot{(2k_1+\frac{1}{2})} \Longrightarrow 4k_2m-4k_1n=n-m</math>. Let <math>n-m=a</math> so the equation becomes <math>4(m(k_2-k_1)+k_1a)=a</math>. Let <math>k_2-k_1=X</math> and <math>k_1=Y</math>, then the equation becomes <math>a=4Ym+4Xa \Longrightarrow \frac{a(1-4X)}{m}=4Y</math>. Note that <math>X</math> and <math>Y</math> can vary accordingly, and <math>4\mid{a}</math>. Next, we do casework on <math>m\pmod{4}</math>:
 +
 
 +
If <math>m\equiv 1\pmod{4}</math>:
 +
 
 +
Once <math>a</math> and <math>m</math> are determined, <math>n</math> is determined, so <math>a+m\leq30</math>. <math>a\in \{4,8,12,\dots,28\}</math> and <math>m\in \{1,5,9,\dots,29\}</math>. Therefore, there are <math>\sum_{i=1}^{7}{i}=28</math> ways for this case such that <math>a+m\leq30</math>.
 +
 
 +
If <math>m\equiv 3\pmod{4}</math>:
 +
 
 +
<math>a\in \{4,8,12,\dots,28\}</math> and <math>m\in \{3,7,11,\dots,27\}</math>. Therefore, there are <math>\sum_{i=1}^{6}{i}=21</math> ways such that <math>a+m\leq30</math>.
 +
 
 +
If <math>m\equiv 2\pmod{4}</math>:
 +
 
 +
Note that <math>8\mid{a}</math> since <math>m</math> in this case will have a factor of 2, which will cancel out a factor of 2 in <math>a</math>, and we need the left hand side to divide 4. Also, <math>1-4X\equiv 1\pmod{4}</math> so it is odd and will therefore never contribute a factor of 2. <math>a\in \{8,16,24\}</math> and <math>m\in \{2,6,10,\dots,30\}</math>. Following the condition <math>a+m\leq30</math>, we conclude that there are <math>6+4+2=12</math> ways for this case.
 +
 
 +
If <math>m\equiv 0\pmod{4}</math>:
 +
 
 +
Adding all the cases up, we obtain <math>28+21+12+2=\boxed{063}</math>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja Magnetoninja]
  
 
==Remark==
 
==Remark==
Line 93: Line 117:
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/O84aJ5OTZ2E
 
https://youtu.be/O84aJ5OTZ2E
 +
 
~mathproblemsolvingskills
 
~mathproblemsolvingskills
−
 
  
 
==Video Solution==
 
==Video Solution==
 
https://www.youtube.com/watch?v=LUkQ7R1DqKo
 
https://www.youtube.com/watch?v=LUkQ7R1DqKo
 +
 
~Mathematical Dexterity
 
~Mathematical Dexterity
  

Latest revision as of 08:12, 3 December 2023

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

It is trivial that the maximum value of $\sin \theta$ is $1$, is achieved at $\theta = \frac{\pi}{2}+2k\pi$ for some integer $k$.

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. (This argument seems to have a logical flaw)

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 $\binom 82 + \binom 72 + \binom 42 + \binom 42 + \binom 22 + \binom 22 = \boxed{063}$ 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 $\{4, 20\}, \{12, 28\}$

Note that $16\mid{a}$ since $m$ will cancel out a factor of 4 from $a$, and $\frac{a}{m}$ must contain a factor of 4. Again, $1-4X$ will never contribute a factor of 2. Simply inspecting, we see two feasible values for $a$ and $m$ such that $a+m\leq30$.

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:

\[\dbinom{8}{2}+\dbinom{7}{2}+\dbinom{4}{2}+\dbinom{4}{2}+\dbinom{2}{2}+\dbinom{2}{2} = 28+21+6+6+1+1 = \boxed{063}\] possible pairs $(m,n)$ that satisfy the conditions.

-KingRavi

Solution 3

We know that the range of sine is between $-1$ and $1$, inclusive.

Thus, the only way for the sum to be $2$ is for $\sin(mx)=\sin(nx)=1$.

Note that $\sin(90+360k)=1$.

Assuming $mx$ and $nx$ 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 $28+21+14 = \boxed{063}$.

Solution 4

The equation implies that $\sin(mx)=\sin(nx)=1$. Therefore, we can write $mx$ as $2{\pi}k_1+\frac{\pi}{2}$ and $nx$ as $2{\pi}k_2+\frac{\pi}{2}$ for integers $k_1$ and $k_2$. Then, $\frac{mx}{nx}=\frac{m}{n}=\frac{2k_1+\frac{1}{2}}{2k_2+\frac{1}{2}}$. Cross multiplying, we get $m\cdot{(2k_2+\frac{1}{2})}=n\cdot{(2k_1+\frac{1}{2})} \Longrightarrow 4k_2m-4k_1n=n-m$. Let $n-m=a$ so the equation becomes $4(m(k_2-k_1)+k_1a)=a$. Let $k_2-k_1=X$ and $k_1=Y$, then the equation becomes $a=4Ym+4Xa \Longrightarrow \frac{a(1-4X)}{m}=4Y$. Note that $X$ and $Y$ can vary accordingly, and $4\mid{a}$. Next, we do casework on $m\pmod{4}$:

If $m\equiv 1\pmod{4}$:

Once $a$ and $m$ are determined, $n$ is determined, so $a+m\leq30$. $a\in \{4,8,12,\dots,28\}$ and $m\in \{1,5,9,\dots,29\}$. Therefore, there are $\sum_{i=1}^{7}{i}=28$ ways for this case such that $a+m\leq30$.

If $m\equiv 3\pmod{4}$:

$a\in \{4,8,12,\dots,28\}$ and $m\in \{3,7,11,\dots,27\}$. Therefore, there are $\sum_{i=1}^{6}{i}=21$ ways such that $a+m\leq30$.

If $m\equiv 2\pmod{4}$:

Note that $8\mid{a}$ since $m$ in this case will have a factor of 2, which will cancel out a factor of 2 in $a$, and we need the left hand side to divide 4. Also, $1-4X\equiv 1\pmod{4}$ so it is odd and will therefore never contribute a factor of 2. $a\in \{8,16,24\}$ and $m\in \{2,6,10,\dots,30\}$. Following the condition $a+m\leq30$, we conclude that there are $6+4+2=12$ ways for this case.

If $m\equiv 0\pmod{4}$:

Adding all the cases up, we obtain $28+21+12+2=\boxed{063}$

~Magnetoninja

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://youtu.be/O84aJ5OTZ2E

~mathproblemsolvingskills

Video Solution

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

~Mathematical Dexterity

See Also

2021 AIME I (Problems • Answer Key • Resources)
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_7&oldid=206407"