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"

(→‎Solution 1)
(→‎Solution 1)
 
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. (This argument seems to have a logical flaw)
+
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 *check note at bottom*)
  
 
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>.
Line 19: Line 19:
  
 
This solution was brought to you by ~Leonard_my_dude~
 
This solution was brought to you by ~Leonard_my_dude~
 +
 +
 +
==Note: Detailed Explanation for <math>m \equiv n \equiv 3 \pmod 4</math>==
 +
Many of the solutions I see here lack an explicit explanation for why every solution to <math>m \equiv n \equiv 3 \pmod 4</math> constitutes a valid solution to the problem, so I will explain it here. The reason why all solutions where <math>m \equiv n \equiv 1 \pmod 4</math> works is pretty intuitive, as <math>\frac{m}{n}</math> would then be able to be expressed as <math>\frac{4a + 1}{4b + 1}</math>. Similarly, note that <math>-(4x + 1) \equiv 3 \pmod 4</math>. Thus, when solving the equation for <math>\frac{m}{n} = \frac{4a + 1}{4b + 1}</math>, every case in which <math>m \equiv n \equiv 3 \pmod 4</math> will produce a valid solution when <math>a</math> and <math>b</math> are both negative. For example, when <math>m = 7</math> and <math>n = 11</math>, <math>a = -2</math> and <math>b = -3</math> gives <math>\frac{4a + 1}{4b + 1} = \frac{-7}{-11} = \frac{7}{11}</math>. Thus, we add <math>\binom 72</math> as every case where <math>m \equiv n \equiv 3 \pmod 4</math> produces a valid solution.
 +
 +
- yangomango
  
 
==Solution 2==
 
==Solution 2==

Latest revision as of 12:18, 27 December 2024

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 *check note at bottom*)

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~


Note: Detailed Explanation for $m \equiv n \equiv 3 \pmod 4$

Many of the solutions I see here lack an explicit explanation for why every solution to $m \equiv n \equiv 3 \pmod 4$ constitutes a valid solution to the problem, so I will explain it here. The reason why all solutions where $m \equiv n \equiv 1 \pmod 4$ works is pretty intuitive, as $\frac{m}{n}$ would then be able to be expressed as $\frac{4a + 1}{4b + 1}$. Similarly, note that $-(4x + 1) \equiv 3 \pmod 4$. Thus, when solving the equation for $\frac{m}{n} = \frac{4a + 1}{4b + 1}$, every case in which $m \equiv n \equiv 3 \pmod 4$ will produce a valid solution when $a$ and $b$ are both negative. For example, when $m = 7$ and $n = 11$, $a = -2$ and $b = -3$ gives $\frac{4a + 1}{4b + 1} = \frac{-7}{-11} = \frac{7}{11}$. Thus, we add $\binom 72$ as every case where $m \equiv n \equiv 3 \pmod 4$ produces a valid solution.

- yangomango

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=238001"