2021 AIME I Problems/Problem 7
Find the number of pairs of positive integers with such that there exists a real number satisfying
The maximum value of is , which is achieved at for some integer . This is left as an exercise to the reader.
This implies that , and that and , for integers .
Taking their ratio, we have It remains to find all that satisfy this equation.
If , then . This corresponds to choosing two elements from the set . There are ways to do so.
If , by multiplying and by the same constant , we have that . Then either , or . 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 . There are ways here. (This argument seems to have a logical flaw)
Finally, if , note that must be an integer. This means that belong to the set , or . Taking casework on , we get the sets . Some sets have been omitted; this is because they were counted in the other cases already. This sums to .
In total, there are pairs of .
This solution was brought to you by ~Leonard_my_dude~
In order for , .
This happens when mod
This means that and for any integers and .
As in Solution 1, take the ratio of the two equations:
Now notice that the numerator and denominator of are both odd, which means that and have the same power of two (the powers of 2 cancel out).
Let the common power be : then , and where and are integers between 1 and 30.
We can now rewrite the equation:
Now it is easy to tell that mod and mod . However, there is another case: that
mod and mod . This is because multiplying both and by will not change the fraction, but each congruence will be changed to mod mod .
From the first set of congruences, we find that and can be two of .
From the second set of congruences, we find that and can be two of .
Now all we have to do is multiply by to get back to and . Let’s organize the solutions in order of increasing values of , keeping in mind that and are bounded between 1 and 30.
For we get .
For we get
For we get
Note that since will cancel out a factor of 4 from , and must contain a factor of 4. Again, will never contribute a factor of 2. Simply inspecting, we see two feasible values for and such that .
If we increase the value of 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 and and then assign them in increasing order. Thus there are:
possible pairs that satisfy the conditions.
We know that the range of sine is between and , inclusive.
Thus, the only way for the sum to be is for .
Note that .
Assuming and are both positive, and could be . There are ways, so .
If both are negative, and could be . There are ways, so .
However, the pair could also be and so on. The same goes for some other pairs.
In total there are of these extra pairs.
The answer is .
The equation implies that . Therefore, we can write as and as for integers and . Then, . Cross multiplying, we get . Let so the equation becomes . Let and , then the equation becomes . Note that and can vary accordingly, and . Next, we do casework on :
Once and are determined, is determined, so . and . Therefore, there are ways for this case such that .
and . Therefore, there are ways such that .
Note that since in this case will have a factor of 2, which will cancel out a factor of 2 in , and we need the left hand side to divide 4. Also, so it is odd and will therefore never contribute a factor of 2. and . Following the condition , we conclude that there are ways for this case.
Adding all the cases up, we obtain
The graphs of and are shown here in Desmos: https://www.desmos.com/calculator/busxadywja
Move the sliders around for and to observe the geometric representation generated by each pair
~MRENTHUSIASM (inspired by TheAMCHub)
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