2021 AIME II Problems/Problem 7
Contents
Problem
Let and be real numbers that satisfy the system of equations There exist relatively prime positive integers and such that Find .
Solution 1
From the fourth equation we get Substitute this into the third equation and you get . Hence . Solving, we get or . From the first and second equation, we get . If , substituting we get . If you try solving this you see that this does not have real solutions in , so must be . So . Since , or . If , then the system and does not give you real solutions. So . Since you already know and , so you can solve for and pretty easily and see that . So the answer is .
~math31415926535 ~minor edit by Mathkiddie
Solution 2
Note that can be rewritten as . Hence, .
Rewriting , we get . Substitute and solving, we get We refer to this as Equation 1.
Note that gives . So, , which implies or We refer to this as Equation 2.
Substituting Equation 2 into Equation 1 gives, .
Solving this quadratic yields that .
Now we just try these two cases:
For substituting in Equation 1 gives a quadratic in which has roots .
Again trying cases, by letting , we get , Hence . We know that , Solving these we get or (doesn't matter due to symmetry in ).
So, this case yields solutions .
Similarly trying other three cases, we get no more solutions, Hence this is the solution for .
Finally, .
Therefore, .
~Arnav Nigam
Solution 3
For simplicity purposes, we number the given equations and in that order.
Rearranging and solving for we have Substituting into and solving for we get Substituting and into and simplifying, we rewrite the left side of in terms of and only: Let from which Multiplying both sides by rearranging, and factoring give Substituting back and completing the squares produce If then combining this with we know that and are the solutions of the quadratic Since the discriminant is negative, neither nor is a real number.
If then combining this with we know that and are the solutions of the quadratic or from which Substituting into and we obtain and respectively. Together, we have so the answer is
~MRENTHUSIASM
Solution 4 (way too long)
Let the four equations from top to bottom be listed 1 through 4 respectively. We factor equation 3 like so: Then we plug in equation 2 to receive . By equation 4 we get . Plugging in, we get . Multiply by on both sides to get the quadratic equation . Solving using the quadratic equation, we receive . So, we have to test which one is correct. We repeat a similar process as we did above for equations 1 and 2. We factor equation 2 to get After plugging in equation 1, we get . Now we convert it into a quadratic to receive . The value of will depend on . So we obtain the discriminant . Let d = -5. Then , so , discriminant is , which makes this a dead end. Thus For , making . This means the discriminant is just , so we obtain two values for as well. We get either or . So, we must AGAIN test which one is correct. We know , and , so we use these values for testing. Let . Then , so . We thus get , which leads to the quadratic . The discriminant for this is . That means this value of is wrong, so . Thus we get polynomial . The discriminant this time is , so we get two values for . Through simple inspection, you may see they are interchangeable, as if you take the value , you get . If you take the value , you get . So it doesn't matter. That means the sum of all their squares is so the answer is ~amcrunner
Solution 5
Let the four equations from top to bottom be listed through respectively. Multiplying both sides of by and factoring some terms gives us . Substituting using equations and gives us , and solving gives us or . Plugging this back into gives us , or using the other solution for gives us . Solving both of these equations gives us when and when .
Multiplying both sides of by and factoring some terms gives us . Testing will give us an imaginary solution for , so therefore and . This gets us . Solving for gives us or . With a bit of testing, we can see that the correct value of is . Now we know and , , and it is obvious that and or the other way around, and therefore, , giving us the answer .
~hihitherethere
Video Solution
https://www.youtube.com/watch?v=2rrX1G7iZqg
Video Solution by Interstigation
~Interstigation
See Also
2021 AIME II (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 |
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