2021 AIME II Problems/Problem 7
Let and be real numbers that satisfy the system of equations There exist relatively prime positive integers and such that Find .
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 .
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 .
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
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