2021 AIME II Problems/Problem 7
Contents
[hide]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
Solution 2 (Easy Algebra)
We can factor out the d out of the last 2 equations. Therefore, it becomes abc + d(bc + ac + ab) = 14. Notice this is just abc -4b since bc + ac + ab = -4. We now have abc -4d = 14 and abcd = 30. We then put d in terms of abc, so abc = . We solve fore d and it is either or -5. We can now try for these two values, and plug the rest in to the equation. Thus, we have 33 + = = = . We have 141 + 4 = and we're done.
~Arcticturn
Solution 2 (Easy Algebra)
can be rewritten as . Hence,
Rewriting , we get . Substitute and solving, we get, call this Equation 1
gives . So, , which implies or call this equation 2.
Substituting Eq 2 in Eq 1 gives,
Solving this quadratic yields that
Now we just try these 2 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 a,b)
So, this case yields solutions
Similarly trying other three cases, we get no more solutions, Hence this is the solution for
Finally,
So,
- Arnav Nigam
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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