Difference between revisions of "2021 AIME II Problems/Problem 7"
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ab\left[\frac{ab+4}{3}\right] + b\left[\frac{ab+4}{3}\right]\left[\frac{90}{ab(ab+4)}\right] + \left[\frac{ab+4}{3}\right]\left[\frac{90}{ab(ab+4)}\right]a + \left[\frac{90}{ab(ab+4)}\right]ab &= 14 \\ | ab\left[\frac{ab+4}{3}\right] + b\left[\frac{ab+4}{3}\right]\left[\frac{90}{ab(ab+4)}\right] + \left[\frac{ab+4}{3}\right]\left[\frac{90}{ab(ab+4)}\right]a + \left[\frac{90}{ab(ab+4)}\right]ab &= 14 \\ | ||
ab\left[\frac{ab+4}{3}\right] + \underbrace{\frac{30}{a} + \frac{30}{b}}_{\text{Group them.}} + \frac{90}{ab+4} &= 14 \\ | ab\left[\frac{ab+4}{3}\right] + \underbrace{\frac{30}{a} + \frac{30}{b}}_{\text{Group them.}} + \frac{90}{ab+4} &= 14 \\ | ||
− | ab\left[\frac{ab+4}{3}\right] + \frac{30(\overbrace{a+b}^{-3})}{ab} + \frac{90}{ab+4} &= 14 \\ | + | ab\left[\frac{ab+4}{3}\right] + \frac{30(\phantom{ }\overbrace{a+b}^{-3}\phantom{ })}{ab} + \frac{90}{ab+4} &= 14 \\ |
ab\left[\frac{ab+4}{3}\right] + \underbrace{\frac{-90}{ab} + \frac{90}{ab+4}}_{\text{Group them.}} &= 14 \\ | ab\left[\frac{ab+4}{3}\right] + \underbrace{\frac{-90}{ab} + \frac{90}{ab+4}}_{\text{Group them.}} &= 14 \\ | ||
ab\left[\frac{ab+4}{3}\right] - \frac{360}{ab(ab+4)}&=14. | ab\left[\frac{ab+4}{3}\right] - \frac{360}{ab(ab+4)}&=14. |
Latest revision as of 14:21, 18 April 2021
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
Solution 2 (Easy Algebra)
We can factor out of the last two equations. Therefore, it becomes . Notice this is just , since . We now have and . We then find in terms of , so . We solve for and find that it is either or . We can now try for these two values, and plug the rest into the equation. Thus, we have . We have and we're done.
~Arcticturn
Solution 3 (Easy and Straightforward 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
Solution 4 (Bash: Two Variables, Two Equations)
Number the given equations and in this 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
Video Solution
https://www.youtube.com/watch?v=2rrX1G7iZqg
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.