2021 AIME II Problems/Problem 7
Problem
Let and
be real numbers that satisfy the system of equations
There exist relatively prime positive integers
and
such that
Find
.
Solution 3(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.
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
. From here you already know
and
, so you can solve for
and
pretty easily and see that
. So the answer is
.
~ math31415926535
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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