Difference between revisions of "2021 AIME II Problems/Problem 7"
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Therefore, <math>m + n = 141 + 4 = \boxed{145}</math>. | Therefore, <math>m + n = 141 + 4 = \boxed{145}</math>. | ||
− | + | ~Arnav Nigam | |
==Solution 3== | ==Solution 3== |
Revision as of 00:27, 17 March 2022
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
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
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.