Difference between revisions of "2018 AMC 10A Problems/Problem 12"
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~OlutosinNGA | ~OlutosinNGA | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/ZWvjqjQTqJw | ||
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+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Revision as of 09:42, 2 December 2020
Contents
Problem
How many ordered pairs of real numbers satisfy the following system of equations?
Solutions
Solution 1
We can solve this by graphing the equations. The second equation looks challenging to graph, but start by graphing it in the first quadrant only (which is easy since the inner absolute value signs can be ignored), then simply reflect that graph into the other quadrants.
The graph looks something like this:
Now, it becomes clear that there are
intersection points.
Solution 2
can be rewritten to
. Substituting
for
in the second equation will give
. Splitting this question into casework for the ranges of
will give us the total number of solutions.
:
will be negative so
Subcase 1:![]()
is positive so
and
and
Subcase 2:![]()
is negative so
.
and so there are no solutions (
can't equal to
)
:
It is fairly clear that
:
will be positive so
Subcase 1:![]()
will be negative so
. There are no solutions (again,
can't equal to
)
Subcase 2:![]()
will be positive so
.
and
.
Thus, the solutions are:
, and the answer is
.
Solution 3
Note that can take on either of four values:
,
,
,
.
Solving the equations (by elimination, either adding the two equations or subtracting),
we obtain the three solutions:
,
,
so the answer is
. One of those equations overlap into
so there's only 3 solutions.
~trumpeter, ccx09 ~minor edit, XxHalo711
Solution 4
Just as in solution , we derive the equation
. If we remove the absolute values, the equation collapses into four different possible values.
,
,
, and
, each equal to either
or
. Remember that if
, then
. Because we have already taken
and
into account, we can eliminate one of the conjugates of each pair, namely
and
, and
and
. Find the values of
when
,
,
and
. We see that
and
give us the same value for
, so the answer is
~Zeric Hang
Solution 5
Just as in solution , we derive the equation
. Squaring both sides in the second equation gives
. Putting
and doing a little calculation gives
. From here we know that
is either positive or negative.
When positive, we get and then,
or
.
When negative, we get
and then,
or
. Clearly, there are
different pairs of values and that gives us
~OlutosinNGA
Video Solution
~savannahsolver
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.