2022 AMC 12A Problems/Problem 17
Contents
[hide]Problem
Suppose is a real number such that the equation
has more than one solution in the interval
. The set of all such
that can be written
in the form
where
and
are real numbers with
. What is
?
Solution 1
We are given that
Using the sine double angle formula combine with the fact that , which can be derived using sine angle addition with
, we have
Since
as it is on the open interval
, we can divide out
from both sides, leaving us with
Now, distributing
and rearranging, we achieve the equation
which is a quadratic in
.
Applying the quadratic formula to solve for , we get
and expanding the terms under the radical, we get
Factoring, since
, we can simplify our expression even further to
Now, solving for our two solutions, and
.
Since yields a solution that is valid for all
, that being
, we must now solve for the case where
yields a valid value.
As ,
, and therefore
, and
.
There is one more case we must consider inside this interval though, the case where , as this would lead to a double root for
, yielding only one valid solution for
. Solving for this case,
.
Therefore, combining this fact with our solution interval, , so the answer is
.
- DavidHovey
Solution 2
We can optimize from the step from in solution 1 by writing
and then get
Now, solving for our two solutions, and
.
Since yields a solution that is valid for all
, that being
, we must now solve for the case where
yields a valid value.
As ,
, and therefore
, and
.
There is one more case we must consider inside this interval though, the case where , as this would lead to a double root for
, yielding only one valid solution for
. Solving for this case,
.
Therefore, combining this fact with our solution interval, , so the answer is
.
- Dan
Solution 3
Use the sum to product formula to obtain . Use the double angle formula on the RHS to obtain
. From here, it is obvious that
is always a solution, and thus we divide by
to get
We wish to find all
such that there is at least one more solution to this equation distinct from
. Letting
, and noting that
, we can rearrange our equation to
The smallest value
where
is
, which is not in our domain so we divide by
to obtain
. By the trivial inequality,
. Furthermore,
, so
. Also, if
, then the solution to this equation would be shared with
, so there would only be one distinct solution. Finally, because
due to the restrictions of a sine wave, and that
due to the restrictions on
, we have
with
. Thus,
, so our final answer is
.
~sigma
Video Solution 1 (Quick and Simple)
~Education, the Study of Everything
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=0gCMvUmZtpI
See Also
2022 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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 |
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