Difference between revisions of "2022 AMC 12A Problems/Problem 17"
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- Dan | - Dan | ||
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+ | ==Video Solution 1 (Quick and Simple)== | ||
+ | https://youtu.be/Tl5hBEkHzbA | ||
+ | |||
+ | ~Education, the Study of Everything | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2022|ab=A|num-b=16|num-a=18}} | {{AMC12 box|year=2022|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:54, 14 November 2022
Problem
Supppose 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
Video Solution 1 (Quick and Simple)
~Education, the Study of Everything
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.