2021 AMC 12A Problems/Problem 19
Contents
[hide]Problem
How many solutions does the equation have in the closed interval
?
Solution 1 (Inverse Trigonometric Functions)
The ranges of and
are both
, which is included in the range of
, so we can use it with no issues.
This only happens at on the interval
, because one of
and
must be
and the other
. Therefore, the answer is
~Tucker
Solution 3 (Graphs and Analysis)
Let and
This problem is equivalent to counting the intersections of the graphs of
and
in the closed interval
We make a table of values, as shown below:
The graph of
in
(from left to right) is the same as the graph of
in
(from right to left). The output is from
to
(from left to right), inclusive, and strictly decreasing.
The graph of in
(from left to right) has two parts:
in
(from left to right). The output is from
to
(from left to right), inclusive, and strictly decreasing.
in
(from right to left). The output is from
to
(from left to right), inclusive, and strictly increasing.
If then
and
So, their graphs do not intersect.
If then
Clearly, the graphs intersect at
and
(at points
and
), but we will prove/disprove that they are the only points of intersection:
Let and
It follows that
Since
we know that
by the cofunction identities:
Applying Solution 2's argument to deduce that and
are the only points of intersection. So, the answer is
Graphs of and
in Desmos: https://www.desmos.com/calculator/9ypgulyzov
~MRENTHUSIASM (credit given to TheAMCHub)
Video Solution by OmegaLearn (Using Triangle Inequality & Trigonometry)
~ pi_is_3.14
See also
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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.