2025 AIME II Problems/Problem 9
Problem
There are 
values of 
in the interval 
where 
. For 
of these values of , the graph of 
is tangent to the -axis. Find 
.
Solution
For 
to happen, whatever is inside the function must be of form 
. We then equate to have
\begin{align*}
7\pi\cdot\sin(5x)=k\pi
\sin(5x)=\frac{k}{7}
\end{align*}
We know that 
, so clearly 
takes all values 
. Since the graph of 
has 5 periods between 
and 
, each of the values 
give 
solutions each. 
give 
solutions each and 
gives 
solutions (to verify this sketch a graph). Thus, 
.
We know that the function is tangent to the x-axis if it retains the same sign on both sides of the function. This is not true for points at 
because one side will be positive and one will be negative. However this will happen if because the sine function 'bounces back' and goes over the same values again, and 
of these values exist.\\
Thus, 
.
See also
| 2025 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
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