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 1
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, 
.
Solution 2 (Calculus)
For 
, we must have 
for some integer . Then 
always satisfies the equation. Notice that on each period of 
, each 
is a 
-value at two distinct points, and each 
is a -value at one point each. Thus each period has 
points satisfying the equation. Since the period is 
and the domain has a length of 
, we find that periods occur in our domain if we include 
. Adding the case where 
, there are a total of 
roots over ![$x\in[0,2\pi]$](http://latex.artofproblemsolving.com/0/1/6/0162b685a82070213f71210ab2a5bd8d897f33a0.png)
. Subtracting the cases at and 
yields 
total roots. This is our .
Next, we take the derivative of 
; using a hideous combination of chain rules we find that
![\[f'(x)=35\pi\cos(5x)\cos(7\pi\sin(5x))=0\]](http://latex.artofproblemsolving.com/b/5/8/b587e5c72b9573f6b2bf022071abd21b92841580.png)
Thus, for a point to be tangent to the -axis, we must have either 
or 
. In the first case, we know that from earlier, so 
. Then 
, so . Recall that over each of the five periods, only one point will satisfy 
, and only one point will satisfy 
. Thus there are 
points in this case.
In the second case, we must have . Substituting yields 
. But this is impossible since 
and 
, so there are no points in this case.
As a result, 
, so 
.
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 | ||
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