2025 AIME II Problems/Problem 9

Problem

There are $n$ values of $x$ in the interval $0<x<2\pi$ where $f(x)=\sin(7\pi\cdot\sin(5x))=0$ . For $t$ of these $n$ values of $x$ , the graph of $y=f(x)$ is tangent to the $x$ -axis. Find $n+t$ .

Solution

Taking the inverse on both sides yields $7\pi\cdot\sin(5x)=k\pi$ for $1\le k\le 7$ . Dividing on both sides and isolating the sine yields $\sin(5x)=\frac{k}{7}$ . For each $1\le k\le 6$ , there will be 10 solutions, and there will be 5 solutions for $\sin(5x)=1$ in the given domain. Thus, $n=65$ .

Basic calculus techniques would tell us that solving the equation $f’(x)=\cos(7\pi\cdot\sin(5x))\cdot35\pi\cos(5x)=0$ would give us the extrema. Solving the equation would give us either $\cos(5x)=0$ which gives $x=\frac{(2k+1)\pi}{10}$ for $0\le k\le 9$ , which are also all roots. Solving the other equation would very quickly tell us that they are all not roots besides being extrema. This tells us that $t=10$ .

Summing gives the final answer, $\boxed{075}$ .

(Please make sure I did not make any mistakes, thanks. If it is verified then please remove this line.)

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2025 AIME II Problems/Problem 9

Problem

Solution