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Difference between revisions of "2025 AIME II Problems/Problem 9"
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There are <math>n</math> values of <math>x</math> in the interval <math>0<x<2\pi</math> where <math>f(x)=\sin(7\pi\cdot\sin(5x))=0</math>. For <math>t</math> of these <math>n</math> values of <math>x</math>, the graph of <math>y=f(x)</math> is tangent to the <math>x</math>-axis. Find <math>n+t</math>.
 
There are <math>n</math> values of <math>x</math> in the interval <math>0<x<2\pi</math> where <math>f(x)=\sin(7\pi\cdot\sin(5x))=0</math>. For <math>t</math> of these <math>n</math> values of <math>x</math>, the graph of <math>y=f(x)</math> is tangent to the <math>x</math>-axis. Find <math>n+t</math>.
  
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== Solution 1 ==
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== Solution ==
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Taking the inverse on both sides yields <math>7\pi\cdot\sin(5x)=k\pi</math> for <math>1\le k\le 7</math>. Dividing on both sides and isolating the sine yields <math>\sin(5x)=\frac{k}{7}</math>. For each <math>1\le k\le 6</math>, there will be 10 solutions, and there will be 5 solutions for <math>\sin(5x)=1</math> in the given domain. Thus, <math>n=65</math>.
 
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Basic calculus techniques would tell us that solving the equation <math>f’(x)=\cos(7\pi\cdot\sin(5x))\cdot35\pi\cos(5x)=0</math> would give us the extrema. Solving the equation would give us either <math>\cos(5x)=0</math> which gives <math>x=\frac{(2k+1)\pi}{10}</math> for <math>0\le k\le 9</math>, 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 <math>t=10</math>.
 
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Summing gives the final answer, <math>\boxed{075}</math>.
 
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(Please make sure I did not make any mistakes, thanks. If it is verified then please remove this line.)
 
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(I think this solution is wrong, I got 149 not 75. See below -lisztepos)
 
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== Solution 2 ==
 
 
For <math>\sin(7\pi\cdot\sin(5x))=0</math> to happen, whatever is inside the function must be of form <math>k\pi</math>. We then equate to have
 
For <math>\sin(7\pi\cdot\sin(5x))=0</math> to happen, whatever is inside the function must be of form <math>k\pi</math>. We then equate to have
 
\begin{align*}
 
\begin{align*}

Revision as of 22:47, 13 February 2025

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

For $\sin(7\pi\cdot\sin(5x))=0$ to happen, whatever is inside the function must be of form $k\pi$. We then equate to have \begin{align*} 7\pi\cdot\sin(5x)=k\pi

\sin(5x)=\frac{k}{7} \end{align*} We know that $-1\le \sin{5x} \ge 1$, so clearly $k$ takes all values $-7\le k \ge 7$. Since the graph of $\sin{5x}$ has 5 periods between $0$ and $360$, each of the values $k=-6,-5,-4...-1,1,2...6$ give $10$ solutions each. $k=-7,7$ give $5$ solutions each and $k=0$ gives $9$ solutions (to verify this sketch a graph). Thus, $n=139$.

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 $k=-6,-5,-4...4,5,6$ because one side will be positive and one will be negative. However this will happen if $k=-7,7$ because the sine function 'bounces back' and goes over the same values again, and $t=10$ of these values exist.\\ Thus, $n+t=\boxed{149}$.

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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