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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 ==
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== Solution 1 ==
 
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*}
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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 <math>k=-6,-5,-4...4,5,6</math> because one side will be positive and one will be negative. However this will happen if <math>k=-7,7</math> because the sine function 'bounces back' and goes over the same values again, and <math>t=10</math> of these values exist.\\
 
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 <math>k=-6,-5,-4...4,5,6</math> because one side will be positive and one will be negative. However this will happen if <math>k=-7,7</math> because the sine function 'bounces back' and goes over the same values again, and <math>t=10</math> of these values exist.\\
 
Thus, <math>n+t=\boxed{149}</math>.
 
Thus, <math>n+t=\boxed{149}</math>.
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== Solution 2 (Calculus) ==
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For <math>f(x)=0</math>, we must have <math>7\pi\cdot\sin(5x)=k\pi</math> for some integer <math>k</math>. Then <math>\sin(5x)=\frac{k}{7}</math> always satisfies the equation. Notice that on each period of <math>\sin(5x)</math>, each <math>k\in\{-6,-5,\ldots,5,6\}</math> is a <math>y</math>-value at two distinct points, and each <math>k=\pm7</math> is a <math>y</math>-value at one point each. Thus each period has <math>13\cdot2+2\cdot1=28</math> points satisfying the equation. Since the period is <math>\frac{2\pi}{5}</math> and the domain has a length of <math>2\pi</math>, we find that <math>5</math> periods occur in our domain if we include <math>x=0,2\pi</math>. Adding the case where <math>x=0</math>, there are a total of <math>28\cdot5+1=141</math> roots over <math>x\in[0,2\pi]</math>. Subtracting the cases at <math>x=0</math> and <math>x=2\pi</math> yields <math>139</math> total roots. This is our <math>n</math>.
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 +
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Next, we take the derivative of <math>f(x)</math>; using a hideous combination of chain rules we find that
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<cmath>f'(x)=35\pi\cos(5x)\cos(7\pi\sin(5x))=0</cmath>
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Thus, for a point to be tangent to the <math>x</math>-axis, we must have either <math>\cos(5x)=0</math> or <math>\cos(7\pi\sin(5x))=0</math>. In the first case, we know that <math>\sin(5x)=\frac{k}{7}</math> from earlier, so <math>\cos(5x)=\sqrt{1-\left(\frac{k}{7}\right)^2}=0</math>. Then <math>\left(\frac{k}{7}\right)^2=1</math>, so <math>k=\pm7</math>. Recall that over each of the five periods, only one point will satisfy <math>k=7</math>, and only one point will satisfy <math>k=-7</math>. Thus there are <math>2\cdot5=10</math> points in this case.
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In the second case, we must have <math>\cos(7\pi\sin(5x))=0</math>. Substituting <math>\sin(5x)=\frac{k}{7}</math> yields <math>\cos(k\pi)=0</math>. But this is impossible since <math>\cos(0)=1</math> and <math>\cos(\pi)=-1</math>, so there are no points in this case.
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As a result, <math>t=10+0=10</math>, so <math>n+t=139+10=\boxed{149}</math>. ~eevee9406
  
 
==See also==
 
==See also==

Revision as of 01:51, 14 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 1

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, $n+t=\boxed{149}$.

Solution 2 (Calculus)

For $f(x)=0$, we must have $7\pi\cdot\sin(5x)=k\pi$ for some integer $k$. Then $\sin(5x)=\frac{k}{7}$ always satisfies the equation. Notice that on each period of $\sin(5x)$, each $k\in\{-6,-5,\ldots,5,6\}$ is a $y$-value at two distinct points, and each $k=\pm7$ is a $y$-value at one point each. Thus each period has $13\cdot2+2\cdot1=28$ points satisfying the equation. Since the period is $\frac{2\pi}{5}$ and the domain has a length of $2\pi$, we find that $5$ periods occur in our domain if we include $x=0,2\pi$. Adding the case where $x=0$, there are a total of $28\cdot5+1=141$ roots over $x\in[0,2\pi]$. Subtracting the cases at $x=0$ and $x=2\pi$ yields $139$ total roots. This is our $n$.


Next, we take the derivative of $f(x)$; using a hideous combination of chain rules we find that

\[f'(x)=35\pi\cos(5x)\cos(7\pi\sin(5x))=0\]

Thus, for a point to be tangent to the $x$-axis, we must have either $\cos(5x)=0$ or $\cos(7\pi\sin(5x))=0$. In the first case, we know that $\sin(5x)=\frac{k}{7}$ from earlier, so $\cos(5x)=\sqrt{1-\left(\frac{k}{7}\right)^2}=0$. Then $\left(\frac{k}{7}\right)^2=1$, so $k=\pm7$. Recall that over each of the five periods, only one point will satisfy $k=7$, and only one point will satisfy $k=-7$. Thus there are $2\cdot5=10$ points in this case.

In the second case, we must have $\cos(7\pi\sin(5x))=0$. Substituting $\sin(5x)=\frac{k}{7}$ yields $\cos(k\pi)=0$. But this is impossible since $\cos(0)=1$ and $\cos(\pi)=-1$, so there are no points in this case.

As a result, $t=10+0=10$, so $n+t=139+10=\boxed{149}$. ~eevee9406

See also

2025 AIME II (Problems • Answer Key • Resources)
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Problem 8 		Followed by
Problem 10
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