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>. | ||
| − | == Solution == | + | == Solution 1 == |
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>. | 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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(Please make sure I did not make any mistakes, thanks. If it is verified then please remove this line.) | (Please make sure I did not make any mistakes, thanks. If it is verified then please remove this line.) | ||
| + | (I think this solution is wrong, I got 149 not 75. See below -lisztepos) | ||
| + | |||
| + | == 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 | ||
| + | \begin{align*} | ||
| + | 7\pi\cdot\sin(5x)=k\pi | ||
| + | \sin(5x)=\frac{k}{7} | ||
| + | \end{align*} | ||
| + | We know that <math>-1\le \sin{5x} \ge 1</math>, so clearly <math>k</math> takes all values <math>-7\le k \ge 7</math>. Since the graph of <math>\sin{5x}</math> has 5 periods between <math>0</math> and <math>360</math>, each of the values <math>k=-6,-5,-4...-1,1,2...6</math> give <math>10</math> solutions each. <math>k=-7,7</math> give <math>5</math> solutions each and <math>k=0</math> gives <math>9</math> solutions (to verify this sketch a graph). Thus, <math>n=139</math>.\\ | ||
| + | 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>. | ||
| + | |||
| + | ==See also== | ||
| + | {{AIME box|year=2025|num-b=8|num-a=10|n=II}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Revision as of 22:24, 13 February 2025
Contents
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
Taking the inverse on both sides yields 
for 
. Dividing on both sides and isolating the sine yields 
. For each 
, there will be 10 solutions, and there will be 5 solutions for 
in the given domain. Thus, 
.
Basic calculus techniques would tell us that solving the equation 
would give us the extrema. Solving the equation would give us either 
which gives 
for 
, 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 
.
Summing gives the final answer, 
.
(Please make sure I did not make any mistakes, thanks. If it is verified then please remove this line.) (I think this solution is wrong, I got 149 not 75. See below -lisztepos)
Solution 2
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
