Difference between revisions of "2021 Fall AMC 12A Problems/Problem 19"
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==Problem 19== | ==Problem 19== | ||
− | Let <math>x</math> be the least real number greater than <math>1</math> such that | + | Let <math>x</math> be the least real number greater than <math>1</math> such that <math>\sin(x)</math>= \sin(x^2)<math>, where the arguments are in degrees. What is </math>x<math> rounded up to the closest integer? |
− | <math>\textbf{(A) } 10 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 14 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 20 | + | </math>\textbf{(A) } 10 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 14 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 20$ |
==Solution 1== | ==Solution 1== |
Revision as of 21:22, 23 November 2021
Problem 19
Let be the least real number greater than such that = \sin(x^2)x\textbf{(A) } 10 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 14 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 20$
Solution 1
The smallest to make would require , but since needs to be greater than , these solutions are not valid.
The next smallest would require , or .
After a bit of guessing and checking, we find that , and , so the solution lies between and , making our answer
Note: One can also solve the quadratic and estimate the radical.
~kingofpineapplz
See Also
2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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