Difference between revisions of "2021 Fall AMC 12A Problems/Problem 19"

(See Also)
(See Also)
Line 16: Line 16:
  
 
==See Also==
 
==See Also==
{{AMC12 box|year=2021 Fall|ab=A|before=First Problem|num-a=19}}
+
{{AMC12 box|year=2021 Fall|ab=A|before=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:41, 23 November 2021

Problem 19

Let $x$ be the least real number greater than $1$ such that sin$(x)$ = sin$(x^2)$, where the arguments are in degrees. What is $x$ rounded up to the closest integer?

$\textbf{(A) } 10 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 14 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 20$

Solution 1

The smallest $x$ to make $\sin(x) = \sin(x^2)$ would require $x=x^2$, but since $x$ needs to be greater than $1$, these solutions are not valid.

The next smallest $x$ would require $x=180-x^2$, or $x^2+x=180$.

After a bit of guessing and checking, we find that $12^2+12=156$, and $13^2+13=182$, so the solution lies between $12{ }$ and $13$, making our answer $\boxed{\textbf{(B) } 13}.$

Note: One can also solve the quadratic and estimate the radical.

~kingofpineapplz

See Also

2021 Fall AMC 12A (ProblemsAnswer KeyResources)
Preceded by
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png