Difference between revisions of "2023 AMC 12B Problems/Problem 20"
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
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| + | {{AMC12 box|year=2023|ab=B|num-b=19|num-a=21}} | ||
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Revision as of 20:34, 15 November 2023
Solution
Denote by 
the position after the 
th jump.
Thus, to fall into the region centered at 
and with radius 1, 
.
Therefore, the probability is
![\[ \frac{2 \cdot 2 \arcsin \frac{1}{4}}{2 \pi} = \boxed{\textbf{(E) } \frac{2 \arcsin \frac{1}{4}}{\pi}}. \]](http://latex.artofproblemsolving.com/6/e/d/6edb5c671433291b14a0e16f099947181f8a4717.png)
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
| 2023 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 19 |
Followed by Problem 21 |
| 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. 
