Difference between revisions of "2023 AMC 12B Problems/Problem 20"
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<math>\textbf{(A)}~\frac{1}{6}\qquad\textbf{(B)}~\frac{1}{5}\qquad\textbf{(C)}~\frac{\sqrt{3}}{8}\qquad\textbf{(D)}~\frac{\arctan \frac{1}{2}}{\pi}\qquad\textbf{(E)}~\frac{2\arcsin \frac{1}{4}}{\pi}</math> | <math>\textbf{(A)}~\frac{1}{6}\qquad\textbf{(B)}~\frac{1}{5}\qquad\textbf{(C)}~\frac{\sqrt{3}}{8}\qquad\textbf{(D)}~\frac{\arctan \frac{1}{2}}{\pi}\qquad\textbf{(E)}~\frac{2\arcsin \frac{1}{4}}{\pi}</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Denote by <math>A_i</math> the position after the <math>i</math>th jump. | Denote by <math>A_i</math> the position after the <math>i</math>th jump. | ||
Thus, to fall into the region centered at <math>A_0</math> and with radius 1, <math>\angle A_2 A_1 A_0 < 2 \arcsin \frac{1/2}{2} = 2 \arcsin \frac{1}{4}</math>. | Thus, to fall into the region centered at <math>A_0</math> and with radius 1, <math>\angle A_2 A_1 A_0 < 2 \arcsin \frac{1/2}{2} = 2 \arcsin \frac{1}{4}</math>. | ||
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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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| + | ==Solution 2== | ||
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| + | (Diagram in progress......) | ||
| + | (Writing in progress......) | ||
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| + | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
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==See Also== | ==See Also== | ||
{{AMC12 box|year=2023|ab=B|num-b=19|num-a=21}} | {{AMC12 box|year=2023|ab=B|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 22:24, 15 November 2023
Contents
Problem
Cyrus the frog jumps 2 units in a direction, then 2 more in another direction. What is the probability that he lands less than 1 unit away from his starting position?
Solution 1
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)
Solution 2
(Diagram in progress......) (Writing in progress......)
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. 
