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Difference between revisions of "2023 AMC 12B Problems/Problem 20"

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==Solution 1==
 
==Solution 1==
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[[File:2023AMC12BP20.png|center|500px]]
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
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==Solution 2==
 
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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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 
  
 
==Solution 3(coord bash)==
 
==Solution 3(coord bash)==

Revision as of 23:25, 15 November 2023

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?

$\textbf{(A)}~\frac{1}{6}\qquad\textbf{(B)}~\frac{1}{5}\qquad\textbf{(C)}~\frac{\sqrt{3}}{8}\qquad\textbf{(D)}~\frac{\arctan \frac{1}{2}}{\pi}\qquad\textbf{(E)}~\frac{2\arcsin \frac{1}{4}}{\pi}$

Solution 1

2023AMC12BP20.png

(Writing in progress......)

~isabelchen

Solution 2

Denote by $A_i$ the position after the $i$th jump. Thus, to fall into the region centered at $A_0$ and with radius 1, $\angle A_2 A_1 A_0 < 2 \arcsin \frac{1/2}{2} = 2 \arcsin \frac{1}{4}$.

Therefore, the probability is \[ \frac{2 \cdot 2 \arcsin \frac{1}{4}}{2 \pi} = \boxed{\textbf{(E) } \frac{2 \arcsin \frac{1}{4}}{\pi}}. \]

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 3(coord bash)

Let the orgin be the starting point of frog. Then, WLOG assume that after the first jump, it is at the point (2,0). Then, the range of all possible places the frog can jump to at its second jump is the circle with equation $(x-2)^2+y^2=2^2$.If it landed $1$ unit within its starting point (the orgin), then it is inside the circle $x^2+y^2=1$. We clearly want the intersection point. So we're tring to solve the system of equations $x^2+y^2=1$ and $(x-2)^2+y^2=2^2$. We have $x=\frac{1}{4}$, so $y=\pm\frac{\sqrt{15}}{4}$. (writing continueing) ~ddk001

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

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