2023 AMC 12B Problems/Problem 20

Revision as of 17:38, 15 November 2023 by Professorchenedu (talk | contribs) (Solution)

Solution

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)