Question

A square is drawn in the Cartesian coordinate plane with vertices at $(2,2),$ $(-2,2),$ $(-2,-2),$ and $(2,-2).$ A particle starts at $(0,0).$ Every second it moves with equal probability to one of the eight lattice points closest to its current position, independently of its previous moves. In other words, the probability is $\dfrac{1}{8}$ that the particle will move from $(x,y)$ to each of $(x,y+1),$ $(x+1,y+1),$ $(x+1,y),$ $(x+1,y-1),$ $(x,y-1),$ $(x-1,y-1),$ $(x-1,y),$ or $(x-1,y+1).$ The particle will eventually hit the square for the first time, either at one of the $4$ corners of the square or at one of the $12$ lattice points in the interior of one of the sides of the square. The probability that it will hit at a corner rather than at an interior point of a side is $\dfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m+n?$

$4$

$5$

$7$

$15$

$39$

Solution:

By symmetry, group the relevant interior points into three types: $C=\{(0,0)\},$ the "axis" points $A=\{(\pm1,0),(0,\pm1)\},$ and the "diagonal" points $I=\{(\pm1,\pm1)\}.$ Let $a,c,i$ be the probabilities of eventually hitting a corner starting from a point of type $A,C,I.$

Reading off the transition probabilities (a point in $A$ goes to $A$ with prob $\tfrac28,$ to $C$ with $\tfrac18,$ to $I$ with $\tfrac28,$ and to a side interior with $\tfrac38,$ etc.) gives $a=\tfrac28 a+\tfrac18 c+\tfrac28 i,\quad c=\tfrac48 a+\tfrac48 i,\quad i=\tfrac28 a+\tfrac18 c+\tfrac18.$

Solving yields $a=\dfrac{1}{14},$ $c=\dfrac{4}{35},$ $i=\dfrac{11}{70}.$ The required probability is $c=\dfrac{4}{35},$ so $m+n=4+35=39.$

Thus, the correct answer is E.

2017 AMC 12A Problem 22

Solution:

Problem 22 in Other Years