2017 AMC 12A Problem 22

Problem 22 of 25HarderCounting & Probability

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 (points with integer coordinates) 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),

(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?

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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,

C

with

\tfrac18,

I

with

\tfrac28,

and to a side interior with

\tfrac38,

etc.) gives

\begin{aligned} &a=\tfrac28 a+\tfrac18 c+\tfrac28 i,\quad \\ &c=\tfrac48 a+\tfrac48 i,\quad \\ &i=\tfrac28 a+\tfrac18 c+\tfrac18. \end{aligned}

Solving yields

a=\dfrac{1}{14},

c=\dfrac{4}{35},

i=\dfrac{11}{70}.

The required probability is

c=\dfrac{4}{35},

m+n=4+35=39.

Thus, the correct answer is E.

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