2017 AMC 12A Problem 22

Attempt Problem 22 of the 2017 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

22.

A square is drawn in the Cartesian coordinate plane with vertices at (2,2),(2,2), (−2,2),(-2,2), (−2,−2),(-2,-2), and (2,−2).(2,-2). A particle starts at (0,0).(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 18\dfrac{1}{8} that the particle will move from (x,y)(x,y) to each of (x,y+1),(x,y+1), (x+1,y+1),(x+1,y+1), (x+1,y),(x+1,y), (x+1,y−1),(x+1,y-1), (x,y−1),(x,y-1), (x−1,y−1),(x-1,y-1), (x−1,y),(x-1,y), or (x−1,y+1).(x-1,y+1). The particle will eventually hit the square for the first time, either at one of the 44 corners of the square or at one of the 1212 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 mn,\dfrac{m}{n}, where mm and nn are relatively prime positive integers. What is m+n?m+n?
44
55
77
1515
3939
Answer: E
Concepts:random walkrecursive probabilitysymmetry
Small hint:
By symmetry, track only three interior states: the center, the four edge-midpoint neighbors, and the four diagonal neighbors
Big hint:
Write hitting-probability equations for each state and solve the linear system for the center’s value
Solution:
By symmetry, group the relevant interior points into three types: C={(0,0)},C=\{(0,0)\}, the “axis” points A={(±1,0),(0,±1)},A=\{(\pm1,0),(0,\pm1)\}, and the “diagonal” points I={(±1,±1)}.I=\{(\pm1,\pm1)\}. Let a,c,ia,c,i be the probabilities of eventually hitting a corner starting from a point of type A,C,I.A,C,I.
Reading off the transition probabilities (a point in AA goes to AA with prob 28,\tfrac28, to CC with 18,\tfrac18, to II with 28,\tfrac28, and to a side interior with 38,\tfrac38, etc.) gives a=28a+18c+28i,c=48a+48i,i=28a+18c+18. \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=114,a=\dfrac{1}{14}, c=435,c=\dfrac{4}{35}, i=1170.i=\dfrac{11}{70}. The required probability is c=435,c=\dfrac{4}{35}, so m+n=4+35=39.m+n=4+35=39.
Thus, the correct answer is E.
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