2021 AIME II Problem 8

Below is the professionally curated solution for Problem 8 of the 2021 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AIME II solutions, or check the answer key.

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Concepts:random walkrecursive counting

Difficulty rating: 2840

8.

An ant makes a sequence of moves on a cube where a move consists of walking from one vertex to an adjacent vertex along an edge of the cube. Initially the ant is at a vertex of the bottom face of the cube and chooses one of the three adjacent vertices to move to as its first move. For all moves after the first move, the ant does not return to its previous vertex, but chooses to move to one of the other two adjacent vertices. All choices are selected at random so that each of the possible moves is equally likely. The probability that after exactly 88 moves that ant is at a vertex of the top face on the cube is mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

Solution:

All 327=3843 \cdot 2^7 = 384 allowed move sequences are equally likely, so we count those ending on the top face. Classify the ant after each move by its face (bottom or top) and by whether its last move was vertical: after a vertical move the two allowed continuations are the two horizontal edges at the new vertex, while after a horizontal move one continuation is horizontal and one is vertical.

Let (Bh,Bv,Th,Tv)(B_h, B_v, T_h, T_v) count sequences ending on the bottom or top with last move horizontal or vertical. Each sequence splits into two, following Bh=Bh+2Bv,Tv=Bh,Th=Th+2Tv,Bv=Th.B_h' = B_h + 2B_v, \qquad T_v' = B_h, \qquad T_h' = T_h + 2T_v, \qquad B_v' = T_h. After move 11 the counts are (2,0,0,1),(2, 0, 0, 1), and iterating gives (2,0,2,2),(2, 0, 2, 2), (2,2,6,2),(2, 2, 6, 2), (6,6,10,2),(6, 6, 10, 2), (18,10,14,6),(18, 10, 14, 6), (38,14,26,18),(38, 14, 26, 18), (66,26,62,38),(66, 26, 62, 38), and after the eighth move Th=62+76=138T_h = 62 + 76 = 138 and Tv=66.T_v = 66.

So 138+66=204138 + 66 = 204 of the 384384 sequences end on the top face, giving probability 204384=1732\frac{204}{384} = \frac{17}{32} and m+n=17+32=49.m + n = 17 + 32 = 49.

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