Problem Analysis #6

<think> Let me set up the transformations as group elements acting on (x,y).

Reflections/operations:

• Origin (point reflection): (x,y) → (-x,-y)
• x-axis: (x,y) → (x,-y)
• line y=x: (x,y) → (y,x)
• line y=-x: (x,y) → (-y,-x)

Starting at (2,1). We apply 3 random operations, each with prob 1/4. Want probability we return to (2,1).

Let me think of these as elements of the dihedral group D4 (symmetries of square). Let's represent each operation as a matrix.

Let a = origin reflection = -I = [[-1,0],[0,-1]] (rotation by 180°) b = x-axis = [[1,0],[0,-1]] c = y=x = [[0,1],[1,0]] d = y=-x = [[0,-1],[-1,0]]

These are 4 specific elements of D4 (order 8 group).

We need sequences of 3 operations (each from {a,b,c,d}) whose product (applied to vector) gives identity on (2,1). Since (2,1) is generic, we need the product to be identity matrix.

Total sequences: 4^3 = 64.

Count sequences g1 g2 g3 (applied as g3(g2(g1(v)))? order matters but product must be identity regardless). Actually applying operation each second: v → g1 v → g2 g1 v → g3 g2 g1 v. Need g3 g2 g1 = I.

Count triples (g1,g2,g3) from {a,b,c,d} with product = I.

Let me compute the group structure. D4 elements:

• I
• r = rotation 90: [[0,-1],[1,0]]
• r2 = 180: -I = a
• r3 = 270: [[0,1],[-1,0]]
• reflections:
  • b = x-axis [[1,0],[0,-1]]
  • y-axis [[-1,0],[0,1]]
  • c = y=x [[0,1],[1,0]]
  • d = y=-x [[0,-1],[-1,0]]

Our set S = {a, b, c, d} = {r2, mx, m_{y=x}, m_{y=-x}}.

Note a = r2 (a rotation), others are reflections.

Products: reflectionreflection = rotation. rotationreflection=reflection.

We need product of 3 elements = I.

Let me denote each element. Let me just brute force using multiplication.

Let me define multiplication. I'll compute products via matrices.

Elements: a = [[-1,0],[0,-1]] b = [[1,0],[0,-1]] c = [[0,1],[1,0]] d = [[0,-1],[-1,0]]

Product g3 g2 g1 must = I.

Let me think: for each pair (g1,g2), product g2 g1 = M, then need g3 = M^{-1}, and g3 must be in S. So count = number of (g1,g2) such that (g2 g1)^{-1} ∈ S, i.e. g2 g1 ∈ S^{-1} = S (since all these are involutions? a is involution (r2^2=I), b,c,d are reflections involutions). So S^{-1}=S.

So we need g2 g1 ∈ S, and then g3 = (g2 g1)^{-1} = g2 g1 (since it's an involution, its inverse is itself... only if g2g1 is an involution). Wait g3 = (g2 g1)^{-1}. For g3 to be in S, need (g2 g1)^{-1} ∈ S. Since S elements are all involutions, S^{-1}=S means inverse of each element of S is in S. But (g2 g1) might not be in S itself; we need its inverse in S.

Let me just count directly: number of triples with product I = number of (g1,g2) with g2 g1 ∈ S^{-1}... let me recompute. Need g3 g2 g1 = I → g3 = (g2 g1)^{-1}. g3 ∈ S. So need (g2 g1)^{-1} ∈ S.

Number of valid (g1,g2,g3) = number of (g1,g2) pairs such that (g2 g1)^{-1} ∈ S, with g3 determined uniquely.

So count pairs (g1,g2) where g2 g1 ∈ S^{-1} = S (since S closed under inverse). Let me verify S closed under inverse: a^{-1}=a (yes a in S), b^{-1}=b, c^{-1}=c, d^{-1}=d. Yes all involutions. So S^{-1}=S.

So need g2 g1 ∈ S.

Compute all products g2 g1 for g1,g2 ∈ {a,b,c,d}, count how many land in S.

Let me build multiplication table. I'll compute g2·g1 (matrix product, g2 first applied... matrix multiplication M=g2*g1).

Let me list matrices. a=[[-1,0],[0,-1]] b=[[1,0],[0,-1]] c=[[0,1],[1,0]] d=[[0,-1],[-1,0]]

Compute products M = X·Y for X (second) , Y (first). I'll compute for each ordered pair (X,Y) meaning X*Y.

Let me compute all 16.

aa = I (since a^2=I). Not in S (I not in S). ab = [[-1,0],[0,-1]][[1,0],[0,-1]] = [[-1,0],[0,1]] = y-axis reflection. Not in S. ac = -c = [[0,-1],[-1,0]] = d. In S! ad = -d = [[0,1],[1,0]] = c. In S! ba = [[1,0],[0,-1]][[-1,0],[0,-1]] = [[-1,0],[0,1]] = y-axis. Not in S. bb = I. Not in S. bc = [[1,0],[0,-1]][[0,1],[1,0]] = [[0,1],[-1,0]] = r3 (rotation). Not in S. bd = [[1,0],[0,-1]][[0,-1],[-1,0]] = [[0,-1],[1,0]] = r (rotation 90). Not in S. ca = [[0,1],[1,0]][[-1,0],[0,-1]] = [[0,-1],[-1,0]] = d. In S! cb = [[0,1],[1,0]][[1,0],[0,-1]] = [[0,-1],[1,0]] = r. Not in S. cc = I. Not in S. cd = [[0,1],[1,0]][[0,-1],[-1,0]] = [[-1,0],[0,-1]] = a. In S! da = [[0,-1],[-1,0]][[-1,0],[0,-1]] = [[0,1],[1,0]] = c. In S! db = [[0,-1],[-1,0]][[1,0],[0,-1]] = [[0,1],[-1,0]] = r3. Not in S. dc = [[0,-1],[-1,0]][[0,1],[1,0]] = [[-1,0],[0,-1]] = a. In S! dd = I. Not in S.

Now count pairs (X,Y) where XY ∈ S: ac, ad, ca, cd, da, d*c → 6 pairs.

So 6 valid configurations. For each, g3 determined as (g2g1)^{-1} ∈ S, valid.

Total favorable = 6, total = 64.

Probability = 6/64 = 3/32.

p+q = 3+32 = 35. </think>

Setting up the operations

The four operations are involutions in the dihedral group:

• Origin: $-I$, x-axis: reflection, $y=x$, $y=-x$.

We need the product of 3 chosen operations to equal the identity (since $(2,1)$ is generic).

Counting

For a triple $(g_1,g_2,g_3)$, we need $g_3 g_2 g_1 = I$, so $g_3 = (g_2 g_1)^{-1}$. Since all four operations are involutions (closed under inverse), $g_3 \in S$ requires $g_2 g_1 \in S$.

Computing all products $g_2 g_1$, the ones landing back in $S=\{a,b,c,d\}$ are: $a c=d,\; a d=c,\; c a=d,\; c d=a,\; d a=c,\; d c=a$

That's 6 valid pairs, each forcing a unique valid $g_3$.

Probability

$\frac{6}{4^3} = \frac{6}{64} = \frac{3}{32}$

So $p+q = 3+32 = 35$.

\boxed{35}