Problem Analysis #6

To solve the problem, we analyze the motion of a point starting at $(2,1)$, which undergoes three successive reflections over one of four transformations:

• A: Reflection over the origin: $(x, y) \mapsto (-x, -y)$
• B: Reflection over the x-axis: $(x, y) \mapsto (x, -y)$
• C: Reflection over the line $y = x$: $(x, y) \mapsto (y, x)$
• D: Reflection over the line $y = -x$: $(x, y) \mapsto (-y, -x)$

Each transformation is selected with equal probability $\frac{1}{4}$, and the process lasts for 3 seconds (i.e., three transformations are applied). The goal is to find how many of these 64 possible sequences result in the final position being again $(2,1)$, and then compute the reduced fraction $\frac{p}{q}$ representing that probability. Finally, we return $p + q$.

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Step 1: Total Number of Sequences

Since each second involves choosing one of the four transformations independently, there are:

$$4^3 = 64 \text{ possible sequences}$$

Each sequence has probability $\left(\frac{1}{4}\right)^3 = \frac{1}{64}$.

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Step 2: Understanding When a Sequence Returns to (2,1)

Let’s denote the three transformations in a sequence as $T_1$, $T_2$, and $T_3$. The final position is:

$$T_3(T_2(T_1(2,1)))$$

We want this to equal $(2,1)$. Since all transformations are linear (specifically orthogonal), the composition corresponds to the matrix product of the individual transformations’ matrices. Importantly, the only orthogonal transformation in the group generated by the set $\{A, B, C, D\}$ that fixes the point $(2,1)$ is the identity transformation.

Why? Because:

• All transformations are orthogonal.
• The only linear transformation fixing a nonzero vector like $(2,1)$ must be the identity or a reflection over a line through the origin containing $(2,1)$.
• However, no such reflection exists within our set of operations (as they correspond to lines with slopes $0$, $\infty$, $1$, and $-1$).
• Moreover, the determinant argument shows that the only sequence yielding determinant $+1$ (required for identity) involves an even number of determinant $-1$ transformations (i.e., from B, C, D).

Thus, a sequence returns to $(2,1)$ if and only if the composition of the three transformations equals the identity transformation.

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Step 3: Count Valid Sequences

We now count how many sequences of three transformations from $\{A, B, C, D\}$ compose to the identity.

Through detailed enumeration and analysis (as shown in the full reasoning), it turns out that only permutations of the set $\{A, C, D\}$ yield the identity transformation when composed in order.

There are $3! = 6$ such permutations:

• A, C, D
• A, D, C
• C, A, D
• C, D, A
• D, A, C
• D, C, A

All of these sequences result in the final position being $(2,1)$.

No other sequences—including those with repeated transformations or involving B—achieve this result.

Hence, there are exactly 6 successful sequences out of 64 total.

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Step 4: Compute the Probability

$$\text{Probability} = \frac{6}{64} = \frac{3}{32}$$

This is already in lowest terms since $\gcd(3, 32) = 1$.

So $p = 3$, $q = 32$, and the requested value is:

$$p + q = 3 + 32 = \boxed{35}$$