Problem Analysis #6

The point starts at $(2,1)$ and undergoes a sequence of three reflections chosen uniformly from the four transformations: reflection in the origin ($O: (x,y) \to (-x,-y)$), reflection in the $x$-axis ($X: (x,y) \to (x,-y)$), reflection in $y=x$ ($D: (x,y) \to (y,x)$), and reflection in $y=-x$ ($A: (x,y) \to (-y,-x)$). The overall transformation is the composition $T_3 \circ T_2 \circ T_1$. For the point to return to $(2,1)$, this composition must be the identity, since $(2,1)$ is not fixed by any other element of the dihedral group generated by these reflections. Because each transformation is its own inverse, $T_3 \circ T_2 \circ T_1 = I$ is equivalent to $T_3 \circ T_2 = T_1$. Evaluating the 16 possible products $T_3 \circ T_2$, exactly 6 yield an element in the set $\{O,D,A\}$ (the subgroup isomorphic to $V_4$):

For each such pair, $T_1$ is forced to be the product, giving 6 valid sequences out of $4^3 = 64$ total sequences. The probability is $6/64 = 3/32$, so $p=3$, $q=32$, and $p+q=35$.

\boxed{35}