= Solution
Replace the two arms by three identical arms of length $N$ attached to one central <qubit>. Label the hub $c$ and the sites on arm $a$ by $(a,j)$, $a=1,2,3$, $j=1,\ldots,N$. Use <XY exchange interactions> with coupling $J_0/\sqrt3$ on each hub-to-first-site link, and coupling $J_j$ on each arm link $(a,j)$ to $(a,j+1)$. Initially excite only the hub: this is the <product state> $|1\rangle_c$ with every arm <qubit> in $|0\rangle$.
In the symmetric <single-excitation subspace>, define $|s_0\rangle=|c\rangle$ and $|s_j\rangle=3^{-1/2}\sum_{a=1}^3|a,j\rangle$. The hub coupling is
$$
H|s_0\rangle=\frac{J_0}{\sqrt3}\sum_{a=1}^3|a,1\rangle=J_0|s_1\rangle,
$$
and the reverse coupling has the same coefficient. All other radial links have coefficients $J_j$, just as in part (c). Thus the <symmetric-arm reduction of an exchange Hamiltonian> is again exactly $H_T$. At time $\pi/2$ its evolution is
$$
|s_0\rangle\longmapsto|s_N\rangle=\frac{|1,N\rangle+|2,N\rangle+|3,N\rangle}{\sqrt3}.
$$
Every nontip <qubit> is zero, so tracing them out leaves the three tips in the pure <W state>
$$
\boxed{|W_3\rangle=\frac{|100\rangle+|010\rangle+|001\rangle}{\sqrt3}.}
$$
The essential change is the additional identical arm and the normalization of each hub coupling by $\sqrt3$ instead of $\sqrt2$.
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