= Solution
Use <one-bit remote preparation of real qubit states>. Alice measures her <qubit> in the <orthonormal basis>
$$
|\alpha_\theta\rangle=\cos\theta|0\rangle+\sin\theta|1\rangle,\qquad|\beta_\theta\rangle=-\sin\theta|0\rangle+\cos\theta|1\rangle.
$$
Because this change of basis is real and orthogonal, the shared <Bell state> has the expansion
$$
|\Phi^+\rangle=\frac{|\alpha_\theta\rangle_A|\alpha_\theta\rangle_B+|\beta_\theta\rangle_A|\beta_\theta\rangle_B}{\sqrt2}.
$$
Her <projective measurement> gives each outcome with probability one half. She sends $m=0$ for the first outcome and $m=1$ for the second. Bob's conditional <pure state> is respectively $|\alpha_\theta\rangle$ or $|\beta_\theta\rangle$.
For $m=0$ Bob does nothing. For $m=1$ he applies the fixed <unitary operator>
$$
U=iY=\begin{pmatrix}0&1\\-1&0\end{pmatrix},\qquad U|\beta_\theta\rangle=|\alpha_\theta\rangle,
$$
where $Y$ is the <Pauli Y gate>. Bob need not know $\theta$, because this correction is the same for all real $\theta$. \b[Both outcomes produce the exact density operator $|\alpha_\theta\rangle\langle\alpha_\theta|$, using one classical bit and only local operations.]
This <remote state preparation> consumes the shared <entangled state>. Before receiving the bit, Bob has the mixture $(|\alpha_\theta\rangle\langle\alpha_\theta|+|\beta_\theta\rangle\langle\beta_\theta|)/2=I/2$, independent of $\theta$, as required by <quantum no-signalling>. The real-state restriction is what permits a fixed unitary correction of the orthogonal outcome.
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