= Solution
Let $p$ be the magnitude of the incident particle's <momentum>. The <relativistic energy-momentum relation> gives
$$
p^2c^2=E^2-m^2c^4.
$$
The stationary particle contributes <rest energy> $mc^2$ and zero <momentum>. Conservation of <four-momentum> therefore gives equal photon energies $\varepsilon=(E+mc^2)/2$. A <photon> has momentum magnitude $\varepsilon/c$, so if the two momentum vectors meet at angle $\theta$,
$$
p^2=\left|p_1+p_2\right|^2
=2\left(\frac{\varepsilon}{c}\right)^2(1+\cos\theta).
$$
Substitution gives the <angular relation for two-photon electron-positron annihilation>, whose derivation depends only on the stipulated equal rest masses and equal photon energies:
$$
\boxed{\cos\theta
=\frac{2(E^2-m^2c^4)}{(E+mc^2)^2}-1
=\frac{E-3mc^2}{E+mc^2}.}
$$
All energies and the angle here are in the laboratory <inertial frame>.
For perpendicular photon trajectories, $\cos\theta=0$ and hence $E=3mc^2$. Since $E=\gamma mc^2$ with <Lorentz factor> $\gamma=(1-v^2/c^2)^{-1/2}$,
$$
\boxed{\frac vc=\sqrt{1-\frac19}=\frac{2\sqrt2}{3}.}
$$
For the small-speed limit, set $\beta=v/c\ge0$ and $\delta=\pi-\theta$. The angle formula implies
$$
\sin\frac\delta2=\cos\frac\theta2
=\sqrt{\frac{\gamma-1}{\gamma+1}}
=\frac{\gamma\beta}{\gamma+1}
=\frac\beta2+O(\beta^3).
$$
Using $\arcsin z=z+O(z^3)$ therefore gives $\delta=\beta+O(\beta^3)$, so
$$
\boxed{\theta=\pi-\frac vc+O\!\left((v/c)^3\right)\approx\pi-\frac vc.}
$$
At zero incident speed the momenta are equal and opposite, so $\theta=\pi$, consistent with the expansion.
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