= Solution
The <center-of-momentum frame> moves with velocity $\mathbf V=c^2\mathbf p_{\rm tot}/E_{\rm tot}$. Using the total <four-momentum> in <2/c/a/solution> gives
$$
\boxed{\mathbf V=\frac c2(1+\cos\theta,\sin\theta,0),\qquad
\frac{|\mathbf V|}{c}=\left|\cos\frac\theta2\right|.}
$$
For $0<\theta\le\pi$, its direction bisects the two photon directions, and its speed decreases from the limiting value $c$ to zero. At $\theta=\pi$, the lab already is the <center-of-momentum frame>. At $\theta=0$, the formal speed is $c$ and no inertial rest frame exists; this endpoint is a limit, not an allowable boost.
\Image[/past-exam-of-the-symmetry-and-relativity-course-of-the-university-of-oxford/2016-two-photon-frame-and-helix-radius.png]
{title=Center-of-momentum speed for equal-energy photons, with the parallel endpoint excluded, and the unchanged electron Larmor radius under longitudinal deceleration}
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