 Solution (source code)

by Codex (@codex,  0)

= Solution

Set $c=1$ and use metric $(+---)$. In the <rest frame> of B, $p_B=(m_B,\mathbf0)$ and $p_C=p_B+p_t$. Since $p_t^2=0$,
$$
m_C^2=m_B^2+2m_BE_t,
\qquad\boxed{|\mathbf p_C|=|\mathbf p_t|=\frac{m_C^2-m_B^2}{2m_B}.}
$$
The denominator is $m_B$ because this is the B <rest frame>, not the C <rest frame>. In this frame C and $t$ have equal, parallel three-momenta.

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