= Solution
Let $\mathbf p_1$ and $\mathbf p_2$ be the incoming and proton <momenta>. The other incoming atom is at rest, so its total <energy> is $m_1c^2$. By <four-momentum conservation>, the tritium atom has
$$
E_3=E_1+m_1c^2-E_2,\qquad\mathbf p_3=\mathbf p_1-\mathbf p_2.
$$
The specified perpendicular emission gives $\mathbf p_1\cdot\mathbf p_2=0$. Therefore the <relativistic energy-momentum relation> implies
$$
\begin{aligned}
m_3^2c^4&=(E_1+m_1c^2-E_2)^2-c^2(|\mathbf p_1|^2+|\mathbf p_2|^2)\\
&=(E_1+m_1c^2-E_2)^2-(E_1^2-m_1^2c^4)-(E_2^2-m_2^2c^4)\\
&=m_2^2c^4+2(m_1c^2+E_1)(m_1c^2-E_2).
\end{aligned}
$$
Dividing by $c^4$ gives
$$
\boxed{m_3^2=m_2^2+2\left(m_1+\frac{E_1}{c^2}\right)\left(m_1-\frac{E_2}{c^2}\right).}
$$
Here $E_1,E_2$ are total <energies>, not just <kinetic energies>. This identity is conditional on the collision admitting the stated emission: the calculated energies must satisfy the positive-energy mass-shell conditions, in particular $E_2\ge m_2c^2$ and $E_3\ge m_3c^2$. No nonrelativistic approximation has been used.
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