= Solution
For the <photopion threshold on a photon background>, write $M=m_pc^2$ and $m=m_\pi c^2$ in <energy> units. For a head-on <photon> of <energy> $\epsilon$ and a highly relativistic <proton> of total <energy> $E$, the invariant squared centre-of-momentum <energy> is approximately $M^2+4E\epsilon$. At threshold the final <proton> and <pion> are at rest relative to one another, so
$$
M^2+4E_{\rm th}\epsilon=(M+m)^2,
\qquad
\boxed{E_{\rm th}=\frac{m(2M+m)}{4\epsilon}.}
$$
Using a typical CMB <photon energy> $\epsilon\simeq2.70k_BT=6.28\times10^{-4}\,\mathrm{eV}$ gives
$$
\boxed{E_{\rm th}\simeq1.1\times10^{20}\,\mathrm{eV}.}
$$
Thus photopion losses suppress the arrival of <protons> of order $10^{20}\,\mathrm{eV}$ from sufficiently distant sources. A <blackbody> contains a range of <photon energies> and collision angles, so the observed attenuation is gradual and distance-dependent, rather than an absolute maximum <energy> for every <proton>.
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