= Solution
Choose the positive $z$ direction along the incident <Electron>. With the neglected lepton <masses> and the paper's <natural units>, the initial <four-momenta> are
$$
p_-=(E_-,0,0,E_-),\qquad p_+=(E_+,0,0,-E_+).
$$
The squared <centre-of-mass energy> is the <Lorentz invariant>
$$
s=(p_-+p_+)^2=(E_-+E_+)^2-(E_--E_+)^2=4E_-E_+.
$$
Producing the <Upsilon meson> as the sole final resonance requires $s=M_\Upsilon^2$. The threshold positron <energy> is therefore
$$
\boxed{E_+^{\min}=\frac{(10.580\,\mathrm{GeV})^2}{4(9.000\,\mathrm{GeV})}
=3.10934\,\mathrm{GeV}\simeq3.11\,\mathrm{GeV}.}
$$
At that beam setting, <four-momentum conservation> gives $E_\Upsilon=E_-+E_+=12.10934\,\mathrm{GeV}$ and $p_{\Upsilon,z}=E_--E_+=5.89066\,\mathrm{GeV}$. Hence
$$
\boxed{\beta_\Upsilon=\frac{p_{\Upsilon,z}}{E_\Upsilon}
=\frac{9-3.10934}{9+3.10934}=0.48646.}
$$
The resonance moves in the incident-<Electron> direction, with <velocity> approximately $0.486c$. The laboratory sum of beam <energies> exceeds the resonance <rest energy> because the asymmetric beams also give it laboratory <kinetic energy>.
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