= Solution
Treat the atmospheric <proton> as stationary in the <Earth> frame, the incident <antiproton> as vertically downward, and neglect atmospheric <energy loss>, scattering and the <Earth>'s curvature initially. Take $m_pc^2=0.9383\,\mathrm{GeV}$. <Conservation of energy> and equal outgoing <energies> give
$$
E_\mu=\frac{35+0.9383}{2}\,\mathrm{GeV}=17.9692\,\mathrm{GeV},\qquad \gamma_\mu=\frac{E_\mu}{0.106\,\mathrm{GeV}}\simeq169.52.
$$
Equal <energies> imply equal <momentum> magnitudes. Their transverse <momenta> cancel, so the two <muons> have equal downward components $p_z=p_{\bar p}/2$, where
$$
p_{\bar p}c=\sqrt{(35\,\mathrm{GeV})^2-(0.9383\,\mathrm{GeV})^2}\simeq34.9874\,\mathrm{GeV}.
$$
Each <muon> therefore makes an <angle> $\theta$ with the downward vertical satisfying
$$
\cos\theta=\frac{p_{\bar p}c/2}{\sqrt{E_\mu^2-m_\mu^2c^4}}\simeq0.9735.
$$
The paths are not both vertical: replacing their path lengths by $20\,\mathrm{km}$ would miss this two-body kinematic correction. Since $v_z=p_zc^2/E_\mu$, the proper flight time over vertical height $H$ is
$$
\Delta\tau_\mu=\frac{H}{\gamma_\mu v_z}=\frac{2Hm_\mu c^2}{c(p_{\bar p}c)}.
$$
For <equal-energy atmospheric muon-pair survival>, independent exponential <particle decays> with mean proper <muon lifetime> $\tau_0=2.2\,\mu\mathrm s$, both survive with <probability>
$$
\boxed{P_{\rm both}=\exp\left[-\frac{4Hm_\mu c^2}{c\tau_0(p_{\bar p}c)}\right]\simeq0.692.}
$$
Equivalently each travels about $20.55\,\mathrm{km}$, with proper flight time about $0.404\,\mu\mathrm s$, and survives with <probability> about $0.832$. The result is decay-only survival to an ideal surface, not a prediction including air interactions or detector efficiency. For a spherical <Earth> of radius $R_E$, the exact path length is $\ell=(R_E+H)\cos\theta-\sqrt{R_E^2-(R_E+H)^2\sin^2\theta}$; inserting $P_{\rm both}=\exp[-2\ell/(\gamma_\mu v_\mu\tau_0)]$ gives a small curvature correction.
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