= Solution
Take unit-curvature <Real projective space> $\mathbb{RP}^n=S^n/\{\pm1\}$ with $n\geq2$, and any point $p=[v]$. It is complete with positive <sectional curvature>. A unit-speed <geodesic> through $p$ lifts to
$$
\widetilde\gamma(t)=\cos t\,v+\sin t\,w,\qquad |v|=|w|=1,\quad\langle v,w\rangle=0.
$$
Since the two representatives of an endpoint are antipodal, <Riemannian distance> in the quotient is the smaller of the two spherical distances:
$$
d([v],[u])=\arccos|\langle v,u\rangle|,\qquad d(p,\gamma(t))=\arccos|\cos t|.
$$
This equals $t$ for $0\leq t\leq\pi/2$ and is strictly smaller just afterwards. Thus the <Riemannian cut point> occurs at time $\pi/2$ in every direction.
Along a curvature-one <geodesic>, every normal <Jacobi field> with initial value zero is $J(t)=\sin t\,P_ta$, where $P_t$ is <parallel transport> and $a\perp\dot\gamma(0)$. Its first further zero is $t=\pi$. The quotient map is a <local isometry>, so the <Jacobi field> equation and conjugacy times are unchanged. Hence
$$
\boxed{t_{\mathrm{cut}}=\pi/2<\pi=t_{\mathrm{conj}}\quad\text{in every direction}.}
$$
This is <cut and conjugate times in round real projective space>.
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