= Solution
Write the cone as $z=r\cot\alpha$, with $r$ the distance from its axis and $0<\alpha<\pi/2$. Its <Euclidean metric> gives
$$
ds^2=dr^2+r^2d\theta^2+dz^2=\csc^2\alpha\,dr^2+r^2d\theta^2.
$$
Consequently a differentiable path expressed as $r=r(\theta)$ has <arc length>
$$
\boxed{S=\int\sqrt{r^2+\csc^2\alpha\,(r')^2}\,d\theta.}
$$
For $L(r,r')=\sqrt{r^2+\csc^2\alpha(r')^2}$, the <Euler-Lagrange equation> has the <Beltrami identity> because $L$ has no explicit $\theta$ dependence:
$$
L-r'\frac{\partial L}{\partial r'}=\frac{r^2}{L}=C.
$$
For a nonradial minimizing path, $C>0$. Rearranging and integrating gives
$$
(r')^2=\sin^2\alpha\,r^2\left(\frac{r^2}{C^2}-1\right),\qquad \boxed{r\cos\bigl((\theta-\theta_0)\sin\alpha\bigr)=C.}
$$
The endpoint conditions determine $C$ and $\theta_0$ on the selected angular branch.
To verify that this stationary path really minimizes <arc length>, use the <local isometry from a circular cone to the plane>. Set $s=r/\sin\alpha$ and $\phi=\theta\sin\alpha$; then $ds^2+s^2d\phi^2$ is the plane <metric>. The displayed path is a straight line $s\cos(\phi-\phi_0)=C/\sin\alpha$, so its segment realizes the plane distance. Choose endpoint angular lifts with $|\Delta\theta|\leq\pi$; then $|\Delta\phi|<\pi$, and the joining segment avoids the apex. Its length is
$$
\frac1{\sin\alpha}\sqrt{r_1^2+r_2^2-2r_1r_2\cos(\sin\alpha\,\Delta\theta)}.
$$
Other lifts with developed angular difference less than $\pi$ give at least this distance; an angular change of at least $\pi$ has infimum at least $s_1+s_2$, attained only if passage through the apex is allowed. This proves the global choice above. Endpoints on a common generator have the radial segment as shortest path, which is the limiting case omitted by the $r(\theta)$ parametrization. If an endpoint is the apex, its shortest path is a generator. Thus \b[shortest paths are straight segments after developing the cone].
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