Shortest path on unit sphere under $\|\cdot\|_\infty$

Let $X$ be $\mathbb{R}^3$ with the sup norm $\|\cdot\|_{\infty}$. Let $Y=\{x\in X: \|x\|_{\infty}=1\}$. For $x,y\in Y$ define $d(x,y)$ to be the arc length of shortest paths on $Y$ joining $x,y$. (It is easy to see that generally there are more than one shortest path and these shortest paths must be a union of line segments on $Y$.) My question is: If $y\neq -x$, is the set $$Y\cap \{\lambda x+\mu y: \lambda\ge 0, \mu\ge 0\}$$ always a shortest path joining $x,y$?

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Somewhat surprisingly (for me), this is not the case.

Consider two points $p_x$ and $p_y$ on the faces $f_x$ and $f_y$ lying on the planes $x=1$ and $y=1$, respectively. On $f_x$, we can’t travel in the $x$ direction, and on $f_y$ we can’t travel in the $y$ direction, so the length of a path connecting the points is at least the sum of the distances of the points from the common edge. On the other hand, the length is also at least the difference in the $z$ coordinates of the two points. If we choose these two bounds to be the same, then there is exactly one shortest path connecting the points, namely the one that runs diagonally on each face in order to cover the same distance along $z$ as towards the edge.

The coordinates of the points in this case can be parametrized as



where $z$ is the $z$ coordinate of the point where the shortest path crosses the edge, and $a,b\ge0$.

Now the path defined in the question lies in the plane

$$\lambda p_x+\mu p_y=\lambda\left[\pmatrix{1\\1\\z}+\pmatrix{0\\-a\\-a}\right]+\mu\left[\pmatrix{1\\1\\z}+\pmatrix{-b\\0\\b}\right]\;.$$

For this to be the shortest path, it would have to contain the point $(1,1,z)$. But setting this expression equal to $(1,1,z)$ yields $\lambda a=\mu b$ from the first two equations, and substituting that into the third leads to $\lambda z+\mu z=z$. Unless $z=0$, this implies $\lambda+\mu=1$, which contradicts the first two equations unless $a=0$ and $b=0$. Thus in general the shortest path does not lie in that plane.