I am trying to solve an exercise on page 13 of the book *Metric structures on Riemannian and non-Riemannian spaces* by Gromov.

Construct a closed, convex surface $X$ in $\mathbb R^3$ such that any two points $a,b\in X$ can be joined by a curve $\gamma\subset X$ of length $$\ell(\gamma)\le c|a-b| \tag1$$ where $c<\pi/2$. Here $c$ is independent of $a,b$.

### Remarks

Here $|a-b|$ is the Euclidean norm of the vector $a-b$. The geometric meaning of inequality (1) is that the surface is not too twisted: a bug crawling from $a$ to $b$ along the surface does not have to travel much further than if it flew directly from $a$ to $b$.

A closed convex surface is precisely the boundary of a convex bounded set.

Gromov calls the smallest value of $c$ for the surface satisfies the above the *distortion* of $X$. Other authors call it the *constant of quasiconvexity*.

### Some ideas

- A sphere has distortion $\pi/2$. Indeed, any curve connecting antipodal points (distance $2r$) has length at least $\pi r$, where $r$ is the radius.
- Ellipsoids are no good; they are distorted more than spheres. Look at the vertices of the shortest axis.
- More generally, every centrally symmetric surface has distortion at least $\pi/2$. Indeed, let $a\in X$ be a nearest point to the center of symmetry, and $b$ its antipode. Any curve connecting $a$ to $b$ stays outside of a ball with diameter $ab$, and therefore has length at least $\frac{\pi}{2}|a-b|$.
- One can consider closed curves instead of surfaces, hoping to get inspiration from there. But the distortion of a closed curve cannot be less than $\pi/2$; proof here. That is, a circle is the least distorted closed curve.
- Among non-symmetric $X$, a natural candidate is the regular tetrahedron, but it does not work. The dihedral angles $\alpha=\cos^{-1}(1/3)$ are too small and difficult to get around: $c$ cannot be less than $1/\sin (\alpha/2) = \sqrt{3}>\frac{\pi}{2}$.
- Minkowski sum of a tetrahedron and a sphere of sufficiently large radius might work, but the length estimates look scary.

Any better ideas?