Convex surface on which any two points $a,b$ can be joined by a curve of length $(\pi/2-\epsilon)|a-b|$ 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?

 A: How about a sharp cone?  Suppose the cone's lateral surface unrolls to a
circular sector of angle $2\theta$ for some small positive $\theta$.  Then:
$\bullet$ the base is flat, so any $a,b$ on the base are joined by
a line of length $|a-b|$.
$\bullet$ if $a$ is on the base and $b$ on the side, then
we can choose $\gamma$ to go straight down from $b$ to the edge
and thence straight to $a$; if these two segments have
lengths $x,y$ then $|a-b| = \sqrt{x^2 - \epsilon(\theta) x y + y^2}$
for some $\epsilon(\theta)$ that approaches zero as $\theta \rightarrow 0$,
so $\ell(\gamma) = x+y \leq (\sqrt 2 + \delta(\theta)) \, |a-b|$ for some small
$\delta(\theta)$ that again tends to zero as $\theta \rightarrow 0$.
$\bullet$ Finally, if $a,b$ are both on the side then the shortest $\gamma$
is a path that unrolls to a straight line on a sector of angle at most $\theta$.
At worst $a$ and $b$ are at the same height, separated by $\psi \leq \theta$
on the unrolled cone, and thus by $(\psi/\theta) \pi$ on a circular
cross-section of the solid cone.  Then
$$
\ell(\gamma) = \frac
 {\mathop{\rm sinc} \frac\psi2}
 {\mathop{\rm sinc} \frac\pi2 \! \frac\psi\theta} 
|a-b| 
$$
where $\mathop{\rm sinc}(x) = \sin(x)/x$.
Since $\mathop{\rm sinc}$ is logarithmically convex upwards, the ratio 
$\mathop{\rm sinc} \frac\psi2 \big/ \mathop{\rm sinc} \frac\pi2 \! \frac\psi\theta$
is an increasing function of $\psi$, so is maximized at $\psi = \theta$,
where it equals $\frac\pi 2\mathop{\rm sinc} \frac\theta2$ $-$ which is still
less than $\pi/2$ for any positive $\theta$, so we've attained
$c < \pi / 2$.
A: For completeness, I add some numerical details to the excellent answer by Noam D. Elkies. Let $C$ be the cone such that the lateral surface unrolls to a circular sector of angle $2\theta$. There are two sources of distortion: 


*

*Between two points on lateral surface (worst case: same level, diametrally opposite). This gives $c\ge (\pi/\theta) \sin(\theta/2)$, blue curve below.

*Between points on lateral surface and on the base (worst case: they are equally distant from the edge). This gives $c\ge \sqrt{2\pi/(\pi-\theta)}$, red curve below. 



The distortion is minimized when the curves intersect, $\theta\approx 0.53378$. This corresponds to the cone of radius $\approx 0.1675$ at height $1$; pretty sharp indeed. 

This least-distorted cone has $c\approx 1.5522$, beating the sphere for which $c=\pi/2 \approx 1.5708$. 
Lower bounds, for comparison


*

*[Gromov] Every subset of $\mathbb R^n$ with $c<\pi/(2\sqrt{2}) \approx 1.11$ is contractible.

*[Pansu] Every subset of $\mathbb R^n$ with $c<\frac{\alpha}{2\sin(\alpha/2)}$, $\alpha=\cos^{-1}(-1/n)$, is contractible. For $n=3$ this evaluates to $\approx 1.17$.


The estimate of Pansu is in Appendix A of the same book. I don't know of any better lower bounds.
