This question was inspired by an answer to the "Magic trick based on deep mathematics" question. I wanted to post it as a comment, but I ran out of characters! I'm sure there must be a collection of standard results related to this question, but I don't know where to start looking.

First, a quick definition. The *diameter* of a set $S \in \mathbb{R}^n$ is $\sup\{d(x, y) \mid x, y \in S\}$.

A sheet of paper is a good physical example of a Riemannian 2-manifold with boundary, and a table is a good physical model of (a subset of) $\mathbb{R}^2$. Embed the paper isometrically in $\mathbb{R}^2$ by laying flat on a table.

Draw the outline of a circular cup on the paper. It seems obvious that no matter how you embed the paper in $\mathbb{R}^2$, the outline of the cup will always be a metric circle, and it will always have the same diameter $D$.

Now, lift the paper into the air, embedding it isometrically in $\mathbb{R}^3$. If you let the paper flop around, the outline of the cup might not be a metric circle anymore... **but will it still have diameter $D$?**

Finally, cut along the outline of the cup, removing an open disk from the sheet of paper. The paper now has a second boundary component, and it's no longer simply connected. The paper has also gained a surprising property: you can bend it around in midair (that is, embed it isometrically in $\mathbb{R}^3$) so that the outline of the cup has diameter greater than $D$! **What's the important property of the paper that we changed to make this possible?**

*Comments*

I don't think you need to cut along the outline of the cup to make this work... you could probably just cut out any disk contained within the outline of the cup. So maybe simply-connectedness is the important property?

My gut tells me that if you draw two dots on the sheet of paper, the distance between the dots is maximized when the paper is flat on the table. When you bend the paper around in midair, the dots can get closer together, but they can never get farther apart. I think this is equivalent to the statement that if $\delta$ is the natural distance function on the paper, $d$ is the distance function in $\mathbb{R}^3$, and $F$ is an isometric embedding of the paper in $\mathbb{R}^3$, $d(Fx, Fy) \le \delta(x, y)$ for all points $x$ and $y$ on the paper.