# Moving Line Segment Problem

So I was contemplating the following problem: Given two circles of radius $r$, which will position vertically so that their centers are separated by a distance of $2r+d$ (and thus, at the nearest points, the circles are $d$ units apart). Suppose we have a line segment which is also positioned vertically so that its endpoints each lie a different circle to the right of those circle's centers. The question is the following: For what lengths $l$ can this line segment exist (easy question), but more importantly can also be moved to the analogous position on to the left of the centers by only moving its endpoints along the circles that they each began on and without changing the line segment's length?

I have reasoned informally that the answer is that any length $l$ that can exist can also be moved to the equivalent position in the required manner, which means that $d<l<4r+d$, but my question is how to actually, formally, prove it? And what of the more general question being alluded to, when can we move a line segment with endpoints on distinct curves to another position with endpoints on the curves by moving said endpoints along the respective curves? (I realize this latter question is really broad, I'm most interested in what, if anything, is known about it.)

Edit: Here is a photo of a free-hand sketch of the situation (apologies for my crappy drawing and handwriting) https://goo.gl/photos/cZnb6znt5Ju8mdfh8

• Having trouble visualizing what you're asking. Can you provide a diagram? Mar 26 '16 at 18:02
• I unfortunately don't have the tools handy to create a good one, but I guess I could draw a picture of it and take a photo w/ my phone and upload it. Would that help? Mar 26 '16 at 18:07
• That only works for $l=2r+d$, also note that it is the endpoints of the line segment that move (along the relevant circles), not the circles themselves. Mar 26 '16 at 18:16

Assume $0<r<1$, and let $$s\mapsto z(s):=(r\sin s,1- r\cos s),\qquad t\mapsto w(t):=(r\sin t,-1+r\cos t)$$ be the two circles. The upper circle is then drawn counterclockwise, the lower clockwise; and $s=0$, resp. $t=0$, produce the two points where these two circles come nearest to each other.

Any choice of a point $(s,t)$ on the Torus ${\mathbb T}=({\mathbb R}/2\pi)^2$ determines a pair of points on the two circles, and the (euclidean) distance between these two points is $\ell(s,t):=\sqrt{f(s,t)}$, whereby $f(s,t)$ computes to $$f(s,t)=|z(s)-w(t)|^2=2\bigl(2+r^2-2r\cos s-2r\cos t+r^2\cos(s+t)\bigr)\ .$$ The following images show contour plots of the function $\ell:{\mathbb T}\to{\mathbb R}$, the first of them having $(0,0)$ as center, and the second of them having $(\pi,\pi)$ as center. We can see that all contour lines with one exception are loops, and these loops are symmetric with respect to $(0,0)$, resp. $(\pi,\pi)$. This implies that one may walk from any point $(s_0,t_0)$ on such a contour line to the opposite point $(-s_0,-t_0)$. But this means that the linkage configuration corresponding to the point $(s_0,t_0)$ can be continuously transformed into the configuration corresponding to $(-s_0,-t_0)$.

The contour line corresponding to the level $\ell=2$ consists of two loops intersecting at the points corresponding to $(0,\pi)$ and $(\pi,0)$. One of these loops appears as straight line $\phi+\psi=\pi\ {\rm mod}\ 2\pi$ and parametrizes the rotary motion of a vertical bar of length $2$.

It will be computationally difficult to convert these pictures into formulas which prove that the level lines for $\ell<2$ and for $\ell>2$ are indeed loops. But it is easily checked that all four critical points $(0,0)$, $(0,\pi)$, $(\pi,0)$, and $(\pi,\pi)$ of $f$ are nondegenerate. Morse theory then tells us that all level lines for $\ell_{\min}<\ell<2$ as well as all level lines for $2<\ell<\ell_{\max}$ have to be loops.

• This is a good approach, but there are a few details to be addressed: First one I noticed is the way you defined distance...I mean distance in the Euclidean sense. I am unsure whether or not what you have corresponds to that (aka that needs to be shown). Next, the level curves need not be circles (indeed they do not appear to be), all that is needed is that there is a path from $(s_0,t_0)$ to $(-s_0,-t_0)$ (this is also something I would like shown if it's not excessively messy/difficult). Apr 2 '16 at 17:13
• @Justin Benfield can the line segments cross through the interior of the circles ? Apr 3 '16 at 9:45
• @mercio They can, and in fact, if $l>2r+d$ it must be the case that line segment crosses the interior of the circle. What must also be the case is that the endpoints of the line segment remain on the boundary of the circle whose boundary they started on. Apr 3 '16 at 22:50

Imagine the point $(0,0)$ is in the center of the line between the two circles. Hence, the center of the top circle is $(0,r+\frac{d}{2})$ and the center of the bottom circle is $(0,-r-\frac{d}{2})$. The line on the right with the length $l$ connect two points, say $A$ from the top and $B$ from the bottom, with coordinates $(x,y_A)$ and $(x,y_B)$. Note that the line on the right is vertical, hence $A$ and $B$ share the same $x$. In addition $y_A=-y_B$.

In this case, the distance between $A$ and $B$ is simply $l = 2\times y_A$. Since the point $A$ lies on the top circle, hence satisfying its equation, we get $$y_A = \pm \sqrt{x^2 - r^2} +r+\frac{d}{2}$$ Hence, $$l = \pm 2\sqrt{x^2 - r^2} +2r+d$$

To maximize, looking at the right side where $x=0 \to r$, we take the derivative of $l$ and we will get $x=0$. Hence, $y_A = \frac{d}{2}$ or $y_A = 2r+\frac{d}{2}$.

Clearly, $y_A=2r+\frac{d}{2}$ is for the maximum and $y_A=\frac{d}{2}$ is for the minimum. Hence, $d\leq l \leq 4r+d$.

• This proves that the only lengths, $l$, of the prescribed line segment are indeed $d<l<4r+d$. But there is more that must be shown: Can you always move this line segment to the equivalent position on the left side, with the constraint that the movements must preserve the length of the line segment and the endpoints must remain on the circle they started on? (both must hold at all times) Apr 3 '16 at 22:55