I think that this question is well known but I cannot remember its name, and now I am interested in it and wanted to look it up, but cannot find anything just based on a description. If anyone knows the name or can find it (or anything similar), that would be very helpful. The question is as follows:

You have a rigid rod of unit length, and some curve in 3d space, which is linear for all but a finite portion of its length. We say the rod is "on the curve" if each of its endpoints are on the curve. Call the "ends of the curve" the 2 linear portions that are infinite. Prove or disprove that for all possible such curves, we can move the rod from one end of the curve to the other while staying on the curve the whole time.

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    $\begingroup$ So the curve is a single infinite line with a middle section cut out and replaced by something else -- or can the two ends of the curve go in independent directions? $\endgroup$ Jul 13, 2016 at 12:31
  • $\begingroup$ It can go in independent directions, although I think the two would be equivalent, no? $\endgroup$ Jul 13, 2016 at 12:55
  • $\begingroup$ x @Hardik: A curve that consists of one narrow 180° turn connected to infinite straight rays parallel to each other cannot be navigated by the ladder -- there'll be no way to turn it around. $\endgroup$ Jul 13, 2016 at 12:59
  • $\begingroup$ Oh yes, youre right. Then I'll require the ends to go in opposite directions. $\endgroup$ Jul 13, 2016 at 13:00
  • $\begingroup$ A precise condition is stated in Theorem 2 quoted below. The beginning and end segments (which need not be infinite) must be isolated in a sense, a sense violated by Henning's example. $\endgroup$ Jul 13, 2016 at 13:08

2 Answers 2


One name in the literature is: The Ladder Movers' Problem:


Goodman, Jacob E., János Pach, and Chee Yap. "Mountain climbing, ladder moving, and the ring-width of a polygon." American Mathematical Monthly 96.6 (1989): 494-510. (ACM link, JSTOR link.)

      Fig.4 from tech report

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    $\begingroup$ Note that requiring the curve to be polygonal rather than just continuous is crucial for the result to hold. $\endgroup$ Jul 13, 2016 at 15:00

The problem seems to be surprisingly dependent on the precise assumptions we make. Consider, for example, this curve

plot of the curve defined by $ \kern-2em\displaystyle \gamma(t) = \begin{cases} \langle \max(t,-1),t \rangle & t \le 0 \\ \langle t,\frac34 t \sin(5 \log(t)) \rangle & 0 < t \le 1 \\ \langle \cos(t-1)),\sin(t-1) \rangle & 1 \le t \le 1+\pi \\ \langle -1, t-1-\pi \rangle & 1+\pi \le t \end{cases} $

This is a continuous curve with a Lipschitz continuous parameterization (the derivative of $t\sin(\log(t))$ is bounded) and is therefore rectifiable. And the ends are in opposite directions as required.

Yet, the ladder cannot be moved continuously along the curve.

Imagine that it moves from high parameter values to low ones. Whenever the front is is at one of the points where the curve touches the upper red line ($y=\frac34x$), the back end will need to be around point $A$ on the circular arc, since that is the only place on the curve behind the front end that has the right distance from it. ($A$ is the center of the radius-$1$ circle that has $y=\frac34x$ as tangent at the origin).

Conversely, when the front end is at one of the points where the curve touches the lower red line, the back end must be around $B$. Since the front oscillates between the two red lines infinitely many times before it reaches the origin, so must the back end oscillate between $A$ and $B$. But that means that the position of the back end cannot tend to a limit while the front end approaches $(0,0)$, so a continuous movement of the ladder is impossible.

(The factors of $\frac34$ and $5$ are just there to allow the diagram to show enough of the wiggles to make it clear what is going on. It would work just as well with $t\sin(\log t)$ as with $\frac 34t\sin(5\log t)$, only not as visibly. It is important to keep the largest of the wiggles from extending beyond the unit circle, though -- otherwise the argument about where the back end of the ladder has to be won't work).

Requiring the curve to be (piecewise) smooth will probably avoid this kind of trouble.

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    $\begingroup$ Umm... the OP didn't ask to answer the quoted question. They asked how is that question called, not answered. $\endgroup$
    – EKons
    Jul 13, 2016 at 16:12
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    $\begingroup$ Still, a cool example illustrating some unintuitive complexities that can occur. The inverse, what family of curves (continuous or not) can be produced, is probably also interesting. $\endgroup$ Jul 13, 2016 at 18:06
  • $\begingroup$ It doesn't seem to require that both ends of the rod must move at once. In your example, if the rod's endpoints are initially $(-1,1)$ and $(-1,0)$ we can slide it to endpoints $(-1,0), (0,0),$ and then keep the end $(0,0)$ fixed while rotating the other end along the semi-circle to $(1,0).$ But the part of the curve sufficiently close to $(1,0), $ and lying below the $x$-axis, has slope $>1.$ So when the rod's endpoints are at $(0,0), (0,1)$ we cannot move the endpoint $(0,0)$ onto the ray $\{(x,x): x<0\}$ without puling the other end off the curve $\endgroup$ Jul 13, 2016 at 19:32
  • $\begingroup$ @user254665: The endpoints don't have to move at once, but they do have to keep a distance of exactly one unit at all times. The point is that for this curve you cannot get it into position $(-1,0),(0,0)$ in the first place. $\endgroup$ Jul 13, 2016 at 19:55
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    $\begingroup$ @TheGreatDuck: That's not a reason to withhold relevant information. In particular, the variant the problem Joseph O'Rourke found a solution for is not exactly the same as the problem the OP described. This answer is to point out that the difference between them is meaningful, such that a reader isn't misled into thinking because the answer to the polynomial variant is "yes", the answer to the general variant is the same. $\endgroup$ Jul 14, 2016 at 7:30

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