In a plane, there is a robotic arm consisting of $n \ge 2$ segments of length 1, like this:

  • The first segment is fastened to a single point ("origin"), but it can rotate freely around that point.
  • All other segments are connected to the previous segment by means of a joint, so they can bend with respect to the previous segment. However, the joints have limited allowance: the segment cannot deviate from the direction of the previous segment by more than some angle $\alpha$ (or, in other words, the angle between no two consecutive segments can be smaller than $\pi-\alpha$). All the joints have the same allowance.

Here's a rough picture of such an arm with $n = 4$ segments:

enter image description here

The blue circles are the joints, the blue lines are the segments. The dashed lines show the possible angles of the first joint.

The question: What is the least allowance $\alpha$ that makes it possible for the robotic arm to reach any point in the disk of radius $n$ around the "origin" (the point where the first segment is fastened)?

Source: I made up this puzzle on my own.

P. S.: As you can see, this is my first puzzle here. Please point out anything that I could have done better. (The solution is pretty nice and while it's nothing terribly complicated, you will need to come up with a bit of elegant reasoning. So I hope that makes it a math puzzle and not just a math problem.)


2 Answers 2


I would say the maximum allowance is $\alpha = \frac{2\pi}n$. If all arms deviate with this angle from the previous, the arms form a regular polygone with $n$ edges, and the robot can touch the origin with the tip of the arm. By widening the angle and rotating around the origin, the robot can then reach every point of distance $n$ to the origin.
To see that this angle is necessary, assume $\alpha < \frac{2\pi}n$. Assume all joints are bent to the right with maximal allowance. Then the joints all lie on one common circle. If seen from the midpoint of the circle, two consecutive joints are an angle of precisely $\alpha$ away. Since there are $n$ segments, the angle between the origin and the tip of the robot arm is (when seen from the center of the circle) equal to $2\pi - n \alpha >0$. Hence the tip of the robot arm cannot reach the origin.

  • 1
    $\begingroup$ You'll need to provide a proof why it isn't possible to reach the origin for any $\alpha < 2\pi/n$. $\endgroup$
    – Magma
    Commented May 7, 2020 at 19:33
  • $\begingroup$ Or (which in my opinion is simpler), rot13(gung va nal cbyltba jvgu nyy fvqrf rdhny, gur yrnfg vafvqr natyr vf ng zbfg $\pi-2\pi/n$). $\endgroup$
    – Ramillies
    Commented May 7, 2020 at 21:15

Partial answer:

First, we can simplify the problem by

moving all the segments so that they all start at the origin

And so the problem reduces to

finding sets of $n$ complex numbers of unit length whose sum is 0 and where the maximal angle between them is minimized.
(Notice that while the initial problem allowed for arms bending in both directions, the shuffle makes it clear that any valid arm corresponds to one with all joints bending in the same direction.)

An obvious answer is

the $n$'th roots of unity, as mentioned already

But this fails to account for the fact that the angle between the first and last segments isn't a joint and thus isn't limited! The problem thus reduces to proving the (non?)existence of

a balanced set with one angle greater than $\frac{2\pi}{n}$ and all others less than $\frac{2\pi}{n}.$

At very least, I believe I can prove that

there is no continuous path from the roots of unity to another solution (i.e. regular polygons are rigid under the angle constraint)

but I don't know what that allows me to show.

  • $\begingroup$ I don't understand your first spoiler... For instance the second segment can never start at the origin. I think you're on the right track, however. $\endgroup$
    – Ramillies
    Commented May 8, 2020 at 13:23
  • $\begingroup$ What I mean is that for any valid arm that exists, I can rot13(cresbez gung bcrengvba naq genafsbez gur nez vagb n frg bs irpgbef qrfpevovat gur qverpgvbaf bs gur frtzragf) which simplifies the problem. $\endgroup$ Commented May 8, 2020 at 13:37
  • $\begingroup$ I see, you essentially meant that rot13(lbh znxr n frcnengr pbbeqvangr flfgrz sbe rnpu frtzrag fb gung vg fgnegf va gur bevtva bs vgf bja flfgrz). That's of course correct and useful. $\endgroup$
    – Ramillies
    Commented May 8, 2020 at 14:04
  • $\begingroup$ Nice how this solution transforms the geometric puzzle into an algebraic one. This approach might still wind up verifying @daw's solution though. $\endgroup$
    – humn
    Commented May 8, 2020 at 16:55
  • $\begingroup$ I'd be very surprised if it didn't, but this is certainly a more rigorous way to go about it. $\endgroup$ Commented May 8, 2020 at 17:59

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