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Take $1$ step forward, turn $90$ degrees to the left, take $1$ step forward, turn $90$ degrees to the left ... and keep going, alternating a step forward and a $90$-degree turn to the left.

Where do you end up walking? It's very easy to see that you end up walking on the perimeter of the same square, with side equal to $1$ step (every $4$ steps you are back at the point where you started), if you are walking on a plane. But what happens if you walk on the surface of a sphere?

The answer obviously depends on how long is your step in relationship to the sphere's radius $R$. If your step length is $\pi R$, for example, you end up walking back and forth over the same lune, formed by two half circumferences at $90$ degrees angle. If your step length is $\pi R/2$, you end up walking on a triangle, formed by quarter circumferences at $90$ degrees angle. I believe one can prove the following:

a) For any $x\geq 0$ and any arbitrarily small $\epsilon>0$, there is a step length $y\in(x,x+\epsilon)$ such that the process above takes you back to the starting point in a finite number of steps.

b) No step length that brings you back to the starting point in a finite number of steps is a rational multiple of the length of the circumference, $2\pi R$, except for integer multiples of $\pi R/2$ (which keep you cycling over the same $1$,$2$, or $3$ points).

c) Any step length that does not bring you back to the starting point in a finite number of steps eventually brings you all over the surface of the sphere, in the sense that for any point $P$ on the sphere's surface and any arbitrarily small $\epsilon>0$, you eventually end up within distance $\epsilon$ of $P$. In fact, for any point $P$ and any angle $\alpha$, you eventually end up within distance $\epsilon$ of $P$ and with an angle between $\alpha$ and $\alpha+\epsilon$ of your starting direction (measured as the angle between the respective great circles)!

Can you prove any or all of the above? Note that a proof of b) would automatically yield an answer to this related question.

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  • $\begingroup$ For 1, picking any $y$ algebraically independent of $\pi$ will get you arbitrarily close to the starting point. $\endgroup$ – Stella Biderman Jan 23 '17 at 18:09
  • $\begingroup$ @StellaBiderman that's not ... quite what a. is about: I want to really go back exactly to the starting point. But it's an interesting observation, that might yield clues about c. Would you care to write it down a little more extensively as an answer? $\endgroup$ – Anonymous Jan 23 '17 at 18:15
  • $\begingroup$ When I get off from work, probably. My lunch break is just about over. $\endgroup$ – Stella Biderman Jan 23 '17 at 18:16
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    $\begingroup$ It seems that a) can be solved as stated by using stereographic projection; that preserves angles. Now draw a square on the plane and center the sphere. Then project the square onto the sphere. now adjust the sphere size to satisfy your $x$ . Have I missed something? $\endgroup$ – rrogers Jan 25 '17 at 13:48
  • $\begingroup$ In my answer below I showed that b) is true, but I don't see how that is related to the question you cite at the end. Just curious. $\endgroup$ – Intelligenti pauca Jun 14 '17 at 9:40
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Let our walk $A_1, A_2, A_3,\ldots$ take place on a unit sphere of center $O$, and let $\theta$ be the step length (of course we move along great circles): $\angle A_1OA_2=\angle A_2OA_3=\ldots=\theta$, while each plane $A_iOA_{i+1}$ is orthogonal to plane $A_{i+1}OA_{i+2}$.

Consider now circle $\Gamma$, passing through $A_1$, $A_2$ and $A_3$, let $C$ be its center and set $\phi=\angle A_1CA_2=\angle A_2CA_3$. A counterclockwise rotation of angle $\phi$ about axis $OC$ carries $A_1$ to $A_2$, $A_2$ to $A_3$ and $A_3$ to some point $B$. But $\angle A_3OB=\theta$ and plane $A_3OB$ is orthogonal to plane $A_2OA_3$. In other words: $B=A_4$, and by the same reasoning all points $A_i$ must belong to $\Gamma$. It follows that proposition c) in the question is false: the walk makes me move only along a well defined circle, depending only on my starting point and on my step length $\theta$.

To compute angle $\phi$ as a function of $\theta$, we can set up a coordinate system such that $A_1=(0,0,1)$ and $A_2=(\sin\theta,0,\cos\theta)$. Point $A_3$ can be obtained by rotating $A_1$ clockwise by $\pi/2$ about line $OA_2$, yielding: $A_3=(\sin\theta\cos\theta,\sin\theta,\cos^2\theta)$. But $\angle A_1A_2A_3=\pi-\phi$, so that: $$ \cos(\pi-\phi)={(A_1-A_2)\cdot(A_3-A_2)\over ||A_1-A_2||^2}, $$ whence: $$ \phi=\pi-\arccos\left({1-\cos\theta\over2}\right). $$

The path returns to its starting point if $n\phi=2m\pi$, with $n$ and $m$ positive integers, which by the previous equation is equivalent to: $$ \cos\theta=1+2\cos\left(2{m\over n}\pi\right). $$ Given $t\in[-1,1]$, the right hand side can be as close to $t$ as one wants, if rational number ${m\over n}$ varies in the range ${1\over4}\le{m\over n}\le{1\over2}$. It follows that proposition a) is true: for any $x\ge0$ and any arbitrarily small $ϵ>0$ one can always find ${m\over n}$ such that the above expression for $\cos\theta$ takes a value comprised between $\cos x$ and $\cos(x+\epsilon)$. The corresponding value for $\theta$ will then give a step length between $x$ and $x+\epsilon$.

Proposition b) could also be true, because such a value of theta is not, in general, a rational multiple of $2\pi$. One should prove that the above equation for $\theta$ as a function of ${m\over n}$ only gives a rational multiple of $2\pi$ when $\theta=k\pi/2$. I don't know if an answer is possible: I asked that in a new question (see EDIT below).

Diagram below is taken from an interactive GeoGebra worksheet, which draws the first eight steps of the path, allowing to set the step length with a slider.

enter image description here

EDIT.

Point b) of the question, as shown above, amounts at proving that equation $$ \cos A-2\cos B=1, $$ where $A$ and $B$ are rational multiples of $2\pi$, has only solutions with $A$ a multiple of $\pi/2$.

It turns out that this is indeed true, see the answer to the question I asked. After that answer was given, I found a nice paper by Conway and Jones which explicitly refers to rational sums of cosines of rational angles: their Theorem 7 can be immediately applied to our case, showing that the only solutions of the above equation have $A$ and $B$ both multiples of $\pi/2$. We can thus confirm that proposition b) is true.

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