I recently was reminded of a puzzle I solved in college and thought I'd give it a shot again. However, being distanced from college math, I am having a harder time remembering how I arrived at the solution.

The problem is as follows: Imagine you are standing in the middle of an open field. You walk forward 16 feet, turn right and walk 8 feet, turn right and walk 4 feet, and so on. This continues indefinitely. When you finally reach an infinite number of turns, how far will you be from your original starting point (as the crow flies)? Generalized, here is what the problem looks like:

enter image description here

I did manage to find the original solution that I came up with for the problem. However, I did not show my work so the process is lost. After attempting to resolve this without any luck, I thought I would toss this out to the community to solve as a fun puzzle.

For reference, this is what I believe to be the solution:

$$\frac{a}{\sqrt{r^2 + 1}}$$

  • $\begingroup$ not hard. separately do the $x$ and $y$ coordinates of the final point, compared with the starting point, as (strictly alternating) series. $\endgroup$ – Will Jagy Jul 20 '14 at 21:07
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    $\begingroup$ Something bugs me about the phrase "When you finally reach an infinite number of turns..." since that will never happen. Perhaps you could edit to "In the limit of this process"? $\endgroup$ – Matthew Leingang Jul 21 '14 at 18:47

The horizontal displacement will be $$L_x=a-r^2a+r^4a-\ldots\;=\frac{a}{1+r^2},$$ and the vertical one $$\;L_y=ra-r^3a+r^5a-\ldots=\frac{ra}{1+r^2}.$$ The result then follows from the Pythagorean theorem: $$L=\sqrt{L_x^2+L_y^2}=\frac{a}{\sqrt{1+r^2}}$$

  • $\begingroup$ This is very similar to what I attempted earlier, I think I was just missing a few things. Do you mind expanding your work on how you arrived at your final displacements? $\endgroup$ – Tyler Murry Jul 20 '14 at 21:14
  • $\begingroup$ @TylerMurry $a$ is the length of the first step, $r^2a$ the length of the third one, $r^4a$ of the fifth, etc. Similarly $ra$ is the length of the second step, $r^3a$ corresponds to the fourth, $\ldots$. I just take them with appropriate signs and then use that $1-r^2+r^4-r^6+\ldots =\frac{1}{1+r^2}$. $\endgroup$ – Start wearing purple Jul 20 '14 at 21:17

We can describe this walk in the complex plane. Starting at $z=0$ we add in sequence $$a,-i r a, -r^2 a, ir^3 a,\cdots, (-i r)^k a,\cdots$$ This is simply the geometric series $$\sum_{k=0}^\infty (-i r)^k a = \frac a{1+ir}$$ which is at a distance of $$\left|\frac a{1+ir}\right|=\frac a{\sqrt{1+r^2}}$$ from the origin.

Additionally :

An immediate generalization is that if we turn an angle of $\phi$ to the right rather than $\dfrac\pi2$, we have the series $$\sum_{k=0}^\infty a\ (r e^{-i\phi})^k = \frac a{1-r e^{-i\phi}}$$ and thus a displacement of $$\frac a{\sqrt{1-2r\cos\phi+r^2}}$$ from the origin.

  • $\begingroup$ One addendum that goes to the symmetry argument mentioned elsewhere in the $\phi=\pi/2$ case: Note that if we group and add pairs of adjacent terms, then the series takes the form $a(1-i r)-r^2 a(a-i r)+\cdots = a(1-i r)(1-r^2+\cdots)$. Consequently the complex aspect of this (i.e. direction) drops out and all that's left is the distance. $\endgroup$ – Semiclassical Jul 21 '14 at 22:18
  • $\begingroup$ I can't believe how complicated will the solution of your generalized problem be without using complex approach. $\endgroup$ – A. Chu Jul 26 '14 at 14:21
  • $\begingroup$ @BrianRushton: Thank you very much for your generous bounty on this old answer of mine. I'm quite pleased that you found it worthy of such. $\endgroup$ – Semiclassical Sep 27 '14 at 13:45

It is immediately obvious that a transformation consisting of a clockwise rotation of $\pi/2$ radians and a contraction by a factor of $r$ about the limiting endpoint of the path will map the black path to a proper subset of itself; therefore, if we enumerate the vertices starting from the outermost vertex, the red line segment is a subset of a single line passing through all odd vertices, and the endpoint is the intersection of this line with the (perpendicular) line passing through all even vertices. Therefore $d$ satisfies $d^2 + (rd)^2 = a^2$, or $d = a/\sqrt{1+r^2}.$

enter image description here

  • $\begingroup$ I love this answer, but it would be even better with a picture! Maybe you could add a modified version of the picture in the OP? $\endgroup$ – André 3000 Jul 22 '14 at 1:13
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    $\begingroup$ @SpamIAm Funny you mention that...I was working on the above animation just as you made that comment. $\endgroup$ – heropup Jul 22 '14 at 1:30
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    $\begingroup$ That is really quite gorgeous, and makes the symmetry of the problem manifest. $\endgroup$ – Semiclassical Jul 22 '14 at 23:55

A symmetry argument not requiring any geometric/infinite series summation.

Consider just one "turnaround": just after he makes an "up" move of $r^3 a$.

He will next make a step of $r^4 a$ to the right, parallel to the first right, as part of the second "turnaround".

He is now at a distance $D_1 = \sqrt{(a-r^2a)^2 + (ra - r^3a)^2} = a(1-r^2)\sqrt{1+r^2}$ from the original point.

Now if for initial step $a$, the total distance moved after infinite "turnarounds" is $f(a)$ then we must have that

$$f(a) = D_1 + f(r^4a) = f(a) + D_1 + r^4f(a) \implies f(a) = \frac{D_1}{1-r^4}$$

This is because if $D$ is the distance for $a$, then $cD$ is the distance for $ca$, and all the "turnaround" points are co-linear.

Thus the distance moved is

$$\frac{a(1-r^2)\sqrt{1+r^2}}{1-r^4} = \frac{a}{\sqrt{1+r^2}} $$


I would calculate horizontal and vertical distances separately and then use Pythagoras theorem to calculate the total distance from the start. The horizontal distance is $a-r^2a+r^4a-...=\frac{a}{1+r^2}$ and the vertical distance is $ra-r^3a+r^5a-...=\frac{ar}{1+r^2} $. The total distance is $\sqrt{\frac{a^2}{(1+r^2)^2}+\frac{(ar)^2}{(1+r^2)^2}}=\frac{a}{\sqrt{1+r^2}}$.


Another, but equivalent, way of viewing this.

The first two steps together with the line joining their endpoints form a right angled triangle T with hypotenuse h = $a\sqrt(1+r^2)$.

The third and fourth steps are each $r^2$ of the length of the original sides so they form two sides of a similar triangle with the hypotenuse colinear with the hypotenuse of T. By similarity, it has length $r^2$h. Therefore this vertex is $h(1-r^2)$ from your starting point. This pattern repeats so the vertex of successive triangles all lie on the same line, with distances based on the partial sums of $h(1 - r^2 + r^4 - ...)$. Therefore the limit, if there is one, must be $h/(1 + r^2) = a/\sqrt(1 + r^2)$. In this case it's geometrically obvious that the limit does exist, even without using any of the formal methods of confirming it.


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