It's a hot day. You are in the middle of a flat sandy plane that stretches as far as you can see. Your hands are shackled and you are surrounded by soldiers of the mighty Pythagorean Brotherhood. Ten steps from you a man, his hands also in shackles, is standing frozen. His name is Hippasus of Metapontum. A Pythagorean soldier yells at Hippasus. "Right-angle turn and move!"

Hippasus makes a 90° left turn and hesitates. A drop of sweat slowly moves down his front head. He then makes a step followed by a 90° right turn. He again hesitates. Another step, and again a 90° left turn. The soldier raises his sword and shouts "Death to the irrational!" Hippasus' body collapses and his head rolls in your direction, coming to a halt just two steps away from your feet.

The soldier turns to you. "Now you go. Make a right angle turn followed by a nonzero number of straight steps in forward direction, and keep repeating this. Each step needs to be of exactly the same size. Each turn needs to be an exact right angle turn and must happen at a position that is a whole number of straight-line steps away from all previous turns. You need to make a total of 100 right angle turns, 50 to the left and 50 to the right. If you fail, you will be killed on the spot, if you succeed, you are a free man, and a witness of the fact that integer distances rule the world. Go!"

You think for a few seconds, and make a left turn. What's next?


4 Answers 4


I'm going to use compass directions to reduce confusion.

You are going to walk in a figure eight, with a width of $c$ all the way up, and with north and south loops of height $a$ and $b$. You start at the south east corner of the figure eight, facing east.
You turn left and walk $b$ steps north, then turn left again and walk $c$ steps west across what will be the middle bar of the figure eight. You are now in the middle of the west side, facing west.
From here you can make four left turns around the south loop, walking $b$ south then $c$ east then $b$ north then $c$ west. Or you can make four right turns around the north loop, walking $a$ north then $c$ east then $a$ south then $c$ west. Each one puts you back at the same spot, facing west. Twelve of each loop gives you 48 right turns and 48 left turns. You started with two left turns, so now you finish with two right turns half-way around the north loop and end up in the north east corner. That's 50 each and 100 total.

Now all we need to do is choose some numbers.

We must have $a,c$ form a pythagorean triangle; $b,c$ must also; and so must $a+b,c$. That covers all 6 diagonals of the figure eight, and the rest are integer sides and the sums of integers. Then every corner of the figure eight is an integer distance from the others.
It feels like overkill to pull this out again, but a=44, b=117, c=240 will do the job. That's 16025 steps; I recommend taking very small steps.

An immediate upvote to anyone who finds smaller numbers, because it is hot out here.

You announce your intentions to the soldiers, hinting that the quicker you finish, the quicker everyone can get into the shade. Being the Pythagorean Guard, of course they are able to help; one calling himself @crazyiman tells you to use:

a = 11, b = 80, c = 60.

This only requires 5275 steps. You refrain from announcing exactly how much faster this will be, and start walking.

  • 6
    $\begingroup$ a=80, b=11, c=60 works. $\endgroup$ Feb 15, 2015 at 13:06

No proper solution yet, but here are 2 degenerate solutions.

Trivial solution: take zero (which is a whole number of) steps on each turn, turning alternately left and right 50 times each. All points are co-located, so distance between any pair of points is 0, which is a whole number.

Solution with a single zero-step: turn left 50 times, taking alternately 3 and 4 steps (going anticlockwise in the same 3x4 rectangle - the ends of each diagonal are 5 steps away from each other). Turn right, take zero steps, then turn right again 49 times, retracing the same rectangle as before, going clockwise this time.

  • 1
    $\begingroup$ Ahhh... have to close this loophole. Thanks! $\endgroup$
    – Johannes
    Feb 15, 2015 at 15:50

I came up with the same solution as Callidus. Adding a few appendix notes.

First, here's some dirty code I drafted up to find the correct dimensions:

#could be in the main program but this is a little easier to understand
def issquare(x):
    for z in range(200):
        if x==z**2:
            return True
    return False

#list of pythagorean triples where the legs are smaller than 100
c = []

#populate c
for x in range(1,100):
    for y in range(x,100):
        if issquare(x**2 + y**2):

ubound = len(c)

#for each pair in c, check if they satisfy the form (x, y), (y, z), (x+z, y)
for x in range(ubound):
    for y in range(x+1,ubound):
        if c[x][1]==c[y][0]:
            if (c[x][0]+c[y][1], c[y][0]) in c:
                print(c[x], c[y], (c[x][0]+c[y][1], c[y][0]))
            if (c[y][0], c[x][0]+c[y][1]) in c:
                print(c[x], c[y], (c[y][0], c[x][0]+c[y][1]))

Second, here's a MSPaint diagram to make the strategy easier to understand:

enter image description here


Earth is not flat. It is not even spherical. But that's not a problem, as long as we assume that the surface is at least topologically "smooth enough" as an embedding of a unit sphere (I think it needs to be $C_1$ but it might even be $C_0$)

Follows a more exact explanation of the ideas in the previous answer.

! By continuity, from any point on earth, take the set of maximum circles. Take as a function of the direction, the difference of the length of the maximum circle in that direction and in the perpendicular direction. This function is either constantly 0 or it takes positive and negative values. In any case there is at least one direction, where the maximum circle has the same length as its perpendicular.

! Now, choose a step length that's $\frac{1}{40000}$ of that length, and walk 40000 steps in that chosen direction. You get to the initial point (so $0$ steps away of it). Now turn right or left at will, and walk 40000 more steps. As the two "meridians" have the same length, you'll get again to the initial point. Turning lefto or right just changes the way you go around the meridian, not its length.

! So you can walk 40000 steps, turn left or right at will, walk another 40000 steps, again at will, and at every single stop you will be at the starting point, that is 0 steps away.

  • $\begingroup$ You conveniently forgot about the requirement that the number of left- and right-turns need to be in balance? :) $\endgroup$
    – Johannes
    Feb 16, 2015 at 2:20
  • $\begingroup$ @Johannes see the more complete explanation. As you can probably see, it doesn't matter at all if you turn left or right, or if you walk+turn two, seven or a million times. $\endgroup$
    – rewritten
    Jan 19, 2016 at 11:45

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