I was recently at an international airport, trying to get from one end of a very long terminal to another. It inspired in me the following simple maths puzzle, which I thought I would share here:
Suppose you are trying to get from one end A of a terminal to the other end B. (For simplicity, assume the terminal is a one-dimensional line segment.) Some portions of the terminal have moving walkways (in both directions); other portions do not. Your walking speed is a constant 
- Suppose you need to pause for some period of time, say to tie your shoe. Is it more efficient to do so while on a walkway, or off the walkway? Assume the period of time required is the same in both cases.
- Suppose you have a limited amount of energy available to run and increase your speed to a higher quantity
(or
, if you are on a walkway). Is it more efficient to run while on a walkway, or off the walkway? Assume that the energy expenditure is the same in both cases.
- Do the answers to the above questions change if one takes into account the various effects of special relativity? (This is of course an academic question rather than a practical one. But presumably it should be the time in the airport frame that one wants to minimise, not time in one’s personal frame.)
It is not too difficult to answer these questions on both a rigorous mathematical level and a physically intuitive level, but ideally one should be able to come up with a satisfying mathematical explanation that also corresponds well with one’s intuition.
[Update, Dec 11: Hints deleted, as they were based on an incorrect calculation of mine.]
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23 October, 2011 at 10:23 pm
lhm
The optimal strategy of pausing on moving walkways and running between walkways can be seen as a consequence of unit-elasticity. The slope of a tangent to the “isodisplacement” curves gives the time change for a unit change in velocity. this is greater at lower speeds than higher ones due to the convex shape.
In the relativistic case one can draw a similar picture by replacing the velocity v with the rapidity r = atanh v/c. The isodisplacement curves are also convex but are relative to half the rapidity r. To see this let v’->v in the equation from TT’s comment so that dT/dv = -T/v√(1-v²/c²) and solve for r to find T tanh r/2 = const.
25 May, 2012 at 2:15 am
Lori
Nice puzzle. I realise this is a late response but I didn’t see an intuitive explanation to all three parts without significant algebra manipulations. One possibility is to use a displacement-time (Minkowski) diagram to compare the time taken to cover a given distance at different speeds as shown below (which I hope comes out OK!). Define:
T1(u;v)=net additional time when speed is decreased to u by an amount v
T2(u;v)=net time reduction when speed is increased from u by an amount v
The first diagram shows the Newtonian case in which times in all frames of reference are equal. The three lines T,T’ and T” represent stationary, slower and faster frames of reference respectively. It’s clear that for fixed v, T1(u;v) is increasing in u and T2(u;v) is decreasing in u which implies the answers to parts 1 & 2 respectively.
In the second diagram, time and displacement are now relative to frame of reference and so the X axis label has been omitted. The line segments u and v are orthogonal to the lines T’ or T” for the corresponding times T1′ or T2” respectively (these two cases are actually distinct but have been combined for economy). The T-intercept represents the time dilation for a unit of time in the moving frame T’ or T”. Changing u and holding v fixed in the reference frame T’ or T” implies the slope of the line segments u and v change orthogonal to the lines T’ or T” but the same relationships apply as in the Newtonian case hence the answer to part 3.
Newton SR T T T' T' | / | / | + T'' |_ X T'' | /| _/ | \_ /| _/ | / | _/ | \_ u / | _/ | / | _/ | \_ / | _/ | / |T2 _/ | \_ / |T2'' _/ | / | _/ | X_ v | _/ | u / V | _/ | /| \_ | _/ |+------+/ | / | \X/ | /| _/ | / | _/ | / | _/ | /T1'|_/ | /T1|_/ | / _X | / _+ | / _/ | / _/ | / _/ | /_/ | /_/ |//___________________ X |//____________________25 May, 2012 at 2:21 am
Lori
Above diagram again (some of the symbols were interpreted as html tags):
Newton SR T T T' T' | / | / | + T'' |_ X T'' | /| _/ | \_ /| _/ | / | _/ | \_ u / | _/ | / | _/ | \_ / | _/ | / |T2 _/ | \_ / |T2'' _/ | / | _/ | X_ v | _/ | u / V | _/ | /| \_ | _/ |-------+------+/ | / | \X/ | /| _/ | / | _/ | / | _/ | /T1'|_/ | /T1|_/ | / _X | / _+ | / _/ | / _/ | / _/ | /_/ | /_/ |//___________________ X |//____________________8 June, 2012 at 7:53 am
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17 July, 2012 at 8:39 pm
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22 December, 2012 at 2:40 pm
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3 May, 2013 at 3:40 pm
ybing81
Such a model seems to be somewhat similar to the charge transportation in semiconductors, where concepts like “trap” and “mean free path” may find analogues in this ingenious puzzle.
9 September, 2013 at 6:20 pm
anthonyquas
I can’t resist mentioning another airport-inspired puzzle that I heard from Isaac Kornfeld. Many airlines have a policy that they will only transport suitcases up to a maximum “linear dimension” (that is height + width + depth). The question is “can one cheat?”
More specifically, do exist suitcases S1 and S2, where the linear dimension of S1 is more than the linear dimension of S2, but still S1 fits completely inside S2 (for example diagonally)???
21 December, 2014 at 8:23 pm
Fan
Assuming a two dimensional suitcase, the answer is no, basically by repeated use of the triangle inequality.
2 February, 2014 at 8:00 pm
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15 September, 2014 at 7:11 am
Frances Winters
Clearly the best choice is to avoid the walkways completely. They will be completely blocked by mathematicians tying their shoelaces, cleaning their teeth or whatever.
27 October, 2014 at 4:30 pm
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25 November, 2014 at 7:41 pm
anonymous
Simple solution to 2:
Let G be the length of the ground, W the length of the walkways in total.
Let T be the amount of time you can run at the higher speed, v’. Let T_G be the portion of that time running on the ground.
Assume that you spend T time running at speed v’ and don’t complete the journey, otherwise obviously you just run at full speed the whole time.
Then the time to complete the journey is simply:
T + (G-v’ * T_G)/v + w – (v’+u)(T-T_G)/(v+u). This is a linear function of T_G and the coefficient of T_G is:
(v’+u)/(v+u) – v’/v which can easily be verified to be negative because v’ > v. Therefore to minimize the time, T_G should be as large as possible (i.e you should run on the ground as long as you can).
27 December, 2014 at 12:37 pm
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30 July, 2015 at 4:37 am
Basil
Moving on platforms provides you with a speed boost(provided the platform is moving your way of course) for as long as the platform lasts.
so it’s an issue of maximising te spent on the platforms therefore your downtime tying your shoelace should be wgile on the platform while spending up your energy to run should happen off the platform since running on the platform would get you less time od the speed boost.
I did happen to have the same thoughts on my last trip to england as well and did in fact postpone tying my shoelace till i got on a platform :) I ran without regard for optimising my route though as they were shouting my name for i was late..
regarding Q3 i’m not familiar with relativity so i cant give an opinion but i’d like to consider how the problem would change given a stamina regeneration factor(although i guess the answer there would be run off platforms, rest on platforms unless youre full rested while on thr platform. also run with a platform within sight to regen stamina on it :p)
22 September, 2015 at 10:03 pm
abhishek
the problem is really simple.its the math that you all invoke is complicated.here a simple rigorous solution:
q1:let the total time taken be T if you dont tie your shoeson your optimal route.let t be the time you take to tie your shoes.
1:if you tie your shoes on ground you take T+t total time.
2:if you tie your shoes on walkway you take T+tv/(u+v).since v/(u+v)<1,therefore T+tv/(u+v)<T+t.therefore tie your shoes on walkway
8 December, 2015 at 12:55 pm
Anonymous
I have a similar problem/puzzle that I think would be interesting – can anyone point me to a place where people pose puzzles (and no, it is not a homework problem I am trying to dupe someone into helping me with :) .
It comes from the world of almond harvesting and uncertainty in nut cart positions!
25 January, 2016 at 8:07 am
Anonymous
Take your shoe off and run the whole way, using the walkways when possible. Tie your shoe on the plane.
26 August, 2016 at 12:24 am
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19 December, 2018 at 2:26 pm
Steven E Landsburg
My blog post here tries to get at why people find this problem surprisingly hard.
5 February, 2019 at 7:57 pm
Evan
I still can’t solve it
9 August, 2019 at 6:28 pm
Skylin Chern
Hello, Terry.
A teleplay,小欢喜(Little Joy) has been released in China recently. One scene in it is about walking up and down an escalator in a shopping mall, and then it mentions “Uniqueness of Terrence Tao’s problem”(陶哲轩问题的唯一性).
I don’t know what the “Terrence Tao’s problem” in it means, but I guess it refers to this puzzle. However, if my guess is true, I don’t understand the uniqueness here–the solution to this puzzle is not unique, right?
Have you raised other problem which is unique or whose solution is unique?
Thanks a lot.
12 November, 2020 at 5:26 am
Grüße an die UNI Tübingen
well, thanks mate, we now have to solve this for my physics I course. still no idea but lets try this!
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17 August, 2021 at 6:34 am
Aditya Guha Roy
I just came across this puzzle and coincidentally I thought of something similar while travelling recently. Let me share my thoughts regarding the shoelace tying puzzle which is almost totally devoid of any calculations. Suppose
is the time required to tie the shoelace. Consider two people 
travelling along the same path where 
ties his shoelace on the walkway, and 
ties his shoelace by halting on the ground outside the walkways. Then we can translate the movements of A and B as follows: imagine that A comes optimally till a point 
located at a distance of 
away from his destination, and then boards a special lift instantaneously which takes it to the destination with a constant velocity 
; this lift is the part of the walkway where 
ties his shoelace. Now for B we take B optimally without tying shoelace till the same point where A boarded the lift. Then we let B tie his shoelace without moving any distance and it takes time 
and then after this additional time 
B is allowed to board a lift which travels the distance of $ut$ with a speed of $u+v$; this corresponds to the additional length of the walkway where B travelled at a higher speed. We see that B takes more time to reach the destination than A.
, thus walkways are more optimal than the ground in the sense that the more the time you spend on ground the slower you reach the destination; this is exactly what the above thought translates to in simple words.
A careful observation would show us that on ground there is a boost of 0 and on the walkway there is a strictly positive boost of
17 August, 2021 at 7:47 am
Aditya Guha Roy
Similarly for part 2 one should consider running outside the walkway in quest to reduce the time spent on ground.
18 August, 2021 at 8:28 pm
Aditya Guha Roy
Reblogged this on Aditya Guha Roy's weblog.
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28 May, 2022 at 12:40 am
Susan Elizabeth Mango
Did you give the solutions someplace, for the less mathematically inclined, like myself?
7 September, 2022 at 4:21 pm
Anonymous
Why would it even be a math problem? It is obvious that one should tie the shoe lace on the walkway. One is still moving while on the walkway, but the velocity will be zero if one stops to tie the shoe lace off the walkway. Isn’t it just common sense?