## Did you solve it? Where’s the best place to tie your shoe in an airport?

The solution to today’s travel teaser.

Earlier today I asked you the following questions:

You are in an airport, and you are walking from security to your gate. The distance includes stretches of travelator, or moving walkway, and stretches of floor. You have a constant walking speed, *u*, and the travelators also have constant speed, *v*. So, when you are on the travelator you are traveling at *u *+ *v*.

*1. You need to stop to tie your shoe. If tying your shoe always takes the same time, will you get to the gate quicker if you tie your shoe on the travelator, off the travelator or does it make no difference?*

*2. You have enough energy to walk at faster speed, w,** for a fixed time. In order to get to the gate quicker, do you increase your speed when you are on the travelator, off the travelator or does it make no difference?*

**The answers:**

- Tie your shoelace on the travelator.
- Speed up off the travelator.

**The explanations (simple):**

Lets simplify the details and suppose that there is a single stretch of floor followed by a single stretch of travelator. There’s a lovely intuitive explanation of why it is better to tie your laces on the travelator.

Imagine two people, say, Angela and Beatrice, start walking from security at the same time. Angela stops to tie her shoe immediately before the travelator, but Beatrice ties her shoe as soon as she steps on the travelator. By the time Angela has finished tying her shoe and steps on the travelator, Beatrice will now be way ahead of her and Angela will never make up the distance.

There is a similar intuitive explanation of why it is better to fast-walk before you get to the travelator. Imagine again that Angela and Beatrice start together. Say, to keep it simple, that Angela fast-walks straight away, which puts her at a distance D in front of Beatrice. When Angela steps on the travelator she is D ahead, but while she is on the travelator, and Beatrice is not, Angela increases her distance from Beatrice to more than D. When Beatrice reaches the travelator, she can start fast-walking, but so long as both her and Angela remain on the travelator she will never make up more than D.

**The history**

The shoelace puzzle was first posed by Terence Toe, I mean Terence Tao. (Couldnt resist, sorry). He is one of the most celebrated mathematicians alive. He featured this problem on his blog in 2008, two years after winning the Fields Medal.

**How well did you do?**

About 15,000 of you submitted answers. The majority – just – got the first one correct. For the second question, however, the correct answer was the one chosen by the fewest people. Interesting.

Q1: where do you tie your shoe?

on the travelator: 47.3%off the travelator: 11.4%

doesnt matter: 41.3%

Q2: where do you speed up?

on the travelator: 28.7%

off the travelator: 17.2%doesnt matter: 54.1%

**The explanations (trickier)**

Here is a more thorough explanation of the answers using some algebra. Again lets assume there is a single stretch of floor and a single stretch of travelator. Lets start by defining the terms.

T = time it takes to tie laces.

u:human speed

v: travelator speedD

_{f}: Floor distanceD

_{tr}: travelator distance

The relevant equation:

Speed = distance/time, and so, time = distance/speed

Question 1: Total time = time on floor + time on travelator + time tying laces = (distance on floor/speed on floor) + (distance on travelator/speed on travelator) + T = D_{f}/*u +* D_{tr}/(*u *+ *v**) + T*

If laces tied on floor, the total time: D_{f}/*u + *T + D_{tr}/(*u *+ *v**)*

If laces tied on the travelator, the total time: (D_{f}/*u) + *T + (D_{tr } T*v)/(**u *+ *v**) *

Since* D _{tr} *> D

_{tr}T

*v,*the total time if laces tied on floor is greater, so you get there quicker if you tie laces on the travelator.

Question 2: Now let T = time spent at higher speed *w.*

Total time if speeding on floor: T + (D_{f} T*w)/u *+ D_{tr}/(*u *+ *v**)*

Total time is speeding on travelator: (D_{f}/*u) + *T + (D_{tr } T(*w*+ *v**))/(**u *+ *v**) *

Since *w* > *u* then *w/u > *(*w* +* v*)/(*u* + *v*) and (I promise) this shows that total time is less when you walk fast on the floor.

Because of the time and space constraints of this blog I havent been able to explain each step in detail. Im sure some commentators will be able to help out below the line if you are still confused. Thanks all!

*I post a puzzle here on a Monday every two weeks. **If you are reading this on the Guardian app, click where it says Follow Alex Bellos and youll get an alert when I publish the next puzzle. **If you want to propose a puzzle for this column, please email me Id love to hear it.*

*Im the author of three popular maths books including Alexs Adventures in Numberland and the maths colouring book Snowflake Seashell Star. **You can check me out on Twitter, Facebook, Google+, my personal website or my Guardian maths blog.*

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