A version of part (b) was asked as an Amazon interview question.

A cable of 80 meters (m) is hanging from the top of two poles that are both 50 m from the ground. What is the distance between the two poles, to one decimal place, if the center of the cable is:

(a) 20 m above the ground?
(b) 10 m above the ground?

I added part (a) since it is interesting mathematically. And I admit I did not solve it, but it was fun so I wanted to share. Watch the video for a solution.

Can You Solve Amazon’s Hanging Cable Interview Question

Or keep reading.
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"All will be well if you use your mind for your decisions, and mind only your decisions." Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

First let’s solve (a) 20 m above the ground. The diagram is like the following.

I didn’t remember how to solve this, and most of the references focused on physics concepts (weight, tension). Luckily I found a nice paper from Neil Chatterjee and Nita Bogdan with equations that made sense to me: The hanging cable problem for practical applications.

The setup is as follows. Due to symmetry, we can consider half the problem from one pole to the center. We can then double this distance for the distance between the two poles.

Also we will use a convenient coordinate system, centered at the lowest point of the hanging cable (which takes the shape of a catenary). We get the following “half diagram”:

The equation for a catenary tangent/touching the ground/x-axis is:

y = a cosh(x/a) – a

arc length = a sinh(x/a)

The parameter a is unknown, and we want to solve for the value of x of the pole.

Note the top of the pole is (x, 30), so we get one equation:

a cosh(x/a) = 30 + a

We also know that half of the cable length is 40 m, so using the equation for arc length of a catenary, we get another equation:

a sinh(x/a) = 40

We divide both equations by a to get:

cosh(x/a) = (30 + a)/a
sinh(x/a) = 40/a

We then use the hyperbolic identity:

cosh2t – sinh2t = 1

Substituting, we get:

(30 + a)2/a2 – (40/a)2 = 1

We can solve this to get:

a = 35/3

Then we use the equation:

(35/3) sinh (x/(35/3)) = 40

x = (35/3) arcsinh(120/35)

x = (35/3) ln(120/35 + √((120/35)2 + 1)) ≈ 22.7

As this is half the hanging chain, the distance between the two poles is then double this value:

2x = (70/3) ln(7) ≈ 45.4 m

Now let’s proceed to the second part.

(b) 10 m above the ground

Now we could try the same half diagram as in case (a)

We then get the equations:

cosh(x/a) = (40 + a)/a
sinh(x/a) = 40/a

We then use the hyperbolic identity and substitute the above results to get:

(40 + a)2/a2 – (40/a)2 = 1

But if you try to solve this there is a problem. This equation has no solution! So what is going on?

Let’s think about this logically.

If the cable is 80 m, then half of it is 40 m. But notice 40 m from the top of a 50 m pole is already 10 m above the ground. The cable therefore is hanging directly downward! The cable has to be doubled back upon itself, and the two poles must be coincident and 0 m apart!

This part is actually a trick question: the two poles are a distance of 0 apart. No physics was required to solve this one, just logical thinking!

Now an employee that uses common sense before wasting time with unnecessary calculations…I’d imagine that is probably an employee you want to hire.

I admit I did not solve this problem as I did not recognize the trick. But I like to learn about these problems, because it’s like the old saying goes.

“Fool me once, shame on you. Fool me twice–you can’t get fooled again.”

Sources

CareerCup
https://www.careercup.com/question?id=7949664

Chatterjee, Neil, and Bogdan G. Nita. “The hanging cable problem for practical applications.” Atlantic Electronic Journal of Mathematics 4.1 (2010).
http://euclid.trentu.ca/aejm/V4N1/Chatterjee.V4N1.pdf

https://mav.vic.edu.au/resources/online-mav-puzzle/item/298-two-flagpoles.html

http://themathmompuzzles.blogspot.com/2014/02/oscar-ropes-puzzle.html

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