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 Around the bend (Posted on 2004-03-13)
What is the longest pole I can swing around the 90° corner of a hallway of unit width?

For the simplicity of this problem, the pole must be kept exactly horizontal, while maneuvering it.

 See The Solution Submitted by SilverKnight Rating: 2.5000 (4 votes)

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 Non-calculus solution | Comment 4 of 8 |

Consider a line of length L that touches both outer sides of the corridor, and the inner corner.  Let the line extend beyond the inner corner by x units on one side, and y on the other.

Then, by similar triangles, y = 1/x.

By Pythagoras, L² = (1 + x)² + (1 + 1/x)² = z² + 2z = (z + 1)² - 1, where z = x + 1/x.

By the Arithmetic Mean-Geometric Mean inequality, the minimum value of z = x + 1/x is 2, when x = 1.  (Or consider (sqrt(x) - 1/sqrt(x))² >= 0.)

Since (z + 1)² - 1 is an increasing function for z >= -1, the minimum value of L² occurs when z = 2. Hence the minimum of L² = 8, and of L = 2×sqrt(2).

This is the shortest line; hence the longest pole.

 Posted by Nick Hobson on 2004-03-13 18:36:53

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