 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  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) Comments: ( Back to comment list | You must be logged in to post comments.) 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, L˛ = (1 + x)˛ + (1 + 1/x)˛ = z˛ + 2z = (z + 1)˛ - 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))˛ >= 0.)

Since (z + 1)˛ - 1 is an increasing function for z >= -1, the minimum value of L˛ occurs when z = 2. Hence the minimum of L˛ = 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 Please log in:

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