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Moving Day (Posted on 2004-05-26) Difficulty: 3 of 5
An easier version of this puzzle is here.

A large panel needs to be moved through a corridor, the panel is tall as the corridor. The corridor is A feet wide before a right angle turn, after the turn, it is B feet wide. What is the maximum length of the panel that can pass through this corner.

Overhead view of the hallway:

+------------+---
|           /   |
|          /    |B ft
|         /     |
|        /+------
|       / |
|      /  |
|     /   |
|    /    |
|   /     |
|  /      |
| /       |
|/        |
+<-A ft-->|

See The Solution Submitted by Brian Smith    
Rating: 3.8333 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution SilverKnight's suggestion | Comment 15 of 31 |
(In reply to Full Solution by SilverKnight)

Just for the sake of it, I solved the problem again, but using SilverKnight's suggestion ("differentiate the length of the panel as a function of the angle it makes..."). As in my previous solution, I turned the diagram round so the outer corner was at (0,0) and the inner corner at (b,a).

A line through the inner corner can be written as (y-a)/(x-b)=m, m being the tangent of the line with respect to the horizontal axis; m should be in (-∞,0).

The intersection with the horizontal axis is found for y=0: x=(bm-a)/m. Similarly, x=0 gets the intersection with the vertical axis: y=a-bm.

We want to minimize x+y= (bm-a)(1/m+1). Differentiating, we get to bm+a=0 so m=-√(b/a). (There's another extreme at m=b/a, but that value is outside the desired range.)

For that m, x and y turn out to be the same values as in the other proof, confirming that result.
Edited on May 27, 2004, 12:51 pm
  Posted by Federico Kereki on 2004-05-27 12:42:19

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