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-->|

(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 x²+y²= (bm-a)²(1/m²+1). Differentiating, we get to bm³+a=0 so m=-³√(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**