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 Moving Day (Posted on 2004-05-26)
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.

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

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

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 re: Full Solution | Comment 6 of 31 |
(In reply to Full Solution by SilverKnight)

I disagree with the 45 degrees opinion. My analytical solution goes like this:

First, let's turn the diagram round so the outer corner is at (0,0) and the inner corner at (b,a). At the extreme position of the panel, it goes from (x,0) to (0,y), and we want to minimize (x²+y²). Given x, a little geometry gives y=ax/(x-b), so now we want to minimize x²+(ax)²/(x-b)².

Differentiating, we eventually get to (x-b)³+a²(x-b)-a²x=0. If we set x=b+z, this turns into z³-a²b=0, so z is ³√(a²b). We now get x=b+³√(a²b) and thus y=a+³√(ab²). Calculating √(x²+y²) we finally get √(a^(2/3)+b^(2/3))³.

Just to make sure we may be right, setting a=b=1 (as in the previous problem) gets √(2³)=2√2, which is correct, so we can have a little more confidence in the result.

By the way, since x is not equal to y, then the angle definitely is not 45 degrees.
Edited on May 26, 2004, 3:04 pm
 Posted by Federico Kereki on 2004-05-26 15:00:20

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