Surprising wine bottles
Thirteen identical wine bottles are placed in a box. There is not room for four bottles at the bottom of the box, but ample room for three. A and C are placed next to the walls, while B is placed anywhere between them. The second layer will have room for only two bottles and the position of B will determine their location. The third layer will have three bottles, the fourth two, and the fifth and last three.
Here is the shocker: Regardless of B’s position the three bottles on the top, i.e. the fifth layer, will be horisontal.
Can you prove it?
Problem creator: Charles Payan, the creator of Cabri. He discovered it while playing with his software.
I found the problem here and here. The illustration is taken from the first link.
May 26th, 2009 at 12:37 am
Following http://www.cs.sdsu.edu/news/fortnight/2007.html label the bottles as follows:
|K L M|
| I J |
|F G H|
| D E |
|A B C|
+———–+
Set up Cartesian coordinates with the x-axis horizontal and A at (0, 0). (Throughout this proof, coordinates refer to the centres of the bottles.) Where D is at (x1, y1), B is at (2×1, 0). Where E is at (2×1+x2, y2), C is at (2×1+2×2, 0). Then, because they rest against the sides of the box, F and H are at (0, 2y1) and (2×1+2×2, 2y2). Vector DG=vector BE, so G is at (x1+x2, y1+y2). Thus FGH is a straight line and G bisects FH. The whole setup has 180-degree rotational symmetry about G; thus because ABC is horizontal, so also is KLM.
May 26th, 2009 at 12:41 am
Argh. Perhaps I should give up trying to avoid WordPress mucking up my comments. I thought that “pre” tags enclosed in angle-brackets gave you a monospaced font. I also find that WordPress will sometimes replace the character x with the symbol × — ugh.