Promising them an increase in their allowance if they get the answer, I offer my two sons, Peter and Paul, the following puzzler:

"I am thinking of a rectangle with integer sides, each of which are greater than one inch. The total perimeter of the rectangle is no greater than eighty inches."

I then whisper the total area to Peter and the total perimeter to Paul. Neither of them are allowed to tell the other what they heard: their job is to work out the rectangle's dimensions.

Their subsequent conversation goes like this:

Peter: Hmmm... I have no idea what the perimeter is.
Paul: I knew you were going to say that. However, I don't know what the area is.
Peter: Still no clue as to the perimeter...
Paul: But now I know what the area is!
Peter: And I know what the perimeter is!

What are the dimensions of the rectangle?

First, I still believe the answer is a 5x6 rectangle.

Second, I made even more mistakes than I mentioned in my "Milind is right..." comment, but like I said I still come up with 5x6. Basically, I was too lazy when I was originally eliminating the areas in the first step. I started with the small numbers and started marking the unique areas. I noticed the pattern with the prime numbered dimensions, and just got lazy and carried that theory over and didn't look for more unique areas. I'll explain more later.

Third, Penny, I don't think the answer can be 4x4, and here's why (I think it's basically a timeline issue). So Paul would have the perimeter of 16, right? As you have mentioned, there are 3 ways to make a rectangle with a perimeter of 16: 4x4, 5x3, and 2x6. As you said, WE can remove 5x3 from the list after Peter's first statement. However, let's look at Pauls first statement. BEFORE Peter says anything, Paul says he KNEW Peter wasn't going to know the perimeter. If Paul really had 16 as the perimeter, he wouldn't have been able to say that. He would have to say "Well, if Peter has the area as 16 or 12 he won't be sure what the perimeter is, but he if had the area as 15 he would know the answer. So I don't know what Peter is going to say." But if Paul had a number like 22, 34, or 46, he could list all the possible rectangles that could create those perimeters, consider the areas Peter would have in all those cases and say "Every single one of the areas that Peter might have would NOT be unique, so no matter what Peter won't know the answer yet" even before Peter says so. Get it?

I think you solved for a problem where they simply say "I don't know", "I don't know", "I don't know", "I know", "I know" starting with Peter. I think the first part of Paul's first statement holds a lot more information than you accounted for.

Fourth, here is a better summary of the full solution. Again, we start with 361 possible rectangles. Peter's first statement eliminates 119, leaving 242 possible rectangles. A big key is that the entire column of rectangles containing 23 as a side has been eliminated. That means that all perimeters of 50 or more are eliminated when Pauls says "I knew you were going to say that." Along with all perimeters of 50 or more, we also eliminate all perimeters except for 22, 34, and 46. Over all we have eliminated 221 more rectangles, leaving only 21. Now we look at Peter's second statement. Eliminating any unique areas in this group of 21 means we eliminate 15 rectangles, leaving only 6. Paul now says he knows the answer. Of those 6 rectangles, only one has a unique perimeter, and that is 22 made by a 5x6 rectangle.

Edited on February 1, 2005, 3:52 pm

Posted by nikki on 2005-02-01 15:37:09