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Rectangular Logic (Posted on 2005-01-28) Difficulty: 4 of 5
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?

No Solution Yet Submitted by Sam    
Rating: 3.7500 (12 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Answer + Solution | Comment 1 of 44

EDIT:  The answer I have posted here is wrong.  My basic logic is correct, but I made an error near the end, and I noticed another error due to laziness.  Please see comments 9 and 15 for my corrections.  Also please see comments 6&7 by Miland, 18 by Charlie and 19 by Penny where they post correct solutions.

FYI: I laid all this out with two tables… one of all the possible rectangles and their perimeters, and one of all the corresponding areas. If I deleted a certain cell (representing a specific rectangle) on one table for any reason, I would also delete it from the other. I hope that made sense.

Ok, so since we are told the perimeter is no greater than 80 inches, we know the largest dimension we could have would be 38 (from 2*(2+x)=80 since 2 is the smallest dimensions allowed, and we get x=38). This leaves us with 361 possible rectangles to start with.

Note: A unique area is created by either two prime numbers (P1*P2=A), or a prime and its square (P1*P1^2=A). Due to the definition of prime numbers, there would be no other way to factor A with integers. In this group of 361 possible rectangles, there are 57 unique areas.

Let's start off by considering Paul before Peter says anything. We know that he already knew Peter wouldn't have the answer right away, even before Peter said so. So let's think about this.

Based on what ever perimeter Paul was told, he can figure out every combination of the rectangles dimensions that would produce that perimeter. If ANY possible rectangle had a unique area, he would have to say to himself "Well, there is at least one rectangle which produces my perimeter such that Peter would know right away what the rectangle was because of the unique area." Therefore, Paul's perimeter must not be able to be created by two prime sides, or a prime and its square.

I don't think I explained that very well. Here's what I mean. Let's pretend Paul has the number 20. There are 4 ways that perimeter could be produced: 2x8, 3x7, 4x6, and 5x5. The 3x7 and 5x5 rectangles, however, have unique areas. No other rectangle out there (regardless of its perimeter) could produce either of those areas, which would mean Peter would know right away what rectangle he was dealing with. But since Paul says he knew for sure that Peter wasn't going to know right away, this means that his perimeter DOESN'T have any possible rectangle with a unique area. So 20 would not be a candidate for what perimeter Paul has.

On the other hand, with a perimeter like 22, the possible rectangeles are : 2x9, 3x8, 4x7, 5x6. In all of those cases, the area is not unique (3x6, 4x6, 2x14, or 3x10 would produce the same areas). Therefore 22 is a possible candidate for what perimeter Paul has.

Taking this into account, there are only a few possible perimeters Paul could be told. 22, 34, 46, 54, 58, 70, and 74.

Ok, so from the given info we had 361 rectangles, from Peter's first statement we dropped it down to 304, and from Paul's first statement we can narrow it down to 79. And remember, Peter is following along and has been able to narrow it down to the same 79 rectangles.

From Peter's second statement we can rule out any areas that are "unique" in this set of 79 rectangles. There are 28 such rectangles, leaving us with 51.
Of the 4 ways to make a perimeter of 22, Peter's statement only rules out 1.
Of the 7 ways to make a perimeter of 34, Peter's statement rules out 6, leaving only 1.
Of the 10 ways to make a perimeter of 46, Peter's statement only rules out 6.
Of the 12 ways to make a perimeter of 54, Peter's statement only rules out 3.
Of the 13 ways to make a perimeter of 58, Peter's statement only rules out 3.
Of the 16 ways to make a perimeter of 70, Peter's statement only rules out 3.
Of the 17 ways to make a perimeter of 74, Peter's statement only rules out 6.
Now Paul says he knows the area. The only way he could know that is if the perimeter was 34. The only rectangle that wasn't eliminated by Peter's second statement was 4 by 13. So he knows the area is 52.

And since Peter was keeping track of all this too, and now that Paul said he knew the area, Peter can figure out the perimeter of his rectangle: 34.

So the rectangle is 4 by 13.

Edited on February 1, 2005, 7:08 pm
  Posted by nikki on 2005-01-28 20:29:26

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