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 Rectangular Logic (Posted on 2005-01-28)
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)

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 my solution and method | Comment 28 of 44 |
i'm new here,  so maybe the answer was given already - i looked only at first and last posts and saw some errors, so here's my take.

since it is a rectangle whose perimeter is <= 80, 2 adjacent sides are <= 40 and the area is the product of those 2 sides, so i opened an excel sheet and had it create all unique combinations of factors whose sum is 40 or less, added column Area (=a*b) and column Perimeter (=2*(a+b))

from peter's first statement we can eliminate any area that has only one perimeter, so i sorted my sheet by Area, and added another column (Y1) that put a Y for any nonunique area (area is same as row above or below) (formula=if(or(c5=c4,c5=c6),"Y","")

from paul's first statement, we can eliminate any unique perimeters (but in fact we don't need to), and also any nonunique perimeters that has any matching area that has a unique perimeter (otherwise paul could not have been sure that peter wouldn't know).  in other words all of the areas that match a given perimeter must have a Y in column Y1.  so i sorted the sheet again, this time by perimeter, and placed a Y in column Y2 when all of te rows for a given perimeter have a Y in column Y1.

here i bring to attention that in nikki's first post i see you assume 58 to still be valid, however it isn't (if the area is 138 or 190, peter would hve known the perimeter because the only factors are 6x23=138, 10x19=190) a few other numbers were also wrong. in fact this leaves us only 22, 34, 46 as valid perimeters.

now, since peter still didn't know the perimeter, the area that he knew must be able to match to at least 2 of these perimeters.  this leaves us only 6 possibilities:
• 5 x 6   area=30 peri=22
• 2 x 15 area=30 peri=34
• 3 x 14 area=42 peri=34
• 2 x 21 area=42 peri=46
• 5 x 12 area=60 peri=34
• 3 x 20 area=60 peri=46
since paul now knew the area, the perimeter must be unique in the above list, therefore the perimeter is 22, area is 30, dimensions are 5 by 6.

joe

 Posted by joseph on 2005-02-21 08:19:21

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