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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)

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No Subject | Comment 40 of 44 |

Solution: Say the length and width are L and W respectively, and Peter knows A = LW, Paul knows S = L+W. Clearly A != p1 p2 (product of two primes) since otherwise Peter would know S immediately. But Paul knows that before Peter makes his first statement, therefore S != p1 + p2(sum of two primes) since if it were, to the best of Paul's knowledge it's possible that A = p1 p2 and Paul could not be sure Peter wouldn't know S right away. So Peter now knows S != p1 + p2 in pass 2, and since 4 <= S <= 40, S is in the set {11, 17, 23, 27, 29, 35, 37} (call these the "permissible" values for S).

Claim that if S = 11, Peter and Paul would have the exact conversation that they do: in pass 2, Paul knows that the possible values for S = L+W are

11 = 2+9 => A = 18 = 2x9 = 6x3 => S = 11, 9 => S = 11 (9 not permissible)

   = 3+8 => A = 24 = 2x12 = 3x8 = 4x6 => S = 14, 11, 10 => S = 11

   = 4+7 => A = 28 = 2x14 = 4x7 => S = 12, 11 => S = 11

   = 5+6 => A = 30 = 2x15 = 3x10 = 6x5 => S = 17, 13, 11 => S = 17, 11

therefore Paul knows A = 30 = 5x6 since in the other three cases S has only one permissible value (so Peter would have known S in pass 2) whereas in the fourth case, Peter cannot decide between 11 and 17 in pass 2. But in pass 3 he can eliminate 17 because

17 = 2+15 => A = 30 = 2x15 = 3x10 = 6x5 => S = 17, 13, 11 => S = 17, 11

   = 3+14 => A = 42 = 2x21 = 3x14 = 6x7 => S = 23, 17, 13 => S = 23, 17

i.e. even though Peter does not know S in pass 2, Paul still would not have been able to decide between A = 30 or 42.

So L = 5, W = 6 is a possible solution to the puzzle, and we can show it is unique by observing that if S is one of 17, 27, 23, 29, 35, or 37, then Paul could not know A in pass 2. We have already proven this in the case S = 17, and the other cases are similar (though somewhat tedious to discover):

23 = 2+21 => A = 42 = 2x21 = 3x14 = 6x7 => S = 23, 17, 13 => S = 23, 17

   = 3+20 => A = 60 = 3x20 = 5x12 => S = 23, 17

27 = 3+24 => A = 72 = 3x24 = 8x9 => S = 27, 17

   = 6+21 => A = 126 = 6x21 = 9x14 => S = 27, 23

29 = 5+24 => A = 120 = 5x24 = 8x15 => S = 29, 23

   = 14+15 => A = 210 = 14x15 = 30x7 => S = 29, 37

35 = 13+22 => A = 286 = 13x22 = 11x26 => S = 35, 37

   = 15+20 => A = 300 = 15x20 = 12x35 => S = 35, 37

37 = 4+33 => A = 132 = 4x33 = 11x12 => S = 37, 23

   = 7+30 => A = 210 = 14x15 = 30x7 => S = 29, 37.


  Posted by togo17 on 2005-12-24 23:41:34
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