For what number of non-overlapping unit squares can a figure be formed whose perimeter is numerically equal to the area?
Each square obviously contributes 1 to the area.
Additionally, each square can be one of four types: a "peninsula" (p) which contributes 3 to the perimeter, a "corner" (c) which contributes 2, a "side" (s) which contributes 1, and an "interior" (i) which contributes 0.
Let's ignore peninsulas for a moment and consider a rectangle formed from unit squares. The area is c + s + i. The perimeter is 2c + s. Simplfying shows that c = i, so we need a rectangle with 4 interior squares. This can be a 6x3 rectangle (area = perimeter = 18) or a 4x4 square (area = perimeter = 16).
If we extend the 6x3 rectangle to be 7x3, we've added 2 more sides and 1 more interior square. So we've added 2 to the perimeter and 3 to the area.
Now consider what happens when we append a peninsula. The net change to the perimeter is +2 since the penisula adds 3 but we lose 1 where it is attached, and the net change to the area is +1.
So adding a row to the rectangle (i.e. changing it from n x 3 to (n+1) x 3), and then appending a peninsula somewhere on a side adds 4 to the perimeter and 4 to the area. So we can make a shape with 18 + 4k squares, where k is any nonnegative integer, such that the area is numerically equal to the perimeter.
Similarly, starting with the 4x4 square, we can append a 2x2 square to the side. This adds 4 to the perimeter and 4 to the area. So we can make a shape with 16 + 4k squares, where k is any nonnegative integer, such that the area is numerically equal to the perimeter.
Combined, this means that we can make a shape satisying the conditions of the puzzle with any even number of squares >= 16.
I'll leave this for now, and think about whether it's possible to make a valid configuration with an odd number, or an even number less than 16.
Posted by tomarken
on 2014-07-09 12:10:07