Fractal Folds Lab Answers

2. First approach In the limiting shape, the bottom, left and right sides become straight lines.

So also the bottom, left and right sides of each piece.

Consequently, in the limiting shape the cube of side length 1 contributes three triangles of base and altitude 1 to the surface.

After accounting for these triangles, the remaining three parts of the of the surface - left, right, and top - each contain three triangles of base and altitude 1/2.

Continuing in this pattern, we see the surface is made of triangles:

step	altitude (= base)	number of triangles	area/triangle
1	1	3	1/2
2	1/2	9 = 32	1/8 = 1/23
3	1/4	27 = 33	1/32 = 1/25
4	1/8	81 = 34	1/128 = 1/27
		...
n	1/2n-1	3n	1/22n-1

Consequently, the area of the surface is

(3/2) + (9/8) + (27/32) + (81/128) + ... =

(3/2)(1 + (3/4) + (3/4)² + (3/4)³ + ...)

The series in parentheses is a geometric series with ratio 3/4, hence converging to 1/(1 - 4/3) = 4. So the area is (3/2)4 = 6.

Second approach After accouting for the three triangles of base and altitude 1, the remainder of the surface consists of three pieces - left, right, and top - each a copy of the whole surface reduced by a factor of 1/2 in each direction.

Denote by x the area of the whole surface. Then the three smaller pieces each have area x/4 and so the area satisfies this equation.

x = 3/2 + 3x/4

Solving for x gives x = 6, agreeing with our earlier calculation.

Caution This calculation works only if x is finite. For example, if the surface bent and folded so much that it was more than 2-dimensional, x would be infinite. The displayed equation is valid for x infinite. Before solving any such equaton, we must be careful to establish that the solution is finite.

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