Theorem 0.1. The minimum length for a piece of paper of thickness t to be folded n times in a single direction is given by

L = πt (2n + 4)(2n - 1). 6
(1)

Proof. We begin by folding a single piece of paper in two. The fold forms a half-circle of radius t, which has perimiter πt. Therefore, πt units of the paper are used in the fold itself. A piece of paper shorter than this cannot be folded even once, because there is not enough paper to form the fold. After folding, the result is a two-layer piece of paper, with total thickness 2t.

To make another fold in this two-layer piece of paper, we begin by folding the first layer onto itself. As before, folding the first layer uses up πt units of paper. We then fold the second layer over the first layer. This second-layer fold has radius 2t, so it uses up 2πt units of paper. The total amount of paper used by the second fold, for both layers, is πt + 2πt = 3πt. The result after two folds is a four-layer piece of paper.

In general, the ith fold begins with 2i-1 layers, and folding the jth layer uses jπt units of paper. Therefore, the total length of paper used for the ith fold is given by

2i-1 ∑ jπt = πt2i-1(2i-1 + 1)= πt (22(i-1) + 2i-1). j=1 2 2
(2)

To get the total length of paper required for n folds, we sum this over i from 1 to n, which gives us

πt ∑n ( 2(i-1) i-1) L = -2 2 + 2 (3) i=(1 ) = πt 22n --1-+ 2n - 1 (4) 2 3 πt (2n n ) = 6 2 + 3⋅2 - 4 (5) πt n n = 6 (2 + 4)(2 - 1). (6)
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