PaperFold

Folding a square sheet of paper

by George Taniwaki

After scribbling notes on a square yellow Post-it note, I folded it in half and was about to put it in my pocket when I made an interesting observation. If I fold a square paper exactly in half while holding an edge toward me, the resulting visible area is a rectangle with an area of one-half the total area (top illustration above). The same is true if rotate the sheet 45 degrees and I fold it along the diagonal to make a triangle (middle illustration above). However, if I fold it in half along any other angle, the area of the top layer is exactly one-half, but two additional triangular ears are visible from under the fold (bottom illustration above).

That made me wonder. What angle produces the maximum area, and what is that area? This is the perfect math question for the origami enthusiast.

Let’s solve it.

Finding the relevant dimensions

The first step in solving this problem is to unfold the sheet, use geometry to find any symmetries, and use trigonometry to calculate the area of the two triangular ears as a function of the rotation angle, θ.

Dimension

Unfolding the sheet

By unfolding the sheet, I discover a surprising symmetry. The two triangular ears are identical. Let’s call the lengths of the two exterior sides A and B. The area of one triangular ear is A * B / 2 and the area of both ears is A * B.

The length of the interior side, which is the hypotenuse of a right triangle, can be called C. The angles opposite each side can be labeled a, b, and c. (See figure above.)

Using trigonometry, this means A = C * cos(b) and B = C * sin(b).

The angle b is twice the rotation angle or 2θ. Substituting gives

A = C * cos(2θ) and B = C * sin(2θ).

The length of one side of the square = 1, so A + B + C = 1. Substituting and rearranging gives

C * cos(2θ) + C * sin(2θ) + C = 1

or

C = 1/(cos(2θ) + sin(2θ) + 1)

and finally substituting this value of C in our original equations for A, B, and A * B gives

A = cos(2θ) / (cos(2θ) + sin(2θ) + 1)
B = sin(2θ) / (cos(2θ) + sin(2θ) + 1)
A * B = cos(2θ) * sin(2θ) / (cos(2θ) + sin(2θ) + 1)2

Numeric solution

There are two ways to solve for the maximum value of the area A * B and the angle θ. We can solve it numerically and analytically. Let’s start with a numeric solution. In Excel, I create a table of values of θ from 0 to 45 degrees with steps of 1 degree increments. I convert degrees to radians and calculate the associated values of A, B, and A *B. Then I plot them. The maximum value occurs at θ somewhere between 22 and 23 degrees. Thus, it seems 22.5 degrees is the solution.

At 22.5 degrees, A = B = 0.2929 and A * B = 0.08579.

AreaByAngle

Plot of A, B, and A*B versus θ

Analytic solution

Now let’s prove the numeric solution is correct by using an analytical method. To find the maximum area, I take the derivative of A * B with respect to θ and set to zero. Then solve for θ. The solution uses basic calculus, including a combination of the product rule and chain rule. First, I define A * B as a group of functions as follows:

let f(θ) = cos(2θ)
g(θ) = sin(2θ)
h(θ) = 1 / (cos(2θ) + sin(2θ) + 1)2 = (cos(2θ) + sin(2θ) + 1)2

and

A * B = f(θ) * g(θ) * h(θ)

The derivatives are as follows:

f'(θ) = -sin(2θ)*2
g'(θ) = cos(2θ)*2
h'(θ) = -2 * (cos(2θ) + sin(2θ) + 1)-3 * (-sin(2θ) + cos(2θ)) * 2 = 4 * (sin(2θ) – cos(2θ)) / (cos(2θ) + sin(2θ) + 1)3

and

(A * B)’ = ( f'(θ) * g(θ) * h(θ)) + (g'(θ) * f(θ) * h(θ)) + (h'(θ) * f(θ) * g(θ))

Now I substitute the values of the 3 functions and their derivatives into the last equation.

(A * B)’ = (-sin(2θ)*2 * sin(2θ) * h(θ)) + (cos(2θ)*2 * cos(2θ) * h(θ)) + 4 * (sin(2θ) – cos(2θ)) / (cos(2θ) + sin(2θ) + 1)3 * cos(2θ) * sin(2θ))

Simplifying and rearranging gets:

(A * B)’ = (cos2(2θ) – sin2(2θ)) * 2 * h(θ) + (sin(2θ) – cos(2θ)) * 4 * cos(2θ) * sin(2θ) / (cos(2θ) + sin(2θ) + 1)3

The solutions are values of where A * B’ = 0. This is only true when both cos2(2θ) – sin2(2θ) = 0 and sin(2θ) – cos(2θ) = 0. The only angle where this condition is met is 45° or θ = 22.5°. This proves that the maximum that I found numerically is a maximum area.

[Update: Completed the analytic solution section.]