Question

A. The figure at the right shows a long rectangular strip of paper, one corner of which has been folded over to meet the opposite edge, thereby creating a 30-degree angle. Given the width of the strip is 12 inches, find the length of the crease.
b. Instead of a 30-degree angle, suppose that the angle has an unspecified size t. Use trigonometry to find the length of the crease, expressed in terms of t.
c. Find the approximate value of t that makes the crease as short as it can be. Restrict your attention to values of t that are less than 45 degrees. Explain your method.

Answer 1

A. We want to know the length of the crease. Let x represent that length. In the picture below, that is segment AD. If t represent m∠EAD, then ED = CD = x*sin(t). Of course DF = 12 -ED.

m∠CDF = m∠CAE = 2t

Putting these relationships together, we have
.. DF/CD = cos(2t) = (12-x*sin(t))/(x*sin(t))
Solving for x, we get
.. x*sin(t)*cos(2t) = 12 -x*sin(t)
.. x*sin(t)*(1 +cos(2t)) = 12
.. x = 12/(sin(t)*(1 +cos(2t))
.. x = 6/(sin(t)*cos(t)^2)
For t = 30°, x = 6/(3/8) = 16

For a 30° angle, the length of the crease in 16 inches.


B. From above,
.. crease length = (6 inches)/(sin(t)*cos(t)^2)


C. Using a graphing calculator, we can find the angle that makes the crease length a minimum. See the second figure. It is about 35.264°.

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An exact solution using derivatives gives arcsin(1/√3) ≈ 35.264390°.