Paper folding

• Aug 16th 2009, 01:38 AM
m_i_k_o
Paper folding
A rectangular, flat sheet of paper, 30 by 20cm is placed on a flat surface. One of the vertices is lifted up and placed on the opposite edge, while the other three vertices are held in their original positions. With all four vertices now held fixed, the paper is smoothed flat as shown in the diagram. Note that AR = RQ, AP = PQ.

http://micko.dyndns.org/math.jpg

Let the length of the crease, PR, be L cm and AP be x cm. Express L as a function of x. Using a spreadsheet, graphics calculator or table of values, draw a graph of this function. Hence, determine the minimum length of the crease, and the size of angle PQB for the minimum length
• Aug 17th 2009, 08:07 AM
Trigonometry
Hello m_i_k_o
Quote:

Originally Posted by m_i_k_o
A rectangular, flat sheet of paper, 30 by 20cm is placed on a flat surface. One of the vertices is lifted up and placed on the opposite edge, while the other three vertices are held in their original positions. With all four vertices now held fixed, the paper is smoothed flat as shown in the diagram. Note that AR = RQ, AP = PQ.

http://micko.dyndns.org/math.jpg

Let the length of the crease, PR, be L cm and AP be x cm. Express L as a function of x. Using a spreadsheet, graphics calculator or table of values, draw a graph of this function. Hence, determine the minimum length of the crease, and the size of angle PQB for the minimum length

Let $\displaystyle \angle PQB = \theta$, where $\displaystyle \theta$ is measured in degrees.

Then $\displaystyle \angle QPB = 90 - \theta$

$\displaystyle \Rightarrow \angle RPQ = \angle RPA = \tfrac12(180 -[90-\theta]) = \tfrac12(\theta +90)$

So, from $\displaystyle \triangle PQR,\, \cos\tfrac12(\theta +90)= \frac{x}{L}$

So, using $\displaystyle \cos2A = 2\cos^2A -1,\, \cos(\theta+90)= \frac{2x^2}{L^2}-1$

$\displaystyle \Rightarrow \cos\theta\cos90 -\sin\theta\sin90=-\sin\theta = \frac{2x^2}{L^2}-1$

$\displaystyle \Rightarrow \sin\theta = 1 -\frac{2x^2}{L^2}$

$\displaystyle \Rightarrow PB = x\sin\theta = x\Big(1-\frac{2x^2}{L^2}\Big)$

Now $\displaystyle AP+PB = 20$ (I assume from your diagram that $\displaystyle PB$ is the shorter side of the rectangle)

$\displaystyle \Rightarrow x +x\Big(1-\frac{2x^2}{L^2}\Big)=20$

$\displaystyle \Rightarrow 1+1-\frac{2x^2}{L^2}=\frac{20}{x}$

$\displaystyle \Rightarrow 1 - \frac{x^2}{L^2}=\frac{10}{x}$

$\displaystyle \Rightarrow 1-\frac{10}{x}= \frac{x^2}{L^2}$

$\displaystyle \Rightarrow \frac{x-10}{x}=\frac{x^2}{L^2}$

$\displaystyle \Rightarrow L^2 = \frac{x^3}{x-10}$

$\displaystyle \Rightarrow L = \sqrt{\frac{x^3}{x-10}}$

This, then, is $\displaystyle L$ as a function of $\displaystyle x$. Plotting this in Excel, I produced the chart as shown. This gives a minimum value of $\displaystyle L$ as $\displaystyle 26.0$ cm (1 d.p.) when $\displaystyle x = 15$. For this value of $\displaystyle x,\, \angle PQB = 19.5^o$