Paper folding

**m_i_k_o** · August 16th 2009, 01:38 AM

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.

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

**Grandad** · August 17th 2009, 08:07 AM

Hello m_i_k_o

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.

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

Then $\angle QPB = 90 - \theta$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Grandad

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