Results 1 to 2 of 2

Math Help - Paper folding

  1. #1
    Junior Member
    Joined
    May 2009
    Posts
    28

    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.



    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1

    Trigonometry

    Hello m_i_k_o
    Quote Originally Posted by m_i_k_o View Post
    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
    Attached Thumbnails Attached Thumbnails Paper folding-untitled.jpg  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Paper Folding Questions
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: August 15th 2009, 05:21 AM
  2. Help with math paper
    Posted in the Calculus Forum
    Replies: 0
    Last Post: September 24th 2008, 01:59 PM
  3. a box of paper
    Posted in the Statistics Forum
    Replies: 5
    Last Post: September 7th 2008, 02:42 PM
  4. Subtracting on paper
    Posted in the Algebra Forum
    Replies: 3
    Last Post: August 1st 2008, 05:14 PM
  5. Folding paper
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: July 18th 2008, 04:41 AM

Search Tags


/mathhelpforum @mathhelpforum