Standard paper

Geometry Level 3

The A-series of paper sizes used in Europe have the following properties:

  • A sheet of A0 paper is a rectangle with an area of precisely one square meter. Its length (the long side) is less than twice its width.

  • When A0 paper is cut in half across its long side, two sheets of A1 paper are formed. Likewise, when A1 paper is cut in half across its long side, A2 paper is formed, and so on.

  • All rectangles in the series (A0, A1, A2, ...) are similar.

What is the length of a sheet of A6 paper, in meters?

2 11 / 4 2^{-11/4} 2 3 2^{-3} 2 13 / 4 2^{-13/4} 2 π 2^{-\pi} 2 7 / 2 2^{-7/2}

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3 solutions

Arjen Vreugdenhil
Apr 29, 2017

Let x ( n ) x(n) be the length and y ( n ) y(n) be the width of a piece of A n n paper. Every time we cut the paper in half, the old width becomes the new length; thus we have x ( n + 1 ) = y ( n ) ; y ( n + 1 ) = x ( n ) 2 . x(n+1) = y(n);\ \ \ y(n+1) = \frac{x(n)}2. The similarity property tells us that x ( n + 1 ) y ( n + 1 ) = x ( n ) y ( n ) 2 y ( n ) x ( n ) = x ( n ) y ( n ) x ( n ) y ( n ) = 2 . \frac{x(n+1)}{y(n+1)} = \frac{x(n)}{y(n)}\ \ \ \ \therefore\ \ \ \ \frac{2y(n)}{x(n)} = \frac{x(n)}{y(n)}\ \ \ \ \therefore\ \ \ \ \frac{x(n)}{y(n)} = \sqrt{2}. Thus for each sheet of A n n paper, the length is 2 \sqrt 2 times as long as its width.

For the area we are told that A ( 0 ) = x ( 0 ) y ( 0 ) = 1 , A(0) = x(0)y(0) = 1, and since each next sheet has half of the area, we have A ( n ) = 2 n A(n) = 2^{-n} . Thus x ( n ) y ( n ) = 2 n and x ( n ) / y ( n ) = 2 1 / 2 . x(n)y(n) = 2^{-n}\ \ \ \ \text{and}\ \ \ \ x(n)/y(n) = 2^{1/2}. If x ( n ) = 2 a x(n) = 2^a and y ( n ) = 2 b y(n) = 2^b , then we have { a + b = n a b = 1 / 2 \begin{cases} a+b = -n \\ a - b = 1/2 \end{cases} with the solutions a , b = n 2 ± 1 4 x ( n ) , y ( n ) = 2 n / 2 ± 1 / 4 . a,b = -\frac n 2 \pm \frac 1 4\ \ \ \ \therefore\ \ \ \ x(n), y(n) = 2^{-n/2\pm 1/4}. WIth n = 6 n = 6 we get x ( 6 ) = 2 6 / 2 + 1 / 4 = 2 11 / 4 . x(6) = 2^{-6/2 + 1/4} = \boxed{2^{-11/4}}. (This evaluates as approximately 0.149 m, of 14.9 cm.)

L e t t h e A 0 r e c t a n g l e b e W a n d L = r W . r > 1. W L = r W 2 = 1 f o r a r e a A 0 . . . . . ( 1 ) F o r a r e a A 1 , s i d e s a r e s h o r t r W 2 , l o n g W . S h a p e b e i n g s i m i l a r W r W 2 = r , r 2 = 2 , r = 2 . B u t f r o m ( 1 ) W 2 = 1 r = 2 1 2 Let~the~ A_0~rectangle ~be~W~ and~ L=r*W.~~~~~~ r>1.\\ \therefore~W*L=r*W^2=1~ for~area~ A_0. ....(1)\\ For~area~A_1,~sides ~are~short~\dfrac{r*W} 2,~~long~W.\\ Shape~being ~similar ~ \implies ~\dfrac W {\dfrac{r*W} 2}~=~r,\\ \therefore~r^2=2,~~ \implies~r=\sqrt2. \\ But~from~~(1)~~W^2=\dfrac 1 r=2^{-\frac 1 2}\\ S o W = 2 1 4 f o r A 0 , l o n g s i d e i s r W = 2 1 2 2 1 4 = 2 1 4 . F o r e a c h h i g h e r n u m b e r t h e l o n g s i d e i s 2 1 a n d s h o r t s i d e b e c o m e s t h e l o n g s i d e . T h i s i t b e c o m e s r ( = 2 1 2 ) t i m e s . T h e n e t r e s u l t i s e a c h s t a g e , l o n g s i d e i s r e d u c e d b y 2 1 2 1 2 = 2 1 2 . f o r s i x s t a g e s f o r A 6 , t h e l o n g s i d e i s W ( 2 1 2 ) 6 = 2 1 4 2 3 = 2 11 4 . A s i m p l e e x p l a n a t i o n f o l l o w s . L e t t h e A 0 r e c t a n g l e b e W a n d L = r W . r > 1. W L = r W 2 = 1 f o r a r e a A 0 . . . . . ( 1 ) F o r a r e a A 1 , s i d e s a r e s h o r t r W 2 , l o n g W . S h a p e b e i n g s i m i l a r W r W 2 = r , r 2 = 2 , r = 2 . B u t f r o m ( 1 ) W 2 = 1 r = 2 1 2 S o W = 2 1 4 F o r A 0 L o n g = r W = 2 1 2 2 1 4 = 2 3 4 S h o r t = W = 2 1 4 F o r A n L o n g S h o r t A 0 2 3 4 2 1 4 . A 1 2 1 4 2 3 4 . A 2 2 3 4 2 5 4 . A 3 2 5 4 2 7 4 . A 4 2 7 4 2 9 4 . A 5 2 9 4 2 11 4 . A 6 2 11 4 2 13 4 . So~W=2^{-\frac 1 4}\\ \implies~ for~A_0,~long~side~is~r*W=2^{\frac 1 2}*2^{-\frac 1 4}=2^{\frac 1 4}.\\ For~each~higher~number~the~long~side~is~2^{-1}~and~short~side~becomes~the~long~side.\\ This~ \implies~ it ~becomes~ r(=2^{\frac 1 2})~times.~~\\ The~net~ result ~is~each~stage,~long~side~is~reduced~by~2^{-1}*2^{\frac 1 2}=2^{-\frac 1 2}.\\ \therefore~for~six~stages~for~A_6,~the~long~side~is~W*\Big(2^{-\frac 1 2}\Big)^6\\ =2^{\frac 1 4}*2^{-3}=2^{-\frac {11} 4}.\\ ~~~~\\ A~simple~explanation~follows.\\ Let~the~ A_0~rectangle ~be~W~ and~ L=r*W.~~~~~~ r>1.\\ \therefore~W*L=r*W^2=1~ for~area~ A_0. ....(1)\\ For~area~A_1,~sides ~are~short~\dfrac{r*W} 2,~~long~W.\\ Shape~being ~similar ~ \implies ~\dfrac W {\dfrac{r*W} 2}~=~r,\\ \therefore~r^2=2,~~ \implies~\color{#3D99F6}{r=\sqrt2.} \\ But~from~~(1)~~W^2=\dfrac 1 r=2^{-\frac 1 2}\\ So~\color{#3D99F6}{W=2^{-\frac 1 4} }\\ For~A_0~~~~Long=r*W=2{\frac 1 2}*2^{\frac 1 4}=2^{\frac 3 4}~~~~~~~Short=W=2^{\frac 1 4} \\ For~A_n~~~~~~~Long~~~~~~~~~~~~Short\\ A_0~~~~~~~~~~~~~~~~~2^{\frac 3 4}~~~~~~~~~2^{\frac 1 4}.\\ A_1~~~~~~~~~~~~~~~~~2^{\frac 1 4}~~~~~~~~~2^{-\frac 3 4}.\\ A_2~~~~~~~~~~~~~~~2^{-\frac 3 4}~~~~~~~~~2^{-\frac 5 4}.\\ A_3~~~~~~~~~~~~~~~2^{-\frac 5 4}~~~~~~~~~2^{-\frac 7 4}.\\ A_4~~~~~~~~~~~~~~~2^{-\frac 7 4}~~~~~~~~~2^{-\frac 9 4}.\\ A_5~~~~~~~~~~~~~~~2^{-\frac 9 4}~~~~~~~~~2^{-\frac {11} 4}.\\ A_6~~~~~~~~~~~~~~~2^{-\frac {11} 4}~~~~~~~~~2^{-\frac {13} 4}.\\

Ujjwal Rane
Jul 11, 2017

If folding a L x W rectangle in half gives a similar rectangle, L W = W L / 2 \frac{L}{W}=\frac{W}{L/2}

This gives L = 2 W L = \sqrt{2}W

Moreover, area of size A 0 = L 0 × W 0 = 1 A_0 = L_0 \times W_0 = 1

Which gives L 0 × L 0 2 = 1 L_0 \times \frac{L_0}{\sqrt{2}} = 1

Thus for size A 0 A_0 , L 0 = 2 1 / 4 L_0 = 2^{1/4}

Every even size A2, A4 etc. halves the length.

So for A6 it will be L 0 8 = 2 1 4 3 = 2 11 4 \frac{L_0}{8} = 2^{\frac{1}{4}-3} = 2^\frac{-11}{4}

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