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An Adventure Starts with Trisecting a Rectangle Origami Style

While looking for something fun to do in Origami in Action by Robert J. Lang, I came across a method for trisecting both the height and the width of any rectangle by drawing a diagonal of the rectangle and another diagonal of opposite slope of one half the rectangle. The diagonals can be created by folding as can dividing the rectangle in half.

This method was given without proof but the proof is easy: you just write the equations for the two lines and solve them for the intersection.

$y = {{2h}/w}x$ and $y = h - {h/w}x$

${{2h}/w}x = h - {h/w}x$

$x = w/3$ and $y = {2h}/3$

But I saw in that image another rectangle trisected:

And the potential for many more:

This technique generalizes to dividing a rectangle into $1/n$ if you already have it divided into $1/{n-1}$. Proof:

$y = {{(n-1)h}/w}x$ and $y = h - {h/w}x$

${{(n-1)h}/w}x = h - {h/w}x$

$x = w/n$ and $y = {(n-1)h}/n$

And so I wrote a PostScript program to draw the various possibilities:

%!PS-Adobe-2.0
%%Creator: Richard Kandarian
%%Title: Nsecting a Rectangle
%%CreationDate: Sun Jan 26 08:29:02 MST 2020
%%Pages: 1
%%DocumentFonts: 
%%BoundingBox:0 0 612 792
%%EndComments

% If you have a rectangle divided by n-1 it is straight forward to divide it
% by n, and this graphically shows that many other rectangles of two 
% types of similarity are also so divided.

/ph 11 25.4 mul def % page height
/pw 8.5 25.4 mul def % page width
/h ph 20 sub def % height
/w pw 20 sub def % width
/ni 5 def % number of iterations
/ic 0 def % iteration counter used for coloring lines uniquely per iteration.
/ns 4 def % the number of divisions
/ws 1 ns 1 sub div def % the width scale for n-1 divisions
/hs ns 2 sub ns 1 sub div def % the height scale for n-1 divisions

/nm1sect{ % divide the rectangle n-1 times
  w ws mul 0  moveto 0 h rlineto
  0 h hs mul moveto w ws mul 0 rlineto
  stroke
}def

/next{
  nm1sect
  w ws mul h translate 180 rotate ws dup scale % set up for the next one
  ic ni div 1 13 16 div sethsbcolor
  /ic ic 1 add def
}def

%%EndProlog

%%Page: 1 1

72 25.4 div dup scale % scale to millimeters
.5 setlinewidth

% set up the page
10 10 translate

% Draw the initial rectangle and the diagonals
0 0 w h rectstroke
.25 setlinewidth
0 0 moveto w ws mul h lineto
0 h moveto w 0 lineto
stroke

% march toward the 1/n division point.
ni{next}repeat
nm1sect % draw the last one
showpage

%%Trailer

You may or may not be interested in the details of that short program, but my attention was drawn to the heart of the matter:

  w ws mul h translate 180 rotate ws dup scale % set up for the next one

and especially the scaling by ws, the width scale factor. This is $1/{n-1}$ and since the rectangles converge on the trisecting point it lead me to investigate how $1/n$ would be represented in base $n-1$ since the scaling is like setting up a fractional number with the radix of the inverse of the scale factor. A little fooling about with the Unix dc:

\$ dc1 20 k 3 o 1 4 /
.020202020202020202020202020202020202020202
\$ dc1 20 k 3 o 3 4 /
.202020202020202020202020202020202020202020
\$ dc1 20 k 4 o 1 5 /
.0303030303030303030303030303030303
\$ dc1 20 k 4 o 4 5 /
.3030303030303030303030303030303030
\$ dc1 20 k 5 o 1 6 /
.04040404040404040404040404034
\$ dc1 20 k 5 o 5 6 /
.40404040404040404040404040403

(The dc calculator is a postfix stack machine (reverse polish) and the k sets the precision and the o sets the output radix.)

So you get a repeating fraction alternating 0 with the largest digit, $n-2$, of the $n-1$ radix and if you multiply by the radix (shift left) you get the complement so that the two added together are unity. So I went here Where I found this formula:

So substituting $k$ with $n$ and $b$ with $n-1$:

$1/n = 1/{n-1} + {(n-1) - n}/{(n-1)^2} + {((n-1) - n)^2}/{(n-1)^3} + {((n-1) - n)^3}/{(n-1)^4} + {((n-1) - n)^4}/{(n-1)^5}…$

and simplifying the numerator:

$(n-1) - n = -1$

to get:

$1/n = 1/{n-1} - 1/{(n-1)^2} + 1/{(n-1)^3} - 1/{(n-1)^4} + 1/{(n-1)^5} - 1/{(n-1)^6}…$

now it's possible to combine terms:

$1/n = {n-1}/{(n-1)^2} - 1/{(n-1)^2} + {n-1}/{(n-1)^4} - 1/{(n-1)^4} + {n-1}/{(n-1)^6} - 1/{(n-1)^6}…$

$1/n = {n-2}/{(n-1)^2} + {n-2}/{(n-1)^4} - {n-2}/{(n-1)^6} + {n-2}/{(n-1)^8}…$

and since the exponents are now all even it looks very much like the results I obtained with dc.

Why does stuff like this happen to me?

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