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While looking for something fun to do in
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
\$ 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 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 Why does stuff like this happen to me? |

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