From - Tue Jun  1 10:49:35 1999
From: Noam D. Elkies <elkies@mathematics.harvard.edu>
Newsgroups: rec.puzzles,sci.math
Subject: Re: Trisecting a line
Date: 29 May 1999 14:20:10 GMT
Organization: Harvard Math Department

Gary Edstrom <GBE@PrimeNet.Com> wrote:
>It is possible to divide a line into ANY number of parts using just the
>traditional tools:

>[standard Euclidean construction with parallel lines deleted]

True enough -- though if you want to actually do it, drawing a
parallel line with only straightedge and compass takes quite a few
operations.  For trisection there's a neat alternative which as it
happens doesn't even require intersecting two circular arcs
(the compass is only used out to lay out equal lengths):

To trisect segment AB, draw arbitrary ray from A, choose point C
on it, then D on the same ray with AC=CD; then connect D to B and
extend that line beyond D, and find E on that line with DB=BE.
Finally connect E to C, meeting AB in F:

D
|\
| \
|  \
|   \
|    \
C     B
| .  / \
|   F   \
|  /  .  \
| /     . \
|/        .\
A-----------E


Then AB,CE are medians of triangle ACE and thus trisect each other.

As a bonus, the same diagram gives an alternative segment bisection
method: to bisect AD, draw an arbitrary ray from A, choose point P
on it, then F,B on the same ray with AP=PF=FB; then connect D to B and
extend that line beyond D, and find E on that line with DB=BE.
Finally connect E to F, extending that line to meet AD in C.
Then C is the midpoint of AD.

[I found this myself some 20 years ago, but imagine that it's a
long-known though generally forgotten bit of elementary geometry.]

ObPuzzle: generalize this to prove that for each n=2,3,4,5,...
one may n-sect a given line segment with straightedge and compass
without ever intersecting two circular arcs.

--Noam D. Elkies (Dept. of Math., Harvard University)
 [change "mathematics" to a 4-letter abbr in e-addr]