From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Straight Line Using Compass and Straightedge
Date: 20 Nov 1995 04:48:48 GMT

In article <48b3f7$1gu@su102w.ess.harris.com>,
Jim Ziarno  <jziarno@harris.com> wrote:
>Is it possible to construct a straight line between two points using
>only a compass and a straightedge ?
>
>Note: Assume the distance between the points is much greater than
>the span of the compass or the length of the straightedge

Yes, it is possible. 

Choose a scale of length so that both the span and the length of the
straightedge are greater than, oh, 2 should be enough.

Call one of the points  O  and fix coordinate axes about  O (in any pair
of perpendicular directions). Clearly it will be sufficient to show how
to draw a line segment to any point in the first quadrant.

For non-negative integers  m, n  let _the  (m,n)-square_ denote the set of
points  (x,y) with  m <= x <= m+1  and  n <= y <= y+1. I will show how to draw
a line segment to each point in the  (m,n)-square; by induction on  m+n, if
you like, we will then have the desired result. The (0,0)-square is trivial
since the length of the straightedge is longer than the diagonal of the square.
So let us fix any other pair  (m,n).

If  (x,y)  is a point in the  (m,n)-square, let  x0=x-m,  y0=y-n. We need to
draw a line from  (x,y)  with slope  (y0+n)/(x0+m). The trick is to note that
this is the same as  [(y0+n)/k]/[(x0+m)/k]  for any  k, say k=max(m,n)+1.

There is a well-known method of dividing lengths with ruler and straightedge.
Mark off  k  equal segments along, say, the  x-axis starting at the origin; 
if the segments are sufficiently small, their union will fit in a single 
square. Mark off a length of  y0  on the  y-axis also starting at the origin.
Join the remaining endpoints of these two segments to form a right triangle.
Construct another perpendicular to the  x-axis at the next-to-last endpoint of
the little segments. This cuts off a similar right triangle whose height is
y0/k. In like manner construct a segment of length  1/k. Now copy  n  of these
and the previous one end to end on the  y-axis to create a segment of
length  (y0+n)/k. Actually it makes a little more sense in our case to
copy this length to a vertical line segment with  (x,y)  as its top endpoint.
Using the same technique we create a horizontal line segment going left
from there and having length (x0+m)/k. The line segment joining this point
to (x,y) has the right slope, i.e., it points to the origin. This line
segment has length  L  at most  sqrt(2).

Now it's easy: draw the line segment connecting these points, slide your
straightedge along the line segment a distance  L/2, say, then draw another
segment of length  L  in the same direction. Repeat as necessary until
you reach the origin  O. (This part of the technique is known to  10-year-old
draftsmen the world over.)


Note that the construction gets a tad more complex as the distance from the
point  O  increases. This is unfortunate in practice, because the last
portion of the algorithm tends to magnify errors more in precisely those cases.
This then suggests a line of inquiry in which not only the compass and
the straightedge are made bounded (rather than ideal) but the pencil itself
is given a non-zero width point.


I suppose it's worth mentioning that others have considered what can be
constructed with tools somewhat more limited than the classical set.
Roughly speaking, a compass alone is enough; so is a straightedge alone plus
one circle. Details upon request.

With a _marked_ straightedge it is possible to trisect an angle. Does
anyone know offhand what the complete set of constructible points is
when using this tool? Any other non-standard but simple tools? 

dave