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 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