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