Line with Small Compass and StraightedgeDate: 10/16/96 at 22:32:16 From: Scott Hills Subject: Line with Small Compass and Straightedge Construct a line segment joining two points using only a compass and a straightedge. The two points are farther apart than either the compass or the straightedge can span. I looked at a few theorems in trying to analyze this problem... Including knowing that the opposite angles of an inscribed quadrilateral are supplementary... I'm just stuck... Any ideas? Date: 10/18/96 at 18:8:15 From: Doctor Wilkinson Subject: Re: Line with Small Compass and Straightedge Nice problem! Here's a sketch of a method. If I can find the point half-way between the two given points, that will be good enough, because I can repeat the procedure to get a sequence of points such that I can join consecutive points with the straightedge. Now let me try to make a picture: M X.----.----.Q / / / / / / / / / / / / / / / .---------.T / / / / / / / / / / / / / / / P.--------.S Suppose P and Q are the given points, and that I can find a way to get from P to X, which is close enought to Q so that I can construct the midpoint M of the segment XQ. Then, as the parallelogram diagram shows, if I could get halfway back from M in the direction from X to P, I would have the midpoint of the segment joining P and Q. How to get from P to a point close to Q? Let me draw a circle with center P and a circle with center Q, with the same radius. If I'm really lucky, they'll intersect. (I'm going to make the radius small enough so that I can span the diameter with my straightedge.) Now I mark off six points on the circle around P using the compass with the same radius. I then use the six points obtained as the centers of six more circles. From the new circles I continue the process, using the intersections with the first circle to get started, and so on outward, until I intersect the circle around Q. (This has to happen eventually). The center of the circle that intersects the circle around Q is my point X. Of course I don't have a straight line from P to X, but I do have a broken line formed by joining the centers of intersecting circles. Now through the point M I can construct a line parallel to the last broken-line segment leading to X, and I can trace backwards using circles of half the original radius. It's only the first line segment that I need to use the parallel line construction on. The remaining segments just veer off at angles of 60 or 120 degrees. (Or continue straight on). -Doctor Wilkinson, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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