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### Line with Small Compass and Straightedge

```
Date: 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/
```
Associated Topics:
High School Constructions
High School Euclidean/Plane Geometry
High School Geometry

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