## 13 November 2013

### Geometry without compasses

Plane geometry admits two instruments only: a straightedge and a pair of compasses. With the former, one constructs a straight line through any two points, with the latter: a circle having any point as its centre and passing through any other point. Both 'instruments' are platonically perfect: infinitely large and infinitely precise. Early in the twentieth century, it was discovered that one can do away with the compasses after a single (even partial) use.

Theorem (Severi and Mordukhai-Boltovskoy). Each point of the plane which can be constructed with straightedge and compasses can be constructed with the sole straightedge, if in that plane one circular arc and its center are given.

To prove this theorem, it suffices that we prove that the points of intersection of straight lines and circles can be obtained with the straightedge alone, given a fixed circular arc and its centre.

All constructions below are done with the sole straightedge. The order of the letters A,B, etc. is also the order in which the points are obtained. Yellow denotes elements given, blue the final result or a result reducing the problem to a problem already solved, and white the intermediate constructions.

For the constructions proper, we provide the few indications that might not be evident. Proofs are to be found here (also here), on the pages cited. Special configurations, requiring an adaptation of the argument, are not considered.

A. Constructions not using the arc

Problem 1. Given a bisected segment, construct the parallel line through an arbitrary given point.

The point E is chosen arbitrarily on the ray from A through D, but outside the segment AD. (Proof: p. 52.)

Problem 2. Bisect any segment, given a line parallel to that segment.

The point D is chosen arbitrarily on the parallel line. (Proof: p. 52.)

B. Constructions using the arc but not its centre

Problem 3. Given a straight line and a circular arc without its centre, construct the tangent in an endpoint of the arc.

The points C,D,E, chosen arbitrarily in the arc, are added to A (counted twice) and B, defining an inscribed hexagon of which the tangent is one of the sides. By Pascal's theorem, H, obtained after F and G, is on the tangent. (Proof: p. 47.)

Problem 4. Given a straight line and a circular arc without its centre, construct the points of intersection of the line with the circle.

AD and ED are tangents to the arc (problem 3). G is chosen arbitrarily on EC. (Proof: p. 69.)

C. Constructions using the arc and its centre

Problem 5. Given a straight line and a circular arc with its centre, construct a parallel to the given line.

The point D is the intersection of the line through A and C with the circle (problem 4), and E likewise. (Proof: trivial.)

Problem 6. Given a point, a straight line and a circular arc with its centre, construct the parallel to the given line through the point.

Construct a line parallel to the given line (Problem 5). Chose any segment on the given line and construct its midpoint, using the two parallel lines (Problem 2). Finally, construct a parallel through the given point, using the bisected segment (Problem 1).

Problem 7. Given a point, a straight line and a circular arc with its centre, construct the perpendicular to the given line through the point.

C and D are the points where a parallel to the given line (Problem 6) intersects the circle, and E is the intersection of the diameter trough C withe the circle (Problem 4, twice). The perpendicular is parallel to DE (Problem 4). (Proof: Thales' s Theorem.)

Problem 8. Given two points, a straight line and a circular arc with its centre, construct the points of intersection of the line with the circle which has the first point as its centre and passes through to the second point

Here and below, dashed circles are perfectly defined, but unavailable for constructions. The point F is chosen arbitrarily on the given line. The points D,H,I are intersection points of straight lines with the circle of the arc (Problem 4). (Proof: p. 65.)

Problem 9. Given four points and a circular arc with its centre, construct the power axis of two circles: with the first (third) point as its centre and passing through to the second (fourth) point.

The arc and its centre are not drawn. E is the midpoint of the segment BD (Problems 5 and 2). F is the intersection of lines perpendicular to AE and CE, respectively (Problem 5). The power axis is perpendicular to AC (Problem 5). (Proof: p. 65.)

Problem 10. Given four points and a circular arc with its centre, construct the points of intersection of two circles: with the first (third) point as its centre and passing through to the second (fourth) point.

Construct their axis of power (Problem 9), then its intersection points with one of the circles (Problem 8).

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