What Is Mathematics? An Elementary Approach to Ideas and Methods, 2nd Edition (1996)

PART II. VARIOUS METHODS FOR PERFORMING CONSTRUCTIONS

§4. GEOMETRICAL TRANSFORMATIONS. INVERSION

1. General Remarks

In the second part of this chapter we shall discuss in a systematic way some general principles that may be applied to construction problems. Many of these problems can be more clearly viewed from the general standpoint of “geometrical transformations”; instead of studying an individual construction, we shall consider simultaneously a whole class of problems connected by certain processes of transformation. The clarifying power of the concept of a class of geometrical transformations is by no means restricted to construction problems, but affects almost everything in geometry. In Chapters IV and V we shall deal with this general aspect of geometrical transformations. Here we shall study a particular type of transformation, the inversion of the plane in a circle, which is a generalization of ordinary reflection in a straight line.

By a transformation, or mapping, of the plane onto itself we mean a rule which assigns to every point P of the plane another point P′, called the image of P under the transformation; the point P is called the antecedent of P′. A simple example of such a transformation is given by thereflection of the plane in a given straight line L as in a mirror: a point P on one side of L has as its image the point P′, on the other side of L, and such that L is the perpendicular bisector of the segment PP’. A transformation may leave certain points of the plane fixed; in the case of a reflection this is true of the points on L.

Fig. 37. Reflection of a point in a line.

Fig. 38. Inversion of a point in a circle.

Other examples of transformations are the rotations of the plane about a fixed point O, the parallel translations, which move every point a distance d in a given direction (such a transformation has no fixed points), and, more generally, the rigid motions of the plane, which may be thought of as compounded of rotations and parallel translations.

The particular class of transformations of interest to us now are the inversions with respect to circles. (These are sometimes known as circular reflections, because to a certain approximation they represent the relation between original and image in reflection by a circular mirror.) In a fixed plane let C be a given circle with center O (called the center of inversion) and radius r. The image of a point P is defined to be the point P′ lying on the line OP on the same side of O as P and such that

(1) OP·OP′ = r².

The points P and P′ are said to be inverse points with respect to C. From this definition it follows that, if P’ is the inverse point of P, then P is the inverse of P′. An inversion interchanges the inside and outside of the circle C, since for OP < r we have OP’ > r, and for OP > r, we have OP′ < r. The only points of the plane that remain fixed under the inversion are the points on the circle C itself.

Rule (1) does not define an image for the center O. It is clear that if a moving point P approaches O, the image P′ will recede farther and farther out in the plane. For this reason we sometimes say that O itself corresponds to the point at infinity under the inversion. The usefulness of this terminology lies in the fact that it enables us to state that an inversion sets up a correspondence between the points of the plane and their images which is biunique without exception: each point of the plane has one and only one image and is itself the image of one and only one point. This property is shared by all the transformations previously considered.

Fig. 39. Inversion of a line L in a circle.

2. Properties of Inversion

The most important property of an inversion is that it transforms straight lines and circles into straight lines and circles. More precisely, we shall show that after an inversion

Statement (a) is obvious, since from the definition of inversion any point on the straight line has as image another point on the same line, so that although the points on the line are interchanged, the line as a whole is transformed into itself.

To prove statement (b), drop a perpendicular from O to the straight line L (Fig. 39). Let A be the point where this perpendicular meets L, and let A’ be the inverse point to A. Mark any point P on L, and let P′ be its inverse point. Since OA’·OA = OP’·OP = r², it follows that

Hence the triangles OP’A’ and OAP are similar and angle OP’A’ is a right angle. From elementary geometry it follows that P′ lies on the circle K with diameter OA’, so that the inverse of L is this circle. This proves (b). Statement (c) now follows from the fact that since the inverse of L is K, the inverse of K is L.

It remains to prove statement (d). Let K be any circle not passing through O, with center M and radius k. To obtain its image, we draw a line through O intersecting K at A and B, and then determine how the images A’, B’ vary when the line through O intersects K in all possible ways. Denote the distances OA, OB, OA’, OB’, OM by a, b, a’, b’, m, and let t be the length of a tangent to K from O. We have aa’ = bb’ = r², by definition of inversion, and ab = t², by an elementary geometrical property of the circle. If we divide the first relations by the second, we get

Fig. 40. Inversion of a circle.

a’/b = b’/a = r²/t² = c²,

where c² is a constant that depends only upon r and t, and is the same for all positions of A and B. Through A’ we draw a line parallel to BM meeting OM at Q. Let OQ = q and A’Q = ρ. Then q/m = a’/b = ρ/k, or

q = ma’/b = mc², ρ = ka’/b = kc².

This means that for all positions of A and B, Q will always be the same point on OM, and the distance A′Q will always have the same value. Likewise B′Q = ρ, since a′/b = b′/a. Thus the images of all points A, B on K are points whose distance from Q is always ρ, i.e. the image of K is a circle. This proves (d).

3. Geometrical Construction of Inverse Points

The following theorem will be useful in Article 4 of this section: The point P′ inverse to a given point P with respect to a circle C may be constructed geometrically by the use of the compass alone. We consider first the case where the given point P is exterior to C. With OP as radius andP as center we describe an arc intersecting C at the points R and S. With these two points as centers we describe arcs with radius r which

Fig. 41. Inversion of an outside point in a circle.

intersect at O and at a point P′ on the line OP. In the isosceles triangles ORP and ORP′,

so that these triangles are similar, and therefore

Hence P′ is the required inverse of P, which was to be constructed.

If the given point P lies inside C the same construction and proof hold, provided that the circle of radius OP about P intersects C in two points. If not, we can reduce the construction of the inverse point P′ to the previous case by the following simple artifice.

First we observe that with the compass alone we can find a point C on the line joining two given points A, 0 and such that AO = OC. Γo do this, we draw a circle about O with radius r = AO, and mark off on this circle, starting from A, the points P, Q, C such that AP = PQ = QC = r. ThenC is the desired point, as is seen from the fact that the triangles AOP, OPQ, OQC are equilateral, so that OA and OC form an angle of 180°, and OC = OQ = AO. By repeating this procedure, we can easily extend AO any desired number of times. Incidentally, since the length of the segmentAQ is as the reader can easily verify, we have at the same time constructed from the unit without using the straightedge.

Now we can find the inverse of any point P inside the circle C. First we find a point R on the line OP whose distance from O is an integral multiple of OP and which lies outside C,

OR = n·OP.

We can do this by successively measuring off the distance OP with the compass until we land outside C. Now we find the point R′ inverse to R by the construction previously given. Then

r² = OR′·OR = OR′·(n·OP) = (n·OR′)·OP.

Therefore the point P′ for which OP′ = n·OR′ is the desired inverse.

Fig. 42. Doubling of a segment.

Fig. 43. Inversion of an inside point in a circle.

4. How to Bisect a Segment and Find the Center of a Circle with the Compass Alone

Now that we have learned how to find the inverse of a given point by using the compass alone, we can perform some interesting constructions. For example, we consider the problem of finding the point midway between two given points A and B by using the compass alone (no straight lines may be drawn!). Here is the solution: Draw the circle with radius AB about B as center, and mark off three arcs with radius AB, starting from A. The final point C will be on the line AB, with AB = BC. Now draw the circle with radius AB and center A, and let C’ be the point inverse to Cwith respect to this circle. Then

AC′·AC = AB²

AC′·2AB = AB²

 2AC′ = AB.

Hence C′ is the desired midpoint

Another compass construction using inverse points is that of finding the center of a circle whose circumference only is given, the center being unknown. We choose any point P on the circumference and about it draw a circle intersecting the given circle in the points R and S. With these as centers we draw arcs with the radii RP = SP, intersecting at the point Q. A comparison with Figure 41 shows that the unknown center, Q′, is inverse to Q with respect to the circle about P, so that Q′ can be constructed by compass alone.

Fig. 44. Finding the midpoint of a segment.

Fig. 45. Finding the center of a circle.