 # Cut The Knot!

An interactive column using Java applets
by Alex Bogomolny

# Inversion: Introduction

February 2003 'I suppose I must have gone round in a circle.' The sergeant again exchanged a knowing glance with the whole personnel of the station. 'A fine circle, that circle of yours!' Jaroslav Hasek The Good Soldier Svejk, Viking Press, 1990, p. 255

Let's consider the following problem [Savchev, p. 146-147; Honsberger, pp. 30-31; Stanley Rabinowitz in Crux Mathematicorum, Problem 1070, 1987, 31] that, perhaps surprisingly (because of its simplicity), has several apparently unrelated solutions. Which one sheds more light on the nature of the problem? The configuration consists of a circle Σ with center O and a straight line ST that cuts from Σ a circular segment STS. Circles are inscribed in the segment and, for each, the points A and B of tangency with the segment are joined by a straight line. Prove that all those lines concur at the midpoint M of the arc ST complementary to the segment.

### Proof 1

Let O' be the center of an inscribed circle. Then ΔAO'B is isosceles. Extend AB beyond A and let it intersect the perpendicular OM to ST at point N (not shown on the diagram.) The two triangles AO'B and NOB are similar. Indeed, they have a common angle at B and, since ON||O'A, their respective angles at O' and O are also equal. ΔNOB is therefore isosceles. OB = ON, which implies N = M. The next proof builds on the observation that the triangles AO'B and MOB are not merely similar but are homothetic with center B.

### Proof 2

All circles are similar and, moreover, homothetic. For each pair of distinct circles there are either two or one homothety that maps one of the circles onto the other. Two touching circles (as in the problem) are related by a single homothety with their common point of tangency as the center. All points related by a homothety are collinear with its center B. In particular this is true of the lowest (A and M in the diagram) points of the two circles. Proofs 1 and 2 are simple and are not exactly ad hoc as both apply to a more general situation of a line and a circle that do not necessarily cross and to the circles tangent to both of them.

### Proof 3

Let's make an inversion with center S and radius SM. The point M remains unmoved as a point on the circle of inversion. The line ST is fixed, although not point-wise. The circle Σ becomes a straight line meeting ST at the image T' of the point T and passing through M. Since inversion preserves angles, the tangency points A and B become the tangency points A' and B' in the inverse image. The circles inscribed in the segment map onto the circles inscribed into the angle vertical to the angle ST'M.

Because of the symmetry in the angle bisector, A'B' is perpendicular to the angle bisector of that angle. For the same reason, any circle through A' and B' is also perpendicular to the bisector. The bisector is the inverse image of the circle with center M and radius MS = MT. (It's a circle through S and T whose tangent at T bisects the angle between ST and the tangent to Σ and is therefore perpendicular to MT.) The line MS, which is fixed under the inversion is obviously perpendicular to that circle; it thus remains perpendicular to the image of the circle, i.e. to the angle bisector of the angle ST'M. It then follows that any circle through S, A', B' is bound to pass through M. But any such circle is the inverse image of the straight line through A, B, and M. The situation is illustrated by the following applet:

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

Proof 3 has the virtue of bringing together elements of various geometries: inversive and affine. But check this out:

• Problem 1. Among all circles inscribed in the segment, the one that is tangent to MT' on the side of T is invariant under the inversion with center S and radius SM.

### Proof 4

Let's make an inversion in center M with radius MS = MT. Then the straight line ST will map onto a circle through the center of inversion M. Both points S and T are fixed under that inversion. Therefore, the line ST maps onto the circle Σ. The points A and B where a line through M crosses the line segment and the arc ST are inverses of each other. Any circle through these two points is perpendicular to the circle of inversion and is mapped onto itself. Because of the angle preservation property, if such a circle is tangent to ST at A, it is bound to touch the arc ST at B. A circular segment is formed by two curves -- a circle and a straight line. What one may learn from the proofs 3 and 4 is that the important fact about those two curves is that they map onto each other under an inversion. For example, since the inverse image of a circle is either a straight line or a circle, we can conclude that the proofs 3 and 4 apply to the case where circles are inscribed into a lune -- the shape formed by two circular arcs. This could be verified with the applet below, where, with the box "Let center move" checked, the point M could be moved from its original location.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

The framework created by Proof 4 suggests more problems. For example,

• Problem 2. Circles inscribed in a circular segment may touch. Prove that when they do, their common tangent passes through M.

• Problem 3. Circles inscribed in a circular segment may cross. Prove that when they do, the line through the two points of intersection passes through M.

• Problem 4. A circle overlaps a circular segment so that the four angles it forms with the boundary of the segment are all equal. Let the points of intersection be A1 and A2 on the linear segment and B1 and B2 on the arc such that A1B2 intersect A2B1 inside the segment. Then A1B1 and A2B2 meet in M.

• Problem 5. A circle with center on ST intersects ST in A1 and A2 and Σ in B1 and B2 (A1 is inside Σ, while B1 is above ST.) Prove that, if the two cricles meet at 90°, then both A1B1 and A2B2 pass through M.

Each of the problems may be tackled in its own right, but inversion, by providing an universal explanation, removes from them the facade of distinctiveness. For these and the original problem inversion provides the right kind of backdrop. This is not to say that individual solutions, like the proofs 1 and 2, have no intrinsic value. Proofs 1 and 2 that do not use inversion make nonetheless a case for one of the rather important properties of the inversion, the property that has been used in proofs 3 and 4, viz., inversion maps straight lines that do not pass through the center of inversion onto the circles that do pass through the center, and vice versa.

The proof of that fact could be built on top of Proof 1. Connect T to B and M. Triangles BTM and MAT are similar. Indeed, the two triangles share an angle at M. Also, ∠BTM equals half the measure of the arc BSM. On the other hand, ∠MAT equals half the measure of the sum of arcs BS and MT, but the latter is equal to MS. Therefore, ∠BTM = ∠MAT.

From the similarity of triangles BTM and MAT, we obtain MB/MT = MT/MA, or

 (1) MA·MB = MT2

which is one of the definitions of inversion. The inversion with center M and radius MT (or center M and power MT2) maps the line ST onto the circle Σ, and the latter back on the line ST.

Sometimes a more general definition is used

 (1') MA·MB = k,

where k may be any nonzero real number. A geometric definition underscores the importance of orthogonal circles in the inversive geometry and provides analogy with the symmetry in a straight line.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

Given a circle λ and a point P not on λ. The circles orthogonal to λ and passing through P all pass through one other point P'. The latter is called the inverse of P in the given circle. Obviously, if P' is the inverse of P, then P is the inverse of P'.

It is easy to see that the two definitions are equivalent. The algebraic one (1) is easier to use. The geometric definition bonds inversion to the symmetry in a straight line. Indeed, for any pair of points that are symmetric images of each other in a straight line, any circle orthogonal to the straight line and passing through one of the points is bound to pass through the other. Circles orthogonal to the circle of inversion (axis of symmetry) are invariant under the inversion (symmetry.) The circle of inversion (axis of symmetry) itself consists of points fixed under the inversion (symmetry.) This explains why inversion is often called symmetry in the circle.

Circles that do not pass through the center of inversion are inverted into circles. Any two circles could be inverted into one another. Indeed, any center of homothety of the two circles could be used as the center of inversion. The points that correspond to each other under homothety are called homologous. The points that correspond to each other under inversion are antihomologous. For example, in the applet below the points in the pairs A, B' and A', B are homologous, whereas A, A' and B, B' are antihomologous.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

Because of this connection between inversion and homothety, inversion, as does homothety, enjoys the angle preservation property: under inversion angles do not change. There are other ways to describe that property. In complex analysis, mappings that preserve angles are called conformal, in geometry they are said to be isogonal.

A caveat is in order. Both definitions of inversion leave out the center of inversion that does not correspond to any point. It is customary to complement the definitions by assigning the point at infinity to the center of inversion (and vice versa, of course.) This is not the same infinity as that shared by all parallel lines, rather every straight line closes on itself at the "new" infinity. The straight lines thus may be (and are in inversive geometry) looked at as circles with center at infinity and an infinite radius. The angle preservation property means in particular that two circles tangent at the center of inversion are mapped onto two parallel lines. Two circles that cross at the center of inversion are mapped onto two intersecting straight lines.

Because of this interplay between circles and straight lines, inversion has been used to produce curious results. Here is a classical example. Construct two tangent circles Σ1 and Σ2 and the line L through their centers. We are going to inscribe into the crescent-shaped space between the circles a chain of pairwise tangent circles. The first one, α0, has its diameter on L. The second, α1, is tangent to α0, and both Σ1 and Σ2, and so on. Let hn and rn denote the distance to L from the center and the radius of the circle αn. Then

 (2) hn = 2n·rn.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

Indeed, an inversion with the center at the common point of tangency of Σ1 and Σ2 maps the two circles onto two parallel lines perpendicular to L. The circles αn map an a chain of equal circles inscribed between the two parallel lines with the diameter of the first one on L. For those circles, the analogue of (2) is obvious: h'n = 2n·r'n, where h'n is the distance from the center of the inverse image of αn to L and r'n is its radius. However, the circles in every inverse pair are homothetic with the center of homothety at the center of inversion. Whence, hn/rn = h'n/r'n = 2n. Lastly, I'll consider another famous example said to be "very dear to Jacob Steiner" [Coolidge, p 31].

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

For two concentric circles, either there exists a closed chain of circles tangent to the given two as well as to their immediate neighbors in the chain, or such a chain does not exist. In the former case, the chain could be started at the arbitrary location in the ring between the two circles.

Jacob Steiner's wonderful theorem says that the same holds true even if the two circles are not concentric. A simple proof depends on the following assertion:

 (3) Any two non-intersecting circles can be inverted into concentric circles.

Let first two circles Σ1 and Σ2 lie outside each other. Their radical axis consists of the points from which the tangents to Σ1 and Σ2 are equal. A circle centered on the radical axis and having radius equal to the common tangent to the circles Σ1 and Σ2 from its center is perpendicular to both circles. It is therefore easy two find two intersecting circles α1 and α2 orthogonal to Σ1 and Σ2. Make an inversion with the center at one of the points of intersection of α1 and α2. α1 and α2 will map onto two straight intersecting lines. Let T' be their point of intersection. Σ1 and Σ2 will map onto two circles orthogonal to those lines and therefore both centered at the point T'.

If one of the given circles is located in the interior of the other, we may first make an inversion that will separate the two. Any inversion with the center in the ring formed by the two circles will serve that purpose. ### References

1. J. L. Coolidge, A Treatise On the Circle and the Sphere, AMS - Chelsea Publishing, 1971
2. H. S. M. Coxeter, Introduction to Geometry, John Wiley & Sons, 1961
3. H. S. M. Coxeter, S. L. Greitzer, Geometry Revisited, MAA, 1967
4. J. Hasek, The Good Soldier Svejk, Viking Press, 1990
5. R. Honsberger, In Polya's Footsteps, MAA, 1997
6. S. Savchev, T. Andreescu, Mathematical Miniatures, MAA, 2003 ### Inversion - Introduction 