Cut The Knot!An interactive column using Java applets
by Alex Bogomolny
Let's consider the following problem 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 with center O and a straight line ST that cuts from 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.
Let O' be the center of an inscribed circle. Then AO'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. NOB is therefore isosceles.
The next proof builds on the observation that the triangles AO'B and MOB are not merely similar but are homothetic with center B.
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.
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 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
The situation is illustrated by the following applet:
Proof 3 has the virtue of bringing together elements of various geometries: inversive and affine. But check this out:
Let's make an inversion in center M with radius
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.
The framework created by Proof 4 suggests more problems. For example,
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 nonetheless make 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,
From the similarity of triangles BTM and MAT, we obtain
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 , and the latter back on the line ST.
Sometimes a more general definition is used
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.
Given a circle and a point P not on . The circles orthogonal to 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.
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 1 and 2 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, 0, has its diameter on L. The second, 1, is tangent to 0, and both 1 and 2, and so on. Let hn and rn denote the distance to L from the center and the radius of the circle n. Then
Indeed, an inversion with the center at the common point of tangency of 1 and 2 maps the two circles onto two parallel lines perpendicular to L. The circles n 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 n 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,
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:
Let first two circles 1 and 2 lie outside each other. Their radical axis consists of the points from which the tangents to 1 and 2 are equal. A circle centered on the radical axis and having radius equal to the common tangent to the circles 1 and 2 from its center is perpendicular to both circles. It is therefore easy two find two intersecting circles 1 and 2 orthogonal to 1 and 2. Make an inversion with the center at one of the points of intersection of 1 and 2. 1 and 2 will map onto two straight intersecting lines. Let T' be their point of intersection. 1 and 2 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.
Alex Bogomolny has started and still maintains a popular Web site Interactive Mathematics Miscellany and Puzzles to which he brought more than 10 years of college instruction and, at least as much, programming experience. In February 2003 the site has welcomed its 6,000,000th visitor. Alex holds M.S. degree in Mathematics from the Moscow State University and Ph.D. in Applied Mathematics from the Hebrew University of Jerusalem. He can be reached at firstname.lastname@example.org
Copyright © 1996-2003 Alexander Bogomolny