12. INVERSION

Simple Cases  

Up until now we have examined various interesting configurations, mostly relating to the circle. Now things get even more interesting as we use our knowledge to develop techniques for finding new results.

Let denote a fixed given circle with centre O and radius a. We have defined points P, P' to be inverse if O, P, P' are collinear and . The circle (Gamma) is called the circle of inversion, and a is the radius of inversion.

Now if point P moves along a certain curve, the inverse point P' will also determine a curve – the inverse of the original curve. Inversion is sometimes called ‘reflection in a circle’. Let’s look at some simple examples.

1. If P traces out a straight line passing through O, then what curve does P' trace out?

2. If P traces out a circle with centre O, describe the inverse curve.  What if the circle is ?

3. If P traces out a circle orthogonal to  , what is the inverse curve?     Check your answers ...

Here is a Java applet that you can expperiment with ...
                                                                                                                                                                                                                                                                                                            
1. P' traces out the same straight line, as P, P' , O are collinear.

2. Since , if OP is constant, so is OP'. Hence P' traces out a circle centre O. If OP < a then OP' > a and conversely. If P traces out , then so does P'.

3. Here P' describes the same orthogonal circle. For if T is a point where and the orthogonal circle meet, = OT 2.


































Circle to Circle

Suppose now that P describes a circle C which does not pass through point O.

Theorem 12.1 The inverse of a circle not through O is a circle not through O.

Proof Let C be the circle described by P,
                 M the inverse of O with respect to circle C,
                 M' the inverse of M with respect to circle .
We show that the inverse P' of P describes a circle with centre M'.

Now (P, P'), (M, M') are inverse with respect to     P, P', M', M are concyclic
      OPM = OM'P' (show; opposite angles of a cyclic quadrangle are supplementary)
       OPM  ~   OM'P' (equal angles)      MP / P'M' = OP / OM'        P'M' = (MP / OP).OM'.

Finally, since M was chosen to be the inverse of O with respect to circle C, by the converse of Apollonius’ Theorem, the ratio MP / OP is constant. Length OM' is also fixed. We deduce that P'M' is constant for all positions of P'. That is, the locus of P' is a circle with centre M'.

This completes the proof of the theorem.
                                                                                                                                                                         


































Circle to Line

Theorem 12.2 The inverse of a circle through O is a straight line not through O, and the inverse of a straight line not through O is a circle through O.

Proof Let C be a circle on OM as diameter, and let M' be the inverse of M with respect to the circle of inversion .

Let line m be the polar of M with respect to ; this line meets OM at right angles at M'.

Since M and M' are inverse, and P and P' are given inverse, we know that points P, P' M', M are concyclic points.

We show that P lies on circle C   P' lies on line m.

Now P lies on C    OPM = 90°   OM'P = 90° P lies on line m.

Jakob Steiner was a famous Swiss geometer who lived from 1796 – 1863. He studied inversion and made many discoveries in projective geometry.                          
                      



























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Summary

We notice that as a transformation, inversion has period 2: if you carry it out twice, you get back to where you started. When applying inversions, we are usually not concerned about the details of the circle of inversion, or the nature of the above proofs. We usually just choose a centre of inversion (the point O), and then use the following summary facts:

A straight line through O maps to a straight line through O.
A straight line not through O maps to a circle through O. [Theorem 12.2]
A circle through O maps to a straight line not through O.   [Theorem 12.2]
A circle not through O maps to a circle not through O.       [Theorem 12.1]

There is one further important property of inversion which we shall need:

The angle between two curves (lines) is preserved under inversion

We look at this property next.
                                                                                                                                                     
                                                                                          































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Angle

The angle between two smooth curves at a point of intersection is usually defined as the angle between the tangents to the curves at that point.  Since the tangent itself is the limit of a set of chords, it will be sufficient for us to show that QPR (at left) is preserved under the inversion transformation.

Suppose that under the inversion, points P, Q, R map to points P', Q', R' as at right.

Then points P, P', Q', Q are concyclic (property of inverse points), so OPQ = OQ'P'. Similarly, points P, P', R', R are concyclic so OPR = OR'P'. (See right diagram.)

Now
                QPR = OPR – OPQ = OR'P' – OQ'P' = Q'P'R'.

This shows that the angle between curves is preserved. Some care is required when the point of contact of the two curves is O itself. Here, for example, two curves touching at O invert into two parallel lines, and two circle intersecting at O invert into two straight lines not through O. In this last case, the angle is of course preserved.

                                                                                                                                                                     
                                                                                          


































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Extensions

To solve these problems, invert the figure with respect to O.

1.  Points P, Q, R lie in order on a straight line, and POR = 90°. Show that the circles PQO, QRO are orthogonal.

2. Find the inverse of a set of coaxial circles having O as one of the limiting points.

3. Points O, A, B, X lie on a circle. A line through O cuts XA, XB in U, V respectively.
     Show that the circles OAU, OVB have a common tangent at O.

4. BOC is a given angle. A circle through O and B has its centre on OC, and a circle through O and C has its centre on OB. The two circles meet again in X. Show that OX is a diameter of circle OBC.

5. (a) By thinking about the properties of inverse points, describe where the image of a circle through O lies under inversion with respect to O.
     (b) Points A, B, C lie on a straight line, and O is a point not on that line. Circles OBC, OCA, OAB have centres U, V, W. Prove that the points O, U, V, W lie on a circle. [Hint: You may find it helpful to draw an accurate diagram. Also, remember the Simson line!]

Isn’t this clever?! If you want to explore further, consider inverting the configurations of various theorems, particularly those involving straight lines, with respect to various points of the figure to obtain interesting new results.

                                                                                Hints and Solutions ...

http://en.wikipedia.org/wiki/Inversive_geometry

                                                                                                                                                          
                                                                                          



































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Hints and Solutions

To solve these problems, invert the figure with respect to O.

1.  Points P, Q, R lie in order on a straight line, and POR = 90°. Show that the circles PQO, QRO are orthogonal.

• Line PQR maps to a circle through O with PR as diameter. The two given circles map to perpendicular segments PO, QR (angle in semicircle). Hence the circles are orthogonal.

2. Find the inverse of a set of coaxial circles having O as one of the limiting points.

• We get concentric circles with centre O. For the orthogonal system maps to straight lines through O.

3. Points O, A, B, X lie on a circle. A line through O cuts XA, XB in U, V respectively.
     Show that the circles OAU, OVB have a common tangent at O.

• Draw the inverse figure, and show that image segments AU, VB are parallel. (Look at the angles subtended by AO, BO.)

4. BOC is a given angle. A circle through O and B has its centre on OC, and a circle through O and C has its centre on OB. The two circles meet again in X. Show that OX is a diameter of circle OBC.

• Invert to get a kite OCXB with right angles at B and C. Now show BC pperpendicular to OX.

5. (a) By thinking about the properties of inverse points, describe where the image of a circle through O lies under inversion with respect to O.
     (b) Points A, B, C lie on a straight line, and O is a point not on that line. Circles OBC, OCA, OAB have centres U, V, W. Prove that the points O, U, V, W lie on a circle. [Hint: You may find it helpful to draw an accurate diagram. Also, remember the Simson line!]

• (a) Let the circle of inversion have centre O, radius 1. If OP is a diameter of the circle to be inverted, and P maps to P' then OP.OP' = 1. If C is the centre of this circle, C maps to C' with OC.OC' = 1. OC = OP/2 implies OC' = 2OP'.