Q: How does a geometer capture a lion in the desert?
A: Build a circular cage in the desert, enter it, and lock it. Now perform an inversion with respect to the cage. Then you are outside and the lion is locked in the cage.
a mathematical joke from before 1938
Definitions. An inversion with respect to a circle Γ is a transformation from the extended plane (the plane with ∞, the point at infinity, added) to itself that takes C, the center of the circle, to ∞ and vice versa, and that takes a point at a distance s from the center to the point on the same ray (from the center) that is at a distance of r2/s from the center, where r is the radius of the circle. See Figure 5. We call (P,P') an inversive pair because (as the reader can check) they are taken to each other by the inversion. The circle Γ is called the circle of inversion.
Note that an inversion takes the inside of the circle to the outside and vice versa and that the inversion takes any line through the center to itself. Because of this, inversion can be thought of as a reflection in the circle.
We strongly suggest that the reader play with inversions by using dynamic geometry software such as Geometers Sketchpad®, Cabri®, or Cinderella®. Using any of these you may construct the image, P′, of P under the inversion through the circle Γ as follows:
If P is inside Γ then draw through P the line perpendicular to the ray CP. Let S and R be the intersections of this line with Γ. Then P′ is the intersection of the lines tangent to Γ at S and R. (To construct the tangents, note that lines tangent to a circle are perpendicular to the radius of the circle.)
If P is outside Γ then draw the two tangent lines from P to Γ. Let S and R be the points of tangency on Γ. Then P′ is the intersection of the line SR with the ray CP. (The points S and R are the intersections of Γ with the circle with diameter CP. Why?)
Click here for dynamic JAVA applets using Cinderella® to implement the constructions and showing the images of
a. Prove that these constructions do construct inversive pairs.
The purpose of this part is to explore and better understand inversion, but it will not be directly used in the other parts of this problem.
b. Show that an inversion takes each circle orthogonal to the circle of inversion to itself.
Two circles are orthogonal if, at each point of intersection, the angle between the tangent lines is 90°. (Note that, at these points, the radius of one circle is tangent to the other circle.)
c. Show that an inversion takes a circle through the center of inversion to a line not through the center, and vice versa. What happens in the special cases when either the circle or the straight line intersects the circle of inversion? Note that the line is parallel to the line tangent to the circle at C.
Look at Figure 7 and prove that ΔCPQ and ΔCQ′P′ are similar triangles.
d. An inversion takes circles not through the center of inversion to circles not through the center. Note: The circumference of a circle inverts to another circle but the centers of these circles are on the same ray from C though not an inversive pair.
Look at Figure 8. If P, Q, X invert to P′, Q′, X′, then show that
P′ X′ Q′ = P X Q = right angle,
by looking for similar triangles. Thus, argue that as X varies around the circle with diameter PQ then X′ varies around the circle with diameter Q′P′.
Problem 1. Circles in the Plane, which explores some geometry of circles.
Problem 2. Inversions in Circles, which explores properties of inversion in circles.
Problem 3. Applications of Inversions to the Peaucellier-Lipkin Linkage, which explores the applications of Problems 1 and 2 to an understanding of the linkage.