Inversion. Chapter Constructing The Inverse of a Point: If P is inside the circle of inversion: (See Figure 7.1)

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1 Chapter 7 Inversion Goal: In this chapter we define inversion, give constructions for inverses of points both inside and outside the circle of inversion, and show how inversion could be done using Geometer s Sketchpad. A cartesian coordinate representation and a number of fascinating applications of inversion are also presented. Definition Let O be the center of a fixed circle of radius r in the Euclidean plane. Let P be any point in the plane other than O. An inversion in circle C(O, r), I(O, r), is a function such that if I (O,r) (P ) = P then P OP and (OP )(OP ) = r 2. Here P is called the inverse of P, O is called the center of inversion, and r is called the radius of inversion, and r 2 is called its power. It follows from the above definition that to each point P of the plane, other than O, there corresponds a unique inverse point P. To make the inversion a transformation of the plane, we add to the plane a single ideal point Ω defined to be the inverse of the center of inversion. The point Ω is considered to lie on every line in the plane. 7.1 Constructing The Inverse of a Point: Given the circle of inversion C (O,r) and a point P, how do you construct the inverse of P? If P is inside the circle of inversion: (See Figure 7.1) Draw the ray OP. Draw a perpendicular to OP at P. This intersects the circle of inversion in two points, label one of them Q. Connect the center of the circle O to Q. Draw a perpendicular to OQ from Q. The intersection of this perpendicular with OP is P, the inverse of P. 52

2 Q O P P. Figure 7.1: Inverse of a Point Inside the Circle of Inversion Note that OP Q OQP. (OQ) 2 = r 2. Hence OQ OP = OP OQ. Therefore, OP OP = If P is outside the circle of inversion: (See Figure 7.2) Let M be the midpoint of the segment OP. Construct a circle centered at M of radius MO. It intersects the circle of inversion C (O,r) in two points, Q and R and goes through O and P. Construct the segment QR. The intersection of QR and OP is the point P, the inverse of P. Q... P O.... M P. R Figure 7.2: Inverse of a point Outside the Circle of Inversion Note that OP Q OQP. (OQ) 2 = r 2. Hence OQ OP = OP OQ. Therefore, OP OP = 53

3 Using Sketchpad: Inverting a Line: To invert a straight line in the circle of inversion C (O,r) follow the following steps: 1. Construct the circle of inversion C and the line l. 2. Construct an arbitrary point P on the line l. 3. Construct the ray OP. 4. Construct the intersection of OP and the circle, call it Q. 5. Construct the segment OP. 6. Mark O as the center of dilation. 7. Mark the ratio r. OP 8. Dilate Q by the ratio r centered at O. The image of Q is P the inverse OP image of P. 9. Hide everything except the circle, its center, the point determining its radius, P and the straight line l and the two point determining it. 10. Select everything and create a tool and call it invcirc. 11. Apply the tool a few times and use arc through three points to determine the image of the line. Inverting a Circle: To invert a circle C 1 in the circle of inversion C (O,r), replace l by C 1 in the steps above. 7.2 Inversion Using Coordinates: Theorem An inversion about x 2 + y 2 = r 2 is given by (x, y) (x, y xr 2 ) = ( x 2 + y, yr 2 2 x 2 + y ) 2 Proof. Since (x, y), (x, y ) and (0, 0) are collinear, we have y = y. Now x x d((0, 0), (x, y)) d((0, 0), (x, y )) = r 2, hence x 2 + y 2 x 2 + y 2 = r 2 and (x 2 +y 2 ) = r4 2 y2 x (x 2 + y 2 ) x 2 x 2 + y 2 (x 2 +y 2 ) = r 4. Hence x 2 = r4 x 2 + y 2 y 2 and x 2 = r4 y 2 (x 2 + y 2 ) x 2 + y 2 = r4 x 2 + y y2 x 2 2 x. Hence 2 x 2 + y2 x 2 = r4 x 2 x 2 + y and x 2 (x 2 + y 2 ) = r4 2 x 2 x 2 + y. Hence 2 x 2 = r 4 x 2 (x 2 + y 2 ) 2 and x = r2 x x 2 + y 2. Similarly, we can show that y = r2 y x 2 + y 2. Exercise 1: What is the image of (x 1) 2 + y 2 x 2 + y 2 = 1? = 1 under an inversion in 54

4 Y X Figure 7.3: Inverse of (x 1) 2 + y 2 = 1 in x 2 + y 2 = 1 Answer. Well, x = x and y = y. Hence x x = = 1 and x 2 +y 2 x 2 +y 2 x 2 +1 (x 1) 2 2 y = y. Hence the image of a circle going through the center of the circle of 2x inversion is a line going through the points of intersection of the two circles. Exercise 2: What is the image of x = 1 2 under an inversion in x2 + y 2 = 1? Y X Figure 7.4: Inverse of x = 1 2 in x2 + y 2 = 1 Answer: (x 1) 2 + y 2 = 1 Exercise: What is the image of x = 1 under an inversion in x 2 + y 2 = 1? Answer: (x 1/2) 2 + y 2 = 1/4. Exercise 3: What is the image of x 2 +2x+y 2 = 0 under an inversion in x 2 +y 2 = 1? Answer: x = 1/2. 55

5 Theorem If two circles are orthogonal, (their tangents at the points of intersection are perpendicular), and if a diameter AB of one circle meets the other circle in the points C and D, then OP 2 = OC OD. P A O C B O D Figure 7.5: Orthogonal Circles are Inverses Proof. OP O is a right triangle, hence (OP ) 2 + (P O ) 2 = (OO ) 2. But OO = OC + CO = OC + P O, hence (OP ) 2 + (P O ) 2 = (OC + P O ) 2. Hence (OP ) 2 = (OC) OC P O, which implies that (OP ) 2 = OC (OC + 2P O ). Hence (OP ) 2 = OC OD. Theorem A circle orthogonal to the circle of inversion inverts into itself, and, a circle through a pair of inverse points is orthogonal to the circle of inversion. T P O P A Figure 7.6: Orthogonal Circles are Inverses Proof. Given the circle of inversion C (O,r) and an orthogonal circle centered at A. Let T be one of the points of intersection of the two circles. Now if a line through O meets this orthogonal circle at P and P then OP OP = OT 2 = R 2. Hence P and P are inverse points. Theorem If P P and Q, Q are pairs of inverse points with respect to some circle C (O,r), Then P Q = P Q r2 OP OQ. Proof. If O, P, Q are noncollinear then, OP = OQ and P OQ = Q OP, OQ OP hence OP Q OQ P. Hence P Q = OQ. Hence P Q = OQ P Q OQ = r2 P Q. P Q OP OP OQ OP OQ 56

6 Q. Q... O P P Figure 7.7: P Q P Q = r2 OP OQ 7.3 Applications of Inversion: Given three non-coaxial concurrent circles, construct a circle C tangent to all three circles. Figure 7.8: A Circle Tangent To Three Non-Coaxial Circles Solution: Invert the circles about a unit circle centered at the point of concurrency of the circles creating a triangle. Now construct the inscribed circle and invert this circle in the circle of inversion to create the required circle. Ptolemy s Theorem: In a cyclic convex quadrilateral, the product of the diagonals is equal to the sum of the products of the two pairs of opposite sides. Proof: Invert the circle and the convex quadrilateral about a circle centered at one of the vertices of the quadrilateral, say A. Now B D = B C + C D. Hence, r 2 BC AB AC + CD AC AD = BD AB AD 57 r2 r2

7 B B A C C D D Figure 7.9: Ptolemy s Theorem Hence, BC AD + CD AB = BD AC. Homework Find the image of the objects below under the specified inversion. (See Figures 7.10 and 7.11 ) 2. Prove that the inverse of the circumcircle C c of a triangle ABC with respect to the incircle C i, as a circle of inversion, is the nine point circle of the triangle XY Z determined by the points of contact of C i with the sides of ABC. 58

8 Figure 7.10: Inversion - HW 59

9 Figure 7.11: Inversion - HW 60

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