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

Chapter 7

Inversion

Goal:

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 7.0.1 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) = P0 then P0 ∈ OP and

(OP)(OP0_{) = r}2_{. Here P}0 _{is called the inverse of P, O is called the center of}

inversion, and r is called the radius of inversion, and r2 _{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 P0_{. 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 inversionC(O,r) and a pointP, how do you construct the inverse of P?

•

If

P

is inside the circle of inversion:

– Draw a perpendicular to OP at P. This intersects the circle of inversion in two points, label one of themQ.

– Connect the center of the circleO toQ.

– Draw a perpendicular toOQfromQ. The intersection of this perpendic-ular with OP is P0_{, the inverse of P.}

... ... ... ... ... ... ... ... ... .... .... .... .... .... .... .... .... .... .... .... . ... ... ... ... ...

O P P0

Q

..._... ..._...

..._...

Figure 7.1: Inverse of a Point Inside the Circle of Inversion

Note that ∆OP Q _∼ ∆OQP0_{. Hence} OQ OP =

OP0

OQ. Therefore, OP · OP

0 ₌

(OQ)2 _=r2_.

•

If

P

is outside the circle of inversion:

– Construct a circle centered atM of radiusM O. It intersects the circle of inversionC(O,r) in two points, Q and R and goes through O and P. – Construct the segmentQR. The intersection of QR and OP is the point

P0_{, the inverse of P.}

.... .... .... .... .... .... .... .... .... .... .... . P P0 M O...

... ... ... ..._... ..._.. ... ..._... ..._... Q R

Figure 7.2: Inverse of a point Outside the Circle of Inversion

Note that ∆OP Q _∼ ∆OQP0_{. Hence} OQ OP =

OP0

OQ. Therefore, OP · OP

0 ₌

•

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 linel. 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. DilateQ by the ratio r

OP centered atO. The image ofQ isP

0 _{the inverse}

image ofP.

9. Hide everything except the circle, its center, the point determining its radius,P0 _{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 C1 in the circle of inversion C(O,r),

replacel byC1 in the steps above.

7.2

Inversion Using Coordinates:

Theorem 7.2.1 An inversion about x2_+y2 _=r2 _{is given by}

(x, y)_→(x0, y0) = ( xr

2

x2_+y2,

yr2

x2_+y2)

Proof. Since (x, y),(x0_{, y}0_{) and (0,0) are collinear, we have} y0

x0 =

y

x. Now

d((0,0),(x, y))_·d((0,0),(x0_{, y}0_{)) =r}2_{, hence}√_x2 _+y2

·√x02+y02 =r2and (x2+y2)· (x02_+y02_{) = r}4_{. Hencex}02 ₌ r4

x2_+y2 −y

02_andx02 ₌ r4−y02(x2+y2)

x2_+y2 =

r4

−x02xy22(x2 +y2)

x2_+y2

= r

4

x2_+y2 −

y2_x02

x2 . Hencex

02₊y2x02

x2 =

r4

x2_+y2 and

x02_(x2_+y2₎

x2 =

r4

x2_+y2. Hence

x02 = r

4_x2

(x2_+y2₎2 and x

0

= r

2_x

x2_+y2. Similarly, we can show that y

0

= r

2_y

6

-_X

Y

Figure 7.3: Inverse of(x−1)2+y2_{= 1in x}2_+y2_{= 1}

Answer. Well, x0 ₌ x

x2_+y2 and y

0 ₌ y

x2_+y2. Hence x

0 ₌ x

x2_+1−(x−1)2 =

1 2 and

y0 ₌ y

2x. Hence the image of a circle going through the center of the circle of inversion is a line going through the points of intersection of the two circles.

Exercise 2:

2 under an inversion in x

2_+y2 _{= 1?}

6

?

-6

X Y

Figure 7.4: Inverse ofx=1

2 inx2+y2= 1

Answer: (x−1)2+y2 = 1

Exercise:

Exercise 3:

Theorem 7.2.2 If two circles are orthogonal, (their tangents at the points of

in-tersection are perpendicular), and if a diameter AB of one circle meets the other

circle in the points C and D, then OP2 _=OC

·OD.

O

P

A C B D

O0

Figure 7.5: Orthogonal Circles are Inverses

Proof. ∆OP O0 _{is a right triangle, hence(OP)}2_{+ (P O}0₎2 _{= (OO}0₎2_{. ButOO}0 ₌

OC +CO0 _{=OC +P O}0_{, hence (OP)}2_{+ (P O}0₎2 _{= (OC +P O}0₎2_{. Hence (OP)}2 ₌ (OC)2 _{+ 2}

·OC ·P O0, which implies that (OP)2 = OC ·(OC + 2P O0). Hence (OP)2 _=OC

·OD.

Theorem 7.2.3 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.

P O

T

1

A P0

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 P0 _{then OP}

·OP0 =OT2 = R2. Hence P and P0 _{are inverse points.}

Theorem 7.2.4 IfP P0 _andQ,Q0 _{are pairs of inverse points with respect to some}

circle C(O,r), Then P0Q0 =P Q r

2

OP OQ.

Proof. If O, P, Q are noncollinear then, OP0

= OQ0

P P0

Q

.. .. .. .

.. .. .. .. .. .. .. .. .. .. .. .. .. .

Q0

O

Figure 7.7: P_{P Q}0Q0 = 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 concur-rency 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

diag-onals 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 B0_D0 _{= B}0_C0_+C0_D0_.

Hence,

BC r

2

AB_·AC +CD

r2

AC_·AD =BD

r2

6

?

A

B

B0

C _C0

D

D0

Figure 7.9: Ptolemy’s Theorem

Hence, BC_·AD+CD_·AB =BD_·AC.

Homework

2. Prove that the inverse of the circumcircle Cc of a triangle ∆ABC with respect to the incircleCi, as a circle of inversion, is the nine point circle of the triangle ∆XY Z determined by the points of contact of Ci with the sides of ∆ABC.

6

? 6

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6

?

-6

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-?

6

6 6

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