# text_inverse

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```					INVERSION 1

Definition

Let k0 be a circle with centre at the point O and radius
r>0.
For an arbitrary point X≢O denote by X the point

which        lies        on       the        ray     OX
and                 or                   .
The point X  is called (invers) image of the
point X under inversion with centre O and
radius r (or inversion with regard to the circle k 0 ).
The circle k0 is called a circle of inversion.
The inversion with centere O and radius r will be
denoted by (O,r), or only by .
For the image      of the point X under the inversion
we shall write X = (X).

Exploration

Let k0 be a circle with center O , X ≢ O and X  =
ι(X).
Study the behavior of the point X  when the point X is
inside, outside or on the circle k0 .
Change the position of point X and investigate how the
position of the point X  depends on the position of X.

Conjectures
The inverse image of a point which is inside the circle of
invesrsion is
……………………………………………………
The inverse image of a point which is outside the circle of
invesrsion is
……………………………………………………...
The inverse image of a point which is on the circle of
invesrsion is
Consequences          from the definition:
   Every point in the plane, except the center O , has a
unique inverse image.
 If OX < r, than OX  > r ;              if OX > r, then
OX                          <                      r.
This means that the images of the interior point are
exterior points and vice versa.
 If X k 0 (OX = r) then X= X  , This means
that each point on the circle of inversion is the own
inverse.
 Let X  = (X) and                  =               

and                . Hence            , i.e.
X                    =                     .
Therefore if     is the inverse of X then X
is the inverse of .

Sketch

is inside the circle of inversion k 0 .
AB is a chord,          AB and AB ⊥ OX.
Construct the tangent to k 0 at the point A .
Denote by Y the intersecting point of the tangent and the
line OX .
Construct the point X = (X) .
Which is the relationship between X and Y ?

is outside the circle of inversion k 0 .
Construct the tangents from the point X to the circle k 0 .
Mark the tangent points A and B , then construct the
segment AB .
Denote by Y the intersecting point of the lines OX and
AB .
Construct the point X = (X) .
Which is the relationship between X and Y ?

The triangles OAX and OX  A are similar (why?).
Then                                               .

The triangle ABX is isosceles ( AX=BX ), and AB⊥
OX. Then the triangles OAX and OXA are similar,
hence                  .

It is evidence from the foregoing that to any figure in the
plane corresponds a second figure, such that the
corresponding points of both figures are mutually inverse.
Our goal is to ascertain which properties of a figure remain
unchanged in the inverse figure. For this purpose we need
the following lemma.

L e mma . If      and are any points in the plane and
and     are their inverses respectively, then the triangles
and           are similar.
Proof. From                              we have
the proportion.
Moreover                                     Hence 

From the Lemma it follows that the distance between two
points and the distance between their inverse images are
connected by the expression                  .
It should be kept clearly in mind that as long as the pairs of
points
(A,A’) and (B,B’) are mutually inverse, the other points
on the line-segments AB and A’B’ are not necessarily
mutually inverse.
INVERSION 2

Exploration
Let k0 be a circle with center O , l - a straight line and X
- an arbitrary point on l .
Study the behavior of X  when X is moving along l in
both directions.
So you are exploring what will be the inverse image of
a straight line.

Sketch
Reset construction and construct three points on the line l .
Construct their inverse images. Construct a circle c passing
through the inverse points.

Investigation
Drag the point X again and compare the trace of the point
X  with the circle c .
Does the circle c pass through the point O ?
Change the position of the line l (moving the point A ) and
repeat the investigation.
Check box to see the line a , passing through the centre of
inversion.
Make the similar investigations.

Conjectures
The inverse image of a straight line, not passing through
the         centre        of          inversion         is
…………………………………………..
………………………………………………………………
…….
The inverse image of a straight line, passing through the
centre                   of                      inversion
…………………………………………………….
………………………………………………………………
…….
Theorem 1.
a) A straight line passing through the center of
inversion is inverse to itself.
b) The inverse of any straight line not passing
through the center of inversion is a circle passing
through that center.
Proof a). The proof is a direct consequence from the
definition.
Proof b). Let a be a line, O ∉ a, X ∈ a and X = (X).
M is the foot of the perpendicular from O to a and M =
(M).
Denote by k the circle with diameter the segment OM .
From the lemma we have ∡OXM = ∡OMX = 90° .
Hence the point X ∈ k .
Conversely, let Y’ k and Y is the point of intersection of
the
lines a and OY’ .
Since ∡OYM = 90° , 
OM.OM' r 2 .
Thus OY' =         =
OY     OY
This means that the inverse of the point Y is the point Y ∈
k , previously chosen.
Both reasoning show that the image of a straight line not
passing through the centre of inversion O is a circle
passing through O .

INVERSION 3

Text 19

Exploration
Let k0 be a circle with center O . Consider another circle
c with centre A . For an arbitrary point X ∈ c , study the
behavior of X  =  (X) when X is moving along c .
In this way are exploring which will be the shape of the
inverse of an arbitrary circle.
Change the position of the circle c (dragging the point A)
and repeat the investigation.
Fig 8

Text 20

Sketch
Here you see a circle k 0 with centre O . Let the circle c
with center A , passes through O and X  c is an arbitrary
point.
Moving the point X around the circle c , study the behavior
of the point
Change the position of the circle c (dragging the point A )
and repeat the investigation. X  =  (X) .

Fig 9
Text 21

Construct two points U and V on the circle c and their
inverse images U’ and V’.
Construct a line l passing through U and V’ .
Move the point X again and compare the trace of the point
X’ with the line l .
Change the position of the circle c ( dragging the point A )
and repeat the investigation.

……………………………………………………………..

Text 22

Here you see a circle k 0 with centre O . Let the circle c with
center A does not pass through O and X  c is an
arbitrary point.
Moving the point X around the circle c , study the behavior
of the point X  =  (X) .
Change the position of the circle c (dragging the point A )
and repeat the investigation.

Fig 10
Text 23

Construct three points U , V and W on circle c and their
inverse images U’, V’, W’ .
Construct a circle d , passing through U’ , V’ and W’ .
Move the point X again and compare the trace of the point
X’ with the circle c’ .
Change the position of the circle c ( dragging A ) and its

Text 24

Conjectures

Text 25

The inverse image of a circle, which is passing through the
center of inversion is
bnbn………………….….………………
The inverse image of a circle, which does not pass through
the center of inversion is
…………………………………………...

Text 26

Theorem 2.
a)    The inverse of any circle passing through the
center of inversion is a straight line not passing
through that center;

b)     The inverse of any circle not passing through
the center of inversion is a circle. The center of
inversion and the centers of the two mutually inverse
circles are collinear.

Text 27

Proof. a) Since X = (X) the statement follows directly
from Theorem 1 b).
Proof. b) Let k be a circle not passing through the center
of inversion O .
Denote AB this diameter of k for which the point O lies
on the straight line AB .
Denote by c the circle with diameter the points A = (A)
and
B = (B) .
For an arbitrary point X ∈ k, X =  (X).
Following the Lemma we have:
(i)    △OXA ∼ △OAX, hence ∡OXA = ∡OAX.
(ii)   △OXB ∼ △OBX, hence ∡OXB = ∡OBX.
Further we have ∡AXB = ∡OXB - ∡OXA = 90°.
Then ∡BXA = ∡OBX - ∡BAX = ∡OXB - ∡OXA =
90°.
This result shows that the point X lies on the circle c.
Now let Y ∈ k. Denote by Y the point for which  (Y) = Y.
Using the fact that
△OYA ∼ △OAY and △OYB ∼ △OBY
we conclude that ∡AYB = 90°. Hence the point Y lies on
the circle k. This means that the inverse of the point Y is the
point
Y  k, previously chosen.

Note, however, that the image of the centre of the initial
circle does not coincide with the centre of the image circle.

Conjectures
The inverse image of a circle, which is passing through the
center of inversion is
bnbn………………….….………………
The inverse image of a circle, which does not pass through
the center of inversion is
…………………………………………...

Fig 11

INVERSION 4

Problem
Let τ (O, r) be an inversion. Construct the inverse of a
circle k passing through O.
Solution. We consider two cases.
1) Let the circle k intersects the circle of inversion k0 in
the points P and Q. By Theorem 2a) the inverse of k is a
straight line k’. Since P and Q lye on k0, they are self-
inverse points and therefore lye on k’. Hence k’
coincides with the line PQ.
2) Let the circles k and k0 have no common points. If OA
is a diameter of k, denote by A’ the inverse of A. Then
the straight line k’ through A’, perpendicular to OA is
the inverse of k.

Fig 12
Text 30

Let line l intersects the circle c in the points A and B and a
and b are the tangents to c in A and B respectively. If a and
b are not parallel, denote by C their point of intersection.
Then AC=BC, i.e. the triangle ABC is isosceles. Hence
ABC=ACB.
So we can give the following definition:
If L be a common point of the line l and the circle c.
Denote by t the tangent to c at the point L.
The angle between l and c - (l, c) - is defined as the
angle between l and t - (l, t).

This is then the defined angle between a straight line and a
circle, L could be either A or B with the same result.
Fig 14

Consider two circles c and k intersecting in the points A and
B. Denote by t1 and t2 the tangents to c at the point A and B.
They intersect in the point C. As we have seen earlier
ABC = BAC. Analogously, t3 and t4 are the tangents to
k at the points A and B with intersection point D and of
CBD =  (t2, t4).
Let L is a common point of the circles c and k and
tc and tk are the respective tangents to the circles in
that                                             point.
The angle between c and k -  (c, k) – is defined as
the angle between tc and tk -  (tc, tk).

We have seen that the inversion changes the shape of
certain figures. Let we have two intersecting figures. In the
point of intersection they form some angle. We shall find
that the images of the two figures form the same angle in
their point of intersection.

Exploration
Let k0 be a circle with center O and radius r . Consider
two objects k and l with common points. Their images with
regard to kO are k and l respectively. Study the connection
between
∡(k, l) and ∡(k, l).
1. k and l are two straight lines;
2. k and l are straight line and circle;
3. k and l are two circles.
Changing the position of k and l compare the angles
between the initial figures and their images.

Conjectures
………………………………………………………
If a circle or a straight line is orthogonal to the circle of
inversion………………………………………………
………………………………………………………..

If a circle or a straight line is orthogonal to the circle of
inversion………………………………………………
………………….….……………… …………………..
………………………………………………………….

Theorem 3.
Let k and l are two figures (straight lines and/or circles)
with at least one common point and k and l are their
inverses respectively.
Then (k,l) = (k ,l) .

The proof of this very impotent theorem is quite difficult and we omit
it.

The following useful theorems indicate what changes and
simplifications of a figure can be effected by inversion.

Investigation
You see a configuration of congruent circles, such that each
one is tangent to their neighbors and the two common
tangent lines to these circles.
The circle k0 with center O is the inverse circle.

Change the position and the radius of k0 and study the
behavior of the inverse images.

INVERSION 5
Exploration
The following useful theorems indicate what changes and
simplifications of a figure can be effected by inversion.
In view of their complicity we are not going to include the rigor proofs
of these theorems, but rather to verify them experimentally.

If a circle or a straight line is ortogonal to the circle
of inversion, it is self-invers.

You see a configuration of congruent circles, such that each
one is tangent to their neighbors and the two common
tangent lines to these circles.

You see two tangent circles and a chain
of tangent circles into crescent-shaped space
between the circles. An inversion with the
center at the common point of tangency of
the two circles maps the two circles onto two
parallel lines and the inscribed circles onto a
chain of equal circles inscribed between the
two parallel lines.

Two or more circles, tangent at a point, can be
inverted into parallel lines if the center of inversion
is at the point of tangency.

Two circles that do not intersect can be inverted
into concentric circles.

Steiner’s porism

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.

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