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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 Write your conjectures, please! 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 OXA 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 Write your 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 ∡OXM = ∡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 ∡OYM = 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 radius and repeat the investigation. Text 24 Conjectures Write your 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 ∼ △OAX, hence ∡OXA = ∡OAX. (ii) △OXB ∼ △OBX, hence ∡OXB = ∡OBX. Further we have ∡AXB = ∡OXB - ∡OXA = 90°. Then ∡BXA = ∡OBX - ∡BAX = ∡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 ∼ △OAY and △OYB ∼ △OBY 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 Write your 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 course ABD = BAD. Hence (t1, t3) = CAD = 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 Write your 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. Right Click on the slider and switch “Animation On”

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