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0CJOJQJ^JaJmH sH "hvdCJOJQJ^JaJmH sH "hU*CJOJQJ^JaJmH sH "hQCJOJQJ^JaJmH sH "h=CJOJQJ^JaJmH sH "hr)\CJOJQJ^JaJmH sH (hEhECJOJQJ^JaJmH sH "h{CJOJQJ^JaJmH sH *v . 7 8 ? F K N Q ^ a p ǣzhǣ"hUPCJOJQJ^JaJmH sH (hEhUPCJOJQJ^JaJmH sH U"hU*CJOJQJ^JaJmH sH "hvdCJOJQJ^JaJmH sH "hYCJOJQJ^JaJmH sH "hO=CJOJQJ^JaJmH sH "hhorCJOJQJ^JaJmH sH (hEhECJOJQJ^JaJmH sH (outside of orb S. The theory of polars allows expanding of the proved also on the case when O lies inside or on surface of orb S. If O lies inside S, there is a plane which is not intersecting an orb and which is named as polar of O (it possesses the property that O lies on the polar to each point of this plane). If the Atransformation center lies on it, all the necessary properties are valid. (If the Atransformation center is not in this plane, Atransformation does not transform geometry "point" to "points"). If O lies in orb S its polar is the plane tangent to S at O. I leave to readers own to prove that Atransformations with the centers in this plane maps "straight lines" of geometry to "straight lines" and "points" to "points". Though in this case point O does not set any Atransformation, we can connect with it the map transforming all points of orb S to point (O(X)=O for all X on the orb). And then it is possible to say that B the Atransformation center lies on the polar to O only when O(X) commutes with B(X).
Also Atransformations keep angles between geometry "straight lines". After all Atransformations is an inversion of orb S (real or imaginary, see paper 4), and the inversion keeps angles between circles. "Straight lines" of geometry is a special case of circles, therefore, Atransformation keeps angles also between them. So Atransformations:
map straight lines of a geometry to straight lines,
keep angles between them,
are involute, A(A(X))=%.
In case of geometries of Riemann or Lobachevsky it is enough first two conditions to state that Atransformations keep distances between points and are motions of plane. In the case of geometry of Euclid geometrical similarity also keeps angles between straight lines, but it does not keep distance between points and, consequently, is not a geometry motion. But the similarity is not involute! With regard to item 3 also in this, the Euclidean case, Atransformation is a motion of a plane (Euclidean). More precisely speaking symmetry.
In what is the symmetry? Let B is an arbitrary point on polar O, symmetry to what sets Atransformation with the center at B? If B is outside of the orb, Atransformation with the center at B determines at S the circle of the tangential cone with the top at B. If a point X lies on this circle, B(X)=%. From the theory of polars it is known, that this circle lies in one plane with O and, consequently, is "straight line" of the geometry. Just the symmetry in this straight line is set by Atransformation with the center at B. If point B lies on polar O and is inside S we consider a straight line (B,O). It intersects orb S at two points, this pair points determines geometry "point". Just the symmetry in this "point" of geometry is set by the Atransformation with the center at point B.
Let's specify more that in the case when O lays outside of orb S (a case of geometry of Lobachevsky) the cut set of orb S by the plane polar to O gives known model of CayleyKlein. In this model a point is an intersection of straight line of the sheaf intersecting orb S with , and a "straight line" the intersection of a plane of the sheaf intersecting the orb with .
In my opinion, the produced model of different geometries is nice both the visualization and "homogeneity" (different geometries turn out as a result of simple movement of point O outside of, on surface and inside the orb). Therefore, I think they are convenient for using at education. The model can be studied further.
Circles in different geometries.
How does within the limits of the produced model the circle look? We define a circle a little unusually. Let point P is the center of the circle and point Q lies on it. I name as circle a set of points where can be transformed Q after symmetries in various straight lines passing through %.
Drawing 1.
(Points P and Q; some straight lines L1, L2, L3 passing through P; images of point Q under symmetry in these straight lines L1(Q), L2(Q), L3(Q))
Let's determine what it means for our model. Point P is represented in our model by two points 1 and 2 lying on one straight line with O. Atransformation leaving pair ( 1, 2) motionless is:
1. either the Atransformations which centers lie on straight line ( 1, 2); then points 1 and 2 are swapped;
2. or the Atransformations which centers lie on a plane tangent to 1 and on a plane tangent to 2, i.e. on intersection of these planes. These Atransformations map point 1 in 1 and 2 in 2.
On the other hand, we interested only in those Atransformations which lie on , polar of O. Just they interest determine symmetries of geometry. It is known from the theory of polars that two planes described in paper 2 are intersected at a straight line lying on , polar of O. Let s designate this straight line as L. Point Q of geometry is represented on an orb by the pair of points Q1 and Q2 (lying on one straight line with O). Draw through L and Q1 a plane, its intersection with S gives a set of points on S where can be mapped Q1 under the influence of Atransformations with the centers lying on L. The intersection of an orb and a plane is a circle. Therefore, the geometry circle is represented by a circle on an orb (more precisely, by a pair of circles, after all we can draw a plane also through L and Q2). We have advanced the reasoning suitable for all three cases of disposition S and O, i.e. for geometries of Riemann, Lobachevsky and Euclid simultaneously. The case described in item 1 corresponds to point symmetry (in point P).
Not to be confused in pairs of points (representing a geometry point), it is possible by some or other convenient mode to consider only a part of orb in which lies one of the representative of the pair. For example, if O is outside of and above the orb we consider only that part of the orb which lies over polar O closer to the O.
This model has a shortcoming. It is threedimensional. But we study flat geometries of Riemann, Euclid and Lobachevsky. Therefore now I give an account of flat model homogeneously enveloping all three cases.
Flat model of various geometries.
The flat model begins so simply that it is even amusing. We take any three circles A, B, C. We name them as "straight lines" of a geometry, and pairs of points of their intersection as geometry "points". Also we name all circles lying in bundles formed by these three as geometry "straight lines". Pair of intersection points from the resulting set we name as geometry "point", any circle which is passing through this pair of points as "straight line" of the geometry, also any circle from a bundle set by two circles from the set is also "straight line" of the geometry. (Exactly circles we include in "straight line" of the geometry, not imaginary inversions.)
What the geometry? It depends on a disposition of initial circles A, B, C. If they are the Riemannian (i.e. one of three circles separates intersection points of two others), the geometry also turns out Riemannian. If they are the Euclidean (i.e. all three intersect at one point), the geometry also turn out Euclidean. The same is for Lobachevsky's three circles (i.e. neither circle does not separate intersection points of two others). (See paper 6 about a circle orthogonal to three given.)
Before all consider the case of Euclidian disposition of three circles A, B, C. Let O is a point in which all of them are intersected. It is easy to see that any circle from the set constructed by the mode specified above passes through . (And vice versa any circle which is passing through O could be got from A, B, C by the described mode.) Let s make any inversion with the center O. Under it all circles of the set ("straight lines" of the geometry) are mapped to straight lines of a usual Euclidean plane. Geometry "points" are the pairs of points (%,O) where X is any point of the plane. Under the inversion O is mapped to infinitely remote point, and X to any other point of the plane. We can operate with this pair as with one point because the second point in all pairs is identical and infinitely far. We have received a usual Euclidean geometry.
What circles intersected at O, represent parallel straight lines? Since parallel straight lines do not have intersection point (except infinitely remote, and it is represented by point O), O should be a unique common point for such circles. It is possible only if they are tangent to each other at O. So, parallel straight lines are represented by circles tangent to each other at point O.
A(O)=(O)=!(O), and any circle which is passing through O leaves O motionless under inversion. Therefore no composition of inversions could map any other point X to O. This is the property of infinitely remote point. No motion could map to it and it is motionless under all motions of a plane.
In the case of the Riemannian disposition of three initial A, B, C any two circles representing "straight lines" of the geometry are intersected, so straight lines of Riemannian geometry behave. If taken A, B, C are Lobachevsky's circles, there are lot of noneintersected "straight lines" of the geometry among the circles representing them, that it is peculiar to Lobachevsky's geometry. In the cases of Riemannian or Lobachevsky dispositions of three initial circles A, B, C there is orthogonal to all of them circle I. In Lobachevsky's case I is the real inversion which has a motionless circle, in the case of Riemannian disposition I is the imaginary inversion. All circles got by the described mode representing "straight lines" of the geometry are orthogonal to I. It follows from the fact that all circles from bundles (A,B), (B,C), (A,C) are orthogonal to I, since I is orthogonal to A, B, C, and pairs of intersection points of the circles representing "straight lines" are conjugated under I. (see paper 6 and paper 2).
If the radius of circle I is very small or we are placed very far from this circle, properties of geometries of Lobachevsky and Riemann are similar to properties of geometry of Euclid. After all very small circle is almost a point. Or, if far to leave from a circle, it looks like a point too.
Connection of flat and space models.
Let's connect this model with constructed before space model. The space model is reduced to getting out of the Atransformation with the center at O of orb S. Then the set of Atransformations commuting with O(X) is considered; they lie on a polar to O. Atransformations are inversions of an orb. Choosing the Atransformation with the center O means simply choosing the inversion; and the Atransformations commuting with this inversion are the inversions commuting with the given point . Therefore two models are equivalent. It would become absolutely obvious, if I begin construction of the flat model so: we name circles orthogonal to the given inversion I as straight lines of the geometry; if I is an imaginary inversion, it is the geometry of Riemann; if real Lobachevsky's geometry. This mode would be worse for two reasons. 1. The geometry of Euclid falls out. 2. The circle I is absolutely not necessary for proof of many theorems.
As well as in space model I consider, what is circle within the limits of the offered model. For this purpose again we use the definition of circle given on fig. 1. "Straight lines" of geometry which are passing through "point" P of the geometry are the circles which are passing through the pair of points 1 and 2; we search how do inversions through all these circles act at a "point" of the geometry. A "Point" is a pair of points Q1 and Q2, but it is quite enough to us to trace an action on one of these points, for example Q1.
Drawing 2.
(Two intersected circles F and , points of their intersection P1 and 2, point Q1, circle " which is passing through 1 and orthogonal to F and H)
The circles which are passing through 1 and 2 form the bundle. We spend through Q1 circle ", orthogonal to any two circles of this bundle, e.g. given F and ; on properties of bundles it is orthogonal to all circles of bundle (F,H), hence, under inversions through circles from the bundle point Q1 moves along circle T. Therefore, circle of the geometry is represented by the circle on the plane. Our reasoning envelops at once all three possible cases (Riemann, Euclid and Lobachevsky).
But, in case of Lobachevsky's geometry not any circle on a plane is a circle of geometry of Lobachevsky. Let two circles representing "straight lines" of the geometry of Lobachevsky are not intersected.
Drawing 3.
(Two noneintersected circles F and , the bundle centers (F,H) points O1 and O2; point Q1; circle " orthogonal to F and and passing through Q1)
Circle " orthogonal to F and does not represent any circle of the geometry of Lobachevsky since it does not have "center" in this geometry. The centers of imaginary bundle (F,H) O1 and O2 lie on I the circles orthogonal to all circles representing "straight lines" the geometry of Lobachevsky. Since " passes through O1 and O2, " necessarily intersects I such circles do not represent circles of the geometry of Lobachevsky.
Let's notice one more good property of this model: the angle between the circles representing "straight lines" coincides with the angle between the straight lines, which these circles represent.
Fruitfulness of flat model.
In my opinion this flat model is rather convenient for proofs of theorems of all geometries Riemann of Euclid, Lobachevsky. Especially the model is fruitful for proofs of theorems of triangles; they are reduced to the proof of certain properties of three circles A, B, C.
The distance between "points" of geometries needs to be studied and defined specially (on the basis of the harmonious ration, paper 4, 5), but angles between "straight lines" simply are angles between circles representing them. Therefore, it is easier to prove theorems about angles, bisectrixes, etc. I mention two examples:
1. The sum of angles of a triangle of different geometries.
2. Triangle bisectrixes intersect each other at one point in all three geometries.
The sum of angles of a triangle or angles in tricircle5.
In geometry of Euclid.
Drawing 4.
(Three circles A, B, C intersected at one point O. The smallest circle, C, lies inside circles A and B; points of intersection A with B O and , B with C O and Q, A with C O and P; the identical angles which form a semicircle at point O are marked)
The second intersection points of circles A, B, C among themselves (it is possible to discard the first point O since all straight lines and circles representing them are intersected at it) are the tops of the Euclidean triangle. Triangle PQH is represented in model by three arcs PQ, QH, and HP. Now we use the fact that circles are intersected at two points and angles are identical in both points. We see that angles between corresponding arcs converge at point O and there supplement each other (without intersections) to the angle of 180 degrees. Hence, the sum of angles between the specified arcs is equal to 180 degrees and the sum of angles of the triangle too. Q.E.D. This proof does not demand carrying out of any auxiliary lines. And at the usual proof it is necessary to draw a straight line through one of the tops parallel to the third leg.
In geometry of Riemann the sum of angles of a triangle always is more than 180 degrees.
Drawing 5.
(Three Riemannian circles A, B, C; circle D passing through those intersection points of A and B with C which lie inside A and B and through intersection point of A with B which lies outside of !)
Let's draw three Riemannian circles A, B, !. As legs of the triangle we take the arcs limiting the area lying inside all circles. We draw circle D through intersection point of A with B and through intersection points of C with A and C with B. Circles A, B, D are intersected at one point, what means on proved earlier that the sum of their angles is equal to 180 degrees. But circle D passes inside the arcs making the triangle and through its tops, therefore, the sum of angles between D, B, A always is less, than the sum of angles between A, B, C, therefore, the sum of the angles formed by arcs , ! and ! is more than 180 degrees. Q.E.D. (In my opinion this reasoning cannot be understood without a drawing, but even in bad drawing it is transparent enough; the main thing not to be mistaken in orientation of angles and not to get confused with basic and complementary angles).
In Lobachevsky's geometry the sum of angles of a triangle always is less than 180 degrees.
Drawing 6.
(Three circles of Lobachevsky A, B and !; B is placed inside A and C; auxiliary circle D passing through intersection point of A with C and intersection points of B with A and B with C which lie more close to another intersection point of A and C)
In this case we consider as a triangle three intersection points A, B, C among themselves which lie further from that intersection point of A with C through which passes D. In this case D does not enter into the triangle though pass through its two tops. Therefore, the sum of angles of tricircle , B, C is less than sum of angles tricircle A, B, D, and the last is equal to 180 degrees since A, B, D are intersected at one point. As well as in the previous case the reasoning is very transparent even at bad drawing.
In the cases of Riemann or Lobachevsky there is a problem of a choice of three points which can be considered as triangles tops. Also it is not clear what arcs to consider as legs. This problem is reduced to the following. A geometry point is represented by a pair of intersection points of circles. How to choose from this pair one point? For this purpose it is convenient to to use the circle I orthogonal to A, B, C. If I is the real circle and we deal with Lobachevsky's geometry, the pair conjugated through it points is separated by this circle. We can choose as the representative the point of pair lying inside circle I. Then all points of the geometry are the points inside circle I. If I is the imaginary inversion we can choose any real circle orthogonal to I. Under the influence of I this circle is turned inside out; it separates conjugated through I points; and, again, we can choose as the geometry points the interior of this circle. However, not in all theorems there is a necessity to use it.
Though theorems becomes easier to prove because the circles are intersected at two points and this increases an amount of equal angles, it is necessary to be careful because of considering orientation of angles and distinguishing direct and additional angles.
Isogonal circles.
Before the proof of the theorem of bisectrixes we study properties of isogonal circles.
Let two circles A and B are taken. Circle C is called as isogonal to A and B if the angle between A and C is equal to the angle between B and C. The particular case of isogonal circles is tangency of C to A and B or orthogonality of C to A and B. Considering of isogonal circles accustoms us to be attentive towards a disposition of angles and to notice their orientations. We will prove that all circles isogonal to two given A and B are divided into two sets. One set is orthogonal to one bisectrix between A and B and the second to another bisectrix. Circles orthogonal to A and B belong to both sets at once. If A and B are tangent, they have only one bisectrix. Isogonal to them circles are also divided into two sets. The first is orthogonal to the unique bisectrix, and the second to all circles which are passing through the point of tangency of A and B. Circles orthogonal to A and B lie in both of these sets simultaneously.
The proof of the fact, that circles from these sets are isogonal to A and B, is trivial. Let I is any bisectrix between A and B, I(A)=B. Let C is orthogonal to I, I(C)=C, then the angle between A and C is equal to the angle between I(A)=B and I(C)=C, i.e. is equal to the angle between B and C, Q.E.D. If A and B are tangent and C passes through the point of tangency A and B:
Drawing 7.
(Two tangent at point P circles A and B; circle C which is passing through P; common tangential straight line at point P to circles A and )
Any circle C which is passing through P intersects A and B at the same angle as their common tangential straight line. It is similar to that any straight line intersects pair of parallel straight lines at one angle. Q.E.D.
It s remained only to prove that any circle isogonal to two given A and B belongs to one of the described sets. At first we consider the case when A and B are not tangent. It is enough to prove, that on every circle C isogonal to A and B there is a pair of points conjugated with each other through any bisectrix between A and B.
If C is tangent to A and B, any circle which is passing through points of tangency is isogonal to A and B
Drawing 8.
(Circles A and B; circle C tangent to A and B; circle !1 which is passing through points of tangency of A with C and B with !)
Since C and A are tangent, !1 is isogonal to C and A; since C and B are tangent, !1 is isogonal to C and B. Hence, !1 is isogonal to A and B. Q.E.D.
Another proof. It has been proved in paper 3 that tangent circles A and B are orthogonal to one of bisectrixes between A and B, tangency points are conjugated through this bisectrix, hence, !1 passing through these points of tangency is orthogonal to this bisectrix. But just now we have proved that circles orthogonal to the bisectrix between A and B are isogonal to A and . Q.E.D.
Orientation and disposition of angles.
Before the advance it is necessary to produce simple properties of orientation (disposition) of angles. We begin with the geometry of straight lines. Let three straight lines , ", K are intersected at point O and we know the angles between 1 and " and between "1 and K. Then the angle between 1 and K is equal either the sum or a difference of these angles, depending on their orientation. I name as orientation a direction of rotation of point O at motion from one straight line to another on a known angle. The situation with circles intersected at one point precisely the same. But in the geometry of circles there is a phenomenon which is not present in the geometry of straight lines. Two circles and " have a second intersection point. And if one considers orientation of the same angle between and " at the second intersection point, it is opposite to the orientation at the first intersection point!
Drawing 9.
(Two circles H and T intersected at points O and O1; arrows show the direction of motion from to " near by the intersection points (arrows go on the smaller angle between and T))
Points near by point O rotate in one direction and near by O1 in another. Certainly, there is also a transitive case when these points move along a straight line, but the straight line does not have orientation! Therefore, speaking about angle orientation in the case of circles is necessary to add in what intersection point we determine it. We notice one more. If the angle between and " is equal to 90 degrees ( and " are orthogonal) there is no possibility to speak about angle orientation because this angle is equal to the complementary.
Let's return to isogonal circles. Let circle C isogonal to A and B intersects circle A at points A1 and A2, circle B at points 1 and 2. We want to prove that C is orthogonal to one of bisectrixes between A and B. We don t take interest in point A2 and consider only the angle orientation between A and C at point A1 and orientations of angles between B and C at points 1 and 2 equal to it. As the last two orientations are opposite each other one of them coincides with the angle orientation between A and C at point A1, and another is the opposite. We choose from points 1 and 2 that one in which angle orientation between B and C is opposite to the angle orientation between A and C at point A1. Let it is the point 1.
Let's draw through A1 and 1 arbitrary circle D and show that it is always isogonal to A and B.
Drawing 10.
(Circles A and B; circle C intersecting them at points A1 and 1 chosen as it has been described above; circle D which is passing through points A1, 1)
Let's consider angles between A and D, D and C, A and C, and their orientations at point A1. Let, for example, the angle orientation between C and D is the same as between A and C. Then the angle between A and D is equal to their sum. Now we consider angles and their orientations at point 1. The angle between B and C according to choice of point 1 has the orientation opposite to orientation of angle between A and C, the angle between C and D has the orientation opposite to those that has at point A1 (since circles C and D are intersected at points A1 and 1 and orientations of this angle are opposite in these points). So, the angle between B and D is equal to the angles between B and C and between C and D. Since the angle between B and C is equal to the angle between A and C, the angle between A and D is equal to the angle between B and D. Q.E.D.
If the angle orientation between C and D at point A1 is opposite to the orientation in the same point of the angle between A and C, the angle between A and D is equal to the angle difference between A and C and the angle between C and D. Corresponding orientations at point 1 both change to the opposite and, consequently, remain opposite each other. The angle between B and D, therefore, is equal to a difference between angles B with D and C with D. Hence, the angle between B and D is equal to the angle between A and D. Q.E.D.
Now we draw through points A1 and 1 circle D1 tangent to A at point A1. On proved D1 forms with B the same angle as with A, i.e. D1 is tangent to circle B at point 1. We have already proved (paper 3) that the circle tangent to A and B is orthogonal to one of their bisectrixes and points of their tangency are conjugated through this bisectrix. Hence, A1 and 1 are conjugated through the bisectrix between A and B, and all circles which are passing through these points are orthogonal to this bisectrix, hence, circle C also is orthogonal to it. Q.E.D.
If A and B are tangent and C is isogonal to them, than if C does not pass through common tangency point our reasoning is transferred without a damage (with small modifications). So, we have proved that circles isogonal to two given are orthogonal to one of bisectrixes or (if the given circles are tangent) pass through the point of their tangency.
The theorem of intersection of bisectrixes of three circles and nonEuclidean geometries.
Let three arbitrary circles A, B, C not tangent to each other are given. D1 is a bisectrix between A and B, and D2 is a bisectrix between B and C. The theorem states that one of two existing bisectrixes between A and C necessarily lies in the bundle formed by D1 and D2.
The proof. Let's consider the bundle of circles orthogonal to D1 and D2. All circles of this bundle are isogonal to A with B and to B with !. Indeed, let O is any circle of this bundle. D1(O)=O, D1(A)=B, hence, the angle between O and A is equal to the angle between O and . D2(O)=O, D2(B)=C, hence, the angle between O and B is equal to the angle between O and C. From these equalities follows that the angle between O and A is equal to the angle between O and C (transitivity of equality). It means that O is isogonal and to A and C. From proved follows that O is orthogonal to any bisectrix between A and C. Lets draw also O1 and O2 from the bundle orthogonal to D1 and D2; each of them is isogonal to A and C and, consequently, orthogonal to some bisectrix between A and C. Since among three circles O, O1, O2 everyone is orthogonal to one of bisectrixes between A and C, one of bisectrixes between A and C is orthogonal to two of these three circles, hence, this bisectrix lies in the same bundle that D1 and D2. Q.E.D. I leave for readers one to consider the case when there are tangent circles among A, B, C.
Let's return to our model of various geometries. According to it three circles A, B, C are straight lines of a geometry (it is not known what geometry Riemann, Lobachevsky or Euclid). Thereby we have proved that in any triangle of any geometry bisectrixes are intersected. However, it is required to establish what bisectrixes!
Having proved earlier that isogonal circles are orthogonal to a bisectrix, we have proved in terms of our model (of any geometry!) that in a triangle with equal angles at one of legs the bisectrix of the opposite angle is orthogonal to this leg. True, here again is required to specify what angles are equal.
So, we see that the offered model makes it possible to prove efficiently theorems of noneEuclidean (and Euclidean) geometries simultaneously.
Paper 8.
End of Appoloniuss problem and other problems on construction.
The paper summary.
In the paper reviewing of Appoloniuss problem comes to the end and singletype problems on construction of orthogonal and tangent circles are considered. The method of fast construction of isogonal circles (see paper 7) is produced.
It is studied when the composition of three inversions lying in one bundle can be the imaginary, and by means of it is found out in various cases how many circles tangent to three given exist. Also geometrical analysis of algebraic outcomes is given.
Returning to Appoloniuss problem (from that place as we have left it in paper 6)
Let's return to the problem about drawing of circle O tangent to three given A, B, C. On the condition O forms an identical (zero) angle with all three circles A, B, C, in other words, is isogonal to them (see paper 7). It follows from here that O is orthogonal to some two bisectrixes: D1 between A and B, and D2 between B and C (and according to the theorem of bisectrixes of paper 7 also to some bisectrix between A and C). For ones who havent read paper 7 I prove here that if O is orthogonal to D1 and D2, the angle between O and A is equal to the angle between O and B and between O and C.
The angle between O and A is equal to the angle between D1(O)=O and D1(A)=B, i.e. to the angle between O and B, the angle between O and B is equal to the angle D2(O)=O and D2(B)=!, i.e. between O and C. Q.E.D.
Circles, orthogonal to D1 and D2 form some bundle. Therefore the problem is reduced to determination in the given bundle the circle tangent to given. Before to solve this not so complicated problem (actually one of alternatives of its solution is already offered in paper 1), we consider a number of singletype problems on construction of tangential or orthogonal circles. Some of them are trivial, and some contain zest. As we are engaged in the geometry of circles initial operations are: inversion, drawing of circles through three points, determination of common points of two circles. We wont draw straight lines and discover the centers of circles for solutions of these problems.
Singletype problems on construction.
I number problems by letters of the Latin alphabet. But before to start the list I remind a solution of absolutely simple problem; so simple and important that it should be remembered separately. It was already written about in papers 3 and 4: drawing the circle which is passing through the given point and lying in the given bundle.
Solution. We take two arbitrary circles of a bundle. Draw two circles orthogonal to them, and then draw through the given point a circle, orthogonal to these two. It lies in the same bundle as two initial circles. Q.E.D. Note that if the initial bundle is the real one, it is simply possible to draw a circle through the given point and intersection points of two arbitrary circles of the bundle (that is, certainly, easier than the described construction), and if the bundle is tangent, it is possible to draw through the given point a circle tangent to two arbitrary circles of the bundle (drawing of tangential circles is not so simple). But if the bundle is the imaginary, the solution offered originally is the most convenient. Besides, it is convenient because solves the problem uniquely for bundles of any types.
So, problems on construction. Many problems from the list were already considered more than once. But it is convenient to collect them at one place.
A. Drawing of a circle orthogonal to given circle O and passing through two given points 1 and 2. Solution: draw the circle through 1, 2, O( 1).
B. The same, but 1 lies on O. Solution: draw the circle through 1, 2, O( 2).
C. Drawing of a circle which is tangent to O at given point 1 and passing through given point 2. Solution: draw circle O1 through 1 and 2, orthogonal to O (see item B), choose point 3 out of it and O and draw circle O2 orthogonal to O through points 3 and 1. Then draw O3 orthogonal to O2 and passing through 1 and 2. O3 is the required circle. Proof: O, O2, O3 pass through 1; O2 is orthogonal to O and O3, hence, O and O3 are tangent at 1. (For similar reason O1 is tangent to O2 and both of them are orthogonal to O.)
Drawing 1.
(All described above circles and points)
D. Drawing of a circle orthogonal to O and passing through 1 and 2 when 1 and 2 both lie on O. Solution. The method of items A and B does not work since now O( 1)= 1, O( 2)= 2 and there is no third point in order to draw circle through it and 1, 2. Therefore we construct at first any circle O1 tangent to O at point 1 (item C) and draw using item B circle O2 orthogonal to O1 through 1 and 2. Since O1 is tangent to O at point 1, O2 is the required. Algorithm: take an arbitrary point 3 which is not lying on O, draw circle K through 1, 3, O( 3); K is orthogonal to O. Take an arbitrary point 4 which is not lying on K and O and draw circle O1 through 4, K( 4), 1 O1 is tangent to O at point 1. Draw circle O2 through points 2, 1, O1( 2), and it is the required.
Constructions in this and previous items have useful analogy in the geometry of symmetries of straight lines: if we need to construct a straight line which is passing through the given point and a parallel to the given straight line, we can take an arbitrary point reflect it in the given straight line, draw through received pair of points a straight line, reflect in it the given point and draw through received pair of points a straight line which is the required line.
Now we pass to reviewing of bundles.
E. The real bundle and circle O are given. Need to draw a circle from this bundle orthogonal to the given circle O. Solution: since the real bundle is a population of the circles which are passing through two given points 1 and 2 the problem is already solved in items , , D. (Depending on, whether points 1 and 2 lie on O).
F. The imaginary bundle which both centers 1 and 2 do not lie on O is given. Need to draw a circle from this bundle orthogonal to O.
Solution.
Drawing 2.
(Circle O, circles O1 and O2, passing through points 1 and 2 and intersecting O; 1 and 2 lie at the same side from O)
We draw as it is specified in paper 6 circle O3 to orthogonal three Lobachevsky's circles O, O1, O2. Since O3 is orthogonal to O1 and O2, it lies in the imaginary bundle with the centers at 1 and 2. On construction it is orthogonal to . O3 is the required circle.
If O separates points 1 and 2, three circles O, O1, O2 are Riemannian circles and the imaginary inversion which does not have motionless circle is orthogonal to all of them. In this case there is a required inversion, but there is no required circle (or it is the imaginary circle).
G. The same, but exactly one of the centers of the imaginary bundle lies on O. Solution: in this case there is neither circle, nor inversion (real or imaginary) from the given bundle orthogonal to . (If not consider as we did in paper 5 center of the bundle as small inversion .) Proof. Let I is the required inversion, and P1 that center of the bundle which lies on by definition of an imaginary bundle I( 1)= 2. I(P1) should lie on O since I under the supposition is orthogonal to O. But on the condition only one center of the bundle lies on O, therefore 2 cannot lie on O. Contradiction. Hence, required I does not exist. Q.E.D.
H. The same, but O passes through 1 and 2. Solution: O is orthogonal to any circle of the given bundle.
I. Tangent bundle of circles and circle O are given. Need to draw a circle from this bundle orthogonal to O. Solution. Let A and B are two circles from the tangent bundle, P is the point of their tangency. The circle which is passing through O(P) and tangent to B at point P is the required (it is orthogonal to O since passes through points symmetric in O: P and O(P)). About construction of such circle see in item !. (If P lies on O, O is either orthogonal to all circles of the given bundle or neither.)
Now we solve problems about circles in bundles tangent to O. The same method will be used: it is discovered circle orthogonal to O and to all circles of the given bundle. Two circles of the given bundle which are passing through intersection points of O and are sought for. Let s recollect that Appolonius s problem brought us exactly to this problem.
J. Need to construct a circle lying in an imaginary bundle with centers P1 and 2 and tangent to the given O.
Solution.
1. Let neither 1 nor 2 do not lie on the O.
Drawing 3.
(Circle O, points 1 and 2, circle which is passing through 1 and 2 and orthogonal to O, intersection points of and O Q1 and Q2)
Circle O1 drawn through Q1 and belonging to the given imaginary bundle is the required one. Also circle O2 passing through Q2 and belonging to the bundle is the required. Thus there are two solutions of the problem.
The proof. Circle is orthogonal to all circles of the imaginary bundle with the centers at 1 and 2 (since passes through the bundle centers), hence O1, a circle of this bundle, also is orthogonal to H. But is orthogonal to O, and all three circles O, O1 and pass through point Q1. Hence, O is tangent to O1. Q.E.D. The same is about O2.
We have proved, that O1 and O2 are the required circles. Now we prove that there are no other circles satisfying to the condition. Let I belongs to the given bundle and tangent to O. Draw circle through the centers of the bundle and the point of tangency of O with I. Since passes through the bundle centers it is orthogonal I, and because are I and O tangent, is orthogonal to O. Hence, I passes through the intersection point of O with the circle orthogonal to O and to all circles of the given bundle. Such circle is unique. Q.E.D.
2. One of the bundle centers lies on . In this case there is only one circle tangent to O and belonging to the given imaginary bundle. I suggest modifying of the proof of item 1 by your own.
3. Both centers lie on O. Then O is orthogonal to all circles of the given bundle, hence there are no the circles in the bundle tangent to O.
The problem solution does not depend on that separates O 1 and 2 or not.
K. Need to draw a circle belonging to the real bundle with centers P1 and 2 and tangent to given circle O. Solution (when exists!). The solution is similar to previous case: we draw circle orthogonal to all circles of the bundle and O intersecting circle O at points Q1 and Q2. Circles O1 and O2, passing accordingly through Q1, P1, P2 and Q2, P1, P2 are the required ones.
1. 1 and 2 lie by one side from O.
lies in the imaginary bundle with centers P1 and 2. We draw using item F. The proof, that turning out O1 and O2 are unique, is similarly to the previous case.
2. 1 and 2 lie by different sides from . In this case the inversion orthogonal to the given bundle and circle O is imaginary (see item F), it does not have motionless points and, consequently, there are no intersection points with O. Problem has no solution. It is trivial to prove differently: if O separates 1 and 2, any circle which is passing through 1 and 2 intersects O and consequently is not tangent to it.
3. One of the bundle centers lies on . In this case we cannot draw required circle H. Nevertheless there is one and only one circle passing through 1 and 2 and tangent to O. See item C.
4. If 1 and 2 both lie on O the required circle does not exist.
L. A bundle of circles tangent at point P is given. Need to discover a circle of the bundle tangent to O. Solution. We draw circle , orthogonal to all circles of the given bundle as follows: through points P, O(P), O1(O(P)), where O1 is an arbitrary circle of the given bundle. It is orthogonal to O since passes through the pair of conjugated through O points, it is orthogonal O1 since passes through the pair of conjugated through O1 points, and since it passes through P and is orthogonal to one of bundle circles, it is orthogonal to all circles of tangent bundle. Further is similar to previous cases.
If the center of bundle P lies on O, then either O itself belongs to this bundle or is not tangent to any circle from it.
Our list is finished.
Construction of circles isogonal to three given and completion of Appoloniuss problem.
Last items settle Appoloniuss problem. Lets repeat: we choose two bisectrixes D1 and D2 between A and B and between B and C. Draw a circle from bundle (D1, D2) tangent to A (either B, or C). Such circles may be two. It is tangent automatically to all remaining since it is isogonal to all three circles. As these two bisectrixes can be chosen by 2x2=4 ways, altogether there are 4E2=8circles tangent to given A, B, C. How to comprehend geometrically the offered method of solution of Appolonius s problem?
Let's begin with simple but an interesting case: we search for a circle tangent to A, B, C which are tangent to each other. In this case there is only one bisectrix between any pair of initial circles. Since any of three circles is isogonal to two remained (is tangent to them) it is orthogonal to the bisectrix between two others. Therefore each bisectrix swaps points of tangency of the opposite circle with two others and leaves motionless the last point of tangency (since simply passes through it). This case of a disposition of points of tangency and bisectrixes is considered in the theorem of triple symmetry (paper 3). It follows from this theorem that all bisectrixes are intersected at the angle of 60 degrees.
It is possible to construct a bisectrix for example between A and B as a circle from the tangential bundle orthogonal to given C using item I of the list. Since the bisectrix is orthogonal itself to the opposite circle, it is enough to discover its intersection points with this circle and to choose from these pairs by a point and to draw two circles O1 and O2 tangent to A, B, C.
Drawing 4.
(Three circles A, B, C tangent to each other; three bisectrixes between them D1, D2, D3, intersected among themselves at two points; circle O orthogonal to A, B, C; intersection points of bisectrixes with circles A, B, C and two circles drawn through these points. These circles are tangent to A, B, C and conjugated through .)
We can search for a circle tangent to A, B and C in bundle (B,C). Such circles in this bundle are B and C. They do not give us new circles, and whether the circle is tangent to itself, is a problem more likely philosophical.
We see that in this case the offered algorithm of a solution of Appoloniuss problem works successfully and conveniently. Successfully it works always, but not always it is possible to name it "convenient". Let, for example, there are noneintersected circles among given. Then the imaginary bisectrixes appear. To discover the centers of the bundle set by imaginary bisectrixes is not very convenient. How it can be made? To discover the center of the imaginary bundle set by inversions (real or imaginary) S and T one can draw two circles orthogonal to S and T. Intersection points of these circles are the centers of bundle (S,T). We construct the orthogonal circle taking an arbitrary point X and drawing the circle through X, S(X), T(X) for this purpose we have no need to know if real or imaginary are S and T and to know the motionless circles of these inversions.
Let now S and T bisectrixes between A and B and between B and C. We choose point X on circle A. Then S(X) and T(X) can be discover by drawing circles O1 and O2 orthogonal accordingly to A and B; B and C through point X. Let O1 intersects B at points 1 and 2, O2 intersects C at points !1 and !2.
Drawing 5.
(Three not intersected circles A, B, C; point X on A; circle O1 which is passing through X and orthogonal to A and B; circle O2 passing through % and orthogonal to A and C; 1 and 2 points of intersection of O1 with B; !1 and !2 points of intersection of O2 with !)
Circles O1 and O2 are tangent to each other since they are orthogonal to circle A and both pass through point X lying on A. It is easy to show (paper 3, 6) that under the action of S point X moves along O1 and S(X) is equal to one of two intersection points of O1 with B 1 or 2. Specifying to what is equal S(X) we choose one of two bisectrixes between A and B (see paper 6). Similarly, specifying "(X) we choose a bisectrix between A and C.
Circle passing through X, 1, !1 is orthogonal to some two bisectrixes (one between A and B, the second between A and C) because the pair of points X, 1 is conjugated through the bisectrix between A and B and the pair X, !1 is conjugated through the bisectrix between A and C. Just so each of three circles: X, 1, !2; X, 2, !1; X, 2, !2 is orthogonal to some two bisectrixes between A and B and between A and C. It follows from here that all these four circles are isogonal to A, B and C (form with all of them equal angles).
To discover the center of the bundle formed by the bisectrixes between A and B and between A and C it is necessary to construct in each of four cases one more circle orthogonal to pair of the bisectrixes. For this purpose we choose point Z on A and do with it similar operations. We get four circles; each of them is orthogonal to some pair of bisectrixes between A and B and between A and C. But it is not always simply to specify what of these fours circles are orthogonal to the same pair. At least it cannot be named convenient.
It is possible to act in another way. Since A, B, C are not intersected there is a circle I orthogonal to all of them. It is orthogonal also to all bisectrixes. See paper 6 for method of constructing it. Intersection points of I with four constructed circles which are passing through X also are the centers of bundles of the corresponding pairs of bisectrixes. If some circle from the four does not intersect I, then the pair of bisectrixes corresponding to this circle is intersected forming the real bundle. So, we discover the centers of the imaginary bundles formed by bisectrixes.
The required circle, tangent to A, B and C, lies in the bundle orthogonal to the bundle of bisectrixes. Since in the considered case the bundle of bisectrixes is imaginary, the tangential circle lies in the real bundle. We use point K.
Small application of the theory of groups leads to the big simplification.
In all our constructions the main role is played by a circle from a bundle of two bisectrixes orthogonal to one of three given circles A, B, C. Now we express algebraically inversion which it realizes. Let S is a bisectrix between A and B, " a bisectrix between A and C. On the theorem of bisectrixes one of two bisectrixes between B and C lies in the same bundle as S and T. Designate it as H. On the theorem of inversions of one bundle any composition of S, " and is again an inversion (and belonging to the same bundle). We consider action of F=T*H*S on circle . S(A)=B, X(B)=C, T(C)=A (by definition of bisectrixes). Hence F(A)=A or FAF=A, i.e. F commutes with A (is orthogonal). Consecutive action of three bisectrixes returns A on the place. It happens, whatever three bisectrixes existing between A, B and C we choose. As we have chosen so that it belongs to bundle (S,T), F is an inversion (real or imaginary). F is orthogonal to A, i.e. if there is a motionless circle of inversion F (i.e. F is the real inversion), this circle is the required circle from the bundle of bisectrixes orthogonal to A; and through its intersection points with A passes tangential to A, B, C circle.
We can already make a useful conclusion: if F=T*H*S is the imaginary inversion, the required circle tangent to A, B, C does not exist because F and A have no intersection points, and F has no motionless points at all. See item K, part 2.
Let F is the real inversion. We use the equality F=T*H*S for determination of intersection points of F and A (using paper 5 terminology: F*A=F=T*H*S*A is the beplet symmetry with the extremities at required points) since F and A are orthogonal inversions. We draw circle I orthogonal to A, B, C (we assume that circles A, B, C are Lobachevsky's circles if they are Riemannian, construction of bisectrixes is not difficult and the centers of their bundles also are easy to discover; see paper 1). I is orthogonal also to all bisectrixes, hence I is orthogonal to F. Let A1 and A2 intersection points of I with A; since I and A are orthogonal to F, F(A1)=A2. Let X an arbitrary point on A. Since F(X)=T(H(S(X))) we can discover F(X) using the reception, similar to drawing 5.
Drawing 6.
(Circle A; circles I and F orthogonal to A and to each other; intersection points of I and A A1 and A2; intersection points of F and A F1 and F2; point X and point F(X); circle W orthogonal to A and passing through X and F(X); circle Z which is passing through points F1 and F2)
Circle W is orthogonal to F since passes through a pair of conjugated through F points. Circles W and I form a bundle orthogonal to F and A, it means that any circle orthogonal to W and I passes through intersection points of F and A. Draw this circle Z and we get required points F1 and F2 through which the circles tangent to A, B, C pass. Then we show how to build point F(X)=T(H(S (X))).
Drawing 7.
(Three not intersected circles A, B, C; point X on circles A; circle O1 orthogonal to A and B, intersecting B at points 1 and 2; circle O2 which is passing through 1 and orthogonal to B and C, it is tangent to circle O1 at point 1. Intersection points of O2 with C !1 and !2)
S (X) is either 1 or 2. We can choose an arbitrarily point from this pair, let it be 1. Draw through 1 circle O2, described in the delineation, orthogonal to B and C, and intersecting C at points !1 and !2. (S(X)) is either !1 or !2 we can choose arbitrarily. We draw now circle W through three chosen points X, (X) and S(H(X)). It is orthogonal to S and , therefore, is orthogonal to both T and F, hence, F(X) lies on W. But F(X) lies and on A, and because of this F(X) is the intersection point of W with A. If F(X)=X, X is the required point; we have discovered a motionless point; W is tangent to A and is the required circle tangent to A, B, C. If W is not tangent to A, the intersection point of W with A other then X is the required F(X). Construction is convenient by not demanding a choice of the third bisectrix and by not demanding as a matter of fact drawing bisectrixes. They are present only in proofs. Besides we have adduced the simple mode of construction of circles isogonal to given A, B, C.
Algorithm for Appoloniuss problem.
We summarize the done work. Describe the algorithm of construction of the circle, tangent to given A, B, C; or rather, the algorithm of discovering tangency of this circle to one of three given, e.g. to A.
1. We take an arbitrary point X on A. Draw through it a circle, orthogonal to A and B. Choose any of two intersection points of this circle with B.
2. From the chosen point draw a circle orthogonal to B and C. Choose any of two intersection points of this circle with C.
3. Draw circle G through X and two chosen in items 1 and 2 points. We designate the second intersection point of this circle with A as Y. (If drawn circle G is tangent to A, it is also tangent to both B and C.)
4. Draw circle I orthogonal to A, B, C. It intersects circle A at points I1 and I2.
5. Draw circle W through I1 and I2 orthogonal to A; draw circle V through X and Y also orthogonal to A.
6. If W and V are intersected, there are no required points (F sets the imaginary inversion). It is necessary to choose differently points from the pairs of intersection D and C with orthogonal it circles (items 1 and 2).
7. If W and V have no common points, the centers of the imaginary bundle formed by W and V are the required.
The note. Drawing of circle I orthogonal to A, B, C can be bulky enough. Besides, it exists only if A, B, C are Lobachevsky's circles. It is possible to do without it: to take one more point X1 on A, to make the construction of items 13. We designate the gotten point as Y1. Then draw circle W not through points I1 and I2 but through X1 and Y1. Items 6 and 7 are left without modifications.
In the offered algorithm it is not said about bisectrixes at all, and this strongly simplifies construction.
The algorithm discovers points of tangency of circle O (tangent to given A, B, C) with A. Or rather it discovers two points of tangency of circle A with circles O1 and O2, each of them is tangent also to circles B and C. O1 and O2 are symmetric in the circle commuting with A, B, !. To construct circles O1 and O2 it is necessary to discover also their points of tangency with circles B and C. The easiest way to do is by inverting the gotten points of tangency through bisectrixes S and H. It can be made, as well as earlier: to draw a circle orthogonal to A and B through the constructed point, etc. It is possible to discover points of tangency with B and C and in another way, simply realizing the offered algorithm on these circles. But here it is important not to be mistaken in bisectrixes in order not to get on B and C points of other circles also tangent to A, B and C.
As it has been told F=T*H*S is an inversion orthogonal to A, and F1=S*T*H orthogonal to B, F2=H*S*T orthogonal to C. Then using these equalities (enough the first, F=T*H*S) we discover when the required circle does not exist. As it has been shown it is not present when inversion F is the imaginary. So, when F is the imaginary inversion?
The theorem of a composition of inversions of one bundle.
Whatever are inversions T, H, S belonging to one bundle their composition is the imaginary inversion when and only when one or three of them are imaginary.
Proof.
1. If composition T*H*S is the imaginary inversion, all these inversions lie in the imaginary bundle (since in the real or tangent bundle there are no imaginary inversions).
2. A composition of three real inversions of one bundle is always the real inversion.
3. A composition of two imaginary inversions is always the real inversion. (It is possible to be convinced of it considering a bundle of concentric circles, see paper 4 when this composition is a similarity).
4. Therefore, if either or S is the imaginary inversions, or T and H are the imaginary inversions and the remained inversion is real, then T*H*S is the real inversion since we can substitute a composition of two imaginary inversions by a composition of two real ones and the whole composition can be reduced to three real inversions which is always real.
5. If T and S are imaginary and X is the real we cannot simply rearrange them in composition T*H*S. But it is possible to use the concept of conjugated elements of group of motions (see paper 5). Lets consider composition =S*T*H. K is the real inversion since S and T are imaginary. Let s map the motionless circle of inversion K by inversion S. We get inversion S*(S*T*H)*S1=T*H*S (we have removed the parentheses and have used the fact that S is involution). Since K is the real inversion, S (K) is also the real inversion. Q.E.D
6. Let s prove that if there is only one imaginary inversion in a composition, e.g. , the outcome is the imaginary inversion. We prove by contradiction. Let outcome is real inversion K. =T*H*S, hence H=T1*K*S1=T*K*S. Right hand side is the composition of three real inversions; therefore, X is also real. The contradiction.
7. The composition of three imaginary inversions due to item 3 is reduced to the composition where only one imaginary inversion and two real. This composition according to previous item is imaginary. Q.E.D.
We have sorted out all cases and were convinced that T*H*S is the imaginary inversion only when there are one or three imaginary bisectrixes.
Geometrical conclusions.
Using the given reasoning we discover the unique nontrivial case when there is no required circle O tangent to three given A, B, C.
It should be such disposition of circles A, B, C that for all cases of a choice of bisectrixes X and S (the remained bisectrix between A and C gets out automatically, it is that which lies in the bundle of first two chosen) composition T*H*S is imaginary. How to choose it? If it actually is not drawn on a delineation and we do not see intersection points of bisectrixes. For this purpose Ill use a simple property: let three bisectrixes form a bundle. We substitute first two of them with the second possible bisectrixes. Then the remained, third bisectrix, belongs to the bundle formed by the substituted. But if we substitute only one (or all three bisectrixes), the turned out three bisectrixes cannot be in one bundle. It is not difficult to prove the property. I designate this replacement rule as (!).
Let's designate as "1 and "2 two possible bisectrixes between A and C, as S1 and S2 two possible bisectrixes between A and B, as 1 and 2 between B and C. If it is known any one case when bisectrixes form a bundle, we can discover all remaining chances using only a rule (!). Let, for example, "1, S1, 1 are in one bundle. Then, on (!) "2, S2, 1 are in one bundle; the same is for "2, S1, 2 and T1, S2, H2. I.e. any two bisectrixes with the code 2 (not belonging to the first bundle) and one bisectrix with the code 1 (belonging to the first bundle) are in one bundle. If we apply the rule (!) to any of enumerated triples of bisectrixes we get the case already enumerated. Triangle bisectrixes are get out in the same way.
First of all well show that if any two circles, for example, B and C are intersected, then regardless of A disposition (except tha case when A is in one bundle with B and C) there is O tangent to A, B, C. The condition that B and C are intersected is equivalent to that their bisectrixes 1 and 2 are both real. We choose from "1, "2 and S1, S2 by one real bisectrix (there is at least one real bisectrix between any circles). Whatever bisectrixes "1 or "2 belongs to the bundle formed by chosen bisectrixes it is real. Therefore we have three real bisectrixes in one bundle, their composition is real, and hence, the required circle O exists.
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In this case the whole plane is divided into 8 parts. Each of these parts can be described, specifying outside or inside of circles D1, D2, D3 it lies (it not true for the case when circles form a straight line; It s not clearly where is an interior and where exterior of a straight line; But this elimination can be neglected in this case). In other words each of eight parts is an intersection of interiors or exteriors of these three circles. There are eight such intersections: from each circle interior or exterior can be taken for intersection. Totally it gives 2E2E2=8 possible combinations. And all from these 8 hypothetically possible cases are realized on a plane. If, for example, we would take four circles there would be 16 cases. But some of them would not be realized, since the corresponding intersection would be empty.
Let's return to calculation of angles. We choose among these eight areas of a plane that which lies inside all circles and draw circle S through tops of this (bounding the area) triarcs: E, H, B. Calculate angles of S with D1, D2, D3. For this purpose use the same method as for the previous case. At points , , B three circles converge and converge in such manner that the sum of angles between these three circles is equal to 180 degrees.
(<D2,S + (<S,D3 + (<D2,D3=180 degrees (at point where D2 and D3 are intersected).
(<D2,S + (<S,D1 + (<D1,D2=180 degrees (at point B where D2 and D1 are intersected).
(<D1,S + (<S,D3 + (<D1,D3=180 degrees (at point X where D1 and D3 are intersected).
Since we consider angles (<D1,D2; (<D2,D3; (<D3,D1 as known we have three equations with three unknowns: (<S,D1; (<S,D2; (<S,D3. Solving this system we get:
(<S,D1=90 +0.5* ((<D2,D3 (<D3,D1 (<D1,D2),
(<S,D2=90 +0.5* ((<D1,D3 (<D2,D3 (<D2,D1),
(<S,D1=90 +0.5* ((<D1,D2 (<D1,D3 (<D2,D3).
Let's consider now the remaining 7 circles constructed on intersection points of D1, D2, D3. As it has been shown in paper 2 and 6 there is the imaginary inversion I commuting with D1, D2, D3. I swaps intersection points of any two circles. It has been proved in paper 6 that two circles conjugated through imaginary inversion cannot have common points. It follows from here immediately that the circle drawn through any three intersection points of D1, D2, D3 cannot have common points with the circle drawn through remained three points (since the first triple of points is conjugated with the second by imaginary inversion I). From this conjugacy follows also that D1, D2, D3 are isogonal to these conjugated circles. In the same way we can trace how arc of circles D1, D2, D3 are transformed by imaginary inversion I.
Let's return to considering of angles. Let s take now those angles are faced inside of eight triarcs. Having calculated angles between S and D1, D2, D3 we have considered those angles between D1, D2, D3 which are faced inside triarcs (intersections of interiors of D1, D2, D3). But we can consider also complementary angles. As the angle between D1 and D2 we can consider an angle lying inside this triarcs as well as lying outside, equal to 180 (<D1,D2. And do the same for each pair of circles. As there are three pairs, it is possible to make 2E2E2=8 combinations of angles, choosing at one moment the basic, at another the complementary angle (I do not consider now the case when there is the right angle equal to it s additional among these angles). On the other hand we have eight triarcs. The thought arises that each of these triarcs just realizes one of eight possible combinations of angles. But it not so!
On drawing 3 we see that, for example, triarc C, E, B differs from triarc , B, by two angles. Angles at points and B are varied to additional, and the angle at point is equal to the angle at point C (I consider the angles faced inside triarc). Similarly with triarcs , A, and , B, F comparing them with , B, we see that two angles (at the common arc) is varied, and the third remained invariable. If we compare these triarcs, e.g. , C, B and A, , among themselves, we see that they differ by two angles. After imaginary inversion through I four considered triarcs pass to four conjugated. And thus angles in conjugated triarcs are identical. Thus, 4 combinations of angles are actually realized. How to discover them? It is possible to use a method similar to that was in considering possible combinations of bisectrixes between three circles. (see paper 8).
It is enough to discover any combination of the basic and the complementary angles faced inside one triarc. Having substituted two angles by the opposite we again get a set of the angles faced inside some triarc. Four possible combinations exhaust all chances of a disposition of angles.
But why only four cases are possible? It has to do with the fact that it is possible to think of an angle as a set of points. Thus, an intersection of interiors of two circles forms the same angle as an intersection of exteriors. And an intersection of interior of a circle with exterior of another circle sets the same angle as the intersection of the exterior of the first with the interior of the second. In other words the same angle exists both outside and inside a circle. The equality of these angles is realized under imaginary inversion.
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