The Apollonius problem
 

The formulation of Apollonius problem is exclusively simple: let the plane is given on three circles. It is necessary to construct two circles tangent to the set of three initial cicrles using a compass and a ruler.

It is possible to define only eight tangent circles (four pairs). It is impossible to construct tangent circles in some positions of initial circles (it means, that all circles are real; while taking in account image geometry operations it is possible to get eight solutions, may be except of cases of degeneration).

The solution is usually found as follows: ones select the circle of among the circle of minimum radius among all circles and reduce them on the value of radius of selected circle. Therefore, the selected circle appears to be like point (the null-radius circle) and others are to become smaller. Then them more simple task is to be solved: construct a circle passing through the point and tangent the same time to the two given straight-lines. After solving this problem resulting circumference is increased by the radius of the selected circle that ultimately leads to the solution of the original problem.

Despite the fact that this solution leads to a correct result, it does not reveal the fundamental properties of the plane, which are found for the formulation of the problem of Apollonius. A full understanding of these properties requires recourse to the concepts of projective geometry and transform of inversion. Let us to explore these properties and find a unified way to solve the problem of Apollonius, which does not require the measurement of distances, and will be based only on the properties of lines and circles of the projective plane. We shall complete the solution approach gradually, while considering methods of work in the system of geometric modeling Simplex.

We proceed to carry out constructions without resorting to measuring the radius of the circles. The approach will allow us to extend a single method of solving the problem not only in the circles selected as input, but also on the point and straight-lines in various combinations. The arguments used in the algorithm for solving the problem will be the same and does not depend on the type of the original objects.

Let's run the Simplex system, create a new project by clicking the button , and begin initial consideration of the problem with the construction of a circle tangent to three given circles, each of which does not lie in the inner area of ​​any other circle. Later, we will show that this restriction is immaterial.
   
1. Let us place three circles d1 , d2  and d3 on the screen . To do this, use the function, "Free circle," which can be accessed by clicking on the button. . Move the cursor into the free space of the window, click the left mouse button and hold it, to outline the first circle. Release the left mouse button to complete the circle input into the program. The center of the circle and its radius will be determined on the basis of the initial and final position of the cursor. The system will automatically generate command constructing a circle with certain coordinates and the radius of the circle and assign it a name d1 .
 

Fig. 1
 
Circle d1 is given with center coords -175 , -152.8 and radius 110 .
 
Circle d2 is given with center coords 26 , 181 and radius 60 ..
 
Circle d3 is given with center coords 220 , -28 and radius 41 .
   
2. Press the spacebar to switch to object selection cursor mode . Hold down the Shift key, successively select all three circles so that their image will be displayed with thickened line.
 

Fig. 2
   
3. Press the keyboard button corresponding uppercase letter D. This will build the command of circle d4 definition, that is perpendicular to three given circles d1 , d2  and d3 . The circle d4 will be selected.
 

Fig. 3
 
Circle d4 , is rectangular on three circles d1 , d2 , d3 .
   
4. Keeping the circle (without any additional clicks of the mouse in the drawing of building windows), press the «Backspace», at the same time to resume the allocation of initial circles d1 , d2  and d3 .
 

Fig. 4
   
5. Press the keyboard Latin letter o. As a result, the program will add several features required to build centers of similarity for pairs of circles d1–d2, d1 –d3 and d2– d3 . It should be noted that all three of the similarity center would actually lie on a straight-line.
 

Fig. 5
 
Line o1 is tangent to circles d2 and d3 in points p1 and p2 .
 
Line o2 is tangent to circles d3 and d2 in points p3 and p4 .
 
Point p5 is given as intersection of lines o1 and o2 .
 
Line o3 is tangent to circles d2 and d1 in points p6 and p7 .
 
Line o4 is tangent to circles d1 and d2 in points p8 and p9 .
 
Point p10 is given as intersection of lines o3 and o4 .
 
Line o5 is tangent to circles d3 and d1 in points p11 and p12 .
 
Line o6 is tangent to circles d1 and d3 in points p13 and p14 .
 
Point p15 is given as intersection of lines o5 and o6 .
 
Line o7 is given with triplet p5 , p10 , p15 .
   
6. Perhaps, during the last action some part of the generated objects will be located outside of the display area of ​​the window in which the drawing is formed. In order to "drag" the image of all objects in the area of ​​the viewport, press the F4 key.
 

Fig. 6
   
7. Now we pick out, holding down the Shift key, the straight-line o7  and the circle d4.
 

Fig. 7
   
8.

Change the color of the display of the objects on red. For this purpose in a palette object attributes you need to expand the list of outline color and select the item that is responsible for the appointment of the selected objects in red.

During operation, the system will ask for confirmation, designate whether the selected attribute of all selected objects, or selectively to perform this assignment. It is recommended response to a question by pressing the "To all" button.

 
 

Fig. 8
   
9. Holding down the Shift key add the circle d1 to the existing allocation.
 

Fig. 9
   
10.

Press the keyboard to uppercase letter D. It is likely that the screen does not display the newly constructed circle d4.

Circle d5 , is rectangular on three circles o7 , d4 , d1 .

This may, because the radius of this circle is very large and the system is unable to display it. In order to display the circle, move the center of the circle d1 down, for example. To do this, select the tool "shaper" from palette, positioning it above the cross hairs that displays the center of the circle, click on the left mouse button and drag the mouse down. When this happens the entire drawing rebuilds, and image of circle d4 appears on the screen.
 

Fig. 10
 

Fig. 11
   
11. Once the left mouse button is released, there appears next drawing on the screen.
 

Fig. 12
   
12. Hold down the Shift key to add a dedicated circle d1  and press the key p on the keyboard. In this case, the program will be added to the function of defining circles d1  and d4  intersection points p16, p17.
 

Fig. 13
  Points p16 and p17 are given as intersections of circles d1 and d5 .
   
13. Deselect all the objects by moving the mouse cursor in the image free of geometrical objects place and performing on it click with the left mouse button. Selection status of the points p16 , p17  will be removed.
 

Fig. 14
   
14. The result of these actions is a circle d5 appearance, but the radius of which is still quite large. If you now press the F4 key on the keyboard, it will lead to the retraction of the screen in the entire drawing, including the d5-circle. d5 , with the result that most of the objects of geometrical construction will be difficult to see.
 

Fig. 15
   
15. To return to previous performance positioning and scale of the drawing, press the keyboard shortcut Alt + Z.
 

Fig. 16
   
16.

At the same time, the drawing shows that some points, which would be desirable to observe, remain outside the display area of ​​the viewport. In order to incorporate the necessary objects in the display area, you can use the following method:

1. Select all the objects of algorithm using the key combination Ctrl + A.
 

Fig. 17
   
17.

2. Hold down the Shift key, point the cursor over the image of d5  circle and press left mouse button. Thus, the circle d5 will be excluded from the selection.
 

Fig. 18
   
18. 3. Type the key combination Shift + F4. Thus, only those objects that were selected, will be zoomed and drawn completely in the screen.
 

Fig. 19
   
19. Deselect all objects again by clicking on the place, which is free of geometrical images.
 

Fig. 20
   
20. Now, hold down the Shift key, select the cursor selection o7 , d4 , d2  Now, hold down the Shift key, select the cursor selection o7 objects, d4, d2, and press the key corresponding to the uppercase letter D. This will enable us to build another circle d6 , which is perpendicular to the source selected objects o7 , d4 , d2.
 

Fig. 21
 
Circle d6, is rectangular on three circles o7, d4, d2 .
 
Points p18 and p19 are given as intersections of circles d6 and d2.
   
21. Hold down the Shift key, to add to the existing selection circle d2  and press p key on the keyboard to get the point of intersection of these circles p18,p19.
 

Fig. 22
   
22. Following the same logic constructs, we input the circle d7 , perpendicular to the object o7, d4, d3 , into the program into the program p20 , p21 of circles d7 , d3 crossing.
 

Fig. 23
 
Circle d7 , is rectangular on three circles o7 , d4 , d3 .
 
Points p20 and p21 are given as intersections of circles d7 and d3 .
   
23. To construct a circle tangent to three given circles, deselect all objects, and then hold down the Shift key, select the points p16 , p19, p20  and press the key on the keyboard with the Latin letter g. As a result, the program will feature introduced d8 , construction of a circle passing through three given points p16 , p19, p20 . A similar action is to perform with points p17, p18, p21 , to get a second tangent circle d9
 

Fig. 24
 
Circle d8 passes points p17 , p19 and p20 .
 
Circle d9 passes points p16 , p18 and p21 .
  Example
   
  Now it is possible to perform additional studies of the resulting model. It is recommended to test its behavior by changing with shaper position of d6, d7, d5 circles centers and their radii. All of their centers are on line o7 , and they have common crossing in two stable points. It is also interesting that in these points, you can draw a straight-line, and the centers of the circles conjugating d8, d9 and d4 circle also will be on the line. These properties are easy to discover and explain from the resulting drawing. However, you can find many other interesting properties in the same drawing, if carried out some additional constructions. Perform some of them.
   
  Connect points p17, p19 with line o9 , select them, and type o button on keyboard. Convert the segment in an endless line, change the attributes by clicking on the button with the symbol ~. Make sure that the line o9 passes through the center of cimilarity of circles d1 , d2 - point p10 . Passing straight-line o10 through the points p16 , p18 will find that it will go through the point p10 . Similar statements are true with respect to other points of contact of the desired circles to the three given circles (it is recommended to build and test them).
 

Fig. 25
   
  The next interesting feature is that the pair obtained through touch points are on the set of circles. Let's construct the circle passing three points p17, p16 , p18 , for example. To do this select the points and type g button on keyboard. Verify that the resulting circle passes through the point p19 .
 

Fig. 26
   
  It is simple to check, that a similar property is shown in relation to points p17 , p16 , p21 , p20 и p20 , p21 , p18 , p19 . Have a look how arranged circumferentially received and formed construction (intersect at a common point) and where the centers users distinguish these circles.
 

Fig. 27
 

The results lead to the idea that the solution to the problem of Apollonius is largely dependent on solving the problem of finding the intersection of two circles and points drawing circles, perpendicular to the given circles. This thought prompts a more detailed study of these basic tasks, even more so given method is not suitable for solving the problem of Apollonius, for example, if one of the starting circles covers two other circles. It is clear that it is impossible to produce a sufficient number of tangent lines to get the similarity axis initial circles in this case. However, this does not mean that the Apollonius problem has no solution - it is. Therefore, we need to deeply explore the properties of geometric objects that define the essence of the fundamental geometrical problem to be solved.

Looking ahead, I want to say that this study will allow us to open a large number of unexplored and perhaps previously unknown properties of the spreading problem of Apollonius in the field of imaginary geometry, let go to the analogies in multidimensional spaces, connect this problem with the geometry of the triangle and see its recursive nature therein.

The next discussion will be devoted to the problems of intersection of circles and constructing the so-called radical images - straight-lines and points induced on pairs and triples of circles. As we shall see soon, that radical images defined on geometry of circles, can be simply connected with straight-lines and points, that will allow us to make some important generalizations about linear and circular object of a plane.