This article is based on one of the sections of the book "Python Programming" by Sedgwick, Wayne and Dondero, already mentioned earlier. This section is called "Systems of iterative functions", and it describes the construction of various images, such as the Sierpinski triangle, the Barnsley fern, and some others, using a fairly simple algorithm, which, moreover, is also easily implemented.
I'll start by describing this algorithm. I will use mathematical terminology, including the one that the authors of the book, in the course of their narrative, do not use. A purely mathematical view of algorithms makes it easier for me to understand them, and it is quite convenient for me to express them using a mathematical language.
So, in order to understand the theoretical part of the article, the reader will need knowledge of some sections of mathematics, which are usually read in technical universities. Namely, it would be useful to get acquainted with the theory of probability and elements of mathematical analysis.
The theoretical part of the article will be followed by a practical one describing the implementation of the algorithm in the C99 language. Since the results of the program will be images, we will use the pgraph graphics library in the program, assuming that the reader, at least in general terms, is familiar with it.
So, let's move on to the theoretical part of our story.
Iterative functions and random sequences
Before describing the scheme by which images will be built, let's talk about sequences whose members are calculated using recurrent formulas.
Let's define 2 sequences, x n n = 1 ∞ and y n n = 1 ∞ , using the following recursive formulas:
X n = f x n - 1 , y n - 1 , n ∈ ℕ , y n = g x n - 1 , y n - 1 , n ∈ ℕ .
Let us explain that x 0 and y 0 are some predefined numbers, and f(x, y) and g(x, y) are some functions of two variables called iterative. The process of calculating the next member of a particular sequence through such functions will be called iterations, and the above set of recurrent formulas is an iterative scheme.
The recursive way of specifying sequences is most likely familiar to the reader if he studied mathematics at a university. Somewhat unusual may seem to be a "cross" method for calculating the members of sequences, in which for calculating n th member of each of the two sequences is needed not only n− 1st member of the same sequence, but also n− 1st member is different.
Now consider a scheme for constructing members of two sequences that uses not just one pair of iterative functions, but m steam. Each of these functions will be linear in both variables and will also contain an additive constant. More specifically, the functions will look like:
F k x , y = a k x + b k y + c k g k x , y = d k x + e k y + h k , k = 0 , 1 , … , m - 1 .
For each n, starting from 1, will randomly select a number from 0 to m− 1, and when calculating x n and y n in recurrent formulas, a pair of iterative functions will be used, the indices of which are equal to a given random number. Note that the random numbers "appearing" before each iteration need not be equally probable. However, for different steps, the probability of occurrence of a particular fixed number is the same.
Let us now formulate what has been said in strict mathematical language. Consider a sequence of discrete random variables T n = 1 ∞ , independent in the totality, distributed according to the same law. Namely: each random variable takes the values 0, 1, ..., m− 1 with corresponding probabilities p 0 , p 1 , …, p m-1 .
Now the sequences x n n = 1 ∞ and y n n = 1 ∞ are defined using the following iterative scheme:
X n = f T n x n - 1 , y n - 1 , n ∈ ℕ , y n = g T n x n - 1 , y n - 1 , n ∈ ℕ .
As earlier, x 0 and y 0 is some predefined numbers.
Thus, each of the sequences is random, that is, its members are random variables. However, each of these sequences can be "realized", i.e., all its members can be calculated (of course, there will be infinitely many such implementations).
Let's take a look at the main question of this section. And what does the images we are going to build have to do with this pair of random sequences? Very simple. Let us construct a realization of these two sequences. For every natural n couple ( x n, y n) can be considered as the coordinates of a point given in a Cartesian rectangular coordinate system on a plane. So, the image corresponding to some pair of implemented sequences is the locus of all such points on the plane.
It would seem that for each implementation of a pair of sequences, we will get our own image, different from the others. However, paradoxical as it may seem, the resulting images will almost coincide each time (i.e., when plotted on a computer, they will be indistinguishable by the human eye). And with an appropriate selection of iterative functions and distribution laws of random variables involved in the formation of sequence members, very interesting patterns can be created.
We add that when constructing images on a computer, we, of course, will perform only a finite (but sufficiently large) number of iterations.
About generating pseudo-random numbers
When writing a program, we will face the need to generate pseudo-random numbers distributed, generally speaking, not evenly, but according to a predetermined law. At the same time, we will only have a software generator of pseudo-random numbers uniformly distributed over the interval . How to get the first one from the second distribution?
Let's translate the problem into a mathematical plane. Let there be a continuous random variable U, distributed evenly on the segment . Let us set ourselves the goal of constructing a discrete random variable T as a function of U, In the way that T took values 0, 1, ..., m− 1 with corresponding probabilities p 0 , p 1 , …, p m-1 .
Solving the problem is very simple. Let us introduce into consideration the sum of probabilities
s k = ∑ i = 0 k - 1 p i , k = 0 , 1 , … , m - 1 .If the upper summation limit over i is less than the lower one, then by definition we will assume such a sum to be equal to 0.
T express through U in the following way:
T = 0 if U ∈ s 0 , s 1 , 1 if U ∈ s 1 , s 2 , 2 if U ∈ s 2 , s 3 , … … … … … … , … … … … … … , m - 1 if U ∈ s m - 1 , 1 .
Obviously the random variable T distributed according to the law we require. Note that, in fact, T- this is the number of the interval in which the random variable falls U(provided that we number the intervals with numbers from 0 to m− 1 in ascending order of their left boundaries).
From a practical point of view, the result obtained allows, at each iteration step, as the number of iterative functions, to take the number of the interval into which the number generated by the generator of pseudo-random numbers uniformly distributed on the interval falls.
Now we can move on to writing the program.
Program structure
The program consists of the main.c file and the files that make up the pgraph graphics library. The content of the main.c file begins with the following directive that includes the graphics library:
#include "pgraph.h"Further, the file contains descriptions of global constant variables and constant arrays. Behind them are the definitions of the get_random_value() and main() functions. The first one generates pseudo-random numbers, and the second one does the main work of building images.
Global constant variables and constant arrays
All the information needed to build a particular image is contained in global constant variables and constant arrays. Of course, for each image, the set of values of constants and elements of constant arrays will be "its own".
The following are descriptions of these constants and arrays.
- n is the number of iterations;
- w - image width in pixels;
- h - image height in pixels;
- xc - abscissa of the beginning of the new coordinate system in the old system;
- yc - ordinate of the beginning of the new coordinate system in the old system;
- l is the length in pixels of a segment parallel to one of the coordinate axes, having unit length in the new coordinate system;
- m is the number of pairs of iterative functions, i.e. the number m;
- s - one-dimensional array of size m containing sums of probabilities of random variables T n (k-th array element contains s k);
- f is a two-dimensional array consisting of m f k(x, yk, 0), (k, 1), (k, 2) contain numbers a k, b k, c k respectively, where 0 ≤ k ≤ m − 1);
- g - two-dimensional array consisting of m"rows" and 3 "columns" containing constants used in functions g k(x, y) (array elements with indices ( k, 0), (k, 1), (k, 2) contain numbers d k, e k, h k respectively, where 0 ≤ k ≤ m − 1).
All variables are of type int , and the base type of all arrays is double .
Let's clarify that the "old" coordinate system means the one defined in the pgraph library. The construction of all images will be carried out in the new system obtained from the old one by parallel transfer (shifts along the abscissa and ordinate axes are equal, respectively x c and y c) and "compression" in l once. Thus, a point that has coordinates in the new system ( x, y), in the old one will have coordinates ( x l + x c, y l + y c). It is superfluous, I think, to explain what kind of storage of numbers x c, y c and l the constant variables xc , yc and l are responsible, respectively.
To store numbers x 0 and y 0 variables are not allocated, since in all cases of imaging, zeros are taken as these numbers.
Pseudo-random number generation: get_random_value() function
The get_random_value() function generates a pseudo-random integer in the range from 0 to m− 1 in accordance with the scheme described earlier. Here is the code for this function:
1. int get_random_value() 2. { 3. double r = (double ) rand() / RAND_MAX; 4. int c = 1; 5. while (s[c]< r && ++c < m) 6. ; 7. return c - 1 ; 8. }Using the standard library function rand(), we get a pseudo-random number in the range from 0 to the value of the RAND_MAX macro, divide the result by this value, and assign the quotient to the variable r (p. 3). Now r stores the number belonging to the interval . It can be approximately considered as the value of a random variable uniformly distributed on this segment.
Let's clarify that the value of the RAND_MAX macro in our case (i.e., in the case of using the MinGW64 compiler version 4.9.2 for 64-bit systems) is 32767.
Now, using a linear search using a while loop, we look for the index of the largest element of the array s that is less than or equal to the value of r plus one, and stores it in the variable c (see page 4-6). Note that if the value of r is zero, the loop is never executed, and the variable c retains a single value (see page 4).
The value returned by the function can be approximately considered as the value of a random variable T described in the section mentioned above.
Image generation: main() function
And here is the main() function code:
1. int main() 2. { 3. image *img = create_image(w, h); 4. double x = 0 , y = 0 ; 5. for (int i = 0 ; i< n; i++) 6. { 7. int r = get_random_value(); 8. double x1 = f[r] * x + f[r] * y + f[r]; 9. double y1 = g[r] * x + g[r] * y + g[r]; 10.x=x1; 11.y=y1; 12. set_color(img, round(x * l) + xc, round(y * l) + yc, BLACK); 13. } 14. save_to_file(img, "out.bmp" ); 15.free(img); 16. return 0 ; 17. }Create an image with the specified dimensions (p. 3). Allocate memory for variables x and y, which will store the current members of the sequences, and initialize them to zero (p. 4). Let me remind you that as numbers x 0 and y 0 , involved in the calculation of the first members of each of the sequences, zeros are taken.
Calculate in the for loop the first n members of each sequence (pages 5-13). We first get a pseudo-random number and write it to r (p. 7). Next, we calculate the current values of the members of both sequences, placing them in temporary variables x1 and y1 (p. 8, 9). When calculating, we use constants that appear in iterative functions and are stored in arrays f and g . The choice of one or another pair of sets of coefficients (and hence, a pair of iterative functions) depends on the value of r , which is used as the first indices of the array elements involved in the calculations.
We rewrite the calculated current values into variables x and y (p. 10, 11). The point coordinates contained in these variables are converted into the coordinates of the original coordinate system, rounded up to integers, and the point with the resulting coordinates is plotted on the image in black (p. 12).
Upon completion of the cycle, save the generated image in the file "out.bmp" (p. 14) and free the memory occupied by the image (p. 15). This completes the function.
Building an image of the Sierpinski triangle
The Sierpinski triangle is a set of points obtained from all points of some initial equilateral triangle as follows. The triangle is divided by three middle lines into 4 triangles, after which the "central" triangle is removed. Further, the same operation is performed with each of the remaining three equilateral triangles. Finally, we do the same with the resulting nine equilateral triangles.
Continuing the described operations to infinity, we finally remove from the original triangle an infinite number of equilateral triangles, the sum of the areas of which is equal to the area of the original one. The remaining points form a line called Sierpinski triangle, which plays an important role in set theory.
The book by Sedgwick and other authors proposes the following method for constructing an image of the Sierpinski triangle. Consider 3 points on the plane that are vertices of an equilateral triangle, for example, points with coordinates 0 , 0 , 0 , 1 , 1 / 2 , 3 / 2 in a Cartesian rectangular coordinate system. We choose at random (with equal probabilities) one of the three vertices of the triangle and build a point dividing the segment connecting the vertex with coordinates 0, 0 and the vertex chosen at random, in half. This is the first point of our image.
The above algorithm can be put into the previously described image construction scheme, which involves random sequences and iterative functions.
We need 3 pairs of iteration functions. Their indices 0, 1, 2 should be chosen with probabilities 1/3, 1/3, 1/3 respectively. The iterative functions themselves are shown below.
F 0 x , y = 1/2 x , g 0 x , y = 1/2 y , f 1 x , y = 1/2 x + 1/2 , g 1 x , y = 1/2 y , f 2 x , y = 1 / 2 x + 1 / 4 , g 2 x , y = 1 / 2 y + 3 / 4 .
Now let's insert into our program the declarations of global constant variables and constant arrays corresponding to given probabilities and given iterative functions. But first, let's define the TRIANGLE macro by placing the following statement in the main.c file after the #include statement
#define TRIANGLEAfter the instructions, paste the following code into the file:
// Sierpinski triangle #ifdef TRIANGLE const int n = 100000 ; //number of iterations const int w = 620 , h = 550 ; //image dimensions const int xc = 10 , yc = 10 ; // coordinates of the beginning of the new coordinate system in the old one const int l = 600 ; //compression ratio const int m = 3 ; //number of pairs of iteration functions const double s = (0 , 0.3333333 , 0.6666667 ); //array of sums of probabilities const double f = ((0.5 , 0.0 , 0.0 ), //array of coefficients for functions f(x,y), {0.5 , 0.0 , 0.5 }, //used to compute x {0.5 , 0.0 , 0.25 }}; const double g = ((0.0 , 0.5 , 0.0 ), //array of coefficients for functions g(x,y), {0.0 , 0.5 , 0.0 }, //used to calculate y {0.0 , 0.5 , 0.4330127 }}; #endifThe above code fragment (without preprocessor directives) will be compiled only if the TRIANGLE macro is defined (and it is defined). Of course, constants that can only be represented using infinite decimal fractions (rational or irrational) were rounded off.
As a result of compiling and executing the program, a graphic file out.bmp appears in the root directory of the executable file, containing the following image:
Building an image of the Barnsley fern
The next image, the construction of which is described in the book by Sedgwick and others, is the image of the Barnsley fern. Now we need 4 pairs of iterative functions. Their indices 0, 1, 2, 3 will be chosen with probabilities 0.01, 0.85, 0.07, 0.07 respectively. And here are the iteration functions themselves:
F 0 x , y = 0 . 5 , g 0 x , y = 0 , 16 y , f 1 x , y = 0 , 85 x + 0 , 04 y + 0 , 075 , g 1 x , y = - 0 , 04 x + 0 , 85 y + 0 , 18 , f 2 x , y = 0 , 2 x - 0 , 26 y + 0 , 4 , g 2 x , y = 0 , 23 x + 0 , 22 y + 0 , 045 , f 3 x , y = - 0 , 15 x + 0 , 28 y + 0 . 575 , g 3 x , y = 0 , 26 x + 0 , 24 y - 0 , 086 .
We now make changes to the program. The #define instruction is replaced by the instruction
#define FERNAnd after the #ifdef -block we place the following piece of code:
// Fern Barnsley #ifdef FERN const int n = 100000 ; const int l = 600 ; const int m = 4 ; const double s = (0 , 0.01 , 0.86 , 0.93 ); const double f = ((0.0 , 0.0 , 0.5 ), {0.85 , 0.04 , 0.075 }, {0.2 , -0.26 , 0.4 }, {-0.15 , 0.28 , 0.575 }}; const double g = ((0.0 , 0.16 , 0.0 ), {-0.04 , 0.85 , 0.18 }, {0.23 , 0.22 , 0.045 }, {0.26 , 0.24 , -0.086 }}; #endifThe result of compiling and running the program is the following image:
Building an image of a tree
Now let's build what is called a "tree" in the book by Sedgwick and others, although what turns out to be depicted is rather like a collection of trees of various sizes. This time, 6 pairs of iterative functions will participate in the iterative process. Their indices 0, 1, 2, 3, 4, 5 will be chosen with probabilities 0.1, 0.1, 0.2, 0.2, 0.2, 0.2 respectively. These are the functions:
F 0 x , y = 0 . 55 , g 0 x , y = 0 . 6 y , f 1 x , y = - 0 . 05 x + 0 . 525 , g 1 x , y = - 0 . 5 x + 0 , 75 , f 2 x , y = 0 , 46 x - 0 , 15 y + 0 , 27 , g 2 x , y = 0 , 39 x + 0 , 38 y + 0 , 105 , f 3 x , y = 0 , 47 x - 0 , 15 y + 0 , 265 , g 3 x , y = 0 , 17 x + 0 , 42 y + 0 , 465 , f 4 x , y = 0 , 43 x + 0 , 26 y + 0 , 29 , g 4 x , y = - 0 , 25 x + 0 , 45 y + 0 , 625 , f 5 x , y = 0 , 42 x + 0 , 26 y + 0 , 29 , g 5 x , y = - 0.35x + 0.31y + 0.525.
#define TREEAfter the last #ifdef block, paste the following code:
//Wood #ifdef TREE const int n = 100000 ; const int w = 620 , h = 620 ; const int xc = 0 , yc = 10 ; const int l = 600 ; const int m = 6 ; const double s = (0 , 0.1 , 0.2 , 0.4 , 0.6 , 0.8 ); const double f = ((0.0 , 0.0 , 0.55 ), {-0.05 , 0.0 , 0.525 }, {0.46 , -0.15 , 0.27 }, {0.47 , -0.15 , 0.265 }, {0.43 , 0.26 , 0.29 }, {0.42 , 0.26 , 0.29 }}; const double g = ((0.0 , 0.6 , 0.0 ), {-0.5 , 0.0 , 0.75 }, {0.39 , 0.38 , 0.105 }, {0.17 , 0.42 , 0.465 }, {-0.25 , 0.45 , 0.625 }, {-0.35 , 0.31 , 0.525 }}; #endifThe output of the compiled program is the image shown below:
The last image that we will build, guided by Sedgwick's book, is the image of a coral. We need 3 pairs of iteration functions. Their indices 0, 1, 2 will be chosen with probabilities 0.4, 0.15, 0.45 respectively. The iterative functions are given below.
F 0 x , y = 0 . 3077 x - 0 . 5315 y + 0 . 8863 , g 0 x , y = - 0 . 4615 x - 0 . x - 0, 0769 y + 0, 2166, g 1 x, y = 0, 1538 x - 0, 4476 y + 0, 3384, f 2 x, y = 0, 5455 y + 0, 0106, g 2 x, y = 0 . 6923 x - 0 . 1958 y + 0 . 3808 .
Replace the #define instruction with the instruction
#define CORALAfter the last #ifdef -block, insert a new block:
//Coral #ifdef CORAL const int n = 100000 ; const int w = 620 , h = 620 ; const int xc = 10 , yc = 10 ; const int l = 600 ; const int m = 3 ; const double s = (0 , 0.4 , 0.55 ); const double f = ((0.3077 , -0.5315 , 0.8863 ), {0.3077 , -0.0769 , 0.2166 }, {0.0 , 0.5455 , 0.0106 }}; const double g = ((-0.4615 , -0.2937 , 1.0962 ), {0.1538 , -0.4476 , 0.3384 }, {0.6923 , -0.1958 , 0.3808 }}; #endifHere is the image we get as a result of compiling and executing the program:
Conclusion
I don't know about you, but it was interesting for me to watch how sets of mathematical formulas "turn" into very funny images. It also surprises me that those who came up with all this were able to choose the probabilities and constants involved in iterative functions in such a way as to achieve such amazing pictures! The method of selecting all these numbers (with the exception of the case of the Sierpinski triangle) is completely incomprehensible to me!
I note that, judging by the images, the Sierpinski triangle and the Barnsley fern are fractals. Most likely, the same can be said about "tree" and "coral", but their fractal nature is perhaps a little less obvious.
Using the link below, as always, you can download the source code of the program discussed in the article. The main.c file has four #define statements, each corresponding to one of the four images. Three of them are commented out. It is clear that in order to move from one image to another, you need to comment out the uncommented instruction and uncomment one of the commented ones. Well, you get the idea...
And with the help of a simple algorithm, you can ensure that the images discussed in the article smoothly "turn" into each other. But this is a topic for a separate article.
The Greek geometers prided themselves on their logical purity; however, as far as physical space is concerned, they were guided by intuition. One aspect of Greek geometry that was particularly influenced by physical considerations was the theory of constructions. Much of the elementary geometry of straight lines and circles can be viewed as a theory of constructions with a ruler and compass. The very name of the subject, lines and circles, reflects the tools that were used to carry them out. And many of the elementary problems of geometry, such as bisecting a line segment or an angle,
building a perpendicular or drawing a circle through three given points can be solved by building with a ruler and a compass.
Once the coordinates are entered, it is easy to show that the points that can be built from points have coordinates in the set of numbers created from the coordinates by the operations and [cf. Muaz (1963) or exercises for section 6.3]. Square roots, of course, appear due to the Pythagorean theorem: if points are plotted, then the distance between them is plotted (Section 1.6 and Figure 2.4). Conversely, the construction is possible for any given length I (Exercise 2.3.2).
Figure 2.4: Building a distance
When viewed from this point of view, the constructions with a ruler and compass look very special and it is unlikely that such numbers will be given, for example, However, the Greeks tried very hard to solve this particular problem, which was known as doubling the cube (so-called because that to double the volume of a cube, one had to multiply the side by Other infamous problems were the trisection of an angle and the squaring of a circle.The last problem was to construct a square equal in area to a given circle, or to construct a number that is equal to the same. they never seem to have abandoned these aims, although they recognized the possibility of a negative solution and allowed solutions by less elementary means.In the following sections we shall see some of them.
The impossibility of solving these problems with straightedge and compass constructions remained unproven until the nineteenth century. As regards the doubling of the cube and the trisection of the angle, the impossibility was shown by Vantzel (1837). The credit for solving these problems, which the best mathematicians have struggled with for 2,000 years, is rarely credited to Wantzel, perhaps because his methods have been superseded by the more powerful Galois theory.
The impossibility of squaring the circle is proved by Lindemann (1882), in a very rigorous way, not only by indefinably rational operations and square roots; it is also transcendental, that is, it is not the root of any polynomial equation with rational coefficients. Like Wantzel's work, it was a rare example of a significant result proven by a minor mathematician. In Lindemann's case, the explanation may be
In that an important step had already been taken when Hermite (1873) proved transcendence. Available evidence for both of these results can be found in Klein (1924). Lindemann's subsequent career was mathematically unremarkable, even embarrassing. Responding to skeptics who thought his success with was a fluke, he set his sights on the most famous unsolved problem in mathematics, Fermat's Last Theorem (for the origin of this problem, see Chapter 11). His efforts ended in failure in a series of unconvincing papers, each correcting an error in the previous one. Fritsch (1984) wrote an interesting biographical article on Lindemann.
I. Introduction.
II. Main part:
the first construction method;
the second method of construction;
the third way to build,
Construction of a segment equal to the product of the other two using a compass and a ruler:
d) the fourth construction method.
2) Construction of a segment equal to the ratio of the other two using a compass and a ruler:
the first construction method;
second construction method.
Conclusion.
Application.
Introduction
Geometric constructions, or the theory of geometric constructions, is a branch of geometry where questions and methods for constructing geometric figures are studied using certain construction elements. Geometric constructions are studied both in the geometry of Euclid and in other geometries, both on the plane and in space. The classical construction tools are compasses and a ruler (one-sided mathematical), however, there are constructions with other tools: only one compass, only one ruler, if a circle and its center are drawn on the plane, only one ruler with parallel edges, etc.
All construction problems are based on construction postulates, that is, on the simplest elementary construction problems, and the problem is considered solved if it is reduced to a finite number of these simplest postulate problems.
Naturally, each instrument has its own constructive force - its own set of postulates. So, it is known that it is impossible to divide a segment using only one ruler into two equal parts, but using a compass, you can.
The art of constructing geometric figures with the help of a compass and ruler was highly developed in ancient Greece. One of the most difficult construction tasks, which they already knew how to perform, was the construction of a circle tangent to three given circles.
At school, they study a number of the simplest constructions with a compass and a ruler (one-sided without divisions): the construction of a straight line passing through a given point and perpendicular or parallel to a given straight line; dividing a given angle in half, dividing a segment into several equal parts using the Thales theorem (in fact, dividing a segment by a natural number); construction of a segment larger than the given one by an integer number of times (essentially, multiplying the segment by a natural number). However, we have never encountered a problem where it would be necessary to multiply a segment by a segment using a compass and a ruler, that is, to construct a segment equal to the product of two given segments, or to divide a segment by a segment, that is, to construct a segment equal to the ratio of the other two segments. This problem seemed very interesting to us, and we decided to investigate it, try to find a solution and the possibility of applying the found solution method to solving other problems, for example, in mathematics and physics.
When solving construction problems, the traditional methodology recommends four stages: analysis, construction, proof, and research. However, the indicated scheme for solving construction problems is considered to be very academic, and it takes a lot of time to implement it, therefore, individual stages of the traditional scheme for solving the problem are often omitted, for example, the stages of proof, research. In our work, as far as possible, we used all four stages, and even then only where there was a need and expediency for this.
And the last thing: the method we found for constructing the above-mentioned segments involves the use, in addition to the compass and straightedge, of an arbitrarily chosen single segment. The introduction of a unit interval is also dictated by the fact that it is necessary at least to confirm the validity of the method we have found for finding a segment on specific particular examples.
GENERAL PROBLEM I
Using a compass and straightedge, construct a segment equal to the product of the other two segments.
Note:
supposed:
The ruler is one-sided, without divisions.
A segment of unit length is given.
Study.
1. Consider the lines y=2x-2 2 and y=3x-3 2 and try to find the coordinates of the point of intersection of these lines by geometric and analytical methods:
a
) geometric method ( Fig.1) showed that the coordinates of the point A of the intersection of these lines: “5” is the abscissa, “6” is the ordinate, i.e. AE=5, AD=6.
b) the analytical method confirms this result, i.e. A (5;6) - the point of intersection of the lines.
Indeed, by solving the system of equations
y=6 А(5;6) - point of intersection of lines.
2. Consider the segment: OB=2, OS=3, AD=6, AE=5.
It can be assumed that BP=OV×OS, because 6=2×3; AE \u003d OB + OS, because 5=2+3 , where
2=OB-slope of the equation y=2x-2 2 , 3=OS - slope of the equation y=3x-3 2 , AD=y A, OD=x A - coordinates of the point A of the intersection of our lines.
We will check our assumption on a general example by the analytical method, i.e. on the equations of lines y=mx-m 2 and y=nx-n 2 (where m≠n) check that the point of intersection of the lines has coordinates:
y=nx-n 2 nx-n 2 =mx-m 2 x=(m 2 -n 2)÷(m-n)=m+n and y=mx-m 2 =m(m+n)-m 2 = mn
coordinates of the point A of the intersection of lines, where m and n are the slopes of these lines, etc.
3. It remains to find a method for constructing a segment. HELL=OB×OC=m∙n=y A - ordinates of point A of the intersection of lines Y=mx-m 2 and Y=nx-n 2, where m≠n and m=OB, n=OC- segments plotted on the axis oh. And for this we must find a method for constructing lines Y=mx-m 2 and Y=nx-n 2 . from the reasoning it is clear that these lines must pass through the points B and C of the segments OB=m and OC=n, which belong to the x-axis.
Remark 1. The above designations of the segments correspond to Fig. 1 "Appendices"
First way constructing a segment AD=mn, where m>1 unit, n>1 unit, m≠n.
single segment
arbitrary segment, m>1ed., n>1ed.
n is an arbitrary segment, where m≠n.
Building (Fig.2)
Let's draw a straight line
On OH we postpone OA 1 = m
On OX we set aside A 1 C 1 \u003d 1 unit
Let's construct C 1 B 1 =m, where C 1 B 1 ┴ OH
Let's draw a straight line A 1 B 1, the equation of which is y=mx-m 2 in the XOU coordinate axes (the scale on the axes is the same).
Note:
Fig.2
Remark 1.
Indeed, the tangent of the slope of this straight line tgά 1 = C 1 B 1 /A 1 C 1 =m/1ed=m, which passes through the point A 1 of the segment OA 1 =m.
Similarly, we build a straight line, the equation of which is Y \u003d nx-n 2.
6. On the OX axis, we set aside OA 2 \u003d n (point A 2 accidentally coincided with point C1).
7. On the OX axis, set aside A 2 C 2 \u003d 1 unit.
8. We build B 2 C 2 \u003d n, where B 2 C 2 ┴ OH.
9. Let's draw a straight line B 2 A 2, the equation of which is Y \u003d nx-n 2.
Remark 2. Indeed, the slope of this straight line tg ά 2 =C 2 B 2 /A 2 C 2 =n/1ed=n, which passes through t. A 2 segment OA 2 =n.
10. We got t.A (m + n; mn) - the point of intersection of the lines Y \u003d mx-m 2 and Y \u003d nx-n 2
11. Let's draw AD perpendicular to x, where D belongs to the x-axis.
12. Segment AD \u003d mn (ordinate of point A), i.e. desired segment.
Remark 3. a) indeed, if in our example, n=4 units, m=3 units, then there should be BP=mn=3 units∙4 units=12 units. This is how it turned out for us: BP = 12 units; b) the line B 1 B 2 was not used in this construction. In B too.
There are at least three different ways of constructing the segment AD=mn.
Second way construction of the segment AD=mn, wherem>1unit,n>1unit,mandn– any.
Analysis
An analysis of the previously constructed drawing (Fig. 2), where using the found method of constructing straight lines Y=mx-m 2 and Y=nx-n 2 found t.A (m+n; mn) (this is the first method), suggests that m.A (m + n; mn) can be found by constructing any of these lines (U \u003d mx-m 2 or U \u003d nx-n 2) and the perpendicular AD, where AD is the perpendicular to OX, AD \u003d mn, D belongs to the axis OH. Then the desired point A (m + n; mn) is the intersection point of any of these lines and the perpendicular AD. It suffices to find the angles of inclination of these straight lines, the tangents of which, according to the slope coefficients, are equal to m and n, i.e. tan ά 1= m and tan ά 2 =n. Considering that tg ά 1 =m/1ed=m and tg ά 2 =n/1ed=n, where 1ed is a unit segment, one can easily construct straight lines whose equations are Y=mx-m 2 and Y=nx-n 2 .
single segment
n n>1 units, m and n are any numbers.
P 
construction (Fig.3)
Fig.3
1. Let's draw a straight line OX.
2. On the OX axis, we set aside the segment OA 1 \u003d m.
3. On the OX axis, we set aside the segment A 1 D \u003d n.
4. On the OX axis, we set aside the segment A 1 C 1 \u003d 1 unit.
5. We build C 1 B 1 \u003d m, where C 1 B 1 ┴ OH.
6. Let's draw a straight line A1B1, the equation of which is Y=mx-m2, in the coordinate axes XOU (the scale on the axes is the same).
7. Restore the perpendicular to OX at point D.
8. We get point A (m + n; mn) - the point of intersection of the line Y \u003d mx-m2 and the perpendicular AD
9. Segment AD=mn, that is, the desired segment.
Output: This second method is more universal than the first method, since it allows you to find the point A (m + n; mn) and when m \u003d n> 1 unit, then the coordinates of this point are A (2m; m 2) and AD \u003d m 2.
In other words, this method allows you to find a segment equal to the square of the given one, the length of which is greater than 1 unit.
Comment: Indeed, if in our example m=3 units, n=5 units, then it should be AD=mn=3 units×5 units=15 units. This is how we did it: AD=15 units.
Third way constructing a segmentAD= mn, wherem>1unit,n>1 unit andm≠ n.
Using Figure No. 2, draw a dashed line straight line B 1 B 2 until it intersects with OX at the point E € OX, and a straight line B 1 B ┴ B 2 C 2, then
B 1 B \u003d C 1 C 2 \u003d OS 2 -OS 1 \u003d (n + 1 unit) - (m + 1 unit) \u003d n-m, and B 2 B \u003d B 2 C 2 -B 1 C 1 \u003d m-n => B 1 В=В 2 В=>∆В 1 ВВ 2 - isosceles, rectangular>∆EC 1 В 1 - isosceles, rectangular => ά=45º
Because OS 1 \u003d m + 1 unit, and EU 1 \u003d B 1 C 1 \u003d m, then OE \u003d OS 1 -EC 1 \u003d m + 1 unit-m \u003d 1 unit.
It follows from the reasoning that the points B 1 and B 2 can be found in a different way, because they are the points of intersection of the straight line EB 1 drawn at an angle ά=45º to the axis ОХ and perpendiculars to ОХ: В 1 С 1 and В 2 С 2, and OE=1 unit. Further, using the previous methods, we will have the following construction method.
Single cut.
n n>1 unit, and m≠n.
Construction (Fig.4)
1. Let's draw a straight line OX.
5.Let's build 
ά \u003d C 1 EV 1 \u003d 45º, where B 1 is the point of intersection of the perpendicular C 1 B 1 with the side ά \u003d 45º.
7. Set aside OA 2 \u003d n, where A 2 € OX.
8. Set aside A 2 C 2 \u003d 1 unit, where C 2 € OH.
9. Restore the perpendicular C 2 B 2 to the OX axis at the point C 2, where B 2 is the point of intersection of the perpendicular with the straight line EB 1.
10. We draw a line A 2 B 2, the equation of which is Y \u003d nx-n 2, until it intersects with the line A 1 B 1 at point A.
11. We lower the perpendicular to OX from point A and get AD equal to mn, where D € OX, since in the coordinate planes of the XOY axes the coordinates of the point A (m + n; mn).
Fig.4
Comment: The disadvantage of this method is the same as that of the first construction method, where construction is possible only under the condition m≠n.
Fourth way constructing a segmentAD= mn, wheremandn- any, greater than a single segment.
Single cut.
n n>1 units, m and n are any.
Construction (Fig.5)
Fig.5
1. Let's draw a straight line OX.
2. Set aside OE = 1 unit, where E € OX.
3. Press EC 1 =m, where C 1 € OH.
4. Restore the perpendicular at point C 1 to the OX axis.
5. Let's build ά=C 1 EV 1 =45º, where B 1 is the point of intersection of the perpendicular C 1 B 1 with the side ά=45º.
6. Postponing OA 1 \u003d m, we draw a straight line A 1 B 1, the equation of which is Y \u003d mx-m 2, A € OH.
7. Set aside A 1 D=n, where D € OX.
8. Restore the perpendicular at point D until it intersects at point A with the line A 1 B 1, the equation of which is Y \u003d mx-m 2.
9. A segment of the perpendicular AD = the product of the segments m and n, that is, AD = mn, since A (m + n; mn).
Comment: This method compares favorably with the first and third methods, where m≠n, since we are dealing with any segments m and n, the unit segment can be less than only one of them involved in the beginning of the construction (we have m> 1 unit).
General Problem II
Using a compass and straightedge, construct a line segment equal to the ratio of the other two line segments.
Note:
the unit segment is less than the divisor segment.
The first way to construct a segmentn= k/ m, wherem>1 unit
Single cut.
Building (Fig.6)
2. On the OU we set aside OM = k.
3. Set aside OA 1 on OX = m.
4. On OH, set aside A 1 C 1 \u003d 1 unit.
5. Let's build С 1 В 1 \u003d m, where С 1 В 1 ┴ ОХ.
6. Draw a straight line A 1 B 1, the equation of which is y=mx-m 2 in the XOU coordinate axes (the scale on the axes is the same, equal to 1 unit).
7. Restore the perpendicular MA at the point M to the axis OY, where A is the point of intersection of MA with the straight line A 1 B 1 (i.e. A € A 1 B 1).
8. Lower the perpendicular from point A to the OX axis until it intersects with the OX axis at point D. The segment AD=OM=k=mn.
9. Segment A 1 D \u003d n - the desired segment, equal to n \u003d k / m.
Fig.6
Proof:
1. The equation of the line A 1 B 1 is really Y=mx-m 2, at Y=0 we have 0=mx-m 2 => x=m=OA 1, and the slope is tg
2. In ∆ADA 1 tg 1 D=AD/A 1 D=B 1 C 1 /A 1 C 1 =>A 1 D=AD×A 1 C 1 /B 1 C 1 =k×1unit/m= mn/m=n, i.e. And 1 D=n=k/m is the desired segment.
Comment. Indeed, if in our example m=3 units, k=15 units, then it should be A 1 D=n=k/m=15 units/3 units=5 units. We did just that.
Second way constructing a segmentn= k/ m, wherem>1 unit
Single cut.
Fig.7
1. We build the XOU coordinate axes.
2. On the OU we set aside OM = k.
3. Set aside OE \u003d 1 unit, where E € OX.
4. Set aside EC 1 \u003d m, where C 1 € OX.
5. Restore the perpendicular at point C 1 to the OX axis.
6. We build C 1 EB 1 \u003d 45º, where B 1 is the point of intersection of the perpendicular C 1 B 1 with the side of the angle C 1 EB 1 \u003d 45º.
7. Set aside OA 1 on OX = m.
8. Draw a straight line A 1 B 1, the equation of which is y=mx-m 2 in the XOU coordinate axes (the scale on the axes is the same, equal to 1 unit).
9. Restore the perpendicular MA at point M to the axis OY, where A is the point of intersection of MA with the straight line A 1 B 1 (i.e. A € A 1 B 1).
10. Lower the perpendicular from point A to the OX axis until it intersects with the OX axis at point D. The segment AD=OM=k=mn.
11. Segment A 1 D=n - the desired segment, equal to n=k/m.
Proof:
1.∆B 1 C 1 E - rectangular and isosceles, since C 1 EB 1 \u003d 45º \u003d\u003e B 1 C 1 \u003d EU 1 \u003d m.
2.A 1 C 1 \u003d OS 1 - OA 1 \u003d (OE + EC1) - OA 1 \u003d 1 unit + m-m \u003d 1 unit.
3. The equation of the straight line A 1 B 1 is really Y=mx-m 2, at Y=0 we have 0=mx-m 2 => x=m=OA 1, and the slope is tg
4.V ∆ADA 1 tg 1 D=AD/A 1 D=B 1 C 1 /A 1 C 1 => A 1 D=AD×A 1 C 1 /B 1 C 1 =k ×1 unit/m= mn/m=n, i.e. And 1 D=n=k/m is the desired segment.
Conclusion
In our work, we found and studied various methods for constructing a segment equal to the product or ratio of two other segments using a compass and a ruler, having previously given our own definition of these operations with segments, since we could not find in any special literature not only the definition of multiplication and division of segments, but even mention of these operations on segments.
Here we have used almost all four stages: analysis, construction, proof and research.
In conclusion, we would like to note the possibility of using the found methods for constructing segments in certain branches of physics and mathematics.
1. If you extend the straight lines y=mx-m 2 and y=nx-n 2 (n>m>0) until they intersect with the OS axis, then you can get segments equal to m 2, n 2, n 2 - m 2 (Fig.8), where OK \u003d m 2, OM \u003d n 2, KM \u003d n 2 - m 2.
R
Fig.8
Proof:
If x=0, then y=0-m 2 => OK=m 2 .
Similarly, it is proved that OM= n 2 =>KM=OM-OK= n 2 - m 2 .
2. Since the product of two segments is the area of a rectangle with sides equal to these segments, then, having found a segment equal to the product of the other two, we thereby represent the area of the rectangle in the form of a segment whose length is numerically equal to this area.
3. In mechanics, thermodynamics, there are physical quantities, for example, work (А=FS, A=PV), numerically equal to the areas of rectangles built in the corresponding coordinate planes, therefore, in tasks where, for example, it is required to compare work by the areas of rectangles, it is very it is simple to do this if these areas are represented as segments numerically equal to the areas of rectangles. And the segments are easy to compare with each other.
4. The considered construction method allows you to build other segments, for example, using the system of equations y=mx-m 3 and y=nx-n 3 , you can build segments with data m and n such as m 2 +mn+n 2 and mn(m+n), since the point A of the intersection of the lines given by this system of equations has coordinates (m 2 +mn+n 2; mn(m+n), and you can also construct segments n 3 , m 3 , and the difference n 3 - m 3 obtained on the OS in the negative region at X=0.
Artworks. ... help compass and rulers. Division algorithm segment AB in half: 1) put the leg compass to point A; 2) install mortar compass equal length segment ...
Biography of Pythagoras
Biography >> Mathematics... building regular geometric shapes help compass and rulers. ... help compass and rulers. More than two ... is equal to b/4+p, one leg is equal to b/4, and another b/2-p. By the Pythagorean theorem we have: (b/4+p)=(b/4)+(b/4-p) or ...
Command is designed for sequential construction of curves and straight lines so that the end of the previous object is the beginning of the next object. Construction of geometry in this way is also possible from the menu Tools → Geometry →
| Parameter | Description |
| This button completes the creation of a chain of geometric elements. In this case, the contour of these elements is closed by connecting the last geometric element with the first point of the chain. This button is active when it is possible to close the chain. For example, a chain will not work if only 2 straight segments are built in series - they can be closed only by 3 straight segments - a triangle (minimal figure) will be obtained. But in the case of a Bezier curve, 2 points are enough to close the contour with the help of the third point | |
| Section | Commands for creating straight lines |
| This button is used to draw an arbitrary straight line segment parallel to the selected straight line. This line may be outside the chain under construction | |
| This button is used to draw a straight line perpendicular to the selected straight line. This line may be outside the chain under construction | |
| This button is used to draw a straight segment tangent to the selected curve. This curve must be outside the chain being built. In some cases, the program may offer several options for constructing tangent segments. To select one of them or all together, use the Previous or Next object buttons or, pointing to each desired option, press the left mouse button. If you specify a specific segment length in the field Length, then it becomes possible to construct a tangent segment, the second point of which may not lie on the selected curve | |
| Arc | Arc Creation Commands |
| This button is used to construct an arbitrary arc by sequentially specifying three points in the graphics window or on the options bar | |
| This button draws an arc tangent to the previous element in the chain | |
| curved curve | Curve Creation Commands |
| This button is used to build a spline based on a series of points. | |
| Spline along the poles | This button is used to create a spline based on a number of limit points. At the same time, you can set Weight points and Order Weight determines the "force of attraction" of the curve to a point on the curve. The greater the weight, the closer the curve is to the point. In fact, this is a parameter of the curvature of the curve (the greater the curvature of the curve, the smaller the radius of the bend, and vice versa). Parameter Order defines the minimum number of points by which the curve will be built. Minimum order 3 - allows you to build a curve from three points |
Building Geometry with the Line Tool
Command Line is designed for sequential construction of straight lines and arcs so that the end of the previous object is the beginning of the next object. The options bar for this command contains a degenerate command menu . Construction of geometry in this way is also possible from the menu Tools → Geometry → Line. The options panel for this button contains the following commands:
| Parameter | Description |
| Section | This button is used to construct an arbitrary straight line segment |
| Arc | This button is used to draw an arc tangent to the previous element in the chain. In this case, the direction of creating an arc is changed by moving the cursor in the opposite direction from the starting point of the arc |
| This button completes the creation of a chain of geometric elements. After that, the program goes into the waiting mode for entering a new chain. | |
| If this button is pressed, then a chain of elements is built. If this button is unchecked, then separate elements (lines or arcs) are drawn. |
Construction of curves and polyline
Creation of curves is possible from the menu Tools → Geometry → Curves. The construction of a polyline is possible from the menu Tools → Geometry → Polyline. The Bezier curve is a special case of the NURBS curve. All of these commands are found on the Geometry toolbar. The ways to build them are listed below:
Button Spline is designed to construct a curve of the same name from a series of points. Buttons Presented in the Options Bar Open object and closed object allow you to build an open and closed curve, respectively, when the first and last points are connected. A closed curve can always be switched to an open curve and vice versa.
The spline has extended editing of characteristic points. This is what the button is for. Edit points on the options panel. Also, this command is automatically called when you double-click the left mouse button on an already constructed curve. In this case, the points of the curve are complemented by tangent segments that pass through the characteristic points of the curve.
The curve can be split into parts using menu commands Split → Curve and Split → Curve into N parts. The first command allows you to split the selected curve into 2 parts at the specified point. The second curve allows you to split the curve into several equal parts. To do this, select the number of parts in the options bar and specify the curve to be split.
By moving the characteristic points (square points) and the ends of the tangent segments (round points) with the mouse, you can control the shape of the curve. You can move these points using the keyboard arrows, to do this, move the cursor to the desired point and press the Enter key. After that, it will be possible to move using the arrows with a step that is a multiple of the current step of the cursor. You can also end the move by pressing the Enter key. There are 3 options for moving characteristic points:
- Move in any direction - if the cursor looks like four diagonal arrows when hovering over a point
- Moving in a limited range of directions - if the cursor looks like four orthogonal arrows when hovering over a point
- Moving the cursor causes the geometry to rotate - if the cursor looks like rotating arrows when hovering over a point.
Curve points can be snapped to other objects and other curve points using global and local snaps. The inclusion of the necessary local snapping in the process of moving a characteristic point is possible by pressing the right mouse button (or pressing SHIFT + F10) and selecting the snapping from the drop-down submenu Binding.
Button Spline along the poles is designed to build a curve - a spline along a series of points. For this type of curve, you can set Weight with points and Order curve in the options bar. Parameter Weight determines the "force of attraction" of the curve to a point on the curve. The greater the weight, the closer the curve is to the point. In fact, this is a parameter of the curvature of the curve (the greater the curvature of the curve, the smaller the radius of the bend and vice versa). Parameter Order defines the minimum number of points by which the curve will be built. Minimum order 3 - allows you to build a curve using three points. The pole spline resembles a regular spline in point edit mode. If the end points of adjacent tangent (tangential) segments in to the spline are connected, then we get a similarity of the spline along the poles. A pole spline is inherently smoother than a regular spline due to the fact that a pole spline provides curvature continuity.
If you build 2 splines along the poles, then you can connect their ends so that continuity (“smoothness”) is ensured at the transition point.
To do this, you need to build an auxiliary line at the transition point with the required slope (for example, a tangent auxiliary line at this transition point) and place the second points from the transition point on this auxiliary line. Now, when moving 3 points and above (when viewed from the transition point), any of these curves will maintain the condition of the curve continuity at the transition point.
You can add a characteristic point by simply clicking the left mouse button on the desired section of the curve.
You can delete a characteristic point using the DEL key when selecting the required point. This will change the shape of the curve.
The interface for working with splines by poles is similar to the interface for working with regular splines. In the options panel, you can also create Open object and a closed object. And with a button Edit points you can also correct the shape of the curve by moving keypoints. In the same way that snaps work with Bezier curves, points are moved and the curve is split into parts.
Button broken line is designed to build a series of interconnected straight lines. A polyline differs from the usual sequence of straight segments in that the shift of any element does not break the line.
The interface for working with broken lines is similar to the interface for working with curves. In the options panel, you can also create Open object, and closed object. And with a button Edit points you can also correct the shape of the polyline by moving the keypoints. In the same way as with curves, snaps work and points are moved. A distinctive feature of a polyline is that it can be broken into separate elements using the menu command Editor → Destroy. After that, individual elements of the polyline can be moved or deleted without affecting other elements.
If it is quite natural that, with the assumption of a greater variety of tools, it turns out to be possible to solve a larger set of construction problems, then one could foresee that, on the contrary, under the restrictions imposed on tools, the class of solvable problems will narrow. All the more remarkable is the discovery made by the Italian Mascheroni (1750-1800): all geometric constructions that can be done with a compass and straightedge can be done with just a compass. It should, of course, be stipulated that it is actually impossible to draw a straight line through two given points without a ruler, so this basic construction is not covered by Mascheroni's theory. Instead, one has to assume that a line is given if two of its points are given. But with the help of a compass alone, it is possible to find the point of intersection of two lines given in this way, or the point of intersection of a line with a circle.
Probably the simplest example of Mascheroni's construction is the doubling of a given segment. The solution was already given on p. 185. Further, on p. 186, we learned how to divide a given segment in half. Now let's see how to bisect an arc of a circle with center O. Here is a description of this construction. With a radius, we draw two arcs with centers From the point O we set aside two such arcs on these arcs and that Then we find the point of intersection of the arc with the center P and the radius and the arc with the center and radius Finally, taking the segment as the radius, we describe the arc with the center P or until the intersection with the arc is the intersection point and is the desired midpoint of the arc. The proof is left to the reader as an exercise.
Rice. 48. Intersection of a circle and a line not passing through the center
It would be impossible to prove Mascheroni's main assertion by showing, for every construction that can be done with a compass and a straightedge, how it can be done with a single compass: after all, there are an infinite number of possible constructions. But we will achieve the same goal if we establish that each of the following basic constructions is feasible with a single compass:
1. Draw a circle if the center and radius are given.
2. Find the intersection points of two circles.
3. Find the points of intersection of the line and the circle.
4. Find the point of intersection of two lines.
Any geometric construction (in the usual sense, with the assumption of a compass and straightedge) is made up of a finite sequence of these elementary constructions. That the first two of them are feasible with a single compass is immediately clear. More difficult constructions 3 and 4 are performed using the inversion properties discussed in the previous paragraph.
Let us turn to construction 3: find the points of intersection of a given circle C with a straight line passing through these points. We draw arcs with centers and radii, respectively, equal to and except for the point O, they intersect at the point P. Then we construct a point reciprocal to the point P with respect to the circle C (see. construction described on page 186). Finally, we draw a circle with a center and radius (it will certainly intersect with C): its intersection points with circle C will be the desired ones. To prove it, it suffices to establish that each of the points is at the same distance from (as for the points, their analogous property immediately follows from the construction). Indeed, it suffices to refer to the circumstance that the point inverse to the point is separated from the points by a distance equal to the radius of the circle C (see p. 184). It is worth noting that the circle passing through the points is the inverse line in inversion with respect to the circle C, since this circle and the line intersect
Rice. 49. Intersection of a circle and a straight line passing through the center
with C at the same points. (When inverted, the base circle points remain fixed.)
The indicated construction is not possible only if the line passes through the center C. But then the intersection points can be found by the construction described on page 188, as obtained when we draw an arbitrary circle with center B intersecting with C at points Method of drawing a circle inverse to a straight line connecting two given points immediately gives a construction that solves Problem 4. Let the lines be given by points (Fig. 50).
Rice. 50. Intersection of two lines
Let us draw an arbitrary circle C and, using the above method, construct circles that are inverse to the lines and These circles intersect at point O and at one more point Point X, the inverse of the point, is the desired intersection point: how to construct it has already been explained above. That X is the desired point is clear from the fact that there is a single point inverse to a point that simultaneously belongs to both lines and, therefore, a point X, the inverse must lie simultaneously on and on
These two constructions complete the proof of the equivalence between Mascheroni's constructions, in which only compasses are allowed, and ordinary geometric constructions with compasses and straightedge.
We did not care about the elegance of solving the individual problems we have considered here, since our goal was to clarify the inner meaning of Mascheroni's constructions. But as an example, we will also indicate the construction of a regular pentagon; more precisely, we are talking about finding some five points on a circle that can serve as the vertices of a regular inscribed pentagon.
Let A be an arbitrary point on the circle K. Since the side of a regular inscribed hexagon is equal to the radius of the circle, it will not be difficult to set aside on K such points that
