Kvant 1971, No. 10 · Pages

Popular science physics and mathematics
quantum
10
1971
magazine
Academy
Sciences of the USSR
And
Academy of Pedagogical
Sciences of the USSR

In the room:

Human language and machine language 1
Geoacoustic exploration of underwater mineral deposits 6
Diffusion in metals 10
About one property of biomial coefficients 16
Spinning Wolf Mechanics 21
Interstellar ships on gravity resources 26
Laboratory «Kvant»
Artificial Mirages 29
Problem «Kvant»
Problems 33
Solutions to problems M64—M69; F75—F78 35
Applicant’s workshop
Systems of algebraic equations 45
Chess and mathematics notes 49
Reviews, bibliography
Collections of author’s errors 53
Not a textbook or a problem 54
Information
Linguistics and mathematics meet at Olympiad 56
The Olympics are held by car 58
Answers, directions, solutions 62
Collector’s Corner
The stamps show nuclear reactors 63
«Kvant» for younger schoolchildren, 3rd cover page
P. S. Guter
O. G. Bespalov, A. I. Nastyukha
B. Kuliti
A. I. Shirtsov
S. A. Krivoshilkov
I. A. Vorobiev
P. Wood
N. B. Vasiliev, I. Sh. Slobodetsky
M. I. Shabunin
E. Ya. Geek
A. F. Khrustalev
S. V. Atsechelov, V. A. Baryshchev
V. V. Raskin
Yu. I. Sorkin, N. V. Shkundina
A. V. Altykis
A. P. Savin

R. S. Guter

Many readers «Kvant» apparently have some idea of ​​what electric computers can do. It is even more interesting to get acquainted with how they do it.

Note that different machines are not required to perform different tasks: electric computers are universal, that is, a new task does not require a new machine, but a new program for it. Programs, methods of composing and recording them will be discussed in this article.

Any electrical computer includes

1. input and output devices;
2. storage devices (in short: memory);
3. arithmetic devices (or simply arithmetic);
4. control device (control).

Input and output are intended for “communication” between a person and a machine. They allow you to “tell” the machine the source data and the method for processing it (that is, the program) and “learn” from the machine the results obtained. The machine’s memory stores all the necessary information: initial data, intermediate and final results, and, of course, the program.

Arithmetic is used to perform operations on numbers that come there from memory (basically these are general arithmetic operations, which explains the term). Finally, the control manages the process of solving the problem itself: it recalls information about the next operation from memory and deciphers it, sends the necessary numbers from memory to arithmetic and sends the result back to memory.

An idea of ​​the connection between the listed devices can be obtained from figure 1, depicting the so-called block diagram of a computer.

The machine’s memory consists of a large number (denoted by N) identical cells, numbered in a row from 0 to N - 1. The cell number is called its address. Each cell contains a specific number

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binary bits, that is, it can contain a set of zeros and ones, the so-called machine word.

The arithmetic device of machines can perform a certain set of elements of operations. The algorithm for solving any problem must be written in the form of a sequence of such elements of operations, which are called programm. The program consists of separate commands, each of which contains information about one elementary operation.

The command (represented as a machine word) is divided into two parts: operational and address. The operational part of the command contains information about the nature of the operation being performed (add, multiply, etc.): the elementary operations of the machines are renumbered and the number of the required operation is written in the bits of the machine memory cell allocated for the operational part. This number is called the operation code (we will simply say code).

The address part of the command contains the addresses of the cells whose contents are involved in the operation. The number of addresses in a command may vary for different machines. We will limit ourselves to considering three-address machines, in the address part of which the addresses of three memory cells are indicated: the first, second and third addresses of the command. For an arithmetic operation (for example, addition), the numbers of cells containing the original numbers (addends) are written in the first and second addresses. The third address indicates the cell number where the result (that is, the amount) should be written. For other elements of operations, the address used may be different (in part, one, two or even all three addresses may turn out to be unnecessary for any command, then the corresponding address cells will be simply filled with zeros).

Let, for example, a system command (a set of elements of operations) of machines contain $64=2^6$ elementary operations.

ny operations. Then six double digits are enough to number them, that is, the operation code can be written in the machine memory as a six-digit double number. It is usually more convenient for the programmer to break this number into two “triads”, each of which is then written with one digit in the octal $(2^3=8)$ number system, thus obtaining a two-digit octal number. If, for example, $N=4096=2^{12}$, then in such a “binary-octal” system, to write any address you will need not a twelve-row, but only a four-row, easier to write and read number.

If, say, code 01 means addition and 04 means division, then the machine word

01 1254 3172 6314

means the command: add the numbers written in cells with addresses 1254 and 3172, and write the sum in cell with address 6314. Similarly, the word

04 4365 2710 5024

it is instructed to divide the number from cell 4365 by the number from cell 2710 and place the part in cell 5024.

The address of the cell from which the control device must extract the next command for execution is recorded in a special counter of commands that are part of the control device. If a certain address I is recorded in the counter, then they say that “the control is in cell I.” When a command corresponding to an arithmetic operation is executed, one is automatically added to the contents of the program counter, so that control is transferred to the cell with the next address. Therefore, if we want the machine to perform any sequence of arithmetic operations, it is necessary to place the corresponding instruction in memory cells in a row.

Let’s write, for example, a program to calculate the value

$$y = \frac{a+bx}{ax+b}.$$

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numbers from cells $a$ and $R_1$ and write the sum in cell $R_1$), which, of course, should not be understood as “equality” $a=0$. By the way, often instead of the = sign, they write something like →; but we will use the arrows for other purposes.

As already mentioned, executing an arithmetic instruction automatically transfers control to the next cell, adding one to the program counter. But very often it is not possible to arrange all the necessary commands “in one column”. Thus, when solving a quadratic control $ax^2+bx+c=0$, depending on the sign of the discriminant $D=b^2-4ac$, it is necessary to calculate either real or minimal roots, which is done, as is known, in different ways.

To change the order of command execution, among the elements of machine operations, special control commands are provided (more precisely, control transfers), which do not perform operations on the contents of memory cells, but change the contents of the program counter.

The simplest command of this type is the unconditional transfer of control command, which always transfers control to the memory cell specified in it. When it is executed, the contents of the program counter are cleared and the address specified in the command is written to it, so that the next command will be called by the control device from the specified cell.

Since for this it is enough to comma one address in the command, and the other two addresses will remain free, they can be used by forcing the machine to perform some other action when executing the same command. Typically, control transfer is accomplished by transferring the contents of one memory cell to another. Assuming, for example, that the third address of the control transfer command is counted in the program counter, you can write it in the meaningful notation cell like this:

$$a \rightarrow b, c$$

(“send the contents of cell $a$ to cell $b$ and transfer control to cell $c$”).

The control transfer condition commands are more complex; they transfer control only when certain acceleration conditions are met. This is also what commands require when programming branching processes. Let’s look at one of these commands:

$$a < b, c.$$

When it is executed, the numbers written in memory cells $a$ and $b$ are compared. If the inequality written in the command is true, then control is transferred to cell $c$, that is, instead of the contents of the program counter, the third address of the command is written to it (as with unconditional transfer of control). If $a \geq b$, then one is added to the contents of the counter, as in arithmetic operations.

Using the described commands, you can already request a solution to some non-computational problems. Suppose, for example, that cells $x_1$ and $x_2$ contain non-toric numbers, from which you need to select the larger one and write it in cell $x$. This can be done using the following program:

1. $x_1 < x_2$, 3,
2. $x_1 \rightarrow x$, 4,
3. $0 + x_2 = x,$
4. $cmon$

Its first command is a conditional transfer of control: if $x_2$ is greater than $x_1$, then control will be transferred to command 3). This is an arithmetic command that will result in the number $x_2$ being written in cell $x$. (Often the number 0 was encountered; it turned out to be convenient to “assign” to the cell numbered 0, which is done in almost all machines.) Then control is transferred to the next cell 4), which contains the command $cmon$.

If $x_1 \geq x_2$, then the command 1) transfers control to the next cell 2), where the unconditional control transfer command is located, sending more than the number $x_1$ to cell $x$ and transferring control immediately to $cmon$.

Instead of specifying the number or condition of the command that is transferring control, it is often just an arrow to the desired command. Here’s what, for example, a similar program looks like, from which

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In search of new sources of raw materials, people are increasingly turning to the countless riches of class minerals hidden by the watery head of the seas and oceans. According to scientists, the reserves of oil, metal ores and other minerals located at the bottom and under the water bodies are several times greater than the reserves of these minerals on land.

But how can you find out where these treasures are hidden underwater? To do this, you need to know the geological structure of the earth’s crust covered by water areas. Using maps of the geological structure, specialists can judge where and what minerals to look for.

For geological studies of the seabed, as on land, geophysical exploration methods are used - seismic exploration and well drilling. But to obtain detailed information about the geology of the area, you need to drill a lot of wells at random, each of which is difficult to drill even on land. And drilling into the seabed is an even more cumbersome and extremely expensive undertaking.

Seismic exploration explores the geological structure of the earth’s crust using explosions. A small hole is drilled in the ground and a charge is detonated in it. The expanding gases formed during the explosion cause deformation of the soil, removing the layer of particles that indefinitely surrounds the explosion site. The elastic forces that arise in the sloped layer transmit pressure to neighboring layers, and from the point of explosion a elastic shock wave (Fig. 1) propagates through the rock in all directions.

Let us assume that under the first fracture of rock I at depth S lies layer II, composed of rock of a different density. The depth of layer II is not difficult to determine (see Fig. 1):

$$S = \sqrt{S_1^2 - \frac{(S_2)^2}{2}},$$

where $S_1 = \frac{1}{2} ABC$ is the path traveled by the direct wave to layer II (the same path travels the reflected wave), $S_2$ is the distance along the surface between the point of explosion and the seeding receiver, recording the arrival power of the reflected wave.

The energy of elastic waves when passing through the earth’s rocks is partially absorbed. The fraction of absorbed energy depends on the frequency of wave oscillations. An explosion produces waves of infrasonic (below 16 hertz) and sound (from 16 hertz to 20,000 hertz) frequencies. Waves of these frequencies are relatively little absorbed by the earth’s rocks, and this determines their use in the sowing receiver.

Seismic exploration has been successfully used to study the structure of marine

At the stern of the ship, a cable goes into the water, at the end of which electrodes are fixed about a meter from the surface of the water. Every 10 seconds, a high voltage pulse is sent to the electrodes from a high-voltage capacitor bank installed on the ship, and a small lightning flashes between the electrodes. A sensitive blow is felt on the hull of the ship, and the sound wave hits the bottom through the water column.

Reflected waves on the sea surface are captured by a receiving device towed behind the ship, which geophysicists call a “scythe”. A scythe is a flexible plastic pipe filled with oil, inside of which there are elements that convert sound.

Fig. 3.

the energy of reflected waves in electrical impulses (piezoelectric elements).

Through the cable connecting the scythe to the corps, electrical impulses go to the amplifier, and after the amplifier the recordings are recorded. The recorder contains a wide strip of electrosensitive paper. There is a “feather” at the top edge of the tape.
When the sound wave from the electric discharge begins its run to the bottom, the pen begins to move across the paper tape to its lower edge, as if repeating the movement of the wave on a smaller scale. And when the reflected waves arrive at the receiving spit and from it through the amplifier to the pen, the pen leaves points on the sensitive paper corresponding to the layers of rock encountered on its way by the sound wave.

Before the next discharge, the pen returns to the top, and the chutchuta paper is pulled out, and the recorder is again ready to record reflected waves.

A few minutes after the start of work, dotted lines on paper form a visual and easy to read figure structure of the geological section under study (Fig. 3).

The geoacoustic location method can be used not only for searching for minerals. With its help, you can solve problems of engineering geology - choose places for hydroelectric dams, for the foundations of port facilities, jetties, tunnels, pipelines, and even search for ancient cities flooded with water. The cost of exploration by this method is 5-10 times cheaper than seismic exploration and exploration by drilling wells.

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Diffusion is a very common phenomenon in nature and is of great practical importance. We are posting here an article by B. Culliti, prepared by the Nauka publishing house for the next issue of the collection “What Physicists Are Thinking About.” The article was published in Scientific American. Translation by V.K. Fedyanin.

Soviet scientists made significant contributions to the study of diffusion. Those who wish to become more familiar with this phenomenon can find a lot of interesting information in the popular science book by Ya. E. Geguzin “Essays on Diffusion in Crystals,” published by the Nauka publishing house in 1970.

The smell of perfume is known to exist at quite a distance. This is explained by the fact that perfume vapor diffuses very easily in the air. Drops of liquid dye in water diffuse easily throughout the vessel. Diffusion in a solid is much more difficult to observe. For this reason, the study of diffusion in solids has become one of the most interesting studies in physics today.

As in many other areas of a person, it reduces effectiveness; in this case, it reduces the precedence of knowledge. For centuries, workers welded metals and produced steel by heating solid iron in a carbon atmosphere, without having the slightest idea of ​​the diffusion processes taking place. Only in 1896 did scientific study of the problem begin.

The English metallurgist Sir William Roberts-Austin measured the diffusion of gold in lead in a simple experiment. He placed a thin disk of gold on the end of an inch-long cylinder of pure lead, placed this cylinder in a furnace where the temperature was maintained at about 200°C, and kept it in the furnace for 10 days. He then cut the cylinder into thin disks and measured the amount of gold that diffused into each lead medium. It turned out that by the “clean” end a completely measurable amount of gold had passed through the entire cylinder; lead diffused in the opposite direction into the depths of the gold disk (Fig. 1). Roberts-Austin discovered that heated metal diffuses into another when they are pressed closely together.

From the point of view of the atomic structure of matter, the permeability of solids is not a country. Nowadays we have a good idea that even the densest solid body is only quite weakly

*) Inch≈2.5 cm.

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connected set of atoms. In the crystals that form a metal, the atoms are arranged in a fixed lattice in certain positions from which they are quite difficult to dislodge.
However, ideal and fully built-up crystal lattices do not exist. They always contain a certain amount of vacancies, or “holes”, in which diffusing atoms can “jump”. When it jumps into a certain vacancy, the atom leaves behind a new vacancy.
A neighboring atom can move into this vacancy; intact, by continuous movements of atoms in the lattice, atom can pass through the crystal (Fig.2).

What makes an atom leave its place in the lattice and “scan” into a vacancy? The fact is that in fact, atoms in lattices are not in fixed positions in an array. As a result of thermal motion, atoms continuously vibrate relative to a certain average or, more precisely, upsaverage position over various possible locations of their location. The amplitude of these vibrations can vary from atom to atom and over time. From time to time, an atom is so far from its equilibrium position that it can jump to an adjacent vacant location. The tendency to this, of course, increases sharply if we heat the body, thereby causing the atoms to vibrate more strongly. Thus, the rate of diffusion increases very quickly with increasing temperature. For example, zinc diffuses into copper at 300°C almost 100 million times faster than at room temperature.

In the Roberts-Austin experiment, gold atoms at 200°C are energetically removed from the sausage to begin jumping into lattice vacancies (the same is the case for lead atoms). After all, if the experiment had been carried out for a long time, the gold atoms would have been evenly distributed throughout the lead cylinder. A simple analysis can be carried out. If you put several black balls into a box of white balls and shake the box for a long time, then the black balls turn out to be evenly distributed among the white ones.

Fig. 1. Figure will encroach on the first Roberts-Austin diffuse experiments. Gold was placed on pure lead (gray cylinder on the left) and the sample was heated. On the right is the same cylinder after diffusion of gold into lead and lead into gold has occurred. This cylinder was cut into thin disks to measure the amount of product-founding gold they contained. The curve shown in the graph on the right shows the change in gold concentration depending on the distance from the top edge of the right cylinder.

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Fig. 2. There may be vacancies in the atomic planes of the crystal. This figure is used in how gold (yellow circle) can diffuse in a lead crystal. The transition of gold atom to the position shown in the lower right figure is possible if there is only one vacancy in the lead lattice.

The diffusion of atoms in a pure metal (called “self-diffusion”) can be altered by using a minute fraction of the element’s radioactive atoms. Let’s say, in order to change the rate of self-diffusion in copper, it is necessary to electron-deposit a small amount of a radioactive isotope of copper onto the end of a cylinder of ordinary copper and heat this cylinder to a certain temperature. After enough time has passed for diffusion to occur, the sample is cut into thin slices and a Geiger counter is used to change the radioactivity of each successive medium. These data allow the diffusion rate to be calculated.

A metal can diffuse into another metal only if there is a vacancy in the crystal lattices, since the atoms of even two different metals have sizes of approximately the same order.

However, atoms of a nonmetallic element, whose dimensions are much smaller than those of a metal atom, can hang between the atoms in a nonmetallic crystal. In this case, diffusion does not depend on the vacancy, therefore, it occurs much faster. However, the most important example of such a case is the diffusion of carbon into iron.

In its directional form, steel is an alloy of carbon and iron. The properties of steel vary depending on the carbon content of the alloy. Steel with a small carbon content (say, on the order of one-quarter percent) is not very strong and does not forge well. Increasing the carbon content makes it stronger but more brittle. A machine part such as a transmission mechanism must have a hard surface to resist wear, and be sufficiently elastic in its internal parts,

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So as not to break due to external impacts. These properties are acquired through a process called surface hardening. A piece of metal is heated and kept in a carbon-rich gas, such as methane. Carbon atoms from the gas diffuse slowly into the steel (Fig. 3). After a few hours they will penetrate to a depth of about one-twelfth of an inch. The surface or “shell” of this piece of steel after this treatment contains about 1.2 percent carbon, while the interior contains only a quarter of a percent (Fig.4).

Knowledge of the processes that occur during the diffusion of atoms into a metal can help us in studying the internal structure of the metal. Generally speaking, a piece of metal is made up of small crystals, or “grains.” The internal structure of an individual grain is highly regular - it is an almost ideal lattice with a certain number of vacancies or other minor deviations from ideality. But taken together, the grains do not form an ordered structure. Their relative position is purely accidental. Very rarely, one grain is adjusted by its edge to another so that the regular arrangement

the atoms in it were, as it were, a continuation of the arrangement of atoms of another grain. Obviously, these grains form a certain skew angle, called the “divergence angle.” Atoms located in the program area between the grains are nestled between crystals and eventually end up inside one of them. As a result, a greater destruction of atoms is created in this area than in grains. Diffusing atoms pass through such region programs much faster than through the regular lattice of the grain itself. It is well known that metal atoms diffuse much faster in a fine-grained metal, which is literally riddled with grain boundaries, than in a metal built from large crystals.

M. Akhter and R. Smoluchowski conducted an experiment to determine the effect of the divergence angle between grains on the diffusion rate. They prepared a cylinder of copper, consisting of large filamentary crystals arranged in such a way that the planes of the lattices made filament angles with each other. An alloy of copper and silver was welded to the end of the copper sample and the whole thing was heated to a temperature of 675 ° C. After 10 days they prepared

Fig. 3. The diffusion of carbon in iron does not depend on the presence of a vacancy. The carbon atoms (black circles) are small enough to fit between the iron atoms (blue circles).

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Fig. 4. Surface hardening of steel occurs as a result of the specific diffusion of carbon atoms into it. The middle part figure shows a micrograph, or structuring, of the distribution of carbon in a piece of steel, the surface of which is in contact with hydrocarbon. The top curve gives the percentage of carbon as a function of the distance to the surface of the piece of steel. The lower curve characterizes the strength of steel in different parts of the piece.

Fig. 4. Surface hardening of steel occurs as a result of the specific diffusion of carbon atoms into it. The middle part figure shows a micrograph, or structuring, of the distribution of carbon in a piece of steel, the surface of which is in contact with hydrocarbon. The top curve gives the percentage of carbon as a function of the distance to the surface of the piece of steel. The lower curve characterizes the strength of steel in different parts of the piece.

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as an experimental error. In order to convince doubters, Kirkendile, in collaboration with A. Smigelax, conducted a new series of experiments. They applied only a layer of copper on opposite sides of the brass stubble, and between the copper and brass on both sides of the brass stubble they placed molybdenum wire so that it would fix the boundary between the copper and brass. After this, the sample was heated to a high temperature. If copper atoms entered the brass rod as quickly as zinc atoms exited it, then the original size of the brass rod would not change. This would be indicated by the stationary position of the molybdenum wire. But in fact the molybdenum wire had shifted, and its dislocation indicated that the brass rod had contracted; that is, zinc atoms came out of stubble faster than copper atoms entered it. Under a microscope, it was possible to smooth out fairly large pores that had formed in the surface as a result of the appearance of a large number of vacancies arising where zinc atoms quickly escaped from the pores.

These results were published in 1947, and what is now called the “Kirkendyle effect” created a real sensation among metallurgists and all researchers who studied the properties of solids.

Old experiments were tested and new ones were carried out.

Since the Kirkendile effect was immediately discovered in many alloys, there was no doubt that it actually existed.

Undoubtedly, the study of diffusion will deepen our understanding of the properties of solids and all those engineering processes in which diffusion plays a major role.

ASCHER’S DRAWING

If a plane is covered with a polygon so that two polygons have either a common side, a common vertex, or have no common points at all, then such a covering is called parquet.

In our magazine (see «Kvant» No. 3, 1970, p. 21) we presented a number of problems related to parquet floors.

For some parquets, it is possible to specify plane transformations that transform the figures that make up the parquet into each other so that the parquet itself does not change.

For the parquet shown on this figure, its transformations also do not change: rotation around the point by 60° and symmetry with respect to straight line l. What other transformations can you point out that do not change this parquet?

You see a completely different kind of parquet on figure by the Dutch artist Escher (see cover). This parquet is replaced by the fact that one of the transformations that does not change it is doubling all distances relative to the center (similarity transformation). Therefore, strictly speaking, figure depicts an infinite number of shrinking lizards. Try to come up with other parquets that do not change their similarity during transformations.

L. Limanov

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A. I. Shirshov

In «Kvant» No. 6 for 1970, an article by D. B. Fuks and M. B. Fuks “Arithmetic of biomial coefficients” was published. In this paper, the authors proposed the following problem: let $p$ be a prime number; find out by what power $p$ the number $C_{p^{n+1}}^{p^n} - C_{p^n}^{p^{n-1}}$ is divisible

This problem has attracted the attention of many of our readers. The editors have received letters from V. G. Gaidalovich (Tashkent), S. P. Krutik (stanitsa Staro-Sherbinskaya, Krasnodar Territory), 9th grade student A. Merkuverev (Leningrad), V. V. Prlensky (Moscow) and other goods who will offer varying degrees of solution to this problem.

We asked the prominent Soviet algebraist, corresponding member of the USSR Academy of Sciences A. I. Shirshov to highlight in detail the solution to the problem.

Schoolchildren are familiar with many popular science books and articles on mathematics. They are written as simply as possible, detailed and accessible, in bright and lively language, with engaging examples and colorful pictures - so as to interest the reader and give him the opportunity to easily understand the material.

How are server-side mathematical articles written in reputable scientific journals? The article by A. I. Shirshov, offered to the attention of readers, gives an idea of ​​the scientific mathematical style of presentation. In terms of its content, it is accessible to high school students, but it is written specifically in the same way as notes are written, say, in the journal “Reports of the USSR Academy of Sciences.”

The scientific and mathematical style is characterized by slope and brevity. As a rule, the presentation is divided into separate steps - lemmas, rising along which we arrive at the final result - the theorem. The initial article always contains a statement of the problem, an explanation of the concepts introduced and notations used, and references to the literature. The proofs are not reproduced in detail - only their idea and the general course of reasoning are outlined. Therefore, you can read such an article only with a pencil and paper in your hands, restoring intermediate calculations; understanding its content requires a certain amount of effort and independent reflection.

A math lover will undoubtedly be interested in learning about the scientific mathematical style of presentation.

In article [1], it was proposed that numbers of the form $C_{p^{k+1}}^{p^k} - C_{p^{k-1}}^{p^{k-2}}$, where $p$ is prime, are divisible by a sufficiently high number $p$. However, this statement has been proven only in the case $p=2$ (Theorem 5 from [1]).

This note is devoted to the proof of a more general theorem.

All numbers $m, k, t, s$, etc., considered below are natural, and $p$ is a prime number. The symbol $(m, n)$, as usual, means the most common divisor of the numbers $m$ and $n$. We will use the symbol $[k, s]_n$ to denote the product of all natural numbers $t$ such that $k \leq t \leq s$ and $(t, n)=1$; on-

Proof. For $n=1$ the equality is trivial. Let’s use the induction method:

$$(\alpha + x_1) \dots (\alpha + x_{n-1}) (\alpha + x_n) - x_1 \dots x_{n-1} x_n = \left\{ x_1 \dots x_{n-1} + \alpha x_1 \dots x_{n-1} \left( \frac{1}{x_1} + \dots + \frac{1}{x_{n-1}} \right) + \alpha^2 N_1 \right\} \times (\alpha + x_n) - x_1 \dots x_n = \alpha x_1 \dots x_{n-1} + \alpha x_1 \dots x_n \left( \frac{1}{x_1} + \dots + \frac{1}{x_{n-1}} \right) + \alpha^2 N_1 + \alpha^2 N_1 x_n = \alpha x_1 \dots x_{n-1} x_n \left( \frac{1}{x_1} + \dots + \frac{1}{x_{n-1}} + \frac{1}{x_n} \right) + \alpha^2 N,$$

Q.E.D.

Lemma 11.

$$[p^k - p^{k-1}, p^k]_p - [1, p^{k-1}]_p = \frac{1}{2} p^{2k-1} (p-1) [1, p^{k-1}]_p \times \left\{ \frac{1}{1 \cdot (p^{k-1})} + \dots + \frac{1}{(p-1) (p^{k-1} - p+1)} + \frac{1}{(p+1) (p^{k-1} - p-1)} + \dots + \frac{1}{(p^{k-1} - 1) \cdot 1} \right\} + p^{4k-2} Q,$$

$p > 2$;

$$[2^k - 2^{k-1}, 2^k]_2 - [1, 2^{k-1}]_2 = 2^{2k-2} [1, 2^{k-1}]_2 \left\{ \frac{1}{1 \cdot (2^{k-1} - 1)} + \frac{1}{3 (2^{k-1} - 3)} + \dots + \frac{1}{(2^{k-1} - 1) \cdot 1} \right\} + 2^{4k-2} Q$$

(in the expressions in curly brackets all numbers that are coprime to $p$ and $2$, respectively, are written down).

Proof. It’s obvious that

$$(p^k - p^{k-1} + s) (p^k - s) = p^{2k-1} (p-1) + s (p^{k-1} - s).$$

Let us now write down the product $[p^k - p^{k-1}, p^k]_p$ and, using this equality, the variables are pairwise co-constituents, equidistant from the ends:

$$[p^k - p^{k-1}, p^k]_p = (p^k - p^{k-1} + 1) (p^k - p^{k-1} + 2) \dots$$

$$\dots (p^k - p^{k-1} + p-1) (p^k - p^{k-1} + p+1) \dots (p^k - 2)(p^k - 1) = \left\{ p^{2k-1} (p-1) + 1 (p^{k-1} - 1) \right\} \left\{ p^{2k-1} (p-1) + 2 (p^{k-1} - 2) \right\} \dots$$

$$\dots \left\{ p^{2k-1} (p-1) + (p-1) (p^{k-1} - p+1) \right\} \left\{ p^{2k-1} (p-1) + (p+1) (p^{k-1} - p-1) \right\} \dots \left\{ p^{2k-1} (p-1) + \frac{1}{2} (p^{k-1} - 1) \frac{1}{2} (p^{k-1} + 1) \right\}.$$

We transform the resulting expression according to Lemma 10, taking $\alpha = p^{2k-1} (p-1)$ and choosing the numbers as $x_i$

$$1 \cdot (p^{k-1} - 1), 2 (p^{k-1} - 2), \dots, \frac{1}{2} (p^{k-1} - 1) \frac{1}{2} (p^{k-1} + 1).$$

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Noticing that
$$1(p^{k-1}-1) \cdot 2(p^{k-1}-2) \dots \frac{1}{2}(p^{k-1}-1) \frac{1}{2}(p^{k-1}+1) = [1, p^{k-1}]_p,$$

we can write:
$$[p^k - p^{k-1}, p^k]_p = [1, p^{k-1}]_p + p^{2k-1}(p-1)[1, p^{k-1}]_p \cdot \frac{1}{2} \left\{ \frac{2}{1(p^{k-1}-1)} + \dots \right\} + p^{4k-2}Q =$$
$$= [1, p^{k-1}]_p + \frac{1}{2} p^{2k-1}(p-1)[1, p^{k-1}]_p \times$$
$$\times \left\{ \frac{1}{1(p^{k-1}-1)} + \dots + \frac{1}{(p^{k-1}-1)1} \right\} + p^{4k-2}Q,$$

which is what lemmy proves at $p > 2$.
The case $p = 2$ is considered according to the same scheme.
Lemma 12.

$$1^2 + 2^2 + \dots + (p-1)^2 + (p+1)^2 + \dots + (p^{k-1}-1)^2 =$$
$$= \frac{1}{6} p^{k-1}(p^{k-1}+1)(2p^{k-1}+1) - p^k L, \quad p > 2$$
(all numbers coprime to $p$ are written down on the left side).
Proof. It is enough to represent the left side in the form
$$\sum_{i=1}^{p^{k-1}} i^2 - p^2 \sum_{i=1}^{p^{k-2}} i^2$$

and calculate both sums according to Lemme 1.
Theorem. The following equality is fair:
$$C_{p^k}^{p^{k-1}} - C_{p^{k-1}}^{p^{k-2}} = p^{3k-1} R, \quad p > 3;$$
$$C_{3k}^{3k-1} - C_{3k-1}^{3k-2} = 3^{3k-2} S;$$
$$C_{2k}^{2k-1} - C_{2k-1}^{2k-2} = 2^{3k-3} T,$$

where $k \geqslant 3$ and $R, S, T$ are integers.
The proof of the theorem is carried out by sequential use of lemme 9, 11, 6, 12, 2.

Comment. From the proof of the theorem, one can additionally see that the numbers $S$ and $T$ indicated in the theorem are not divisible by 3 and 2, respectively. The author does not know how things stand in the general case - whether the number $R$ is divisible or not divisible by $p$.

Literature Cited

1. Fuchs D. B., Fuchs M. B., Arthmetic of biomp coefficients. «Kvant» No. 6, 1970, pp. 17-25.
2. Sominsky I. S., Method of mathematical induction. M., Fizmatgiz, 1961.
3. Egorov A. A., Comparisons by modulus and arithmetic of remainders. «Kvant» No. 5, 1970, pp. 27-33.

Editor’s note. As reader Yu. Trakhtman (Dnepropetrovsk) told us, he managed to obtain certain numbers for which the numbers $R, S, T$ appearing in the theorem are not divisible by $p, 3, 2$, respectively. However, this result is formulated in terms unfamiliar to schoolchildren and is proven by non-trivial means.

24

Leno to the axis; therefore, due to friction, the wolf’s axis tends to take a vertical position. This is easy to verify if you run the fibers obliquely. After some time, its axis becomes vertical. The moment of friction forces is directed towards the vertical (the gimlet rule), therefore, the change in the angular momentum of the wolf is also directed towards the vertical, and the axis of the wolf tends to become perpendicular to the plane of rotation.

So, two moments of force act on the inclined fiber: the moment of a couple of forces - the reaction of support and gravity, and the moment of friction. The movement of a wolf will always occur in the presence of these two moments.

Let’s now return to Thomson’s wolf and try to explain his behavior. Since the Thomson fiber consists of a sphere with a middle apex, its center of gravity is located below the geometric center of the ball from which it is made. When spinning the wolf, we will involuntarily deviate its axis from the vertical position. But since the drag has a spherical shape, its support point changes as a result of such deviation. The axis of rotation of the wolf remains the same - vertical and will not coincide with its geometric axis. And since the center of gravity of the wolf lies below the geometric center of the ball, as a result of this

declination, its center of gravity will no longer lie on the axis of rotation (Fig. 7). He will take position $O'$ and will spin around the vertical axis together with the wolf. When rotating at a high angular velocity, the center of gravity of the wolf will rise in the same way as a ball on a thread rises, if the thread is untwisted as shown in figure 8.

The fiber does not hold it in a lateral position, but by inertia jumps over it, touching its lower plane, on which it rotates (Fig. 2). As soon as this happens, the fulcrum of the wolf moves from point $A$ to point $B$ (Fig. 9), and the drag, rotating around its axis, begins to presuppose near axis $BB'$. In other words, the Thomson wire will behave like an “ordinary” wire. Under the influence of the frictional moment, it will align its axis with the vertical axis $BB'$ and will continue to rotate upward with the ball.

From the above reasoning it is clear that we owe the columnar behavior of Thomson’s wolf to the force of friction. Indeed, if there were no friction force, then the drag, having hit the plane, would return to the lateral position and rotate as long as it had a sufficiently high angular velocity. And then, under the influence of the moment of gravity, it would return to its original position (Fig. 2).

25

Fig. 10.

You can conduct an interesting experiment with Thomson’s top. If you run a top over a surface covered with a thin layer of powder, the powder will leave a mark on the surface of the top - the trajectory of the point of contact of the top with the plane. This trace is shown at figure 10. The line on the top twists into a spiral, but at the equator of the top ball it begins to spin in the opposite direction. Why is this happening? The law of conservation of angular momentum requires that the top rotate in the same direction, both in its original position and in its inverted position. Let us launch the top clockwise (if you look at it from above). If the top were to turn over without ceasing to spin around its axis, then in the inverted state it would spin counterclockwise. Therefore, in order for the law of conservation of angular momentum to be satisfied, the top at some point in time must stop rotating around its axis, passing through its leg, and then begin to rotate in the opposite direction. Judging by figure 10, this is what happens at the moment when the top lies on its side.

As for calculating the size of the top, it can be anything, but such that its center of mass does not coincide with the geometric center of the sphere from which it is made.

TASKS
ABOUT CHESS
TOURNAMENT

1. 8 chess players took part in the tournament*). They all scored a different number of points, and the second prize-winner scored the same amount as all the chess players who took places from 5th to 8th. How did the chess players who took 3rd and 5th places play among themselves?

2. Two 7th grade students and several 8th grade students played in the tournament. The seventh-graders scored 8 points together, and all the eighth-graders scored equally. How many eighth-graders could participate in the tournament?

3. n chess players—grandmasters and masters—participated in the tournament. After the end of the tournament, it turned out that each student gained half of his points in the game against the masters. Prove that $\sqrt{n}$ is an integer.

4. A chess player plays at least one game a day for training and, in order not to overwork, no more than 12 a week. Prove that it is possible to find several consecutive days during which he plays exactly 20 games.

5. Every chess player in the tournament scored half of their points in matches with the participants who took the last three places. How many people took part in the tournament?

6. Prove that at the end of the tournament its participants can be re-determined in such a way that it turns out that not a single participant lost to the next one indefinitely after him.

*) We assume that chess players play one game at a time with each other, that is, the tournament is held in one round.

26

Even the closest stars are very far away. There is still a long way to go to other galaxies. From the Andromeda galaxy, light takes about a million years to reach us.

Is it conceivable to get there in, say, a week? According to Earth time, it does not add up to Andromeda so soon. The light goes on for years and years. But the ship cannot move faster than light. Another thing is the time of the interstellar traveler himself, if his ship flies at a sufficiently high speed. When moving at high speeds, all processes are replaced by the same number of times. The higher the speed, the greater the replacement.

How should we fly if we want to travel to the Andromeda Nebula in a week? A distance of 10⁶ light years must be covered in $\frac{1}{52}$ years. Calculations using the formula of the relative theory result: it is necessary to fly with constant acceleration $a=700\text{ g}$! (g is the acceleration of free fall).

But what about overloads? With an acceleration of 700 g, our weight will increase 700 times! It’s like driving over a starlet with an asphalt roller. Our blood also becomes 700 times thicker. And it must circulate throughout the body. The heart cannot cope with such a significant load. Pilots sometimes go blind during sharp turns - the blood does not rise to the brain, although here the acceleration $a$ does not exceed $5\div 10\text{ g}$. You can lie down to reduce the load on the heart - then you need to raise the blood to a lower height. But even a tenfold overload in this position can only be withstood for three to five minutes.

Is it possible to come up with a better way that allows you to move at any acceleration without any overload?

Is it possible to propose a design for a ship depicting an astronaut under any kind of overload?

Before rushing at an enemy, it is useful to study it.

Where does the weight come from?

If you stand up the external scale, the needle will deviate from the zero position. The earth attracts the body, and since we are motionless relative to the scales, this force must be balanced by the elastic force of the arms. The arrow will show us the weight $P=mg(m-\text{ масса})$. Let us change the weight (readings of external scales) in an elevator rising with acceleration. The arrow will show $m(a+g)$, a re-

27

load: elastic force not only balances the Earth’s gravity, but also gives the body acceleration $a$. If the elevator falls freely, then the weight will be zero - all bodies fall with the same acceleration, our movement will be exactly the same as the movement of the spring, and the spring will not be aware of it. Taking an apple, we will not feel its weight; it does not press on our hand, but falls freely along with our hand.

So, the weight $P=m(a+g)$, if $a$ is directed in the same direction as the acceleration of free fall $g$ and $|a|=|g|$ (free fall), a state of weightlessness occurs. This condition is experienced by astronauts when flying with their engines turned off, or when they are moving with the acceleration of gravity.

It is important that weight is due to both acceleration and gravitational attraction. It became possible to compensate one thing for another. It is possible to create weightlessness because the acceleration of gravity does not depend on either the mass or the substance of the falling bodies. This property was discovered by Galileo. But even now, researchers sometimes cannot resist the temptation to test this property. In 1964, Roll, Krotkov and Dicke established the equality of acceleration for gold and aluminum with an accuracy of 0.00000000001!

We will meet weightlessness in the cabin of a free-fall elevator. There are no overloads. Let us take a massive body created by the acceleration of gravity $a$, equal in magnitude to the desired acceleration of travel. Let’s install an engine on it that can provide just this acceleration. The traveler is accommodated with all comforts in a separate cabin, freely falling onto this body. The cabin falls with acceleration $a$, but the engine is running and the body leaves with the same acceleration.

At zero initial speed, the distance between the cabin and the massive body will remain unchanged. In a freely falling cabin there is weightlessness, but at the same time it rushes with the necessary acceleration.

But before the halfway point, you need to start slowing down. The coordination of the acceleration of the body and the free fall acceleration of the cabin will be disrupted, and the cabin will crash into a massive body. In addition, there is a danger of standing up or falling on the body with any change in the acceleration of the ship (massive body). This means that we need to better think through the design of the ship.

Let’s drill a shaft in the direction of acceleration through the center of a massive body, which has been shaped into a ball. The acceleration of gravity will vary depending on the distance to the center of the ball. Let us denote the acceleration on the surface of the ball by $a_0$ and the radius of the ball by $r_0$. Let’s plot the acceleration $a$ at a distance $r$ from the center of the ball (see figure *).

At distances from $-r_0$ to $r_0$ (that is, inside the ball), the acceleration of gravity changes according to the law $a=a_0 \frac{r}{r_0}$, and at $r<-r_0$ and $r>r_0$

*) This problem is discussed in detail in the article by Ya. A. Smorodinsky “Similar movements”. The dependence of the force of attraction of a body on the distance to the center of the ball was found with the solution problem F43 («Kvant» No. 5, 1971).

28

(outside the ball) acceleration $a = -a_0 \left( \frac{r}{r_0} \right)^2$.

The negative sign of the acceleration means that it is always directed towards the center of the ball.

When a ball moves without acceleration, its center is the point of equilibrium. For small deviations, a restoring force occurs. When accelerating the ship $w$, we change $a_0$, there is always a cabin position at which the free fall acceleration is equal to $w$. To find this position on the graph, draw a straight line below the $r$ axis at a distance $w$. Two intersection points will appear, their coordinates give the required distances. The point inside the ball corresponds to the source equilibrium. With a small change in the acceleration of the ship, the cabin will oscillate around the equilibrium point, itself monitoring the consistency of the accelerations, and all this time the cabin will remain inert.

When braking with the same acceleration, the equilibrium point will be on the other side of the center of the ball.

We have proposed a radical way to combat overloads. What “ifs” stand in the way of realizing this fantastic project?

It does not contradict the laws of nature. The main difficulties lie in the power source: this is a common difficulty of any intercity ship project. But where did the massive body come from? You can either take a body of very large volume or use a substance of very high density. To make it easier to think about this, understand the exercises below.

Exercises

1. If a jet plane dives with an acceleration of 1.5 g, what is the weight of the pilot and where is the force of the weight directed?
2. Find the relative radius and mass of a ball of the same density as the Earth to the radius and mass of the Earth if the acceleration on the surface of the ball is 1000 g.
3. What if the density of the ball is 10^6 times greater than that of the earth? (Under some physical conditions, for example, on stars called white dwarfs, the density of matter significantly exceeds our values.)

29

Professor Everett, discussing the phenomenon of mirage (Nature, November 19, 1874), showed that for the formation of sharply successive mirages, the presence of a horizontal plane with a maximum refractive index is necessary. As you move up and down from this plane, the optical density decreases in proportion to the distance from it.

A horizontal or almost horizontal beam is deflected towards the plane with maximum density and intersects it *). Then it changes its curvature and again deviates towards the plane, which it crosses again and again, passing along a trajectory close to a sinusone.

Observing the trajectory of a light beam in a vessel filled with a salt solution (the optical density of the solution increased with depth), I thought:

*) The nature of the mirage is described in the article by G. Ya. Myakishev “Fermat’s principle and the laws of geometric optics” («Kvant» No. 11, 1970).

that by changing the refractive index of the liquid, it would be possible to make the beam move along a sinusone.

The result is quite beautiful experiments that can be shown in the audience. Apparently, it is worth publishing them, although, as I learned later, similar experiments had already been described by Wiener.

I made a vessel from thick sheet glass approximately 50 cm long, 10 cm high and 2 cm wide. I filled it 3 cm with a concentrated alum solution, and on top I placed a 3 cm thick layer of aqueous alcohol solution (~10%) (The addition of alcohol increases the refractive index of water.) You need to carefully add water to the vessel using a pipette, the shape of which is shown in figure 1.

As a liquid with a high refractive index, the removed weight of which is greater than the remote weight of water, but less than the remote weight of alum solution, I used a mixture

31

Fig. 5.

beats from heating. I have had great success using gypsum plates cast from alabaster (fired gypsum) on thick sheet glass, although these plates are fragile and short-lived. Perhaps slate (roofing slate) slats would be very good, as they can withstand quite a lot of heat and can be made flat and smooth.

The plates only need to be sprinkled with a layer of sand to establish reflection from the surface. The surface of the sand should appear completely robust. To successfully carry out the experiment, you need a completely flat desert; so don’t feel like you’ve paid too much attention to adjusting the records. At one end of the desert we need to create an artificial sky. It is best to use a large mirror mounted on a window and reflecting the sky.

Between the sky and the desert you need to install a small chain of mountains cut out of cardboard. Individual peaks should have a height of 1 to 2 cm, and the valleys between them should rise only slightly above the desert level. The general arrangement is shown at figure 5.

The plates are then heated using gas burners, which need to be moved from time to time to ensure even heating.

If we now look along the desert a little above the level of the sand, we see the mountains clearly standing out against the sky: as the temperature rises, axero begins to form in front of the mountain range, and

Fig. 6.

Fig. 7.

If you place a small marble ball at the far end of the desert, then when the position of the eye is determined, the steep contour will turn into an ellipse, and if you lower the eye, the image will become a segment and eventually disappear. The increase in size in this case is caused by the fact that the direct and reflected images are observed simultaneously and are shifted relative to each other. I observed a similar situation on a lake: when the water is warm and the air is cold, patches of snow on the opposite shore (too small to be distinguished by the eye from a height of a few meters above the lake level) become clearly visible if you go down to the edge of the water.

The atmospheric conditions necessary for mirages to occur can also cause the dust wings often seen in the desert, and on a large scale, death. These deaths, like mirages, can be reproduced on a small scale.

I sprinkled one of the metal plates with quartz sand and heated it with small burners. After a few minutes, very beautiful little deaths began to move across the surface, twisting the thin sand into watery wings; sometimes it lasted 10-15 seconds.

However, vortices created in this way cannot be filled in a large audience. So I looked for ways to create them on a larger scale. To do this, after removing the sand, he heated the plate well and sprinkled it with adhesive: a dense cloud of white steam immediately rose from the hot surface and soon a real death minator about 2 cm high appeared in the center. By placing the plate in the beam of a flashlight in a dark room, you can demonstrate the wings in a larger audience. I find it best to first sprinkle ammonia on the plate and then heat it: then the wings appear almost continuously and often persist for quite a long time.

figure 7 shows a snapshot of one of these deaths, taken in bright sunlight.

This method of creating atmospheric wings is preferable to the old method of creating them using a quick-release drum with transverse partitions *), since here wings arise for the same reasons and under the same conditions as in nature.

*) See the article by N. E. Zhukovsky “Fundamentals of the theory of winga” («Kvant» No. 4, 1971).

33

TASKS

In this issue, “Problem Book «Kvant»” is compiled from problems proposed at the final exams of the first All-Union Olympiad in Mathematics and Physics.

M106. Prove that if the numbers $p_1, p_2, q_1, q_2$ satisfy the inequality
$(q_1 - q_2)^2 + (p_1 - p_2)(p_1q_2 - p_2q_1) < 0,$
then square trinomials
$x^2 + p_1x + q_1$ and $x^2 + p_2x + q_2$
have real roots and between two roots of each of them lies the root of the other.

I. F. Sharyn

M107. a) Given a convex polygon $A_1A_2 \dots A_n$. (Fig. 1). On side $A_1A_2$ points $B_1$ and $D_2$ are taken, on side $A_2A_3$ - points $B_2$ and $D_3$, …, on side $A_nA_1$ points $B_n$ and $D_1$ so that if we construct parallelograms $A_1B_1C_1D_1, A_2B_2C_2D_2, \dots, A_nB_nC_nD_n,$ then straight $A_1C_1, A_2C_2, \dots, A_nC_n$ will intersect at one point. Prove that
$A_1B_1 \cdot A_2B_2 \cdot \dots \cdot A_nB_n = A_1D_1 \cdot A_2D_2 \cdot \dots \cdot A_nD_n$.

6. Prove that the converse statement is also true for a triangle: let points $B_1$ and $D_2$ be chosen on side $A_1A_2$, points $B_2$ and $D_3$ on side $A_2A_3$, points $B_3$ and $D_1$ on side $A_3A_1$ so that if we construct parallelograms $A_1B_1C_1D_1, A_2B_2C_2D_2, \dots, A_nB_nC_nD_n,$ then the lines $A_1C_1, A_2C_2, \dots, A_nC_n$ will intersect at one point.

V. L. Gutemacher

M108. a) Prove that the line dividing the given triangle into two polygons of equal area and perimeter passes through the center of the circle inscribed in the triangle.

6. Prove a similar statement for an arbitrary polygon in which a circle can be inscribed.

Yu. I. Ionin

M109. a) At the vertex $A_1$ of the regular 12-gon $A_1A_2A_3 \dots A_{12}$ there is a minus sign, and at the rest there are plots. It is allowed to simultaneously change the sign to the opposite one at any six consecutive vertices of the polygon. Prove that in several such operations it is impossible to achieve that at the vertex

not $A_2$ turned out to be a minus sign, and in the remaining vertices there were plots.

6. Prove the same statement if it is allowed to simultaneously change signs not at six, but at four consecutive vertices of the polygon.

c) Prove the same statement if it is allowed to simultaneously change signs at three consecutive vertices of a polygon.

M110. On an endless sheet of glued paper, $N$ grids are painted black. Prove that a finite number of squares can be cut from a leaf so that two conditions are satisfied:

1. all the black squares will lie in the cut out squares;
2. in any cut square $K$ the area of ​​black cells is at least $\frac{1}{5}$ and no more than $\frac{4}{5}$ of the area $K$.

G. A. Rosenblom

F118. A mass $m$ of hydrogen and the amount of oxygen required for complete combustion enter the combustion chamber of a jet engine per second. Nozzle output $S$. The pressure in this section is $P$, the absolute temperature is $T$. Determine the traction force of the engine. (9-10 grades)

G. L. Komkin

Fig. 2. Fig. 3.

F119. A matchbox $AB$ with a ball of mass $M$ is suspended from a matchbox $BC$ with a ball of mass $m$ (Fig. 2). Point $A$ oscillates in the horizontal direction with a period of $T$. Find the length of thread $BC$ if it is known that thread $AB$ remains vertical all the time. (10 grades)

G. L. Komkin

F120. A speed skater on an icy road tries to make a turn as close to the inside edge as possible. A cyclist on a cycle track takes a turn as far as possible from the inner edge. How to enforce this difference in the movement of a skater and a cyclist on a turn? The track profile is shown on figure 3. (8 cells)

M. M. Balatsov

F121. In a hermetically sealed vessel, a piece of ice of mass $M$ floats in water, in which a lead pellet of mass $m$ is frozen. How much heat does it take to make the pellet sink? The density of lead is $11,3 \text{ e/cm}^3$, the density of ice is $0,9 \text{ e/cm}^3$, the heat of fusion of ice is $80 \text{ кал/е}$. The temperature of the water in the vessel is $0^\circ \text{C}$. (8 grades)

F122. Three bodies with masses $m_1$, $m_2$ and $m_3$ can slide along a horizontal plane without friction (Fig. 4), and $m_1 \gg m_2$ and $m_2 \ll m_3$. Determine the maximum velocities that two extreme bodies can acquire if at the initial moment of time

Fig. 4.

neither they accumulated, but the average body had a speed of $v$. The impacts are considered to be absolutely elastic. (9 grades)

E. Kuznetsov

36

It remains to solve the following problem: find a point M on line l for which angle PMQ is the simplest: a circle passing through points P, Q and M touches line l. To construct a point M, you need (if the segment PQ is not parallel to the line l) from the point C of the intersection of PQ and l on the line l, plot the segments CM₁ = CM₂ = √(CP·CQ), and then from the resulting points M₁ and M₂ select the one for which the angle MᵢCP is acute (or straight). The proof that it is for this point that the angle PMQ is maximum is almost obvious and is left to the reader *) (Fig. 2).

Oddly enough, many readers who sent us letters thought about one of the possibilities 1) and 2). The greatest solution was sent by O. Kim, G. Levin, A. Pukhalsky, V. Luzzin, Yu. Obolonkov, A. Kostyurin.

M65

a) Let E, F, G be points on sides AB, BC and CA of triangle ABC for which $\frac{AE}{EB} = \frac{BF}{FC} = \frac{CG}{GA} = k$, where 0 < k < 1.

Find the ratio of the area of ​​triangle KLM formed by lines AF, BG and CE to the area of ​​triangle ABC (Fig. 3a).

6. Cut the triangle with the sixth straight line into pieces from which seven equal triangles can be folded.

a) Let us denote the area of ​​the triangle with vertices X, Y, Z by s(XYZ). Using the fact that the ratio of triangles with the same heights (or with the same bases) is equal to the ratio of the bases (corresponding to the ratio of the heights), it is not difficult to prove that

$$s(ACK) = \frac{1}{k} s(ABK) = \frac{k+1}{k^2} s(AEK) = \frac{k+1}{k^2+k+1} s(AEC) = \frac{k}{k^2+k+1} s(ABC).$$

In the same way it can be proven that

$$s(BLA) = s(CMB) = \frac{k}{k^2+k+1} s(ABC).$$

Pozny

$$\frac{s(KLM)}{s(ABC)} = 1 - \frac{3k}{k^2+k+1} = \frac{(1-k)^2}{k^2+k+1} = \frac{(1-k)^3}{1-k^3}.$$

6. Note that for k = $\frac{1}{2}$ this ratio is equal to $\frac{1}{7}$ and each of the three segments AF, BG, CE is divided by two others in the ratio 3:3:1, so that if points M, K, L are drawn, three more lines corresponding to parallel to these segments, then each of the sides of the triangle ABC will be divided in the ratio 2:1:1:2.

*). For more information about this, see the book by N. B. Vasiliev and V. L. Gutenmacher “Straights and Curves”, series “Library of the Physics and Mathematics School”. “Science”, 1969, pp. 76-78.

Now it is not difficult to assemble 7 equal triangles from the 13 pieces into which triangle ABC is cut (one of them consists of a single piece KLM; Fig. 36).

A similar solution to the problem of adding A. Alekseev from Leningrad, A. Alyaev from the Penza region, E. Turkevich, A. Chernyak, Yu. Obolenkov.

M66

Here are some examples where the sum of the squares $k$ of consecutive natural numbers is equal to the sum of the squares $(k-1)$ of the following natural numbers:
$$3^2+4^2=5^2, 36^2+37^2+38^2+39^2+40^2=41^2+42^2+43^2+44^2,$$
$$55^2+56^2+57^2+58^2+59^2+60^2=61^2+62^2+63^2+64^2+65^2.$$
Find a general formula covering all such cases.

Let us prove that for every natural number $k$ there is exactly one natural number $n$ such that
$$(n-k+1)^2+(n-k+2)^2+\ldots+(n-1)^2+n^2=(n+1)^2+$$
$$+(n+2)^2+\ldots+(n+k-1)^2.$$

(Note that on the right side of the equality $(k-1)$ is a summand, and on the left side there is $k$). Equality (1) is equivalent to the following:
$$n^2=[(n+1)^2-(n-k+1)^2]+[(n+2)^2-(n-k+2)^2]+\ldots$$
$$\ldots+[(n+k-1)^2-(n-1)^2]$$
$$n^2=k(2n-k+2)+k(2n-k+4)+\ldots+k(2n+k-2),$$
$$n^2=2k(k-1)n.$$

Thus, the stated statement is proven: equality (1) for natural $k$ and $n$ is satisfied if and only if $n=2k^2-2k$. (When moving from (3) to (4), we took advantage of the fact that the sum of the $(k-1)$ terms of the arithmetic progression $(2n-k+2)+(2n-k+4)+\ldots+(2n+k-2)$ is equal to $(k-1)\frac{(2n-k+2)(2n+k-2)}{2}=2(k-1)n$. Thus, we obtained an exclusion formula. It can be written, for example, like this:
$$(2k^2-3k+1)^2+(2k^2-3k+2)^2+\ldots+(2k^2-2k)^2$$
$$=(2k^2-2k+1)^2+(2k^2-2k+2)^2+\ldots+(2k^2-k-1)^2,$$
where $k$ is an arbitrary natural number. The examples given in the conditions are obtained from it with $k$ equal to 2, 4 and 5.

This formula (or others equivalent to it) has been found by a lot of people - most of them either give a proof close to ours, or use the formula for the sum of squares of natural numbers in their calculations (see Lemma 1, on page 17).

M67

The jeweler was ordered a gold ring of width $h$, which has the shape of a body, the bounded surface of a ball with center $O$ and the surface of a cylinder of radius $r$, the axis of which passes through the point $O$ (Fig. 4). The master made such a ring, but chose $r$ too small. How much gold should be added if $r$ needs to be increased $k$ times, but the width $h$ remains the same?

It is not difficult to prove that the volume of a spherical ring - the body obtained by expressing the segment $AB$ around the diameter $MN$, which does not intersect this segment - is equal to $$\frac{\pi}{6} AB^2 \cdot A_1B_1,$$ where $A_1B_1$ is the projection of the chord $AB$ onto the diameter $MN$. This formula is given, for example, in the book of J. Adam “Elementary Geometry”, part II, chapter. IV, 500.

Fig. 4.

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The volume of the ring is equal to $\frac{\pi}{6} h^3$, that is, it does not depend on the radius $r$, so gold is added to the master.

M68

The grid of lines depicted on figure 5 (see also the cover of “Kvat” No. 2, 1971) consists of concentric circles of radii 1, 2, 3, 4, … with a center at the point $O$, a line $l$ passing through the point $O$, and all possible tangents to the circle parallel to the line $l$. The entire plane is divided by these lines into cells, which are colored in a checkerboard pattern. In the chain of pink dots shown in figure, every two adjacent dots are opposite vertices of a dark cell. Prove that all points of such an infinite chain lie on the same parabola (the pose weight figure is, as it were, woven from yellow and dark parabolas).

Taking the straight line $l$ as the axis $Ox$, and the point $O$ as the origin of coordinates of the rectilinear system, as shown in figure 5. The scale unit is already specified in the conditions. The grid of lines discussed in the conditions is composed of circles of radius $n$ with center $O$ (the equation of such a circle $x^2 + y^2 = n^2$) and straight lines $y = m$, where $m$ and $n$ are all possible integers; $n > 0$. Thus, each point of intersection (or tangency) of these lines—each “node” $(x, y)$ of our grid—corresponds to a certain pair of values ​​$n = \sqrt{x^2 + y^2}$ and $m = y$. Note that when we move from one of the pink dots to the adjacent pink dot, both $m$ and $n$ change by one, so their difference $n - m = \sqrt{x^2 + y^2} - y$ remains constant; this difference for all pink dots is 6.

The equation $\sqrt{x^2 + y^2} - y = 6$ is equivalent to the following: $x^2 + y^2 = (y + 6)^2$; $y = \frac{x^2}{12} - 3$. Thus, all pink points lie on the parabola $y = \frac{x^2}{12} - 3$.

In another similar chain of points, which we drew with the head color, when moving from a point to the neighboring one, $\sqrt{x^2 + y^2}$ and $y$ change by one so that their sum remains constant: $\sqrt{x^2 + y^2} + y = 4$. This equation, after transforming the points, becomes the equation of the parabola $y = -\frac{x^2}{6} + \frac{3}{2}$.

In the same way, it can be shown that any infinite chain of points in which neighboring points are opposite vertices of the same cell (or one segment) lies on the parabola $y = -\frac{x^2}{2p} + \frac{p}{2}$, where $p$ is some non-zero integer. Conversely, any of these parabolas contains such an infinite chain of points (these are the points of intersection of the parabola with all possible lines $y = m$).

39

Note that if $p < 0$, that is, if the branches of parabolaOLS go upward, then if $p$ is even, the parabola goes along the dark cells, and if $p$ is odd, along the yellow ones; if $p > 0$, then vice versa. Moreover, through each intersection of lines and circles of our grid (excluding tangent points lying on line $x = 0$), two parabolas pass; one goes along the yellow cells, the other goes along the dark ones; one is directed with branches up, the other - down.

The solution to this problem is the addition of many honors. Some people use the following geometric definition of parabola in their solution: a parabola is a set of points equally distant from a given point and a given line. Essentially, the solution to problem M68 is very close to proving the equivalence of this definition of parabola and the one known from the school course (a parabola is the graph of a square triangle). If we choose a coordinate system so that a given point is the nacal coordinate, and a given line is the equation $y = c$, then the condition that the point $(x, y)$ is equally distant from the given point and the given line is written by the equation $\sqrt{x^2 + y^2} = |y - c|$, which is equivalent to the equation $y = -\frac{x^2}{2c} + \frac{c}{2}$.

M69

The number 76 has this curious property: the last two digits of the number $76^2 = 5776$ give the number 76 again.

a) Are there any other two-digit numbers like this?

b) Find all three-digit numbers $A$ such that the last three digits of $A^2$ make up the number $A$.

b) Is there an infinite sequence of digits $a_1, a_2, a_3, \dots$ such that for any $n$ the square of the number “$a_n a_{n-1} \dots a_2 a_1$” is “…$a_n a_{n-1} \dots a_2 a_1$”? (The quotation marks here denote the decimal notation of the number. We exclude the obvious answer $a_i = O$ for all $i > 1$.)

The answer to this problem follows: for any $n > 1$ there are exactly two such $n$-digit numbers $n > 1$, the square of which is $A$ (not counting the very long ones: 0 … 0000 and 0 … 0001), - for $n = 2$ these are 76 and 25, for $n = 3$ - 376 and 625, with $n = 4$ - 9376 and 0625, with $n = 5$ - 09376 and 90625, etc., so there are two infinite sequences of numbers discussed in point c): one starting with the numbers 6, 7, 3, 9, 0, 1, …, the other - with the numbers 5, 2, 6, 0, 9, 8, …*).

Let’s start with task a). The easiest way to find the two-digit numbers we need, which indicate pluralities, is as follows. If $A^2$ turns out to be the same two digits as $A$, then $A^2 - A = A$ (A - 1) is divisible by 100 = 25 4, and since the numbers $A$ and $(A - 1)$ are coprime (do not have common factors greater than 1), then one of them must be divisible by 25, and the other by 4. Let’s try to see if each of the numbers is suitable 25, 50 and 75 for role $A$ or for role $A - 1$ (both of these numbers are double digits). To do this, you just need to check which of its neighboring numbers are divisible by 4. This will only be 76 and 24, so $A$ can only equal 25 and 76.

Problem 6 can be solved similarly), but we must immediately understand: if the square of the number $B = \langle a_{n-1} a_{n-2} \dots a_1 \rangle$ turns out to be the digits “… $a_{n-1} a_{n-2} \dots a_1$”, then it is possible, in a unique way, to choose the digit $a_n$ so that the square of the number $A = \langle a_n a_{n-1} \dots a_1 \rangle$ turns out to be “… $a_n a_{n-1} \dots a_1$”.

Let $B^2 = \langle ... b_n a_{n-1} a_{n-2} \dots a_1 \rangle$. Then $(n \geqslant 2 = > 2n > n + 1)$

$$A^2 = (10^n a_n + B)^2 = 10^{2n} a_n + 2 \cdot 10^n a_n B + B^2 = \langle ... c_n a_{n-1} a_{n-2} \dots a_1 \rangle,$$

where $c_n$ is the last digit of the number $2a_n a_1 + b_n$. We must choose $a_n$ so that $c_n$ equals $a_n$, that is, so that $2a_n a_1 + b_n = a_n = (2a_1 - 1)a_n + b_n$ is divisible by 10. Obviously, this can be done in a unique way. In the functions $a_1 = 5$ and $a_1 = 6$ that interest us, the $b_n \rightarrow a_n$ functions that we must indicate are especially simple: if $a_1 = 5$, then $a_n = b_n$, and if $n = 6$, then $a_n = 10 - b_n$ with $b_n > 0$ and $a_n = 0$ with $b_n = 0$.

This approach to solving problem M68 is discussed in more detail in the book by Ya. I. Perel - on “Entertaining Algebra”, pp. 81-85.

*) We do not exclude from consideration the numbers “$a_n a_{n-1} \dots a_1$”, in which the first digit $a_n$ (or several first digits) is equal to zero. Readers who have not considered such numbers should naturally only go up to the four-digit number 9376 and the three-digit number 625.

41

A heavy hoop with lightweight knitting needles is located in a vertical plane and can be rotated around a horizontal axis passing through the center. A small ball is fixed in the thickness of its rim, having the same mass as the hoop itself. What will be the period of the wheel hoop around the axis? How will it change if: a) this hoop is moved to the Moon? b) place it in a liquid in which he will marvel without tremium?

Will engage in analogies. Readers can get acquainted with him by reading the article by Ya. A. Smorodinsky “Similar movements” in the magazine for this year. Let’s use the analogy with a mathematical matrix of length $R$.

If the maximum angle of deviation of the matrix and the ball attached to the hoop is $\alpha_0$ (Fig. 7), then they are at a height of $h=R(1-\cos \alpha_0)$ from the equilibrium position. At the deflection angle $\alpha$, the kinetic energy of the matrix (and the hoop with the ball) is equal to the difference between the potential energies at angle $\alpha_0$ and angle $\alpha$. For the mathematical matrix we get

$$\frac{mv^2}{2} = mgR(\cos \alpha - \cos \alpha_0).$$

Therefore the speed matrix

$$v = \sqrt{2gR(\cos \alpha - \cos \alpha_0)}.$$

Just like the matrix, the potential energy of the hoop with the ball changes. The potential energy of the hoop itself does not change, but the kinetic energy in this case is equal to the sum of the kinetic energies of the ball $E_{\text{ш}} = \frac{mv^2}{2}$ and the hoop $E_0 = \frac{mv^2}{2}$ (for the kinetic energy of the hoop, see the article by A.K. Kikon “Rotational motion of bodies”, «Kvant» No. 1, 1971). Therefore, in the case of a hoop with a ball, it follows from the law of conservation of energy that

$$\frac{mv^2}{2} + \frac{mv^2}{2} = mgR(\cos \alpha - \cos \alpha_0),$$

and ball speed

$$v = \sqrt{gR(\cos \alpha - \cos \alpha_0)}.$$

At any angle of deviation from the equilibrium position, the speed of the ball is $\sqrt{2}$ times less than the speed of the mathematical matrix. Hence,

42

and the period of oscillation of the hoop with a ball is $\sqrt{2}$ times greater than the period of oscillation of the matrix and is equal to

$$T = 2\pi \sqrt{\frac{2R}{g}}.$$

You can approach the solution of this problem anyway. If the mathematical matrix is ​​deviated from the equilibrium position by an angle $\alpha$, then it is returned to the equilibrium position by the component of the gravity force, tangent to the trajectory of the matrix (Fig. 8):

$$F = mg \sin \alpha.$$

Since the angle $\alpha$ is small, then $\sin \alpha \approx \alpha \approx \frac{x}{R}$, where $x$ is the displacement of the matrix from the equilibrium position. From here

$$F \approx mg \frac{x}{R}.$$

Therefore, for the tangential component of the matrix acceleration we can write the equation

$$mg \frac{x}{R} = ma$$ or $a = \frac{g}{R} x = cx,$

where $c = \frac{g}{R}$ is some constant coefficient for a given matrix. The matrix oscillation period is equal to $2\pi \sqrt{\frac{R}{g}}$, that is, $2\pi/\sqrt{c}$.

We will obtain a similar result for a weight attached to a circle and oscillating along a horizontal plane. The equation of motion of such a weight is $kx = ma$, and the period of oscillation is $T = 2\pi \sqrt{\frac{m}{k}}$.

In our case, the force returning the ball to the equilibrium position is equal, as in the case of the mathematical matrix, $mg \frac{x}{R}$, and the acceleration $a$ is imparted to both the ball and the hoop, that is, mass $2m$. Therefore $mg \frac{x}{R} = 2ma$; this means $c = \frac{g}{2R}$ and the period of oscillation of the system is $T = 2\pi \sqrt{\frac{2R}{g}}$.

If a hoop with a ball is moved to the Moon or placed in a vessel with liquid, then its period of oscillation will change in exactly the same way as the period of oscillation of a mathematical matrix: in the first case it will increase by $\sqrt{6}$ times ($g_{\text{л}} = \frac{1}{6} g_3$) due to a change in the acceleration of free fall, and in the second case it will decrease by $\sqrt{1 - \frac{\rho}{\rho_T}}$ times ($\rho$ - the density of water and $\rho_T$ - the density ball materials) due to the influence of the Archimedean buoyancy force.

The correct solution was sent by S. Ivanov (Moscow), G. Zaitsev (Gagra), A. Proskuryakov (Moscow).

F77

A rectangular block, the height of which significantly exceeds its length and width, stands on a horizontal surface. How to determine the coefficient of friction between the block and the surface, having only one measuring device - a ruler?

Having made a loop from the thread, we throw it over the block and pull it in the horizontal direction (Fig. 9). If the thread is draped over a block close to the surface

43

If necessary, the block will slide. If we lift the thread along the block, then when it is at a certain height $h_0$ from the surface, the block will no longer slide, but will begin to turn over. In this case, the force $F$ with which we pull the block is equal to the friction force:

$$F = kmg,$$

and the moment of force $F$ relative to the axis passing through the point $A$ is equal to the moment of gravitational force:

$$Fh_0 = mg \frac{a}{2},$$

where $a$ is the side of the square at the base of the block.

Solving these two equations together, we find

$$k = \frac{a}{2h_0}.$$

By measuring $a$ and $h_0$ using a ruler, we find the coefficient of friction of the block on the surface.

This solution was added by I. Sidorov (Moscow), A. Petrov (Asbest of the Sverdlovsk Region), M. Rozov (Minsk), S. Yudin (Volgograd), D. Shapiro (Chelyabinsk), A. Alexandrov (Glazov Udm. Autonomous Soviet Socialist Republic), G. Simonenko (Kaunas).

An interesting solution by A. Kalik from Baku and S. Sobolev from Moscow. Here it is.

Having tilted the block and holding it at a certain angle to the vertical, we apply a certain force $F$ to the block, directed along the block (Fig. 10). In this case, the reaction force $N = mg + F \cos \alpha$ and the friction force $F_{trp}$ will also act on the block. If the friction force does not reach the maximum value equal to $kN = k (mg + F \cos \alpha)$, then the end of the block will not slide. In this case, the friction force is equal to $F \sin \alpha$. Thus, the end of the bar does not slip when

$$F \sin \alpha < kmg + kF \cos \alpha,$$

that is, when

$$F < \frac{kmg}{\sin \alpha - k \cos \alpha}.$$

It is easy to see that at $k > tg \alpha$ the block cannot be made to move, no matter how large the force $F$. Let us find the angle $\alpha_0$ at which the block begins. Using a ruler, measure the height from the horizontal plane to the edge of the tilted block and the width of the block and calculate $tg \alpha_0$, we find the friction coefficient $k$.

44

20 helium contained in a cylinder under the piston is very slowly transferred from state I ($P_1 = 4,1$ amm, $V_1 = 32$ l) to state II ($P_2 = 15,5$ amm, $V_2 = 9$ l). Each gas with the highest temperature access is simpler if the graph of pressure versus volume is a straight line (Fig. !i)?

Let us plot the isotherm grid on the graph—hyperbola $PV = \text{const} \frac{m}{\mu} RT$. The higher the gas temperature, the farther the vertex of the hyperbola is from the origin. Therefore, it is clear that during process I-II the gas reaches such a high temperature at which the corresponding hyperbola does not intersect the line I-II, but only touches it.

Since the point with coordinates $P$ and $V$, corresponding to the maximum temperature, lies on both the hyperbola and the straight line, its coordinates must satisfy two equations: the equation of the hyperbola

$$PV = \frac{m}{\mu} RT$$

and the equation of the line

$$P = \alpha V + \beta$$

(α and β are constants. We will define them a little later). Substituting $P$ from the second equation into the first, we obtain a quadratic equation for $V$:

$$\alpha V^2 + \beta V - \frac{m}{\mu} RT = 0.$$

Since a hyperbola and a straight line must have only one common point, this equation can only have one root, that is, its determinate must be equal to zero:

$$\beta^2 + 4\alpha \frac{m}{\mu} RT = 0.$$

Solving the last equation for $T$, we find

$$T = -\frac{\beta^2 \mu}{4\alpha m R}.$$

In this expression you need to substitute the values ​​of the coefficients $\alpha$ and $\beta$. Since points I and II belong to the same line $P = \alpha V + \beta$, then $P_1 = \alpha V_1 + \beta$ and $P_2 = \alpha V_2 + \beta$. Solving this system of equations, we find

$$\alpha = -\frac{P_2 - P_1}{V_1 - V_2}, \quad \beta = \frac{P_2 V_1 - P_1 V_2}{V_1 - V_2}.$$

Pozny

$$T = \frac{(P_2 V_1 - P_1 V_2)^2 \mu}{4 (P_2 - P_1) (V_1 - V_2) m R}.$$

Substituting the numerical values ​​of the quantities included in this expression, we obtain

$$T \approx 490^\circ \text{ K}.$$

The correct solution with additives M. Preger (Tomsk), V. Rudko (Kyiv), N. Fedin (Omsk), G. Zaitsev (Gagra), L. Knizherman (Moscow), O. Maslyuk (Novomirgorod, Kirovograd region), V. Trunkov (Gomel), E. Rozov (Minsk), A. Petrov (Asbest Sverdlovsk region), L. Tsiferblam (Lvov), I. Vidorovich (Simferopol), V. Tyryshkin (Kuibyshev), V. Martivov (Ioshkar-Ola), V. Klimov (Solnechnogorsk, Moscow region), A. Tovbi (Odessa), G. Fast (Karaganda), M. Glikman (Chisinau), A. Zisman (Pskov), A. Esin (Liggul Tashkent region), A. Tarnopolsky (Nikolaev), M. Sarkisyan (village of Agaran, Megri district of the Armenian USSR).

I. Sh. Slovodechkiy

45

Systems of algebraic equations

M. I. Shabunin

At university entrance exams, systems of algebraic equations are first proposed; in particular, such systems arise in problems of composing equations. Exam experience shows that many students who come in solve a system of equations cope unsatisfactorily.

The fact is that usually when studying, the question of finding solutions to a system is understood simply as the formal implementation of a series of “usual” algebraic transformations. At the same time, it should be kept in mind that some of the transformation examples may lead to the appearance of extraneous solutions or to the loss of the solution of the original system.

Without appropriate justification for the solution of the system, the logical equations are incomplete. In particular, we may not replace the losses of some solutions. Unfortunately, well-founded, well-presented decisions of the system in the work of applicants are extremely rare.

A detailed presentation of all the important issues relating to the system of equations would require too much space *). Therefore, we will limit ourselves to presenting the basic theoretical principles concerning the solution of three algebraic equations with unknowns, and then consider several examples.

First, let’s recall some definitions.

A system of three equations with unknowns can be written in the form

$$\begin{cases} f_1(x, y, z) = 0, \\ f_2(x, y, z) = 0, \\ f_3(x, y, z) = 0 \end{cases}$$

or, omitting the unknowns for brevity, in the form

$$\begin{cases} f_1 = 0, \\ f_2 = 0, \\ f_3 = 0. \end{cases}$$

We will consider only algebraic systems, that is, such

*) The reader can be recommended to study, for example, the book by V. G. Boltyansky, Yu. V. Sidorov and M. I. Shabunin “Lectures and problems in elementary mathematics”, M., “Nauka”, 1971.

systems in which the left side of each equation of the system is a polynomial in unknowns.

A solution to system (1) is a triple of numbers $(x_0, y_0, z_0)$, when substituting the corresponding one instead of $x, y, z$, each equation of system (1) becomes a true numerical equality. Solve a system introduces you to find all its solutions.

Two systems of equations

$$\begin{cases} f_1 = 0, \\ f_2 = 0, \\ f_3 = 0. \end{cases}$$

And

$$\begin{cases} g_1 = 0, \\ g_2 = 0, \\ g_3 = 0. \end{cases}$$

46

are called equivalent (equivalent) if they have the same solutions, that is, if the solution to system (2) is the solution to system (3) and, conversely, if the solution to system (3) is the solution to system (2). To indicate the equivalence of systems (2) and (3), the notation “(2) ⟨=⟩ (3)” is used.

Note that if we find all the solutions to system (3), equivalent to system (2), then we have also found all the solutions to system (2). In this case, there is no need to check, that is, there is no need to substitute the values ​​of the unknowns found when solving system (3) into the original system (2).

Equation
$$F(x, y, z) = 0$$
(4)
Let’s call it a consequence of system (1) if each solution of system (1) satisfies equation (4). The entry “(1) ⟨=⟩ (4)” will mean that equation (4) is a consequence of system (1).

We will say that system (1) is equivalent to the totality of the system
$$\begin{cases} F_1 = 0, \\ F_2 = 0, \\ F_3 = 0 \end{cases}$$
And
$$\begin{cases} G_1 = 0, \\ G_2 = 0, \\ G_3 = 0, \end{cases}$$
if the following conditions are met:
a) each solution to system (1) is a solution to at least one of systems (5), (6);
b) every solution to any of the systems (5), (6) is a solution to system (1).

The notation “(1) ⟨=⟩ [(5) ∨ (6)]” will mean that system (1) is equivalent to the set of systems (5) and (6).

If system (1) is equivalent to a combination of systems (5) and (6), then it is possible to solve systems (5), (6) and combine the sets of solutions to these systems; this is the union and dact of all solutions of system (1) (the totality of the system appears when considering various cases of type $x > 0$, $x < 0$, $x = 0$).

Let us now formulate several properties that are most often used when solving a system of equations. We will not prove these properties; Using the given definitions, the reader can easily verify the necessary justification on his own.

I. Let one of the equations of system (1), to determine the first, “split” into two equations, for example, represented in the form
$$g_1(x, y, z) g_2(x, y, z) = 0,$$
where $g_1$ and $g_2$ are polynomials in $x, y, z$. Then system (1) is equivalent to the combination of the following two systems:
$$\begin{cases} g_1 = 0, \\ f_2 = 0, \\ f_3 = 0 \end{cases} \quad \text{и} \quad \begin{cases} g_2 = 0, \\ f_2 = 0, \\ f_3 = 0. \end{cases}$$

This statement can easily be generalized to the case when two or three equations of system (1) “break up” into several equations.

II. If equation (4) is a consequence of system (1), then, when connecting equation (4) to system (1), we obtain a system {{1}, {4}} of four equations, equivalent to the original one.

III. Let equation (4), a consequence of system (1), “break down” into two equations:
$$F_1(x, y, z) \cdot F_2(x, y, z) = 0,$$
where $F_1$, $F_2$ are polynomials in $x, y, z$. Then system (1) is equivalent to the combination of the following two systems:
$$\begin{cases} f_1 = 0, \\ f_2 = 0, \\ f_3 = 0, \\ F_1 = 0 \end{cases} \quad \text{и} \quad \begin{cases} f_1 = 0, \\ f_2 = 0, \\ f_3 = 0, \\ F_2 = 0. \end{cases}$$

IV. System of equation (1) is equivalent to each of the following systems:
$$\begin{cases} f_1 - f_2 = 0, \\ f_2 - f_3 = 0, \\ f_3 - f_4 = 0; \end{cases} \quad \text{и} \quad \begin{cases} f_1 - f_2 = 0, \\ f_2 - f_3 = 0, \\ f_3 - f_4 = 0; \end{cases}$$

$$\begin{cases} f_1 + f_2 = 0, \\ f_1 + f_3 = 0, \\ f_2 + f_3 = 0 \end{cases}$$

(the ability to add to one equation of the system another number multiplied by the product).

V. Let one of the equations of system (1), for example, its first equation $f_1(x, y, z) = 0$, be equivalent to the equation $x = \varphi(y, z)$ - a polynomial in $y$ and $z$. Then system (1) is equivalent to system
$$\begin{cases} x = \varphi(y, z), \\ f_2[\varphi(y, z), y, z] = 0, \\ f_3[\varphi(y, z), y, z] = 0. \end{cases}$$