Kvant 1972, No. 1 · Pages

KEANM
1
JANUARY
1972
POPULAR SCIENTIFIC PHYSICS AND MATHEMATICS JOURNAL
USSR ACADEMY OF SCIENCES AND USSR ACADEMY OF PEDAGOGICAL SCIENCES

Editor-in-Chief - Academician I. K. Kikoin,
First Deputy Editor-in-Chief - Academician A. N. Kolmogorov.

Editorial collection:
L. A. Artsimovich, M. I. Bashmakov, V. G. Boltyansky, I. N. Bronshtein, N. B. Vasiliev, I. F. Ginzburg, V. G. Zubov, P. L. Kapitsa, V. A. Kirillin, V. A. Leshkovtsev (deputy editor-in-chief), A. I. Markushevich, M. D. Millonshikov, N. A. Patrikeeva, N. Kh. Rozov, A. P. Savin, I. Sh. Slobodetsky, M. L. Smolyansky (deputy editor-in-chief), Ya. A. Smorodinsky, V. A. Fabrikant.

On the cover of our magazine there is a figure whose outline represents one of the wonderful curves - the cardiond. Read more about cardiond on page 32.

2

S. G. Gindikin

At school, in geometry lessons, we consider the problem of constructing regular polygons using a compass and a ruler. It is easy to construct a regular square quadrilateral. It is quite easy to construct a regular triangle and an almost equally regular hexagon (its side is equal to the radius of the circle circumscribed around it). A trickier thing is to build a regular pentagon.

Having learned how to construct the indicated polygons, it is easy to move on to constructing regular octagons, 10-gons, 12-gons, 16-gons, and 20-gons. It is somewhat more difficult to construct a regular 15-gon.

All these problems were solved back in ancient Greece. However, despite all the successful efforts of the creative Greek geometers, none of them succeeded in constructing either a regular heptagon or a regular pentagon; It was not possible to construct a regular p-gon for any prime p other than 3 and 5. For two thousand years nothing could be achieved in solving this problem. In 1796, Carl Friedrich Gauss (1777–1855) proved the possibility of constructing a regular 17-gon using a compass and ruler.

Over the next few years, Gauss completely solved the problem of constructing regular n-gons with compasses and a ruler, finding out, in particular, that such a construction is impossible for n = 7, 9, 11, 13.

This article talks about constructing a regular 17-sided triangle. The main difficulty, the reader’s expectations, is not related to the fact that he will need some new information (all of which does not go beyond the scope of a school textbook), but to the fact that Gaus’s construction uses a number of deep ideas. Those readers who take the trouble to understand them will be returned with one of the most astonishing discoveries in the history of mathematics.

2

The tenacity with which Gauss followed the path he had chosen, the stormy youthful drive with which he every time, no matter what, overcame the largest ascents leading to the goal, all these difficult trials strengthened his strength and made him capable, after defeating the obstacles already eliminated by others, to move forward uncontrollably, ahead of them. To this praise of creative initiative I must add others: praise of youth. All I want to say by this is that the development of a mathematical genius is subject to the same laws as the development of any other creative ability. For a brilliantly gifted individual, the years of youth, the period when the process of physical growth is just completed, is an era of great discoveries, in abundance; It is during these years that the brilliantly gifted two create those new values ​​that belong to him alone, which they will subsequently present to the world.

F. Klein *)

On June 1, 1796, a note with the following content appeared in the newspaper Jenen-ser Intelligenzbbatt.

Every novice geometer knows that it is possible with geometric tools (that is, with a compass and a ruler) to construct various regular polygons, namely, a triangle, a pentagon, a decagon, and those that are obtained from each of them by successively doubling the number of its sides. This was known in the time of Euclid, and since then it seems that the conviction has been widespread that the field of elementary geometry does not extend further: at least I do not know of a successful attempt to extend it in this direction…

Moreover, the discovery that, in addition to these regular polygons, many others can be geometrically constructed, for example, a seventeen-sided polygon, seems worthy of attention to me.

The note is signed:
K. F. Gauss from Braunschweig, student of mathematics in Göttingen.

This is the first report of Gauss’s discovery. Before in detail

*) Felix Klein (1849-1925) is one of the greatest mathematicians who worked at the turn of the 19th and 20th centuries. His own research is closely related to the work of Gauss. F. Klein was one of the greatest experts on Gauss’s legacy; for many years he supervised the publication of the collected works of Gauss.

talk about it, let’s refresh our memory of what “every novice geometer knows.”

About construction with compasses and ruler

A segment of unit length is assumed to be given. Then, using a compass and a ruler, you can construct new segments, the lengths of which are obtained from the lengths of existing segments using the following operations: addition (Fig. 1a), subtraction (Fig. 1b), multiplication (Fig. 1b), division (Fig. 12) and square root extraction (Fig. 1d).

Having carried out these operations sequentially, using a compass and a ruler, you can construct any segment, the length of which is expressed through a single number of addition, subtraction, multiplication, division and square root operations. Such numbers are called quadratic irrational. It can be proven that no other segments can be constructed using a compass and ruler - but this is a topic for a special discussion.

Construction of regular n-gons

The problem of constructing a regular n-gon, as is easy to understand,

Fig. 1. a) b) c) d) A]

4

Equivalent to the problem of dividing a circle of radius 1 into $n$ equal parts. The chords of the arcs into which the circle is divided are the sides of a regular $n$-gon and the length of each of them is $2\sin \frac{\pi}{n}$ (Fig. 2).

Consequently, for those $n$ for which $\sin \frac{\pi}{n}$ is a square irrationality, one can construct regular $n$-gons with a circlel and a ruler. This condition is satisfied, for example, by the values ​​$n=3, 4, 5, 6, 10$. For $n=3, 4, 6$ this is well known.

Let us show that $\sin \frac{\pi}{10}$ is a square irrationality. Consider an equilateral triangle $ABC$ (Fig. 3), the vertex angle of which is $\frac{\pi}{5}=36^\circ$, length $AB$ is 1; let $AD$ be the bisector of angle $A$. Then $x=AC=AD=BD=2\sin \frac{\pi}{10}$.

We have:

$$\frac{BD}{DC}=\frac{AB}{AC}; \frac{x}{1-x}=\frac{1}{x}, x=\frac{\sqrt{5}-1}{2}.$$

This number is a square irrationality; thus we can construct the side of a regular decagon.

Further, from the possibility of dividing a circle into $p_1 p_2$ equal parts

It follows, of course, that it can be divided into $p_1$ equal parts (in part, it is possible to construct a regular pentagon). The converse statement is, generally speaking, false. Let us indicate two special cases when it is still true.

1. From the possibility of dividing a circle into $p$ equal parts it follows the possibility of dividing it into $2^k p$ equal parts for any $k$.

2. If we reduce a circle into $p_1$ equal parts and $p_2$ equal parts, where $p_1$ and $p_2$ are mutually prime (for example, $p_1, p_2$ are distinct primes), then the circle can be divided into $p_1 p_2$ equal parts.

A few words about complex numbers

We need to know very little about complex numbers: operations on them and geometric interpretation; All this is contained, for example, in the Kochetkovs’ school textbook. Let us say that the complex number $z=a+ib$ is associated with a point with coordinates $(a, b)$ (Fig. 4) and a vector with an end at this point and a beginning at $(0, 0)$. The length of the vector $r=\sqrt{a^2+b^2}$ is called the modulus of the given number. The complex number $z$ can be written in trigonometric form: $z=a+bi=r(\cos \varphi+i\sin \varphi)$; the angle $\varphi$ is called the argument of the number $z$.

6

equations: first, equation (4) is solved, the roots of which are the sums $\varepsilon_1 + \varepsilon_4$ and $\varepsilon_2 + \varepsilon_3$ of the symmetric (see Fig. 6!) roots of equation (3), and then the roots of equation (3) themselves are found from equation (5).

It was in this way that Gauss managed to construct a regular 17-gon: here, too, groups of roots are distinguished, the sum of which is found sequentially from quadratic equations. But how do you find these “good” bands? Gauss has an amazing way to answer this question…

Construction of a regular 17-gon

March 30, 1796 was the day of creative baptism for him (Gauss)… Gauss had already been engaged for some time in grouping roots from units on the basis of his theory of “primitive” roots. And then one morning, when we woke up, he clearly and distinctly realized that the construction of a 17-agon follows from his theory… This event was a turning point in Gauss’s life. He decides to devote himself not to philology, but exclusively to mathematics.

F. Klein

In order to reveal the hidden “symmetries” found by Gauss in the set of roots of the 17th power of units and, using immy, to divide the roots into the necessary groups, we introduce a new numbering of the roots. We will raise 3 to successive powers 0, 1, 2, … and each time take the remainder of dividing the resulting number by 17. We will spare the reader from carrying out these calculations and present the final results in the table. The first line contains the exponents $k$, and below it the remainders from dividing $3^k$ by 17.

Notice that the bottom line contains all the numbers from 1 to 16; then $3^{16}$ gives remainder 1 and then periodic remainders are repeated (prove it!)

The pattern noticed by Gauss is a special case of the following theorem: for every prime $p$ there is a number $l$, called a primitive root, such that among the remainders of the division of $l^k$ by $p$ all the numbers 1, 2, …, $p-1$ are found. This fact was first noted by Euler (1707-1783), but only Legendre (1752-1833) was able to prove it; Other proofs were obtained by Gauss, but, most likely, in 1796 he did not yet have a general theorem, but discovered the given empirical fact by carrying out calculations for specific numbers. This is a very important circumstance, without taking into account the fundamental one, it is difficult to correctly understand the nature of Gauss’s early works.

Let us assign the root $\varepsilon_h$, $k=3^l$, a new number, namely $l$, which we

Table

| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |
| :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: |
| 1 | 3 | 9 | 10 | 13 | 5 | 15 | 11 | 16 | 14 | 8 | 7 | 4 | 12 | 2 | 6 | 1 |

9

Appendix 1.

Now we will prove the square irrationality of the 17th roots of ones. Note that $\varepsilon_k \varepsilon_l = \varepsilon_k + l$ (if $k + l \geq 17$, then $k + l$ is replaced by the remainder of its division by 17), $\varepsilon_k = (\varepsilon_1)^k$. Let’s renumber the roots of units using our table, namely, we will write $\varepsilon_{[l]}$ together with $\varepsilon_k$ (see formula (1)), where $l$ is the number above $k$ in the table. Let’s divide the roots into groups, as shown in figure 8. We will find the sums of roots whose numbers appear in one rectangle (Fig. 8), moving from a large rectangle to a smaller one. First of all, we note that

$$\varepsilon_1 + \varepsilon_2 + \ldots + \varepsilon_{16} = \varepsilon_0 + \varepsilon_1 + \ldots + \varepsilon_{15} = -1.$$

(You can verify this, for example, by considering this expression as the sum of a geometric progression.) So, we have found the sum of the roots in the largest rectangle (No. 0).

Let us denote by $\sigma_m, r$ the sum of $\varepsilon_k$ with those $k$ that give remainder $r$ when divided by $m$. Let us consider the sums of the roots corresponding to the next rectangle (No. 1).

We get

$$\sigma_{2,0} = \varepsilon_0 + \varepsilon_2 + \varepsilon_4 + \ldots + \varepsilon_{14},$$

$$\sigma_{2,1} = \varepsilon_1 + \varepsilon_3 + \varepsilon_5 + \ldots + \varepsilon_{15}.$$

It’s clear that

$$\sigma_{2,0} + \sigma_{2,1} = \varepsilon_0 + \varepsilon_1 + \ldots + \varepsilon_{15} = -1.$$

It can be shown that

$$\sigma_{2,0} \cdot \sigma_{2,1} = 4 (\varepsilon_0 + \varepsilon_1 + \ldots + \varepsilon_{15}) = -4^*.$$

Now, using Vieta’s theorem, we can create a quadratic equation whose roots are $\sigma_{2,0}, \sigma_{2,1}$:

$$x^2 + x - 4 = 0; \quad x_{1,2} = \frac{-1 \pm \sqrt{17}}{2}.$$

To distinguish the roots, use Figure 7. In each of the sums, the roots are included along with their conjugates. It is clear that $\sigma_{2,0} > \sigma_{2,1}$ (in the first case, you need to add and double the real parts of the roots $\varepsilon_1, \varepsilon_2, \varepsilon_4, \varepsilon_8$, in the second - $\varepsilon_3, \varepsilon_5, \varepsilon_6, \varepsilon_7$). So,

$$\sigma_{2,0} = \frac{\sqrt{17} - 1}{2}, \quad \sigma_{2,1} = \frac{-\sqrt{17} - 1}{2}.$$

Let us consider the sums of roots corresponding to the following rectangle (No. 2) on figure 8:

$$\sigma_{4,0} = \varepsilon_0 + \varepsilon_4 + \varepsilon_8 + \varepsilon_{12},$$

$$\sigma_{4,1} = \varepsilon_1 + \varepsilon_5 + \varepsilon_9 + \varepsilon_{13},$$

$$\sigma_{4,2} = \varepsilon_2 + \varepsilon_6 + \varepsilon_{10} + \varepsilon_{14},$$

$$\sigma_{4,3} = \varepsilon_3 + \varepsilon_7 + \varepsilon_{11} + \varepsilon_{15}.$$

*) You can verify this, the wires are direct changes, taking into account that $\varepsilon_k \cdot \varepsilon_l = \varepsilon_k + l$, and it is convenient to use Figure 7. However, Appendix 2 will indicate a way to avoid these tedious calculations.

With respect to all transformations $T_k$, only the set of all roots $\{\varepsilon_1, \dots, \varepsilon_{16}\}$ is invariant.

The cardinal insight is that the group of roots is “better” the greater the number of transformations that leave this group invariant.

Let us introduce another numbering $T_{[l]}$ for $T_k$, as was done for $\varepsilon_k: T_{[l]} = T_k$, $k = 3^l$. In new notations

$$T_{[k]}^{\varepsilon_l} = \varepsilon_{[k+l]}, \quad T_{[m]}(T_{[k]}^{\varepsilon_l}) = T_{[m+k]}^{\varepsilon_l}$$

(the sum in square brackets must be modulo 16). The reader will, of course, discover an analogy with the transition to logarithms, which is not surprising, since $\varepsilon_{[l]} = \varepsilon_{3l}$.

Problem 3. Prove that if a certain set of roots is invariant under some $T_{[k]}$, where $k$ is odd, then this set is invariant under all transformations $T_{[m]}$, that is, if it is not empty, then it will coincide with the set of all roots.

Note. Show that if $k$ is odd, then there is a $m$ such that $km$ gives remainder 1 when divided by 16.

On the other hand, there are two groups of roots invariant under all $T_{[k]}$ with four $k$: roots $\varepsilon_l$ with four $l$ and four roots $l$. We denoted their amounts by $\sigma_{2,0}, \sigma_{2,1}$.

It is clear that $\sigma_{2,0} + \sigma_{2,1} = -1$. Let’s explore $\sigma_{2,0} \cdot \sigma_{2,1}$. This product is the sum of pairwise products $\varepsilon_{[k]}^{\varepsilon_l}$, where $k$ is even, $l$ is odd, each of which is some root $\varepsilon_m$, and there are 64 terms in total. We will show that among them, each of the roots $\varepsilon_{[0]}, \varepsilon_{[1]}, \dots, \varepsilon_{[15]}$ occurs the same number of times (four times), resulting in $\sigma_{2,0} \cdot \sigma_{2,1} = -4$. Let us take advantage of the fact that the transformations $T_{[k]}$ preserve groups of roots when $k$ is even and transform them into one another when $k$ is odd. Each term in $\sigma_{2,0} \cdot \sigma_{2,1}$ is uniquely transformable into the form $\varepsilon_{[m]}^{\varepsilon_{[m+r]}}$, where $0 \leq m \leq 15$, $r = 1, 3, 5, 7$ (prove it!). Let’s group terms with the same $r$. Popular sums will be of the form

$$\varepsilon_{[0]}^{\varepsilon_r} + \varepsilon_{[1]}^{\varepsilon_r} + \varepsilon_{[2]}^{\varepsilon_r} + \dots + \varepsilon_{[15]}^{\varepsilon_r} = T_{[0]}^{\varepsilon_{[0]}^{\varepsilon_r}} + T_{[1]}^{\varepsilon_{[0]}^{\varepsilon_r}} + \dots + T_{[15]}^{\varepsilon_{[0]}^{\varepsilon_r}} = \varepsilon_{[0]} + \dots + \varepsilon_{[15]} = -1,$$

where $\varepsilon_{[k]} = \varepsilon_{[0]}^{\varepsilon_{[r]}}$. We will use the fact that $T_{[m]}^{\varepsilon_{[k]}} \cdot T_{[m]}^{\varepsilon_{[l]}} = T_{[m]}(\varepsilon_{[k]}^{\varepsilon_{[l]}})$ and the already mentioned property $T_{[m]}$.

The values ​​of $\sigma_{2,0}, \sigma_{2,1}$ are found in Appendix 1.

Let’s move on to the next step. We want to introduce into consideration new, smaller groups of roots that are invariant under some $T_{[k]}$. By analogy with Problem 3, it can be shown that in this case $k$ must necessarily be divisible by 4. Therefore, there are four groups of roots that are invariant with respect to all $T_{[4l]}$ and smaller than those already considered; Let’s write down the sums of the roots in each group: $\sigma_{4,0}, \sigma_{4,1}, \sigma_{4,2}, \sigma_{4,3}$. We have already noted that

$$\sigma_{4,0} + \sigma_{4,2} = \sigma_{2,0}; \quad \sigma_{4,1} + \sigma_{4,3} = \sigma_{2,1}.$$

Let’s calculate the product $\sigma_{4,0} \cdot \sigma_{4,2}$; it is represented as a sum of 16 terms of the form $\varepsilon_{[4k]}^{\varepsilon_{[4l+2]}}$, which such term is uniquely written in the form $\varepsilon_{[2m]}^{\varepsilon_{[2m+2r]}}, r = 1, 3, m = 0, 1, 2, 3, 4, 5, 6, 7$.

Let us group the terms with one $r$ and replace that $\varepsilon_{[0]}^{\varepsilon_{[2]}} = \varepsilon_{1} \varepsilon_{9} = \varepsilon_{10} = \varepsilon_{3}, \varepsilon_{[0]}^{\varepsilon_{6}} = \varepsilon_{1} \varepsilon_{15} = \varepsilon_{16} = \varepsilon_{8}$. For $r = 1$ we get the sum

$$T_{[0]}^{\varepsilon_{[3]}} + T_{[2]}^{\varepsilon_{[3]}} + \dots + T_{[14]}^{\varepsilon_{[3]}} = \sigma_{2,1};$$

at $r = 3$ — sum

$$\sum_{k} T_{[2k]}^{\varepsilon_{[8]}} = \sigma_{2,0},$$

that is, $\sigma_{4,0} \cdot \sigma_{4,2} = \sigma_{2,0} + \sigma_{2,1} = -1$. By solving quadratic equations, we found $\sigma_{4,0}, \sigma_{4,2}$.

In the last step we will consider groups of roots that are invariant under $T_{[8]}$; we will use them $\varepsilon_{[0]}^{\varepsilon_{[r+1]}} + \varepsilon_{[1]}^{\varepsilon_{[r+1]}} + \varepsilon_{[2]}^{\varepsilon_{[r+1]}} + \dots + \varepsilon_{[15]}^{\varepsilon_{[r+1]}} = T_{[0]}^{\varepsilon_{[0]}^{\varepsilon_{[r+1]}}} + T_{[1]}^{\varepsilon_{[0]}^{\varepsilon_{[r+1]}}} + \dots + T_{[15]}^{\varepsilon_{[0]}^{\varepsilon_{[r+1]}}} = \varepsilon_{[0]} + \dots + \varepsilon_{[15]} = -1$. This allows us to find $\sigma_{8,0} = 2 \cos \frac{2\pi}{17}$ and thereby complete the solution.

12

No one knows exactly when people learned to create fire and artificially heat bodies to temperatures above ambient. In any case, this happened ten thousand years ago, in an era known as the Stone Age.

But the data on obtaining artificial cold is known quite accurately. Let us conventionally assume that low temperatures are temperatures below those that naturally occur when the weather changes. Faraday first obtained such a low temperature (163°K, or -110°C) in 1840. This is much colder than the bitter cold of Antarctica (the lowest recorded temperature in Antarctica is about -90°C).

Why have people been able to artificially create high temperatures since time immemorial, but they managed to create artificial cold only quite recently?

On the one hand, this is due to the fact that people needed high temperatures more than low ones.

On the other hand, the possible reason for the “lag” in the technology for obtaining low temperatures is too obstructive that high (no need) temperatures are not easier to obtain than low ones. In fact, it is enough to strike a match to get a temperature higher than room temperature. And Faraday, in order to obtain a temperature only half as low as room temperature, had to create something more than just a device.

This “inequality” of methods for obtaining high and low temperatures is not accidental. It reflects one of the basic laws of nature, the so-called second law of thermodynamics (recall that the first law of thermodynamics is the law of conservation of energy).

At first glance, it seems that cooling a body is very simple: you need to bring this body into contact with other bodies of lower temperature. Then the body will naturally move from the cooled body to the cold one, which will lead to the required cooling; If you have ice, it’s easy to make ice cream!

But what if the body needs to be cooled to such a low temperature that there is no suitable other cold body to which the body could naturally transfer? Then, obviously, such a cold body, or, as they say, a refrigerant agent needs to be created, a manufactured refrigerant also needs to provide heat transfer from it to some other body, but not a colder one, because there is no such body, but a warmer one, for example, the environment (air, water, etc.).

First of all, let’s find out what the difficulty is in this.

13

Order and disorder in matter

Let us remember that the temperature of a body is determined by the energies of the random thermal motion of its atoms and molecules. To lower the temperature means to reduce the disorder that reigns among the molecules.

What is disorder? If every item in your apartment is always in a certain place assigned to it and it is always easy to find, then you maintain order in your apartment. If things are classified anywhere, then they cannot be found. Your apartment is a mess.

The larger the size of your apartment, the more time it will take you to find this or that thing; if things are evenly scattered, the more disorder there is in your apartment. Imagine that things in the room can move on their own, changing their position. Then it will become even more difficult to find them, and the more difficult it is, the higher the speed of their movement.

This everyday example helps us understand how to determine what order and disorder are in the molecular world. The degree of disorder is determined by the probability of finding molecules (or another particle) in a certain state, for example, at a certain point in space.

If there is a given number of particles uniformly distributed in a given volume, then the probability of detecting a particle at a given point will be greater, the slower the particles move, the slower they change their position in a given volume. But the speeds of molecules are determined by body temperature. Therefore, the higher the temperature of a body, the greater the degree of disorder among the molecules of which it consists. It is also clear from the above that the greater the volume provided by the particles, the greater the degree of molecular disorder.

Thus, the degree of disorder depends on two factors - the temperature of the body and its volume. An increase in temperature at a given volume, just like an increase in volume at a constant temperature, leads to an increase in the degree of disorder among the molecules.

It turns out that all the processes that occur in nature, in a closed system left to itself (that is, in a system consisting of particles interacting with each other, but not with other bodies), take place in such a way that the degree of disorder increases. And when this degree of disorder reaches its highest value, all processes stop by themselves. A state occurs called the state of equilibrium.

Imagine balls of two colors arranged in a certain order. It took quite a few pages to arrange the balls in the correct sequence. But a slight push is enough for the painstakingly created order to be disrupted and the balls to be randomly mixed. And it is impossible to restore the original order by any means. The probability of such an event means nothing.

And here is an example from the atomic-molecular world. By throwing a piece of sugar into water, we can easily obtain a solution in which sugar molecules are distributed throughout the entire volume of the vessel. But no matter how much we covet, the solvent will not separate into water and sugar by itself. In order to obtain sugar and water separately, you need to expend some energy by heating the solution and evaporating the water.

The physical meaning of the second law of thermodynamics lies in the statement that order and disorder in nature are “unequal”, that all self-inducing processes of establishing equilibrium are accompanied by an increase in disorder. This is the reason why it is so high

14

temperatures are easier to obtain than low temperatures. As body temperature increases (while maintaining its volume), disorder increases. To cool bodies means to increase order, that is, to act, so to speak, “against nature.”

Nevertheless, it turns out that it is still possible to cool bodies. It is therefore possible to make the body move not from a warm body to a cold one, but, for example, from a cooled body to warmer ones. This is how refrigerants are created.

Disorder is a physical quantity

The words “order” and “disorder” that we used here seem at first glance to be somehow unfamiliar, taken from everyday life. In fact, behind these words lies the physical concept of the present, which is important, that it is expressed by a special quantity. This quantity, which expresses the degree of disorder or order (order is simply the absence of disorder), is called entropy. It is denoted by the letter $S$, and it is defined as follows: if a certain amount of the body $\Delta Q$ is supplied to or removed from the body, and the absolute temperature of the body is equal to $T$, then the change in entropy $\Delta S$ is determined by the equality

$$\Delta S = \frac{\Delta Q}{T}.$$

In all processes that happen to bodies, it is not entropy that is important, but rather the change in entropy. (In this respect, entropy is similar to the potential energy of a body. In all problems, it is not the potential energy itself that is important, but its change.)

Of particular interest to us is the case when the process occurs in such a way that the body is not supplied to or removed from the body: $\Delta Q = 0$. Such conditions can be ensured by a well-heat-conducting body or by quickly carrying out the process, when the body does not have time to move from one body to another. Such processes are called adnabate. From expression $\Delta S = \frac{\Delta Q}{T}$ it follows that during adnabate processes the entropy does not change.

Adnabate change in volume

In the technique of artificially producing cold, we are mainly interested in the processes of changing the volume of the body, the processes of their expansion and rolling. They play a particularly important role here.

Let’s imagine that we have a body that we expand or compress adnabatally, that is, without supplying or withdrawing the body. We already know that increasing the volume of a body with a given number of particles leads to an increase in the degree of disorder, that is, to an increase in entropy. But our process is adnabate, in which $\Delta Q = 0$ and entropy do not change. How to ensure the constancy of entropy, provided that it returns a noble increased volume? If the second law of thermodynamics is correct, then this is only possible if, when the volume of a body changes, some other process occurs, due to which entropy decreases by exactly the same amount as it increases due to an increase in volume. We know that entropy changes when body temperature changes. Therefore, we can expect that with an adnabate increase in the volume of a body, its temperature will decrease. This is what actually happens.

Thus, we immediately have a way to obtain body temperature. It is necessary to expand this body adnabatally, that is, to increase its volume. Adnabate expansion is precisely one of the main ways to obtain cold.

One type of expansion is the evaporation of liquid. Even the ancient Egyptians kept drinks in porous vessels. Pores increase the surface of the evaporating liquid and thereby increase

15

the rate of its evaporation. Evaporation leads to cooling of the liquid. The poor thermal conductivity of materials and the high evaporation rate ensure that the evaporation process is adnabate.

Thus, the question of artificial cooling methods comes down to the question of what to expand and how to expand.

What to expand?

It is clear that solids are unsuitable for expansion - they cannot change their volume in any noticeable way. For the same reasons, liquids are also unsuitable. Although the increase in the volume of liquid during their evaporation is used to cool them, especially low temperatures cannot be obtained in this way, because, as it cools, the liquid eventually solidifies. So the lowest temperature that can be obtained when a liquid evaporates is the temperature at which it solidifies. And for substances that are liquid at room temperature, the solidification temperatures are not so low.

It is best to use gases. Gases are capable of unlimited expansion, and if they are also previously at high pressure, then their mass is quite large and therefore the heat capacity of this gas is also quite large. Gases are also characterized by the greatest disorder in the movement of particles. And since our task is to reduce disorder, it is clear that the best ones are to start with a substance in which the disorder is great, so that there is something to reduce!

So, the most suitable substance for our task is gas. On the other hand, the final product, that is, the finished refrigerant, is best in the form of a liquid: a liquid always creates good thermal contact with the body immersed in it. Therefore, the technical part (but not exclusively) of the liquefaction properties of a particular gas.

As you know, any gas can be converted into a liquid. But for each gas there is a certain temperature, called the criterion $T_h$, above which it cannot be formed into a liquid state. In order for gas to be liquefied, its temperature must be below $T_h$. Then to burn it you only need to rock it. And to what extent it depends on how much its temperature is below $T_h$: the lower it is, the less pressure is needed for combustion. You can cool a gas so much that it becomes liquid at a pressure of 1 atmosphere.

To obtain moderate cold, gases with high $T_h$ values ​​are used. Often used, for example, ammonia, which has $T_h = 132,4^\circ\text{C}$. Recently, everyone has been using so-called freons - gases obtained from hydrocarbons by mixing hydrogen with fluorine, chlorine or bromine. In home refrigerators, in particular, freon-12 ($\text{CF}_2\text{Cl}_2$), which has $T_h = 112,04^\circ\text{C}$, is used.

The lowest temperatures for the production of gases, the critical temperatures of which are much lower than room temperature. These include: oxygen, which has $T_h = -118,4^\circ\text{C}$, nitrogen with $T_h = -146,9^\circ\text{C}$, hydrogen with $T_h = -239,9^\circ\text{C}$, and finally helium, which has the lowest critical temperature in nature. He has $T_h = -267,91^\circ\text{C}$.

Burning these gases is not so easy. To do this, they need to be very cool.

How to expand?

Gas can be expanded in different ways. Two methods of evaporation are more “popular” and most often used.

With the first method. Expand the best compressed gas. Therefore, with any method of evaporation, they begin by burning the gas using a special machine - a compressor - to a pressure of tens, and sometimes hundreds of atmospheres. In this case, the gas heats up for the same reason

17

Throttling leads to gas cooling. This is explained by the fact that gas molecules interact with each other. When, during the expansion of a gas, the molecules move away from each other, the arbitrary work against the molecular forces connecting the molecules, and the kinetic energy of the molecules, and, therefore, the temperature of the gas, will decrease. We can say that here too, work is done, but not against external forces, as in detand, but against the internal forces of cohesion between molecules.

The cooling effect of gas during throttling is called the Joule-Thomson effect.

For the Joule-Thomson efficiency, it is important that the gas is not ideal, that is, its molecules interact with each other at a distance, and not just in a direct “head-on” collision. An ideal gas would not cool when throttling. It could be cooled, but it uses detaner.

As already indicated, the cooling during throttling is stronger, the lower the temperature. At room temperature, which is where you start, it is not great. In order, despite this, to achieve significant cooling, resort to a constructive technique, the existence of which is clear from figure 2.

The pipe through which gas flows to the throttle is placed in another, wider pipe. When the gas cools slightly after throttling, it returns from the refrigeration chamber to the compressor through a wide pipe, cooling the oncoming gas “on the way.” Therefore, the next portion of gas will suit the throttle colder. Having passed through the throttle, the gas will cool even more, and returning through the outer pipe, it will cool the oncoming gas even more. After some time, the gas will approach the throttle so cold that throttling will lead to its liquefaction, and the first portion of liquid will appear in chamber X. Now the cold vapor of this liquid will escape through a wide pipe, continuing to cool the oncoming gas.

The cold liquid accumulated in the refrigeration chamber can be immediately used to cool other bodies. Sometimes this “other” body is a salt solution, which then goes to pipes installed in store counters, under an artificial skating rink, etc. But you can do it differently: drain the compressed gas through tap B into special vessels prepared for storing and transporting cold liquids (Dewar vessels). The liquid thus obtained can be used to cool other bodies anywhere and at any time. Common machines that use cooled gas or liquid as a refrigerant in the refrigerating chamber are called refrigerators. A machine designed to dispense a liquid product is called a cooling machine.

Throttling is the most commonly used cooling method. It is, in particular, used in home refrigerators.

One important addition needs to be made here. The fact is that throttling can lead

18

not only to cooling, but also to heating of gas. And for each gas there is a certain temperature, the so-called effective Joule-Thomson inversion temperature ($T_i$), above which heating occurs. This means that if it is desirable to cool the gas by further expansion, care must be taken to ensure that the gas is pre-cooled below the inversion temperature. For most gases, including nitrogen and oxygen, the inversion temperature is much higher than room temperature, so no difficulties arise. But for hydrogen and gel it is significantly lower than normal temperatures: for hydrogen $T_i$ is 204°K, for gel 40°K. Consequently, when compressing these gases, they require pre-cooling before re-settling can begin. To do this, you can use the detaner method. In many installations for compressing hydrogen and gel, both methods are used at once: in the detaner the gas is cooled below $T_i$; further cooling, up to compression, is achieved by further settling. A combination of these methods is used for compression of other gases.

Even closer to absolute zero

The cooling methods described above make it possible to convert any gases, including gels, into liquid. Helicium boils at atmospheric pressure at a temperature of 4.2°K. By intensively pumping out the evaporating gel using a pump, you can lower the vapor pressure above the liquid gel and thereby lower the boiling point to 0.7°K, and if you use a lighter isotope of the gel, its temperature can be raised to 0.3°K. Lower temperatures cannot be obtained in this way.

How to get to temperatures of thousandths of a degree?

Recall that lowering the temperature means reducing that part of the disorder that depends on temperature. To do this, we use a gas whose disorder is high. But at temperatures close to absolute zero, the degree of disorder is already so small that it is very difficult to reduce it. At such temperatures, all substances (except the gel) are solid, so one cannot count on an adabate change in volume. This means that we need to find substances in which, even at these temperatures, disorder still exists; it depends not only on temperature and volume. And physicists also found substances. These are some complex parametric substances.

A paramagnetic substance is a substance containing particles (atoms, ions, molecules) that behave like small magnets. Due to thermal movement, the magnets are oriented in a completely random manner. True, at low temperatures, when thermal movement is very weakened, the interaction forces between the magnets can force them to position themselves correctly. But if the magnets are located at significant distances from one another, then there is no interaction force, and the magnetic disorder persists to the lowest temperatures. This is the case in parametric salts that are very complex in composition, in which, in addition to magnetic particles, there are many other non-magnetic ones, for example, in the salt 2Ce ($NO_3$)₃·3Mg ($NO_3$)₂·24H₂O (cerium-magnetic nitrate). In this substance, only cerium Ce nonones are “magnetic”. But for every nonentity there are about 60 other, non-magnetic atoms. “Magnetics” are so strongly “diluted” with non-magnetic particles that disorder in Ce orientaions persists even at temperatures below 1°K. This is the substance we need: the disorder in it is great and high at absolute zero!

Disorder in a parametric substance can change not only when its temperature changes, but also when its magnetic state changes. And the magnetic state of matter can be controlled in exactly the same way as

19

Fig. 3.

and the volume of the substance. This can be done using a magnetic field. If a parameter salt is placed in a constant magnetic field, then all the magnets will line up parallel to the confluent line of the magnetic field. We will bring order to the orientation of the ions. By removing the field, we will create chaos again. This is completely analogous to the rolling and expansion of gas.

To obtain low temperatures proceed in the following way. A paramagnetic salt in an ampoule $A$ filled with a gaseous gel and located in a vessel with a liquid gel, the temperature of which is about $1^\circ\text{K}$, is placed in a magnetic field between the electromagnetite strips (Fig. 3, $a$). The helium gas in the ampoule provides thermal contact with the liquid gel, so that the temperature of the magnetized cols matches the temperature of the surrounding liquid helium. Helium gas is then pumped out of the ampoule, providing heat-conducting salt from the environment (Fig. 3, $\delta$). After this, the magnetic field is turned off, that is, adiabatic gas formation cols (Fig. 3, $\theta$) is produced. In this case, misorientation of the magnets occurs -

innovations, which should lead to an increase in entropy. And since the demagnetization process is adnabate, the entropy as a whole cannot change and the salt cools.

For almost forty years now, the described method has been used to obtain so-called ultra-low temperatures, that is, temperatures below those that can be obtained using only liquid helium. The lowest temperatures obtained in this way are thousandths of a degree.

Not only atoms, but also atoms cores are magnetic, albeit 1000 times weaker. Nevertheless, if you start not from $1^\circ\text{K}$, but from $0,01^\circ\text{K}$, you can, by adiabatic demagnetization of not atoms, but nuclei, obtain a temperature of about a millionth of a degree. But this is a topic for a separate article.

Exercises

1. Why can expansion in the part and in the process be considered adnabate processes?
2. Is it possible to use ferromagnetic rather than parametric substances in the magnetic cooling method?
3. In what units is entropy measured? What other physical quantity is measured in such units?

20

G. D. Balk, M. B. Balk

Both the mathematics major and the schoolchild come across mathematical propositions, the truth of which he doubts. Otherwise, you need to “prove or disprove” any statement or choose the only correct one from several sentences of answers to a problem.

If the proof of a statement runs into cumbersome calculations, then it is often useful not to start with attempts at proof, but to make a separate report on whether this hypothesis is correct and whether it does not lead to clearly erroneous or inaccurate conclusions.

Let us dwell on some of the advantages of this kind of “check for legal completion” *).

This is what Bramagupta said…

Generalizing Heron’s well-known formula for the planes of a triangle, the Indian mathematician and astronomer Bramagupta (7th century AD) made a statement that can be formulated in the current notation as follows:

*) The article is based on the stepogram of the class of the 9th grade mathematics and mechanics club at school No. 7 in Smolensk (January 17, 1971).

the area S of any quadrilateral (Fig. 1) is expressed through its sides a, b, c, d according to the formula

$$S = \sqrt{(p-a)(p-b)(p-c)(p-d)}$$

here p, as usual, is the meter of the quadrilateral:

$$p = \frac{1}{2}(a+b+c+d).$$

Is Bramagupta’s statement true?

Before they rush to prove this statement (if it is true, then the proof will probably be cumbersome), it makes sense to use it to complement the truth. Let’s use the simplest method: testing for partial cases.

Consider the case when the given quadrilateral is a square; arbitrary rectangle; a rhombus other than a square.

In the first two cases, formula (1) leads to the correct result, in the third it leads to an error: according to the formula $S=a^2$, although it is clear that in this case $S<a^2$.

Therefore, Bramagupta’s statement is wrong.

Let us note in passing that this failure should not force us to abandon

25

It is interesting that the American mathematician M. Goldberg managed to discover in 1967 that Malfatti circles for any triangle do not provide a solution to the “original Malfatti problem” (Problem II).

Used in the consideration of the Malfatti problem, let us assume that “recourse to the limiting case” is often and successfully used to test mathematical propositions for verisimilitude.

Is it just for verification? Is it only in mathematics?

For about six years now, it seems to have turned out that “testing for verisimilitude” can only lead to the verification of dubious hypotheses. But is it not possible to arrive at the result of a “positive” characteristic in this way? Was it possible to obtain a new discovery using these methods? What about other sciences: physics, chemistry, biology? After all, any experiment, strictly speaking, can either refute a false hypothesis or confirm its veracity, but in no way prove it! And yet, we know many “laws of science” obtained experimentally (as they say, “inductively”).

All this, of course, requires a special discussion. In order not to make such a conversation dependent on the continuation of this article, we recommend to the reader the excellent books of György Pólya:

1. Mathematics and truthful reasoning (IL, 1957);
2. How to solve a problem (Uchrediz, 1959);
3. Mathematical discovery (“Science”, 1970).

The reader can also find individual examples related to the topic of this article in our book “Mathematics after Lessons” (“Enlightenment”, 1971).

Number triangle

| | | | | | | 65 | 50 | 66 | 37 | 51 | 67 |
| :— | :— | :— | :— | :— | :— | :— | :— | :— | :— | :— | :— |
| | | | | | | | | | | | | |
| | | | | | | | | | | | | |
| | | | | | | | | | | | | |
| | | | | | | | | | | | | |
| | | | | | | | | | | | | |
| | | | | | | | | | | | | |
| | | | | | | | | | | | | |
| | | | | | | | | | | | | |
| | | | | | | | | | | | | |
| | | | | | | | | | | | | |
| | | | | | | | | | | | | |
| | | | | | | | | | | | | |
| | | | | | | | | | | | | |
| | | | | | | | | | | | | |
| | | | | | | | | | | | | |
| | | | | | | | | | | | | |
| | | | | | | | | | | | | |
| | | | | | | | | | | | | |

Here you do not see the entire number triangle, but only its “left” part. The law of triangle designation is very visible, and to come in this way to the result of a “positive” characteristic? Was it possible to obtain a new discovery using these methods? What about other sciences: physics, chemistry, biology? After all, any experiment, strictly speaking, can either refute a false hypothesis or confirm its veracity, but in no way prove it! And yet, we know many “laws of science” obtained experimentally (as they say, “inductively”).

All this, of course, requires a special discussion. In order not to make such a conversation dependent on the continuation of this article, we recommend to the reader the excellent books of György Pólya:

1. Mathematics and truthful reasoning (IL, 1957);
2. How to solve a problem (Uchrediz, 1959);
3. Mathematical discovery (“Science”, 1970).

The reader can also find individual examples related to the topic of this article in our book “Mathematics after Lessons” (“Enlightenment”, 1971).

V. Berezin

26

LABORATORY “QUANTA”
with EXPERIENCES
PENDULUM
G.L.Kotkin

It is difficult to find physical devices that were as simple in design and reliable in use as matnikki. Meanwhile, experiments with them can lead to interesting results, which are extremely important not only in mechanics, but also in radio engineering and electrical engineering. Try to do the suggested experiments yourself and explain the effects that you will observe. Although the experiments themselves are very simple, however, it is not easy to fully understand the increase in matniki: for this you need to know the basic laws of harmonic vibrations and have experience in trigonometric transformations. Apparently, only tenth graders will be able to fully understand this article, but schoolchildren in grades 8-9 can carry out these experiments and fully explain them, at least qualitatively.

Matnik with suspension division

Make a mat with a split hanger, shown at figure 1. The length of the mat $l$ should be much longer than the length of the top divided part of the hanger.

Push the mat at an angle to the figure plane. You will see that at first the matnik will be qualitatively in the direction only, but soon its movement will turn into rotation in the circle of the vertical axis. Then the mat will begin to oscillate again in the plane, but no longer in the one in which it was pushed. After this, the rotation will begin again, but in the other direction. Gradually it will be replaced by movement in the direction of the first oscillatory shock. Then everything will repeat again from the beginning.

What is the reason for this behavior of the matnik?

Related matniki

a) Two identical mats (for example, nuts on a thread) will be hung on a not very tight stretched thread $AB$ (Fig. 2). Reject one of the matniks and release.

Fig. 1. Fig. 2.

27

Try changing the weights of the weights, changing the lengths of the threads of the mats, and the tension of the upper thread.

6. Attach two matrices to the rigid frame. Then tie the threads of the matniks with a thread at a certain height (Fig. 3). Reject one of the mats and lower it. Look how the matniki will move.

Double matnik

The structure of the double matnik is shown in figure 4. The upper weight is much heavier than the lower one, and the long threads are the same. Push the bottom weight and watch it move.

Let’s now understand the behavior of these matniks.

1. Let’s start with a matnik with an unlinked suspension.

The rather complex behavior of the matnik that you observe can be explained by the fact that its movement consists of two oscillatory movements (Fig. 5): parallel to the plane $zy$ and perpendicular to it. These oscillations occur independently of each other (if the deviations of the mat from the vertical are not too large), and the periods of their corresponding ones are equal

$$T_1 = 2\pi \sqrt{\frac{l}{g}} \quad T_2 = 2\pi \sqrt{\frac{L}{g}}.$$

First of all, let’s learn how to describe the movement of a weight using coordinates.

Let the point $A$ move along a circle of radius $a$ with an angular velocity $\omega$ counterclockwise (Fig. 6a) or clockwise (Fig. 66). Then the projection $B$ of the point $A$ onto the axis $x$ performs a harmonic oscillation*) with an angular frequency $\omega$ equal to $\frac{2\pi}{T}$ and an amplitude $a$.

The projection $C$ of the same point $A$ onto the axis $y$ makes similar oscillations, but reaches its maximum deviation from point $O$ a quarter of a period later - after the time during which the radius $OA$ makes $\frac{1}{4}$ revolutions around the point $O$. It is easy to verify this by following the movements of points $C$ and $B$ when rotating the radius $OA$. At the moment when the projection of the point $A$ onto the axis $x$ is maximally deviated from the point $O$, its projection onto the axis $y$ is just

*) See, for example, the article by Ya. A. Smorodinsky “Similar movements”, «Kvant» No. 9, 1971.

29

direction $OD$ (Fig. 7) to a distance $a$ and then release, then the coordinates of the weight will change according to the formulas:

$$\begin{cases} x = a \cos \omega_1 t, \\ y = a \cos \omega_2 t, \end{cases}$$

Where

$$\omega_1 = \frac{2\pi}{T_1} = \sqrt{\frac{g}{l}},$$

$$\omega_2 = \frac{2\pi}{T_2} = \sqrt{\frac{g}{L}},$$

$$a = \frac{b}{\sqrt{2}}.$$

Recall that $l$ and $L$ have very little difference from each other, that is, the frequencies of $\omega_1$ and $\omega_2$ are almost the same. Therefore, at first, oscillations along the $x$ and $y$ axis will occur almost in phase.

Since $\omega_1 > \omega_2$, over time the oscillation along the $y$ axis lags behind the oscillation along the $x$ axis more and more. But, as we have seen, when a point moves counterclockwise around a circle, the oscillations of its projections on the $x$ and $y$ axes lag behind each other by the fourth period. It follows that when the lag of the oscillations of the weight along the axes $x$ and $y$ reaches the fourth period, it will move just along the circle counterclockwise. When the oscillations of the weight lag along the axes $x$ and $y$ on the field of motion, it will move along the straight line $E'D'$. Then the lag of the oscillation along the $y$ axis will reach the $\frac{3}{4}$ period - the weight moves around the circle clockwise. Finally, when there is a delay of an entire period, the weight again oscillates along the straight line $ED$. This is what you observed.

It is convenient to “retell” these wonders using formulas. System (5) can be represented as

$$\begin{cases} x = a \cos \omega_1 t, \\ y = a \cos (\omega_1 t - \phi), \end{cases}$$

where $\phi = (\omega_1 - \omega_2) t$. Over two or three periods, $\phi$ remains almost unchanged, but over many periods it increases steadily. For $\phi = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$, formulas (6) coincide in sequence with formulas (3), (1), (4), (2).

Fig. 9.

At $\phi = 2\pi$ the weight returns to motion in a straight line $ED$.

The stated theory can be tested experimentally. After all, we can calculate how long it will take for the weight to move through the entire cycle of changes. If it is time $T_0$, then $\phi_0 = (\omega_1 - \omega_2) T_0 = 2\pi$, opening $\frac{2\pi}{T_0} = \omega_1 - \omega_2$, or

$$\frac{1}{T_0} = \frac{1}{T_1} - \frac{1}{T_2}.$$

Now you need to measure the oscillation periods $T_1$ and $T_2$ of the pendulum, then in the $yz$ plane, another time in the $xz$ plane. Time $T_0$ is measured independently. After measuring $T_1$, $T_2$ and $T_0$, check the correctness of formula (7). To measure $T_1$ and $T_2$ it is better to determine the time, say, in 10-15 vibrations. It is immediately revealed what reasons lead to deviations from the calculation. Apparently, the damping of oscillations caused by air resistance is primarily affected. Well, the weights shouldn’t be too light.

In reality, the movements will not occur in a straight line or circle, but along a complex trajectory. This trajectory (Fig. 9) more or less uniformly “fills” the entire square.

If the initial angle with the symmetry plane is different from $\frac{\pi}{4}$, then the characteristic “instantaneous” trajectories of the weight are the diagonals of the right angle and the ellipses inscribed in it.

The fact that the movement of the pendulum consists of independent oscillations is expressed by saying that the superposition principle is valid for its oscillations *).

2. Let’s move on to explaining the movement of the connected pendants (Fig. 2 and 3). If we deflect and then release one of them, we will see that the other will gradually begin to move.

*) Reader «Kvant» have already encountered the principle of superposition when explaining the type of wave on the surface of water («Kvant» No. 1, 1971, p. 32).

30

This is explained by the fact that when the first matrix is ​​deflected by the bond, their thread is deformed and the elastic force acts on the other matrix, imparting acceleration to it. The first matrix will oscillate with less and less amplitude and finally stop. At this time, the amplitude of vibrations of the second matrix will be the most magnificent. Then the first matrix will roll out, and the second will stop, and so on.

The complex oscillatory motion of a system consisting of two matrices can be represented as the result of the addition (superposition) of two oscillations with frequencies $\omega_1$ and $\omega_2$. One of the frequencies is equal to the frequency with which both matrices will oscillate if they are simultaneously deflected in one direction and then released (Fig. 10). Let’s denote this frequency by $\omega_1$. Another frequency is $\omega_2$. This is the frequency with which both matrices will vibrate if they are first simultaneously deflected in opposite directions and then released (Fig. 11).

Note that the angular frequency $\omega_2$ is greater than the angular frequency $\omega_1$. These two vibrations are called normal vibrations of two coupled matrices.

Let’s write the equation of normal vibrations. First

$$\begin{cases} x_1' = a \cos \omega_1 t, \\ x_2' = -a \cos \omega_1 t, \end{cases}$$

second

$$\begin{cases} x_1'' = a \cos \omega_2 t, \\ x_2'' = a \cos \omega_2 t. \end{cases}$$

Let’s add the mixtures $x_1'$ with $x_1''$ and $x_2'$ with $x_2''$. We get:

$$\begin{cases} x_1 = x_1' + x_1'' = a \cos \omega_1 t + a \cos \omega_2 t, \\ x_2 = x_2' + x_2'' = -a \cos \omega_1 t + a \cos \omega_2 t. \end{cases}$$

Using the formula for adding cosines, we get:

$$\begin{cases} x_1 = A \cos \Omega t, \quad \Omega = \frac{\omega_1 + \omega_2}{2}, \\ A = 2a \cos \left( \frac{\omega_1 - \omega_2}{2} t \right). \end{cases}$$

Due to the fact that $\omega_1$ differs little from $\omega_2$, the expression $\cos \left( \frac{\omega_1 - \omega_2}{2} t \right)$ is close to 1 (if $\omega_1$ is equal to $\omega_2$, then this expression is equal to 1). Therefore, it has little effect on the value of $x_1$. Therefore, the factor $2a \cos \left( \frac{\omega_1 - \omega_2}{2} t \right)$ in the expression for $x_1$ can be considered as slowly changing the amplitude (red point on Fig. 12, a). The situation is similar

31

Deal with $x_2$. The oscillations of weights corresponding to these formulas are presented graphically at figure 12, $a$, $b$. Movements of this kind are called beats (or modified oscillations). It is easy to find the period of beats. It is equal to the half-period of the sine (dotted curve) and is determined by the formula

$$\frac{\omega_1 - \omega_2}{2} T_0 = \pi$$ or $$\frac{1}{T_0} = \frac{1}{T_2} - \frac{1}{T_1},$$

where $T_1$ and $T_2$ are periods of normal oscillations. The frequency of the beat is equal to the frequency of normal oscillations. The beat period is greater, the closer $T_1$ and $T_2$ are to each other.

Of course, with a very large beat period, the motion picture will be confused due to the damping of the oscillations.

3. When the double matrix moves, beats are observed. The light matnik either oscillates with great swing, then almost stops. The movement of a heavy matrix can be almost imperceptible. Such beats are explained by the addition of two normal vibrations, shown in figure 13. Do the explanation yourself.

Exercises

1. Derive a formula for the period of oscillation of a mathematical matrix, considering its movements as rotation around each axis.
2. Select the length of the expanded suspension of the matrix so that the weight can be described with eight hands.
3. Select the lengths of two independent matrices so that with their help you can determine time intervals of 10 seconds, 20 seconds, not counting the oscillation “piece by piece”. Such matrices can be used to measure heart rate after charging. (Galileo randomly measured the periods of vibration of lamps in a church, using the pulse beat of a clock).

MIPT crossword

There are many types of student folkcore: songs, jokes, fairy tales. Students of the Moscow Institute of Physics, in addition to those listed, have one more thing - crosswords.

In order to solve such a crossword puzzle, you must first of all have a sense of numbers and imagination. Encyclopedias and reference books will not help here. Among MIPT teachers, individual halves of the crossword words are considered excellent access; among students, the figures are slightly higher. If the crossword puzzle below does not work for you, then look at page 38, where it unscrambles the words horizontally, and write them down, then try to debug the words vertically on your own. The answers to these are given on page 65.

Horizontally: 1. Computer center of a research institute for the organization of individual workers of a production and training plant. 6. The devil is in the nose. 7. Far away. 8. Natural compound. 11. A character from Russian folk tales who appears primarily from behind the mountains. 13. Musical instrument without sound. 14. Srednezamatsky catrist, philosopher, composer, and therefore singer. 17. Patient Cossack. 20. Son of the East. 21. A device for stopping a train by pressing a button. 22. Trick with light.

Vertically: 1. A bad person and also a drinker. 2. Subtlety. 3. Work by P. I. Tchaikovsky. 4. Type of coal. 5. Spectacled kite. 9. Lonely inhabitant of the wild world. 10. The force of attraction to the Earth. 12. Very large wardrobe. 15. Man with a shovel. 16. Legendary downsizing. 18. No goals. 19. Several identical letters.

32

CARDIOID

The figure depicted on the first page of the cover represents one of the replacement curves - the cardioid. The word “cardioid” means “heart-shaped” in Greek.

The properties of this curve were first studied in 1674 by the French academician Ole Roemer.

Cardionda has many “relatives”. These are, first of all, various cyclones, epicyclones and hypocyclondes.

If a circle rolls along a straight line, then a point fixed on this circle leaves a trace on the plane - a curve called a cyclonde (Fig. 1).

If a movable circle rolls along the inside of a fixed circle, then a point fixed on the first of them describes a curve on the plane, which is called an epicyclonde (Fig. 2).

If a movable circle rolls on the outside of a fixed circle, then the curve described by a point lying on the movable circle is called an epicyclonde (Fig. 3).

A cardioid is an epicyclone whose movable and stationary circles are of the same radius (see Fig. 4).

You can read about the features of all these cyclones, including the cardionde, in G. N. Berman’s book “Cycloid” (Gostekhidat, 1954).

figure 5 shows one of the possible constructions of the cardionda. The fixed circle is divided into sections by points. From each such point, as from the center, circles are drawn that also pass through one fixed point on the same circle. Try to prove that with this construction you really get a cardiond (circumscribed around all these circles).

In conclusion, a fun interpretation of the cardionda. Let’s consider a non-slippery, stationary waist, round in cross section. A hula hoop with twice the radius is placed next to it. Then each point of the hoop is described near the waist of the cardiond.

M. L. Smolyansky

33

When solving many mathematical problems, the so-called Bunyakovsky-Cauchy inequality is used

$$\sum_{i=1}^{n} a_i b_i \leq \sqrt{\sum_{i=1}^{n} a_i^2} \cdot \sqrt{\sum_{i=1}^{n} b_i^2} \quad (a_i > 0, b_i > 0).$$

Let us first give a small but very simple proof of this inequality. Let $$\sqrt{\sum_{i=1}^{n} a_i^2} = A, \quad \sqrt{\sum_{i=1}^{n} b_i^2} = B.$$ Then we need to prove that

$$\frac{\sum_{i=1}^{n} a_i b_i}{AB} \leq 1.$$

Using the inequality $xy \leq \frac{1}{2}(x^2+y^2)$, we write the following integer:

$$\frac{\sum_{i=1}^{n} a_i b_i}{AB} = \sum_{i=1}^{n} \left(\frac{a_i}{A} \cdot \frac{b_i}{B}\right) \leq$$

$$\leq \sum_{i=1}^{n} \frac{1}{2} \left[ \left(\frac{a_i}{A}\right)^2 + \left(\frac{b_i}{B}\right)^2 \right] = \frac{1}{2} \left( \frac{1}{A^2} \sum_{i=1}^{n} a_i^2 + \frac{1}{B^2} \sum_{i=1}^{n} b_i^2 \right) = 1.$$

This implies inequality (*). As is known, equality in it is achieved when

$$\frac{a_1}{A} = \frac{b_1}{B}, \quad \frac{a_2}{A} = \frac{b_2}{B}, \dots, \quad \frac{a_n}{A} = \frac{b_n}{B},$$

that is, when

$$\frac{a_1}{b_1} = \frac{a_2}{b_2} = \dots = \frac{a_n}{b_n}.$$

Let us now show how inequality (*) is applied.

34

1. Find the most value of a function
   $$y = a \sin x + b \cos x$$
   $$\left(a > 0, b > 0, 0 < x < \frac{\pi}{2}\right).$$

Solution. Due to inequality (*) we have
$$y = a \sin x + b \cos x \leqslant \sqrt{a^2 + b^2} \cdot \sqrt{\sin^2 x + \cos^2 x} = \sqrt{a^2 + b^2}.$$
The maximum is achieved under the condition
$$\frac{a}{b} = \frac{\sin x}{\cos x},$$ that is, when
$$tg x = \frac{a}{b}.$$

2. A variable type transformer must be designed so that its cross-shaped iron core fills as much of the internal cavity of the round winding as possible. What should be the cross-sectional dimensions of the core shown at figure if the radius of the coil is $R$?
   Solution. Let $\Rightarrow AOB = \alpha$, then
   $$\frac{1}{4} S = y^2 - (y - x)^2 = 2xy - x^2 = R^2 \left(2 \sin \alpha \cos \alpha - \sin^2 \alpha\right).$$
   Hence $S = 2R^2 \left(2 \sin 2\alpha + \cos 2\alpha - 1\right)$. Let’s use inequality (*):
   $$S \leqslant 2R^2 \left(\sqrt{2^2 + 1^2} \cdot \sqrt{\sin^2 2\alpha + \cos^2 2\alpha} - 1\right) = 2R^2 \left(\sqrt{5} - 1\right).$$
   Therefore, $S_{\max} = 2R^2 \left(\sqrt{5} - 1\right)$ is achieved at
   $$\frac{\sin 2\alpha}{\cos 2\alpha} = \frac{2}{1},$$ from where $\alpha = \frac{1}{2}$ arctg 2.

3. Consider three pairs of real numbers $(a_1, b_1), (a_2, b_2), (a_3, b_3)$. Prove that there is a triangle with side $a = \sqrt{(a_2 - a_3)^2 + (b_2 - b_3)^2}, \quad b = \sqrt{(a_3 - a_1)^2 + (b_3 - b_1)^2}, \quad c = \sqrt{(a_1 - a_2)^2 + (b_1 - b_2)^2}$.
   Solution. In order for $a, b$ and $c$ to be sides of a triangle, it is necessary that the inequalities $a + b > c, b + c > a, a + c > b$ be satisfied, each of which is equivalent to the Bunyakovsky-Cauchy inequality. Indeed, the correspondence $a + b \geqslant c$ is equivalent to the inequality $2ab \geqslant c^2 - a^2 - b^2$ or $2 \sqrt{(a_2 - a_3)^2 + (b_2 - b_3)^2} \cdot \sqrt{(a_3 - a_1)^2 + (b_3 - b_1)^2} \geqslant (a_1 - a_2)^2 + (b_1 - b_2)^2 - (a_2 - a_3)^2 - (b_2 - b_3)^2 - (a_3 - a_1)^2 - (b_3 - b_1)^2 = 2 \left[ (a_3 - a_1) (a_2 - a_3) + (b_3 - b_1) (b_2 - b_3) \right]$, reducing by 2 we obtain inequality (*). Equality is achieved only if the numbers $a_3 - a_1, b_3 - b_1$ are proportional to the numbers $a_2 - a_3, b_2 - b_3$; check that in this case the vertices of the triangle must lie on the same straight line.
   It can be proved in a similar way that $a + c \geqslant b, b + c \geqslant a$. This means that a triangle with sides $a, b, c$ exists, although it can degenerate into a segment. You can construct this triangle geometrically - in the Cartesian coordinate system, its vertices must be placed at points with coordinates $(a_1, b_1), (a_2, b_2), (a_3, b_3)$.

35

Exercises

1. Find the most value of a function
   a) $y = a \sin x + b \sin \left( \frac{\pi}{3} + x \right) + c \sin \left( \frac{\pi}{3} - x \right) \left( a > 0, b > 0, c > 0, 0 < x < \frac{\pi}{3} \right)$;
   b) $z = a \sin (x + y) + b \sin (x - y) + c \sqrt{\cos 2x \cdot \cos 2y}$
   $(a > 0, b > 0, c > 0, 0 < x < \frac{\pi}{4}, 0 < y < \frac{\pi}{4})$;
   c) $u = a \cos x + b \cos y + c \cos z + d \sqrt{2 \cos x \cos y \cos z}$, where $a, b, c, d$ are positive numbers, $x, y, z$ are the remaining angles of the triangle;
   d) $u = a \sin x + b \sin y + c \sin z + d \sin t + e \sqrt{2 \cos (x + y) \cdot \cos (y + z) \cdot \cos (z + x)}$, where $a, b, c, d, e$ are positive numbers, $x + y + z + t = 2\pi$;
   e) $y = \frac{(a_1 x_1 + a_2 x_2 + \cdots + a_n x_n)^2}{x_1^2 + x_2^2 + \cdots + x_n^2}$ $(a_i > 0, x_i > 0)$.

2. Find the most and most values ​​of the function $u = \sum_{i=1}^{n} x_i y_i$, if it is known that $\sum_{i=1}^{n} x_i^2 \leq a^2$, $\sum_{i=1}^{n} y_i^2 \leq b^2$ and $a$ and $b$ are positive numbers.

3. Prove inequality
   $$\sum_{i=1}^{n} \sqrt{x_i y_i} \leq \sqrt{\sum_{i=1}^{n} x_i} \sqrt{\sum_{i=1}^{n} y_i},$$
   where all $x_i$ and $y_i$ are positive numbers.

4. Prove inequality
   $$\left( \sum_{i=1}^{n} \sqrt{a_i b_i} \right)^2 \leq \sum_{i=1}^{n} a_i x_i \sum_{i=1}^{n} \frac{b_i}{x_i},$$
   where all $a_i, b_i, x_i$ are positive numbers.

5. Prove inequality
   $$\sum_{i=1}^{n} x_i y_i \sum_{i=1}^{n} \frac{x_i}{y_i} \geq \left( \sum_{i=1}^{n} x_i \right)^2,$$
   where all $x_i, y_i$ are positive numbers.

6. A ball of radius $R$ has a cylinder with the most complete surface. What are its radius and height?

7. Inside a triangle, a certain point is located such that the sum of the squares of the distances from it to the side of the triangle is the smallest. Express these distances in terms of the lengths of the sides of the triangle and find this point.
   8*) The areas of the triangle’s faces are $S_1, S_2, S_3$ and $S_4$. Point $M$ lies inside the triangle at distances $h_1, h_2, h_3$, and $h_4$ (respectively) from the faces.
   a) At what position of the point $M$ the sum
   $$\frac{S_1}{h_1} + \frac{S_2}{h_2} + \frac{S_3}{h_3} + \frac{S_4}{h_4}$$

takes the most meaning?
6) Prove that any triangle has at least one point $M$ for which the sum (*) is minimal. Find how many such points there can be.

*) This problem was sent by our reader G. Hovhannitsan from Yerevan.

PROBLEM WORKER
QUANTA

This section of the magazine is maintained from issue to issue. In it, along with relatively easy problems, problems consisting of several, gradually developing questions are published. The most difficult tasks are marked with asterisks. They are often difficult even for specialists in mathematics and physics. Therefore, do not check in if you cannot cope with this or that task. Try to get back to it again over time.

But a solution has been found. One should not, however, rest on this. Think about how you can generalize the task, clarify or strengthen its result.

In the second part of the section we will post the solution to the problems (approximately six months after the publication of the conditions of the problems) taking into account the letters sent by readers. We recommend that even those who do not try to solve these problems on their own get acquainted with the solutions.

In most offices, we indicate the names of the readers who sent the most successful solutions.

Schoolchildren who regularly submit especially original and complete solutions will receive the right to participate in the regional rounds of the All-Union Olympiad for Schoolchildren along with the winners of regional Olympiads. In addition, the editors of «Kvant» have established several special awards for the authors of the most interesting solutions.

And one last thing. Coming up with a problem is usually more difficult than solving it. It’s more difficult, but it’s also more interesting. Try to come up with problems and send them to us. We will publish the best of them in the “Problem Book «Kvant»”.

Send your solutions and new problems to the address: 117071, Moscow, V-71, Leninsky Prospekt 15, editorial office of the magazine «Kvant», “Problem book «Kvant». On the envelope, after the address, indicate the solutions to which problems you are sending (for example: M122, M124, F146). The solution to each problem (if you send it immediately unpenned on a separate sheet(s), the pages are numbered.

At the end, indicate your last name, first name, patronymic, six-digit code, address, as well as class and school. Decisions are considered only if they are received no later than one and a half months after the publication of the corresponding issue of the journal.

37

3GOALS

M121. Prove that for any $n$ real numbers $a_1, a_2, \dots, a_n$ there is a natural number $k \leqslant n$ such that each of the $k$ numbers

$$a_k, \frac{a_{k-1}+a_k}{2}, \frac{a_{k-2}+a_{k-1}+a_k}{3}, \dots, \frac{a_1+a_2+\dots+a_k}{k}$$ does not exceed

$$\frac{a_1+a_2+\dots+a_n}{n}.$$

M122. The pentagon $ABCDE$ is inscribed in a circle. It is known that the distances from the point $E$ to the lines $AB, BC$ and $CD$ are equal to the correspondence $p, q, r$. Find the distance from point $E$ to line $AD$.

M123. Find all natural numbers $m$ for which

$$1! \cdot 3! \cdot 5! \cdot \dots \cdot (2m-1)! = \left( \frac{m(m+1)}{2} \right)!$$

(the product $1 \cdot 2 \cdot \dots \cdot n$ is denoted by $n!$).

V. P. Beshkarev

M124. Given a triangle $ABC$. Find a point $O$ inside it that has the following property: for any line passing through the point $O$ and intersecting the sides of the triangle $AB$ at the point $K$ and $BC$ at the point $L$, the equality

$$\frac{AK}{KB} + \frac{CL}{LB} = 1.$$

In general, prove that if $p$ and $q$ are arbitrary given positive numbers, then inside the triangle $ABC$ one can specify a point $O$ such that for any line $KL$ passing through this point ($K$ lies on $AB$, $L$ on $BC$),

$$p \frac{AK}{KB} + q \frac{CL}{LB} = 1.$$

G. Nomen (10th grade student)

M125. a) Is there an infinite sequence of natural numbers with the following property:

None of these numbers is divisible by the other, but among each three numbers you can choose two whose sum is divisible by the third?

6. If not, how many numbers can there be in a set that have this property?

b) Solve the same problem with the additional condition: only odd numbers are allowed to be included in the set.

Here is one example of such a set of four numbers: 3, 5, 7, 107. Here, among the three numbers 3, 5, 7, the sum 5+7 is divisible by 3; in the triple 5, 7, 107, the sum 107+5 is divided by 7; in the triple 3, 7, 107, the sum 7+107 is divided by 3; finally, in the triple 3, 5, 107, the sum 3+107 is divided by 5.

Yu. I. Nonin

F133*. A closed metal chain is connected by a thread to the centering axis of the machine and rotates with an angular velocity $\omega$ (Fig. 1). In this case, the thread makes an angle $\alpha$ with the vertical.

Fig. 1.

38

Find the distance from the center of the heavy chain to the axis of rotation.

F134. A heavy plate moves with a speed $v$ in a direction perpendicular to its plane. At an angle $\alpha$ to the direction of motion of the plate, a light ball moves at a speed $u$. Determine the magnitude and direction of the ball’s velocity vector after its elastic collision with the plate.

a) Distance when the ball speed is directed towards and away from the plate.

6. How does the answer change if the plate moves in a direction making an angle $\beta$ with its plane?

F135. A mine lying on the ground explodes from detonation. The mine fragments begin to move symmetrically in all directions with the same speeds $v$. The sizes of all osokols are the same. What fraction of the sedges will fall outside the circle of radius $R$ centered at the point of explosion?

G. L. Kstkin

F136. A parallel plate capacitor has a capacitance of $C$. Charge $+q$ was placed on one of the capacitor plates, and charge $+4q$ was placed on the other. Determine the potential difference between the plates of the capacitor.

V. E. Belonukkin

F137 *. Infrared radiation of the defining wavelength is absorbed by methane ($\text{CH}_4$). Under normal conditions, a 1 cm thick layer of pure methane absorbs 98% of the radiation energy. How many times will such radiation be weakened when passing vertically through the Earth’s atmosphere?

When calculating, take the weight content of methane in the atmosphere equal to $1,4 \cdot 10^{-6}$.

S. M. Kozel

Answers to the crossword (see page 31)

Horizontally: 1. VTSNIIOT-PUSK. 6. Obscurantist. 7. Telephone 8. Joint. 11. Egor. 18. Sax. 14. Akbin. 17. Ataman. 20. Ogly. 21. Stopkop. 22. Presentation.

Constructing a correct 17-gon

Richmond gave a simple construction of the seventeenth triangle $P_0P_1 \dots P_{16}$ (see figure).

Let’s connect point $P_0$ with point $J$, lying on radius $OB$ at a distance $1/4OB$ from the center. On the diameter passing through the point $P_0$, we choose points $E$ and $F$ so that $\Rightarrow OJE$ is equal to four times the angle $OJP_0$, and $\Rightarrow FJE$ is equal to $45^\circ$. Let a circle built on $FP_0$ as a diameter intersect

$OB$ at point $K$, and let a circle with center $E$ and radius $EK$ intersect $OP_0$ at points $N_3$ (between $O$ and $P_0$) and $N_5$. Let us restore the perpendiculars to $OP_0$ at these two points until they intersect with the original circle at points $P_3$ and $P_5$. Then the arc $P_3P_5$ (and the equal arc $P_1P_3$) is equal to the $2/17$ circle. The proof uses several times the fact that the roots of the equation

$x^2+2x\text{ctg}, 2\alpha-1=0$ are equal to tg$\alpha$ and - ctg$\alpha$.

(G. S. M. Coxter, Introduction to Geometry, M., “Hayka”, 1966, p. 49.)

39

Solutions

In this issue we publish solutions to problems M77—M83

M77

The lengths of the two sides of the triangle are 10 and 15. Prove that the bisector of the angle between them is not greater than 12.

We received a lot of different solutions to this problem from readers. Let’s give just three.

I. Let’s carry out $DE \parallel BC$ (Fig. 1). Then triangle $CDE$ is equilateral, therefore $CE=ED=x$, and $x$ one can find similarities of triangles $ABC$ and $ADE$:

$$\frac{15-x}{15} = \frac{x}{10}, \text{ поэтому } x=6 \text{ и } CD < CE + ED = 2x=12.$$

II. By the property of the bisector $BD/DA = BC/CA = \frac{2}{3}$. Draw $BK \perp CD$ and $DL \perp CD$ until they intersect with side $AC$ (Fig. 2). Then $AK=AC - KC=AC - BC=5$, and $KL/LA=BD/DA = \frac{2}{3}$, therefore $KL=2$, and $CD < CL=12.$

III. (In two stats “Area Method” «Kvant» No. 12, 1971, p. 41). Let $\Rightarrow BCD = \Rightarrow ACD = \alpha$, then (Fig. 3) ​​$S_{\triangle ACD} + S_{\triangle BCD} = S_{\triangle ABC} \langle = \rangle al \sin \alpha + bl \sin \alpha = ab \sin 2\alpha,$

Fig. 1. Fig. 2. Fig. 3.

40

because

$$l = \frac{2ab}{a+b} \cos \alpha = \frac{2 \cdot 10 \cdot 15}{25} \cos \alpha = 12 \cos \alpha < 12.$$

(Many people also used other formulas for the bisector:

$$l^2 = ab - mn = ab - \frac{abc^2}{(a+b)^2} = \frac{4abp(p-c)}{(a+b)^2}.$$

Either of these methods can be used to prove that the bisector $l$ of a triangle enclosed between the sides $a$ and $b$ does not exceed $\frac{2ab}{a+b}$ (the harmonic mean of the numbers $a$ and $b$). This estimate is unimprovable: for any $l < \frac{2ab}{a+b}$ one can construct a triangle with sides $a$ and $b$ and a bisector $l$ between them (make sure of this!).

Try to prove the following two statements that generalize the problem:

1. If in a triangle $ABC$, where $BC = a, AC = b$, the point $D$ divides side $AB$ in relation to $BD : DA = m : n$, then $\frac{|mb - na|}{m + n} < CD < \frac{mb + na}{m + n}$. (In our problem $m : n = a : b$).

2. If the segment $CD$ divides the angle $BCA$ in relation to $\Rightarrow BCD : \Rightarrow DCA = p : q$, then $CD < \frac{(p+q)ab}{pa + qb}$. (In our problem $p : q = 1$; here for the proof we can use the following lemma: if $0 < \alpha < \beta < \pi$, then $\frac{\sin \alpha}{\alpha} > \frac{\sin \beta}{\beta}$).

M78

Prove that every non-negative integer can be represented as $\frac{(x+y)^2 + 3x + y}{2}$, where $x$ and $y$ are non-negative integers, and what is the singular representation.

Let us denote the sum $x+y$ by $s$ and rewrite this formula as follows:

$$n = \frac{(x+y)^2 + 3x + y}{2} = \frac{s^2 + s}{2} + s.$$ (1)

The conditions that $x$ and $y$ are integers, $x \geq 0$ and $y \geq 0$ are equivalent to the following: $x$ and $s$ are integers, $x \geq 0$, $s \geq x$. For a given $s$, the value $x$ can take the following values: $0, 1, \ldots, s$; the corresponding value $n$, defined by formula (1), takes the following values: $\frac{s^2 + s}{2}$, $\frac{s^2 + s}{2} + 1$, $\ldots$, $\frac{s^2 + s}{2} + s$. So, each $s = 0, 1, 2, \ldots$ corresponds to a segment of $s+1$ non-negative integers $n$. Note that the last number of the segment corresponding to $s$ is exactly one less than the first number of the segment corresponding to $s+1$:

$$\left( \frac{s^2 + s}{2} + s + 1 \right) = \frac{(s+1)^2 + (s+1)}{2}.$$

Therefore, the segments will also contain all non-negative integers $n$, and each $n$ falls into only one segment, that is, it will correspond to exactly one pair of values ​​$s$ and $x$ (Fig. 4).

Solution problem. We have proven that formula (1) allows us to “inline” all points $(x, y)$ with non-negative integer coordinates with numbers $n=0, 1, 2, \ldots$. figure 5 shows how these numbers are arranged in order.

The correct solution to the addition A. Abdulaev, M. Bukhanov, A. Gritsoryan, A. Zaslavsky, M. Illarionov, N. Kortsenko,