Kvant 1971, No. 9 · Pages

Popular science physics and mathematics
quantum
9
1971
magazine
Academy
Sciences of the USSR
And
Academy of Pedagogical
Sciences of the USSR

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In the room:

Computers and number systems 1
R. S. Guter

Conversation about the causes of uncertainty 7
M. Ya. Azbel

M. V. Ostrogadsky 11
I. A. Maron

“Similar” movements 15
Y. A. Smorodinsky

Moon roads 18
L. G. Aslamazov

Glitter in nature, or why a cat’s eyes glow 20
S. A. Kheifets

Can the machine translate? 25
V. V. Raskin

Problem «Kvant»
Problems 31

Awards «Kvant» 33

Solutions to problems M61—M63 and F72—F74 34
N. B. Vasiliev
I. Sh. Slobodetsky

Applicant’s workshop
Square trinomial 41
M. S. Atamukas

Kinematika 47
I. A. Zaitsev

Chess and mathematics notes 52
E. Ya. Gik

History of the invention of the electric capacitor 56
A. K. Kikoin

Reviews. Bibliography
How they see the invisible 57
V. A. Leitskovtsev

Collector’s Corner
Stamp and label dedicated to M. V. Ostrogadsky 59
N. N. Kolesnikov

Logic cubes 60
G. I. Kopylov

Answers, directions, solutions 62

«Kvant» for younger students

3rd cover page

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Fig. 1.

physical devices that measure the corresponding quantities.

A number appears completely differently in a digital machine. Here it is written in its general numerical form and, therefore, requires the use of some specific system of simplification.

To represent each digit of a number in a digital machine, a special element is allocated, which has a certain number of different sources of sources of sources. For example, in ordinary Russian abacus (which are closer relatives of modern computers than the slide rule), a knitting needle with ten bones put on it is used to depict the number of each digit. The accuracy of recording a number on an abacus is determined by the number of digits of the number, which means the number of knitting needles available on the abacus.

Digital machines also include arithmeters and keyboard computers. In them, to represent each digit of a number, a six with a variable number of teeth is used, which can be located in one of ten different source sources, depending on the angle of rotation of the gear on the shaft, or a stepped roller, also with ten possible sources.

In all cases mentioned there is a blanket and the condition of each element. It is easy to say that this is associated with the decimal system: in each positional system, the number of different signs (digits) required to represent a number is equal to the base of the system. We see here that the power system predetermines its requirements for the design of the computer. It is not difficult to satisfy these requirements: you can put any number of teeth on a counter spoke or cut any desired number of teeth on a gear.

But it also happens that the machine exceeds its requirements for the power system. For example, for electronic, relay and electronic elements, the presence of two stable sources is characteristic. For example, an electron tube conducts or does not conduct current (open or locked), a ferrite core is magnetized in one or the opposite direction, etc. The use of such elements requires access to a numeric system in which there are enough numbers. This is the double system of strength: a positional system with a base of two.

In such a system, the unit of each subsequent rank is twice as large as the unit of the previous one. Therefore, the number is represented as 10, the number “four” as 100, and the number “eight” as 1000.

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Here’s what some numbers look like in the dual system:

| Decimal numbers | Binary numbers | Decimal numbers |
| :— | :— | :— |
| 1 | 1 | 110 |
| 2 | 10 | 7 | 111 |
| 3 | 11 | 8 | 1 000 |
| 4 | 100 | 50 | 110 010 |
| 5 | 101 | 100 | 1 100 100 |

The advantages of the dual system for computers do not end with the fact that it allows the direct use of elements with two sources. To build a computer, the total number of different sources of all elements is important. And in this respect, the double numerical system turns out to be more advantageous than the decimal one. Indeed, to write, for example, integers in the interval 1≤N≤999 in the decimal system, three elements (one for each digit), with ten sources for each, are sufficient. In total we will get 30 possible sources. The binary system will require ten elements, since $2^9=512<999$ and $2^{10}=1024>999$, but there are only two sources each, that is, a total of 20 possible sources.

Another advantage of the dual system is associated with the peculiarities of arithmetic operations. In all positional systems, numerical arithmetic operations on multi-digit numbers are related to actions on single-digit numbers, so to perform them it is enough to know the tables of addition and multiplication of single-digit numbers. In the dual system the tables are also extremely simple:

$$0+0=0 \quad 0 \times 0=0$$
$$0+1=1+0=1 \quad 0 \times 1=1 \times 0=0$$
$$1+1=10 \quad 1 \times 1=1$$

Since there are no other digits here except zero and ones, then multiplying multi-digit numbers only involves shifting numbers and addition. Thus, the basis of the arithmetic device of a computer operating in a double number system is a multi-digit double adder that adds two multi-digit double numbers. With its help, all other arithmetic operations are performed. In turn, a multi-bit adder can be composed of single-digit double adders, one of which is designed to add elements of the same bit.

Such a single-valued double adder must have three inputs: two of them receive the values ​​of the corresponding bits of the two terms, and the third is the ability to transfer to this bit from the least significant one. This adder should have two outputs: one of them is the amount that should be received in a given digit, and the second is a possible transfer from this digit to the highest one. A binary single-digit adder processes circuits as shown in figure 2. A multi-bit double adder is easy to construct from the required number of single-digit adders (Fig. 3).

Each digit applied to any input of a single-digit double adder can be either a zero or a one. Since there are three inputs, everything is possible in all (2^3) different combinations. For any combination, the adder must

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give certain output numbers. In this case, all inputs are equal and the output values ​​are determined not by what is supplied to one or another input, but by the total number of units supplied to the inputs.

For example, if two ones and one zero are supplied to the input, then at the output the adder must give a zero as the digit of the sum in a given digit and a one carried to the most significant digit (1+1+0=10), regardless of whether the zero to the input is a carryover from the low-order digit or a digit of one of the addends. Therefore, the operation of a double single-digit adder can be described by the table below. In it, A and B mean the inputs to which the digits of the corresponding digit of the first and second terms are supplied, C is the output containing the digit of this digit in the sum, and Z1 and Z2 are transfers of the corresponding one from the low-order digit to the given one (input) and from this digit to the highest one (output). Check the accuracy of the results yourself.

Single-digit double adder operation table

| Inputs | Inputs | Inputs | Inputs | Inputs |
| :— | :— | :— | :— | :— |
| A | B | Z1 | C | Z2 |
| 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 |
| 1 | 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 | 1 |

Before considering the circuits that make it possible to implement the operation of the adder in accordance with the table given, let’s get acquainted with the so-called logical operations *), defined on a set of two elements: 0 and 1.

1. The “or” operation (also called logical addition). The result of this operation is equal to one if at least one term is equal to one. Denoting this operation with the sign √, we get the table:

$$0\vee0=0, 0\vee1=1\vee0=1, 1\vee1=1.$$

2. Operation “and” (otherwise known as logical multiplication). Its result is one if both components are ones. This operation is denoted by ∧:

$$0\wedge0=0, 0\wedge1=1\wedge0=0, 1\wedge1=1.$$

3. The operation of “negation” is denoted by the sign ∗. It converts one to zero and vice versa:

$$\neg0=1, \neg1=0.$$

*) See, for example, «Kvant» No. 4, -1971, p. 14.

Fig. 4.

To implement the operations C=A\veeB and C=A\wedgeB it is easy to construct simple concatential circuits (Fig. 4). Here the value 1 for A and B is expressed as a closed contact, and 0 as an open contact. For C, the value 1 corresponds to the presence of voltage at the specified location in the circuit, and 0 corresponds to the absence of voltage. It is clear that for the “and” operation, the current through point C appears at the indicated location in the circuit, and 0 is the absence of voltage. It is clear that for the “or” operation - when at least one is closed.

Common schemes and other types. For example, for logical addition and multiplication it is easy to create diode circuits in which the value 1 indicates the presence of a high potential relative to ground at the corresponding point, and the value 0 indicates the absence of potential.

figure 5 shows a diode matching circuit that performs the “and” operation, while figure 6 shows its own circuit that performs the “or” operation. They use series diodes that conduct current in only one direction, shown by the direction of the side triangle. Instead, one could use an electron tube - a diode with two electrodes, which passes current only from the anode to the cathode.

If we apply a high positive potential at points A and B of the coincidence circuit (Fig. 5), then both diodes are blocked and at point C we get a high potential +E relative to ground. This matches

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Fig. 9.

Computing machines operating in the positional number system can be said to imitate the behavior of a person counting with a “column”. In essence, all machines began with an imitation of what was seen, alive: the first “self-propelled vehicles” had “legs” (Fig. 9), and the first flying machines were planned to be made with flapping wings. It is possible that in the future, more advanced computers will work with modern number systems. It is still difficult to say what such systems should be, but attempts to look for new principles are already being made, and we will now talk about one of them.

The disadvantage of positional number systems (when used in machines) is inter-bit connections that delay the execution of operations: to obtain the final result in any digit, it is necessary that the previous (low-order) digit has already been processed. Non-positional systems (in other respects, as we know, are less convenient) allow you to work with each “digit” independently and separately.

One possible way of representing natural numbers, which can be considered a “non-positional number system”, is the system of residual classes. In the system of residual classes, a number is represented by its remainders from division by the selected system of bases (“modules”).

When choosing one model $p$ different remainders from division by $p$ have $p$ integers taken in a row from the natural series. If there are several models, then the possibility of beginning numbers having representations expands.

Taking the corresponding remainders as symbols of the representation of the number, we obtain a non-positional number system, the digits of which are the remainders for each module depending on

efficiency. The numbers in each digit are different remainders, less than the modulus. The more a given module is selected, the larger the range of represented numbers and the more possible values ​​of the remainders (digits in a given digit). At the same time, all digits of a number in such a number system are completely independent of each other.

Arithmetic operations in such a system are very simple. For example, the remainder of dividing the sum of two numbers by a given module $p$ is equal to the sum of the remainders formed by the individual terms, if this sum is less than $p$, or $p$ less than this sum. A similar rule is true for the multiplication operation.

Consider as an example a system of residual classes defined by three modules $p_1=7, p_2=9, p_3=10$. The number $n$ in this system will be written in the form $(a, b, c)$, where $a, b, c$ is the correspondence of the remainder of dividing $n$ by $p_1, p_2, p_3$. This representation defines unique natural numbers in the range $0 \leq n \leq 629$, with possible values ​​of $a, b, c$ enclosed within $0 \leq a \leq 6, 0 \leq b \leq 8, 0 \leq c \leq 9$. For example:

$$5=(5, 5, 5); 20=(6, 2, 0);$$

Arithmetic operations:

$$(3, 2, 1)+(1, 4, 4)=(4, 6, 5);$$
$$(2, 5, 3)\times(2, 7, 6)=(4, 8, 8).$$

(In the decimal system, this equality goes to the corresponding one)

We do not at all claim, of course, that it is arithmetic that will still find application in the creation of new computers (if only for the reason that with this system it is difficult to compare numbers, which is necessary for many purposes). But it seems to be a simple and respectable example of the connection between “abstract” (and somewhat “old-fashioned” and quite traditional) theoretical-arithmetic laws with “numerical technical” problems of perfect machine mathematics.

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M. Ya. Azbel

on the uncertainty principle

In the second issue of our magazine for 1971, an article by M. Ya. Azbel “Dialogue on Temperature” was published. Stamy attracted many open readers. All the letters said that the article was integran. However, there were many typos in the article, for which the editors and the author apologize. The letters from readers raised very integrated questions. Therefore, we decided not to simply give a list of typos, but to ask the author to continue the conversation with the reader.

Author. I never expected that the editors would help me turn an imaginary dialogue with the reader into a real one, and not even a dialogue, but a conversation. This is what reader Volodya Margulis from Saransk writes.

V. Margulis. It seems to me that your article “Dialogue on Temperature” contains many typos. The Bolshman and Planck constants are overestimated by two orders of magnitude, the electron charge by an order of magnitude, and so on.

Author. You are absolutely right. Let me tell you a professional secret: like many theoretical physicists, I remember only one system of units: CGSE. All estimates were carried out in it and the constants in my article were given: Planck’s constant $h \approx 6.5 \times 10^{-27}$ ere·sec, Bolshman’s constant $k \approx 10^{-16}$ ere/erad, electron charge $e \approx 5 \cdot 10^{-10}$ units. CGSE. (Remember that we are only interested in the order of magnitude!)

However, in an article published in the journal, all calculations turned out to be in the SI system and, moreover, in a “deformed form.” This, of course, is unpleasant, but there is no big problem here. Every student reader who benefits from knowing systems of units will benefit from continuing the calculations and returning them to the correct form. Helpful and respectful. The fact is that typos and typos in scientific articles are not uncommon, and one must learn to treat them without worry, correcting them as necessary.

V. Margulis. I did just that, but the correct calculation for electrons is that the potential energy is two orders of magnitude less than the kinetic energy. And for the gel, the potential energy turned out to be much greater than the kinetic energy! *)

*) For those who have not read the article “Dialogue about temperature”, I will explain what we are talking about. From the so-called uncertainty principle it follows that the smallest possible value of the momentum $p_{\min}$ of a particle concentrated in a region whose dimensions are of the order of $a$, $p_{\min} \sim \frac{h}{a}$. Hence the lowest possible kinetic energy is $E_{\min} \sim \frac{p_{\min}^2}{2m} \sim \frac{h^2}{2ma^2}$, where $m$ is the mass of the particle.

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Author. Another typo has failed you. In all formulas of the article, it is more correct to write Planck’s constant $h$ rather than $h = \frac{h}{2\pi} \cdot \text{Позетому кинетическая энергия нулевых колебаний в гелини порядка}$

$$\frac{h^2}{2ma^2} \sim 2 \cdot 10^{-15} \text{эр}e \sim 20^\circ\text{K}$$

(since $m \sim 10^{-23} e$, $a \sim 3 \cdot 10^{-8} cm$). This value is only 5 times higher than the actual value, despite the roughness of the estimate.

I did not freely calculate the potential energy of interaction between the atoms of the helium, since this cannot be done based on general considerations.

V. Margulis. It is clear to me that this also applies to electron gas. According to the formula $E_{\text{min}} \sim \frac{h^2}{2ma^2}$ with $a \sim 3 \cdot 10^{-8} cm$ it turns out about $2 \cdot 10^{5^\circ}\text{K}$. And since $a$ can change, the estimate only makes sense in order of magnitude. Electronic gas really has an energy of tens, even hundreds of thousands of degrees.

Author. Absolutely right. I must conclude by saying that I really enjoyed your letter. I wish you success in your progress towards science. And one more thing - I apologize to both you and all the other participants in our conversation for turning their letters into replicas.

Reader M. Mokhnatkin (Saratov). But you haven’t convinced me yet. After all, according to the uncertainty principle

$$\Delta p_x \cdot \Delta x = h$$

and not $\Delta p \cdot \Delta x = h$, as written in the “Dialogue”. This means that since $p^2 = p_x^2 + p_y^2 + p_z^2$, the given value of the minimum energy should be tripled!

Author. If uncertainty corresponded with an equals sign, you would be absolutely right. But it is very important to remember: we are talking only about the order of magnitude. Let us assume, for definiteness, that the most important of $\Delta p_x$, $\Delta p_y$, $\Delta p_z$ is $\Delta p_x$. Then for $\Delta p =$

$$\sqrt{(\Delta p_x)^2 + (\Delta p_y)^2 + (\Delta p_z)^2}$$

we get $\Delta p_x \leqslant \Delta p \leqslant \Delta p_x \sqrt{3}$, so $\Delta p \sim \Delta p_x$ and $\Delta p \Delta x \sim h$ are complete.

When calculating orders of magnitude, it is important not to exceed the accuracy of the calculation, “refining” the coefficients - this can only worsen the matter. Such calculations are very useful, but they cannot give more than what they are designed for - an error of ten times is natural.

In particular, for a free particle “locked” in a spherical box of diameter $a$, the exact calculation gives just

$$E_{\text{min}} = h^2/2ma^2$$

Reader V. Rubin (Moscow). Why do you talk about Coulomb interaction in the case of electron gas? After all, electrons by themselves could not form a crystal - they repel each other!

Author. Certainly. That’s why I talked about electrons in metal. There they are not allowed to scatter positive nonsense. To interact with them and relative calculation. And the Coulomb interaction was taken for the following reason. The electrostatic interaction itself is classical. Let there be one free electron for each atom, so that the only characteristic distance of the problem is the mechanical (more precisely, mechanical) distance $a$, and we are only interested in distances of this order. Then the only quantity with the dimension of energy that can be constructed from $a$ and the electron charge $e$ is $\frac{e^2}{a}$.

Of course, such a notation is correct only in order of magnitude and only at such distances. At large distances, different charges shield each other, and the interaction quickly kills with distance.

V. Rubin. One more question. Why, if you have a distance for an electron in a box that exceeds the speed of light, do you still use non-relativistic formulas?

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Author. Because I wanted to show: in non-relativistic quantum mechanics, the presence of “locked” electrons in the nucleus would lead to troubles. And this suggests that, they believe, they cannot be locked there. However, you are, of course, right: such reasoning is not proof.

Since we are talking about this, let’s make a more accurate calculation. Using the uncertainty principles again, we obtain the order of the minimum impulse $p_{\min}$ of an electron “locked” in a nucleus of radius $R$, $p_{\min} \sim \frac{h}{R}$. According to the theory of relativity, $p = mv = \frac{m_0 v}{\sqrt{1 - v^2/c^2}}$ ($m_0$ is the rest mass of the particle, $v$ is its speed). “Our” impulse $p_{\min} \gg m_0 c$. This means that $v \approx c$ and $p \approx mc$ (m is the mass of the relativistic electron). This means that for an electron “locked” in the nucleus, $m \sim \frac{h}{R_c}$, which gives values ​​on the order of the mass of a proton at rest. Consequently, the proximity of the mass of the nucleus to the sum of the masses of only protons and neutrons means the absence of electrons “locked” in the nucleus: once they get into the nucleus, they are easily pulled out of there.

You can see the size of the region in which the particles are concentrated. This region can either strictly limit their movement, like the nucleus - the movement of protons and neutrons or atom - the movement of electrons, or be characteristic only from the point of view of the uncertainty principle.

Reader A. Petrov (Leningrad). If the average distance between particles is $a$, then $\Delta x \approx a$ and possibly even $\Delta x = L$, where $L$ is the size of the system. So, $p_{\min} \sim \frac{h}{L}$!

Author. Absolutely right. We estimate the characteristic energy, and it is determined by the average mass $a$. The smallest possible energy is $\frac{h^2}{2mL^2}$, and intermediate energy values ​​corresponding to $a < \Delta x < L$ are also possible.

Let me explain this remark in more detail. The average distance between electrons in a metal $a$ determines the characteristic distance in the uncertainty principle and the characteristic energy of zero-point vibrations, which we estimated. However, in principle, individual electrons can be in rare states, being at a distance from each other on the order of the size of the system $L$. In accordance with this, the minimum kinetic energy (zero-point energy), which has a small number of electrons, is of the order of $\frac{h^2}{2mL^2}$, that is, extremely small.

And here the cardial difference between Fermi and Bohemian particles is already evident.

Since basic particles can accumulate in unlimited quantities in any state, at very low temperatures they begin to accumulate precisely in a state with low energy - the so-called Bose-Einstein condensation begins in energy space.

What has been said is enough to obtain the basic characteristics of the Bose gas. Let’s start with the distance of thermal radiation concentrated in the cavity of a heated body (and in equilibrium with this body). Since thermal radiation consists of quanta of electromagnetic waves (photons), the number of which per unit volume is determined by the temperature of the heated body emitting them, the only energy characteristic is temperature. Therefore, the average photon energy is on the order of $kT$. Here we are not given a distance stretching, but a warm stretching of energy. Let us adapt the uncertainty principle in a slightly different form by introducing the characteristic time $\Delta t$:

$$h \sim \Delta p \Delta x \sim \frac{\Delta p}{\Delta t} \Delta x \Delta t.$$

But the change in momentum over a given time is the product of the force and the time of its action: $\Delta p = F \Delta t$. This means that $\frac{\Delta p}{\Delta t} \Delta x = F \Delta x$ is the work of force on the path $\Delta x$, that is, according to the law of conservation of energy, the change in energy $\Delta E$. Thus,

$$\Delta E \Delta t \sim h$$

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(a very important form of the uncertainty principle).

For photons, $\Delta E \sim \varepsilon$, and the value $\Delta t$ is a puree of time $\tau$, which characterizes the period of photon oscillations. Therefore $\varepsilon \sim hv\left(v = \frac{1}{\tau}\right)$.

From the resulting equality, taking into account that $\varepsilon \sim kT$, we find the oscillation period $\tau = \frac{1}{v} \sim \frac{h}{kT}$ and the wavelength $\lambda$ (the distance traveled by a wave moving at the speed of light $c$ during the oscillation period $\tau$):

$$\lambda = c\tau \sim \frac{ch}{kT}$$

The wavelength determines the characteristic distance “arriving at the quantum” $^*$). The constituent volume is $\lambda^3$, so the number of photons per unit volume is of the order of

$$\lambda^{-3} \sim \left(\frac{kT}{ch}\right)^3$$

Their total energy per unit volume, that is, the energy density of thermal radiation,

$$E_1 \sim kT \cdot \lambda^3 \sim \frac{(kT)^4}{(ch)^3}$$

We have obtained the significant Stefan-Boltzmann formula! The exact formula differs by the coefficient $-\frac{\pi^2}{20} \approx \frac{1}{2}$.

Now let’s move from the macrocosm of gases to the microscopic world of nuclei. Let’s try to estimate the order of magnitude of nuclear forces, knowing nothing about them, except that the radius of the nucleus is $R \sim 10^{-12} \sim 10^{-13}$ cm, and it consists of protons and neutrons interacting with each other. Kinetic energy of order $\frac{h^2}{MR^2}$. And since the nucleus does not fall apart and is not removed, but is kept in balance by nuclear forces, the potential energy of interaction must be on the order of kinetic!

So we have added to the energy of thousands of billions ($10^{12}$ degrees. Just don’t forget that this is energy measured in degrees,

*) Of course, this statement does not look very convincing, but it turns out to be correct, so just take it on faith.

but not the temperature of the part in the core!

And the difference in the masses of the nuclear parts - proton and neutron - is of the order of the electron mass $m_e$. (This is not accidental, since a neutron can turn into a proton and an electron.) This means that the energy associated with the charge is of the order of $m_e c^2$, that is, “only” of the order of $10^{10}$ degrees and significantly less than the nuclear interaction energy. Therefore, nuclear forces practically do not depend on charge, and it makes no sense to talk about neutrons and protons in the nucleus as separate parts, and they are simply called “nucleons.”

What is the speed of the core parts?

In non-relativistic mechanics $\frac{Mv^2}{2} \sim \frac{h^2}{MR^2}$ and $v \sim \frac{h}{MR}$. The speed is on the order of the speed of light! A more accurate estimate leads to $v \sim 0,1$ $c$, so in many cases for the part in the nucleus, in a rough approximation, it is possible to restrict ourselves to a nonrelativistic theory.

As you can see, understanding functional physical laws gives functional physical consequences, of which we have touched only a small part.

Exercises

1. Determine the dependence of pressure on the density of a part (their number per unit volume) in an ideal Fermi gas $^*$ at low temperatures. Obtain the equation of state of such a gas (up to a numerical factor). Compare it with the equation of state of a classical gas.

2. Find a criterion for the quantumness of an ideal Fermi gas.

3. Determine the temperature dependences of the heat capacity of a weightless gas at low temperatures.

4. Find the criterion for a classical gas.

5. Estimate the size of hydrogen atoms.

6. Given that the connection of atoms in liquids and crystals is determined by the basic electrical system interaction, estimate the distance between atoms.

*) See “Dialogue about temperature”.

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arrives there in May 1822. At the political school, the Sorbonne, the College de France, he listened to lectures by famous scientists: Cauchy, Fourier, Laplace, Monge, Poisson, Legacdre, Sturam, Poncelet, Binet and others, who were pioneering new paths in mathematical analysis, mathematical physics and mechanics. Having studied the conditions and results achieved by the French mathematical school, Ostrogradsky himself began to deal with important and pressing issues of that time, often the work of his Parisian colleagues. In 1826, the Russian scientist presented his first scientific work to the Paris Academy of Sciences - “Metoir on the propagation of waves in a cylindrical pool”, highly appreciated by Ksshi and published in the proceedings of the Academy. The scientific significance of this work can be judged by the fact that back in 1816 the Academy announced a special competition to solve it.

In 1824-1827, Ostrogradsky presented several more themes. These works enhanced the scientific reputation of the young scientist and won him the friendship and respect of many French mathematicians.

But Mikhail Vasilyevich is inexorably drawn to his homeland, where his successes were well known. It is not for nothing that young people who went to study abroad were given parting words by their relatives and friends with the words: “Become Ostrograd.”

In 1828 he left for Russia. This trip was difficult. On the way, he was robbed, and he had to walk from Frankfurt am Main to St. Petersburg. The “Russian pedestrian”, which was also acquired from abroad, looked very suspicious, and the momentary authorities, which I am currently restoring, established secret police surveillance over it. Ostrogradsky probably did not know about this until the end of his days.

Immediately after Ostrogradsky’s arrival in St. Petersburg, his fruitful work at the Academy of Sciences and numerous teaching activities began. The Academy of Sciences highly appreciated Ostrogradsky’s scientific work: in August 1830 he was elected extraordinary, and a year later - ordinary academician in applied mathematics. From that time on, his life was full of creative successes, and his activities were marked by the conferment of a number of honorary scientific titles. Thus, in 1834 he was elected a member of the American Academy of Sciences, in 1841 - a member of the Turin Academy, in 1853 - a member of the Roman Lynch Academy and in 1856 - a corresponding member of the Paris Academy.

Ostrogradsky’s scientific interests were determined early, even before leaving for Paris. In an announcement to the council of Kharkov University, Ostrogradsky wrote back in 1820 that he wanted to “improve himself in the sciences, applied mathematics.” And indeed, he devoted much of his work to mathematical physics and mechanics, laying the foundation of these sciences together with Laplace, Poisson, Fourier, Cauchy, Jacobi, and Hamilton.

Ostrogradsky wrote 15 works on mathematical physics. Most of them relate to problems of body propagation, control theory and hydrodynamics. His work on the theory of tellots was of greatest scientific importance. These studies, in addition to containing important results, relate directly to the theory of body propagation, and have enormous general mathematical significance, since, on the one hand, they laid the foundations for a number of important theories that are developing in our time, and on the other hand, they obtained theorems that are one of the central results of mathematical analysis.

Ostrogradsky was the first Russian scientist to study analytical mechanics. His works on mechanics, including Lectures on Analysis

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tical mechanics" and “Course of celestial mechanics” were the foundation on which the Russian school in the field of mechanics was built and developed. Ostrogradsky’s works on mathematical analysis are in most cases caused by his research in mathematical physics and mechanics: they provide solutions to mathematical questions posed by the theoretical natural sciences of that time. Thus, in connection with his studies of the propagation of a body in a solid, he obtained a significant formula, which is now included in all textbooks of mathematical analysis under the name of the Ostrogradsky-Green formula. Currently, this formula plays a huge role in mathematical physics, vector analysis and other branches of mathematics and its applications. It would not be an exaggeration to say that Ostrogradsky made outstanding contributions in the field of mathematical analysis. His results have entered modern mathematics as an essential and integral part of it and represent that necessary weapon without which mathematics can no longer do.

Ostrogradsky’s interests also included algebra, number theory, and probability theory. According to N. E. Zhukovsky, “in the works of M. V. Ostrogradsky we will be attracted by the society of analysis, the main idea, as broad as the expanse of his native fields.”

Ostrogradsky rendered an invaluable service to Russian science by educating a whole galaxy of talented students who later became outstanding representatives of Russian science. Among his students were: I. A. Vyshnegradsky - the founder of the theory of automatic regulation; N.P. Petrov is the creator of the hydrodynamic theory of lubrication and the author of classical studies on the theory of mechanisms; A. N. Tikhomandrikiy, E. I. Beyer, D. M. Delarue, E. F. Sabinin are famous professors of mathematics, and many other mathematicians and outstanding engineers.

The remarkable thing about the leading representatives of the Russian natural science community and mathematicians in particular was that they were not only innovators in the sciences, bold in their thoughts, but also teachers in the broad sense of the word; They did not confine themselves to the circle of their scientific interests, but devoted a significant part of their time and energy to the cause of education.

The progressive significance of the pedagogical activities of Russian mathematicians. It lies not only in their personal teaching, but in the fact that they took an active part in the general pedagogical movement and had a significant influence on the development of mathematical education and culture in Russia. The pedagogical activity of M. V. Ostrogradsky is especially characteristic in this regard.

A man of his time, a brave thinker and a brilliant lecturer, he worked a lot and fruitfully in the pedagogical field. In development, he taught in the Office classes at the Naval Cadet Corps, opened on the initiative of the famous navigator and scientist I. F. Krusenstern; was a professor at the Institute of the Corps of Railway Engineers

16

Thus, we have obtained all the characteristics of harmonic motion. From this we can conclude that a projected point on a circle can be replaced by a real particle, the movement of which will be described by the formulas obtained above.

Let us recall that the quantities $F$ and $a$ included in Newton’s equation are vectors. They, therefore, can be projected in any direction, and the dependencies between the projections will be described by Newton’s equation.

Tunnels in the Earth

Let us show that a harmonic force also acts on a point located inside the Earth (a force $F = \gamma \frac{Mm}{R^2}$ acts on a point outside the Earth).

Let the material point of mass $m$ be located at a distance $r$ from the center of the Earth. The action that the Earth has on a point can be divided into two parts (Fig. 2): the action of the inner sphere (red) and the action of the outer spherical layer (yellow).

As is known, a spherical layer does not create a gravitational field inside itself (if the density $\rho$ in it is constant) *). Therefore, only the “red part”, whose mass will act on the point

$$M = \frac{4}{3} \pi r^3 \rho.$$ (7)

The force with which the “red part” acts on the point is equal to

$$F = -\gamma \frac{Mm}{r^2} = -\gamma \frac{4\pi\rho m}{3} \cdot r.$$ (8)

Consequently, inside the Earth (in the tunnel) a harmonic force acts on a point, proportional to the distance from the center of the Earth. The movement in the tunnel turns out to be comparable to the movement of the body suspended in the circle - the elastic force of the spring is also proportional to its stretching.

Now we can solve the following problem: to dig through the center of the Earth

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light being at the same level above the water surface (Fig. 1).

A small horizontal mirror will cast light into the eyes of observed $O$ only if the distances from it to the observed and to the source are the same (at point $M$). If the mirror is tilted at an angle $\alpha$ to the observed side, then in order for the reflected light to enter the eyes, it must be slightly away from the observed one (point $N$). The mirror, inclined at an angle $\alpha$ in the opposite direction, should be at point $N'$.

The inclined positions of the mirror are similar to the extreme positions of the waves, at which the light reflected from them still enters our eye. The distance between $N$ and $N'$ therefore determines the length of the light path. At all points between $N$ and $N'$ we will find a section of waves that are inclined enough to reflect the rays into our eye.

Let us now consider the angles between the rays of light. From the drawing it is clear that $\beta + \alpha = \gamma + \delta$, $\beta - \alpha = \varepsilon = \delta$, hence $\gamma = \alpha + \beta - (\beta - \alpha) = 2\alpha$. Thus we come to the conclusion that the angle at which we see the large spot of light is simply equal to the angle between the two observed slopes. It is not difficult to calculate the linear size of the major axis of the spot $NN'$.

The short axis of the reflected light spot is easily found in a similar way. If you move the mirror from the point $M$ in a direction perpendicular to $NN'$, then in order for the reflected light to enter the eyes, the mirror must be rotated through a certain angle around an axis parallel to $NN'$ (Fig. 2). Assuming that the limiting angle of rotation of the mirror is still equal to $\alpha$, we find that the width of the light strip is $PP' = 2htg\alpha$, and, therefore, the short axis subtends the angle $\beta = \frac{2htg\alpha}{\sqrt{l^2 + h^2}}$.

The ratio of the two visible stripes of the spot will be equal to $\beta/2\alpha$ or, assuming that the spot is not large and the angle $\alpha$ is small, equal to $\beta/2\alpha = \sin\omega$, where $\omega$ is the angle at which we look at the water.

The smaller this angle, the more elongated the spot. If the gaze glides over the surface, the spot of light will stretch and narrow indefinitely.

When observing light paths on the surface of the sea, the angle $\omega$ is usually small - the light paths reach the horizontal (see photograph on figure 3), so we can only talk about the width of the path. And although the formulas we obtained are not literally applied in this case, using them, you can not only qualitatively determine the origin of the roads, but also understand the dependence of their width on the wind force and the height of the Sun above the horizon: with an increase in $\alpha$ and $h$, the width of the road returns.

20

S. A. Kheifets

The craving for something shiny is hidden in the depths of consciousness. A child who sees the light for the first time is drawn to the bright; the natives trade both food and water for shiny trinkets. People call a gifted, deep mind “brilliant.” We will not be interested in whether this attention to the shiny is connected with atavistic memories of the burning fire or with hypnosis, we will try to figure out why the shine appears - be it the shine of gold, amaz or cat’s eyes in the dark.

We judge the rage of an object by the amount of light that was reflected from it and into our eyes *). It is clear that this amount depends on the proportion of reflected light and its direction. A white sheet of paper reflects better than black, and a mirror surface reflects better than a matte one.

The direction of light reflected by a surface is determined by the well-known law: the angle of incidence is equal to the angle of reflection. And the energy of reflected light also depends on the refractive index of the medium $n$. If light is incident normally on the boundary of media with refractive indices $n_1$ and $n_2$, then the fraction of the reflected

*) Brightness is determined by the luminous flux hitting the retina.

energy is equal to $\eta = \left( \frac{n_1 - n_2}{n_1 + n_2} \right)^2$. For glass $n=1,52$, for air $n=1$, so only 4.3% of the incident light is reflected. For a glass plate, the proportion of reflected energy will be approximately twice as large, since part of the transmitted light will be reflected from the second surface of the glass and return back. If the rays do not fall vertically, but at an angle, they are reflected better (Fig. 1). At $60^\circ$, not 4.3%, but 9% are reflected from the glass; at $80^\circ$ already 39%. When the angle of incidence approaches $90^\circ$—a grazing ray—almost all the light is reflected (the reflection coefficient approaches 100%). Further, very black bodies, such as graphite, soot, velvet, begin to shine if you look at them at a small angle to the surface.

When light is normally incident on a glass plate, the fraction of reflected energy is close to 8.6%. Is it a little or a lot? It turns out, not much. Fresh snow reflects 80-85% of the incident light, and clouds reflect the same amount.

However, the sedge of glass on the snow is easy to notice, as well as the sparkles of individual snowflakes. This is explained not so much by the reflective properties of the plate, but by the parallelism of the sun’s rays - the

22

Fig. 3. Reflection from an air bubble.

differed in direction. The exception was “sliding” rays. Then the shine was a consequence of almost 100% reflection of the incident light. A similar situation is observed if the beam hits the interface between two media from a denser medium. If the light is incident normally, then the fraction of reflected light energy is determined by the same formula as before. For water, this means that 98% of the light escapes into the air. As the angle of incidence β increases, the intensity of the refracted beam decreases, and at $\sin \beta = \frac{1}{n}$ the refracted beam disappears. In this case, all the incident light is reflected from the boundaries - the so-called total internal reflection occurs. For glass $n = 1,52$, so that the beam, induzione at an angle to the normal $\beta \ge 36^\circ$, does not go out. For water $n = 1,33$, so that the fish can only see the sky through the “hole” with an opening angle of about 45°

Fig. 4. The path of rays in a drop of water.

(Fig. 2). The rest of the water surface (from the point of view of the fish) is bleibt and reflects only up to the reservoir. The interface with total internal reflection is an ideal mirror.

Another interesting application of this effect is light pipes. If you send a beam inside the fiberglass, then you can gently bend these fibers, even spin the nodes, and the light beam will obediently spin, following the law of total internal reflection.

Dip a leaf of silver neva into the water. The reverse side of the sheet, covered with many bubbles, begins to serbrista. And you’ve probably seen the sparkle of a dew drop. In both cases, the matter is in total internal reflection: in the first case - from the external, in the second - from the internal surface of an air bubble or dew drop (Fig. 3, 4). The angle between the rays that entered the drop and those that exited is easy to determine if we take into account that the radius is perpendicular to the surface of the ball. This angle is $\varphi = 4\gamma - 2\alpha$. Descartes was the first to make such a calculation. It is more difficult to analyze the dependence of the deviation angle φ on the impact parameter $d$.

If the incident rays are parallel, then, having passed through the drop, they begin to diverge. However, there is a minimum deflection angle. It is close to 42°. Near this direction, the refracted rays are thicker, that is, the bulk of the refracted rays go precisely in this direction. The dewdrop observed from this angle is quickly especially bright. At the same angle to the sun’s rays a rainbow is visible.

Wherever there is a bright source of light (the Sun, a bright sky), you can always observe its spark-image in the dewdrop (Fig. 5). Rays reflected from the surface of the drop can enter the eye, even if the angle $\alpha$ is close to 180°. In this sense, a drop of water actually reflects the scales of the world.

Speaking of glitter, one cannot help but think of precious stones. Generally

Fig. 7. The course of some rays in the cryliant.

what’s the matter, you need to “get” inside the metal. The distance arising would require a separate article. We will only point out that the peculiarities of metals are declared by the weak connection of some electrons with atoms and the easily mobile “gas” of free electrons formed due to this. Therefore, the movement of electrons in the field of an electromagnetic wave in metals is completely different from the movement in dielectrics, where each electron “knows” its atom. Free electrons easily absorb and reflect back most of the light falling on the surface of the metal. This is why metals reflect radio waves and even visible light well. Typically, electromagnetic waves passing inside the metal are strongly absorbed. It is interesting that sometimes the opposite picture is observed: substances that absorb light at Fig. 8. The difference in the path of the LAN between beams 1 and 2 should be an odd number of half-points.

in a certain part of the spectrum, acquire a metallic sheen (for example, dried filotte ink).

Oxidized metals conduct electricity much worse, so you should expect that they conduct electricity worse than pure ones.

Of course, things are most often beautiful, but sometimes you have to fight with brilliance. If the lenses of binoculars or cameras are biased, this means that light is reaching the eye or photographic plate: part of it is reflected. They remove shine by applying a very thin and transparent film to the surface of the lens. Its thickness and refractive index are selected so that the rays reflected from the two surfaces of the film (Fig. 8) are out of phase by 180° and, interfering, cancel each other out. This cannot be achieved for the entire region of visible light, but it can be achieved for that wavelength to which photographic film or the eye is most sensitive.

We have not considered such an interesting phenomenon as the sparkle of happiness in the eyes - but, apparently, it cannot be described by physical laws alone, and it is unlikely that this is what we should strive for.

Exercises

1. How does the brightness of the Moon change for an astronaut adjacent to it?
2. The reflected wave disappears completely if the intensities of rays 1 and 2 (Fig. 8) are the same. Prove what it must be for this to happen
   $n_{пленки} = \sqrt{n}$.
3. Light reflection can be reduced by smoothing the transition from $n=1$ to $n>1$. Compare the amount of reflected light from the air-glass boundary and from the glass on which a layer of water is poured (pretenbrets absorption in water).
4. Is it possible (and how) to increase the amount of reflected light by applying a transparent film to the glass?
5. Some fabrics show a sheen at certain angles. Give an explanation for this effect.

nogo. After all, think: if there were only 100 languages ​​in the world, then to translate from each language to any other (and vice versa) it would take 100 × 99/2 = 4950 translators. But there are many more languages, and not every pair of languages ​​can use one translator. It is clear that with the advent of powerful electronic computing technology, the hope began to shift the burden of translation onto the machine.

The beginning of this country’s history

The idea of ​​automatic (machine) translation captured the masses (we are talking, of course, about the masses of scientists) in a consistent period, at the dawn of cybernetics, in the first years of the electronic era.

1946 There is a discussion between the American scientists W. Weaver and A. D. Booth about the limits of the use of automatic digital computers. The first believes that the basic “elements of language” can be discovered using decryption devices developed during the war to reveal the secret codes of the enemy, and machine translation can be built on the basis of these “elements”. The other is more insular: it speaks of the fundamental possibility of only the last translation, and even then with a very small dictionary - the amount of computer memory was still very small. (The following year, Booth and his colleague Britten developed a program for such a translation.) But Weaver is optimistic. He shares correspondence with the “father of cybernetics” Norbert Wiener. Wiener was wary of Weaver’s idea and pointed out the great difficulties that stand in the way of exploration. Weaver is not discouraged. His credo: “A sentence in one language is simply an encrypted phrase in another language.” Let’s start decoding the machine. However, Weaver did not pay attention to one extremely important circumstance: with the help of formal decryption operations, easily carried out in a machine, it is possible to reach only the first analogous text in the original language. But the machine cannot understand the text in this language: after all, you need to know this language. During the First World War, codebreakers who famously deciphered all the most secret codes of American radio operators broke their teeth on one such code, which turned out to be simply plain text in an Indian language unknown to them. Thus, scientists have not yet had any information that natural human language can be “deciphered.”

Below is a problem that tells you what languages ​​the ciphertexts are written in, and decryption in this case turns out to be quite easy.

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How far can you travel by car, or a story about how instant success can turn into protracted failure

What can a machine do? She knows how to perform simple, mechanical operations: compare the contents of two cells of her memory, add, subtract, multiply, etc. And she does not care whether she is implementing a translation program or reading the trajectory of a satellite. But a machine can do a lot of such operations, and by combining them in different ways, the programmer can force the machine to perform some kind of operations, purposeful operations.

What operations must a machine perform to translate text from one language to another? First, she must “read” it. It’s not that difficult to do. The text is entered in encoded form into memory cells, and the machine can look into its own memory. The machine must then recognize the words. This is done like this: each word (from one space to another) is compared with a dictionary that must be stored in the machine’s memory. The machine must determine what form each word is in (morphological analysis) in order to then understand the relationships of all the words in a sentence in relation to each other (syntactic analysis). What then? Then you need to translate. Actually, the human translator continues all the described operations instantly and unconsciously, and the translation process itself begins from this moment. It is necessary to find words in the language in which the text is translated (this language is called the output language), equivalent to the words of the language from which the translation is carried out (input language). Then put these words into appropriate forms, that is, arrange them according to the law of the morphological and syntactic structure of the target language. Every step here brings difficulties. We’ll have to figure them out.

The first machine translation enthusiasts expect a generally capable translation (well, maybe with a few exceptions). Let’s take an English phrase:

I know his daughter.

The literal translation of this phrase turns out to be correct:

I know his daughter.

True, the English verb know is conjugated differently than the Russian verb to know: currently only in the third person singular it has the ending -s, and in all other persons it has the general form - know. How then can the machine understand that in the Russian form this verb should be given the form know, but not know and not knowing? This cannot be revealed from the dictionary entry for know, no matter what is written down: one form of know or all possible forms: know, know, knowe, we know, know and, finally, know (me). The clue, however, is contained in the English word I - i, which directly indicates the first person singular. This means that you need to put rules into the machine, according to which in order to translate the verb know (as well as other English verbs at present) the correct form of the Russian verb, it must take into account the subject, that is, translating each word separately using only a dictionary between language correspondences is not enough for correct translation. What if

29

was there an article in the English phrase that is not in the Russian language? What should be written in the word structure of the article? And anyway, what should I write in the dictionary? In Russian, words can appear in different forms: names are declined, verbs are conjugated. Indicate all possible forms of a Russian word with the corresponding English? But this is to separate words, and we must save the machine from memory. And then - the main thing! — how to guess which form to use? After all, in the English language, words have their own forms - which one should you write in the dictionary?

We managed to find answers to these questions quite quickly. It is necessary to create separate dictionaries of stems and endings. Then, for example, the machine will recognize the Latin verb amat as follows: first, it will look in the dictionary of stems for the length of the stems contained there, which is included in this word - it will not find the stem amat, ama, too, but the stem am should be there, and it corresponds to the concept love; it is the machine that remembers and draws the boundary: am/at; at is an ending, and you need to look it up in the dictionary of endings; it corresponds to the third person singular. So, we translate it like this: the base is love, the ending um is we love. Right! But we were lucky: the Latin endings more or less correspond to the Russian ones. However, things are not always smooth with Latin endings. In Latin there is the shortest sentence in the world, it consists of one letter. This is I - Go! In addition, i can represent 8 more different endings. How to guess which of them is used in the text? Again we need to look at the context. Moreover, the solution may not lie in the next word, but through words, after two. How to find out which one? And the linguist must foresee everything and equip the machine with the rules for any case.

Syntax is an even tougher nut to crack. You can see this for yourself.

Task 2. Establish grammatical connections between words in the phrase:
Uncover detailed giants in the land. (The assignment belongs to A.E. Kibrik.)

(A solution is considered good if at least 9 methods are proposed - and there are about 20 in total).

This phrase was chosen specifically. But there are many phrases in the language in which syntactic analysis can be done in more than one way. And this changes the meaning of the sentence. Take the phrase: “Mother loves daughter.” Who loves whom (in other words, what is the subject and what is the object)? When such a phrase gets into the machine, then, even if the machine correctly establishes the forms of all the words, it finds itself in a great difficulty: in order to correctly determine the syntactic structure, it is necessary to go beyond the boundaries of this sentence, turn to a broader context, to understand what is being said. How would it be?

Perhaps the intimidated reader will timidly suggest that we write down in the dictionary and translate not individual words, but entire phrases? After all, one always comes to refer to the context. But if you translate whole phrases, maybe it will be easier? Please! Let’s try it. Let the language have 1000 words (how many are there really?!). The phrase “whatever” has five words. Let’s write down all five-word phrases in the dictionary (without repeating words). Theoretically, their number is $N=1000 \times 999 \times 998 \times 997 \times 996 \approx 10^{15}$.

Of course, there are many fewer meaningful phrases, but still

30

their number will exceed the memory capacity of the most memory-rich of all currently operating machines. In addition, the machine will spend a long time searching for each phrase in such a dictionary.

For many years, linguists have been looking for a way out of this difficult situation. They came up with new, very cunning algorithms, witty models of language, and pictorial methods of describing it. We figured out how to simplify the processing process and how to improve the quality of translation. We met, discussed, argued. But let me ask, the reader, if the matter turns out to be so complicated, then why were the first experiments so successful?

The secret is simple. These experiments were carried out on very small material. And it is clear that for a small and very specific text there are easily all the exceptions and variants; stock up on the necessary rules for all cases that the text contains. And the text is translated. You can see this for yourself using the following problem as an example.

Task 3. Using a system of rules, sequences of Arabic numerals are translated into sequences of Roman numerals. At each step, one rule is applied, replacing one Arabic numeral with one Roman numeral. The following four types of rules are used:

non-contest:
for example, 2→III replace 2 with III;
single-contest singles:
a) 2(3)→III replace 2 with III,
6) (3) 2→III if after (before) 2 there is 3;
two-contest:
(3) 2(4)→III (replace 2 with III if there is 3 before 2 and 4 after)

Each sequence of Arabic numerals is corrected in the following order: first all double-contest, then single-contest double, then single and finally non-contest. Any set of rules of these types is called programming. The program may not contain all types of rules.

Numbered in Arabic numerals from 1 to 6 the following English (French, German) words:

lake, a, this, what, beautiful, is;
champ, un, ce, que, vert, sera;
Feld, das, was, grün, grüne, ist.

Task 1. Make a list of all sentences consisting of these words.
Task 2. Make a minimum list of Russian words necessary for translation from task 1 into Russian, and prove them in Roman numerals.
Task 3. Create a program for translating numerical sequences in such a way that each sequence of Arabic numerals corresponding to any sentence II from task 1 is translated into a sequence of Roman numerals corresponding to a Russian sentence, which is the correct translation of sentence II.

If this task is solved correctly, then a machine translation of all the sentences from task 1 into Russian has been carried out: all that remains is to write the proven words into the corresponding cells of machine memory and enter the program from task 3 into the machine. But this is not enough. It is necessary to come up with such rules, create such a processing program and compile a dictionary so that the machine can translate any text. And this is where all those difficulties emerge, a very small and not the most “difficult” part of which we managed to talk about.

F113. The vessel is half filled with water, in which a piece of ice floats. Kerosene is poured on top of the ice, the upper level of which is set at a height of $h$ from the bottom of the vessel. How will this height change when the ice melts?

Leningrad City Olympiad in Physics, 1971

F114. Two balls with masses $m_1$ and $m_2$ can oscillate on springs

Fig. 2.

equal rigidity of a rod attached to a block with mass $M$ (Fig. 2). The block lies on a horizontal plane. At the initial moment, hide the balls towards each other using a string, the pulling force of which is equal to $T$. The thread is burned out. At what minimum coefficient of friction between the block and the plane will the block not move?

G. L. Komkin

F115. A “black box” - a box with an unknown internal circuit - has two outputs. Resistance $R = 4 \text{ om}$ is connected in series with the box and then this target is connected to a source with e. d.s. $E_1 = 5 \text{ в}$ (Fig. 3). In this case, current $I_1 = 1 \text{ a}$ flows through the circuit. If the target connects to a source with e. d.s. $E_2 = 20 \text{ в}$, then current $I_2 = 2 \text{ a}$ will flow through it. What circuit is inside the black box?

Fig. 3.

The internal resistance of the sources of claims is small.

Y. A. Smorodinsky

F116. If the temperature of the air in the cylinder shown in figure 4 is $t_0 (t_0 > 0^\circ \text{C})$, then it remains to temperature $\frac{t_0}{2}$ in about 30 seconds. The piston begins to move in and out with some frequency. In what case does more ice grow around the cylinder in 50 piston strokes if they are made a) in 1 minute; b) in 1 hour; c) in 30 days?

G. L. Komkin

Fig. 4.

F117. People usually walk along a sticky road, taking small steps. What length of step should a person walk without fear of falling, if its length is 1 meter, and the coefficient of friction of the soles of shoes on the road is 0.1?

I. Sh. Slobodetsky

The editors of the magazine received more than 2000 letters with solutions to problems placed in the “Problem Book «Kvant»” section. Among the authors of these letters, the editorial board selected schoolchildren who regularly sent particularly original and successful solutions. They are loaded with an annual subscription to the magazine «Kvant» for 1972.

Here are the names of the winner of our competition:

Dmitry Grigoriev - Leningrad, boarding school No. 45 at Leningrad State University;
Yuri Obolonkov - Voronezh, school No. 58;
Eduard Turkevich - Chernivtsi, school No. 9;
Mikhail Prager - Tomsk, school No. 51;
Arkady Chernyak - Minsk, school No. 50;
Leonid Braginsky - Frunze, school No. 61;
Viktor Krivitsky - village. Kletnoye, Borisvsky district, Minsk region;
Alexander Grigoryan - Baku, school No. 211;
Grigory Zaitsev - Gagra, school No. 2;
Ivan Grichik - David-Gorodok, Brest region, school No. 1;
Alexander Zhukov - Kirovsk, Khartsyzsky district, Donetsk region, school No. 11;
Anatoly Anishchenko - Sarapul, school No. 13 of the Udmurt Autonomous Soviet Socialist Republic;
Evgeny Gubarev - Moscow, school No. 144.

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To prove statement 1), we will indicate a strategy for the first player that does not depend on the moves of the second player and guarantees that the difference between the remaining two numbers will be at least 32. This strategy describes a very simple one: with each move, cross out numbers after one, that is, cross out the second, fourth, … number from the remaining ones (we assume that the numbers are arranged in ascending order). then after the 1st move the difference between any adjacent of the remaining numbers will be at least 2, after the 2nd - at least 4, after the 3rd - at least 8, after the 4th - at least 16 and after the 5th - at least 32.

To prove statement 2), it is enough to indicate the strategy of the second player, which does not depend on the moves of the first and will allow him to win no more than 32. It is as follows. With his first move, he crosses out all numbers less than 512, or all numbers greater than 512 (it is clear that there are no more than 256 of either of them left). After this, the difference between the outermost remaining numbers will be no more than 512. Similarly, with the second move, he can ensure that all the remaining numbers will be in only one of the segments [0.256]; [256, 512]; [512, 768]; [768, 1024], that is, reduce the difference between the extreme numbers to at least 256. In the same way, on the 3rd move he can reduce this difference to 128, on the 4th to 64 and on the 5th to 32.

I will add a solution to this problem from Yu. Obolonkov from Voronezh and V. Luzey from Moscow.

Of course, it was similarly possible to prove that if a game consists of n moves and starts with the numbers 0, 1, 2, … 2^n, then the “game price” is equal to 2^n. But a detailed study of such a game, starting with an arbitrary set of numbers (not necessarily an arithmetic progression) is a much more difficult task.

M62

Prove that for any odd number a there is a natural number b such that 2^b - 1 is divisible by a.

Here is the shortest proof.

Let’s consider the numbers 2^0 - 1, 2^1 - 1, …, 2^a - 1. There are (a+1) of these numbers. Some two of them give the same remainders when divided by a, because there are only a different such remainders (this reasoning is called “Dirichlet propill,” its applications are described in the article by A. Orlov in «Kvant» No. 7). Let, say, the numbers 2^k - 1 and 2^m - 1 give the same remainders when divided by a and k < m. Then the number (2^m - 1) - (2^k - 1) = 2^k (2^m - k - 1) is divisible by a and, since a is odd, 2^m - k - 1 is divisible by a.

A more general fact is proved in the same way: if natural numbers a and c are coprime, then there is a natural number b such that c^b - 1 is divisible by a. Many readers have noticed that the statement of the problem follows from the following Euler theorem*): for any natural numbers a and c, the number c^φ(a)+1 - c is divided by a, where φ(a) is the number of natural numbers less than a and relatively prime to it, and they even give a formula for calculating the “Euler function”

$$\varphi\left(p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_r^{\alpha_r}\right) = \left(p_1^{\alpha_1} - p_1^{\alpha_1}-1\right) \times \left(p_2^{\alpha_2} - p_2^{\alpha_2}-1\right) \cdots \left(p_r^{\alpha_r} - p_r^{\alpha_r}-1\right).$$

Correct solutions were added by A. Chernyak from Minsk, E. Turkevich, Sh. Slepoy from Chernivtsi, G. Levin from Kuibyshev, N. Demidenko from the village of Kolbovka, Gomel region, A. Pukhalsky from Moscow, Yu. Semenok from the village of Mirny, Krasnodar Territory and other readers.

M63

Is it possible to lay out a square out of 18 tiles of size 1x2 so that there is not a single rectangle of the sides of the square and there is no overlap at the edges of the tiles? (For example, a tile arrangement like figure 2 would not work because there is an AB seam.)

Let’s assume that we managed to lay out a 6x6 square with 1x2 tiles without a “seam”. Then each of the ten straight lines cutting this square

*) See, for example, the book by I.M. Vinogradov “Fundamentals of Number Theory” GITTL, M., 1953 or another textbook on number theory.

36

Fig. 2.
Fig. 3.
Fig. 4.
Fig. 5.

Rat on 1x1 squares, must intersect at least one 1x2 tile. It is not difficult to prove that each such straight line intersects the obligatory even number of bars. Indeed, it cuts off a 6×k straight line from a 6×6 square, consisting of an even number of 1×1 cells; this straight line contains a certain number of whole 1×2 tiles and some even number of 1×1 cells—halves of ingots intersected by a straight line (Fig. 3).

Not if each straight line intersects at least two ingots, then the total number of ingots must be at least 2·10=20, and there are only 18 of them.

Thus, we have proven that it is impossible to cut a 6x6 square into ingots in the required manner.

The correct solution to this problem was added by A. Kostyurin from the Moscow region, I. Kashanov from Ufa, V. Schwartz from Leningrad and others. Most of the attached letters discuss the general question: when can a straight m×n be cut into 1×2 ingots without seams. It is easy to prove that a straight line 2×n, 3×n and 4×n cannot be cut in this way. If m≥5, n≥5 and mn is even (the last condition, of course, is necessary), then in all cases, except for the 6×6 considered above, the required partition exists. To prove it, it is enough to come up with a partition of straight lines 5×6 and 6×8 and then that from a partition of a straight line m×n it is easy to obtain a partition of a straight line (m+2)×n (Fig. 4 and 5).

N. B. Vasiliev

F72

Why does day length change faster in March than in December?

The change in the length of the day is associated with the angle of inclination of the earth’s axis to the plane of the earth’s orbit around the sun (≈23.5°). If the earth’s axis were perpendicular to the orbital plane, the length of the day would remain virtually unchanged.

The tilt of the axis in space does not change, and since the Earth rotates around the Sun, it is visible from the Sun in different ways (Fig. 6) - The Sun illuminates parts of the Earth’s surface at different times, and the length of the day is not constant. It is obvious that in December and June, when in the rural hemisphere the length of the day is minimal or maximum, respectively, and the northern stripe is inclined away from or towards the Sun, the position of the earth’s axis relative to the Sun changes very slowly. Slow changes and long days. At the same time, in March and September, the position of the axis changes quickly - the length of the day also changes quickly.

It is interesting to write an approximate graph of the dependence of the length of the day on the time of year. Roughly, this is a sine wave (Fig. 7). The rate of change during the day, very quickly, is proportional to the tangent

38

Well, the pressure exerted by the molecules of the dissolved substance. This pressure is called osmotic pressure. The condition for equilibrium of liquid in vessels is not equality of hydrostatic pressures on both sides of the partition, but equality, roughly speaking, of the pressure of solvent molecules.

The molecules of the dissolved substance behave like gas molecules. They strive to occupy as much volume as possible by moving the partition (they would succeed if the partition were movable). But since the partition is fixed, it is not the partition that will mix, but the level of the mixture.

The analogy between the surface of gas molecules and molecules of a dissolved substance is complete. If the solution is weak, then the molecules of the dissolved substance are farther from each other and the weak interact with each other - just like the molecules of an ideal gas (they interact with each other only by standing). Therefore, the osmotic pressure $p$ must be related to the volume of the solution $V$, the temperature $T$ and the number $n$ of moles of the dissolved substance by the same formula as the pressure of an ideal gas:

$$pV=nRT$$

(R is the gas constant).

This law was discovered and proven experimentally by Van’t Hoffform. Osmotic pressure does not depend on the nature of the dissolved substance and is determined only by the concentration of the solution. For strong solutions (those with high concentrations), you can write an equation similar to the van der Waals equation for the realization of gas. If we have different substances in the solution, then, by analogy with gases, the osmotic pressure is equal to the sum of the osmotic partial pressures of the individual dissolved substances. A raster in a field of gravity will behave similarly to an ideal gas: the distribution of the number of molecules in height will be exactly the same. This allowed Perrin

“weigh” the atom (see the article by M.P. Bronstein “How the atom was weighed”, «Kvant» No. 2, 1970).

The energy for raising the liquid is taken from the thermal energy of the movement of particles of the dissolved substance. Just like a real gas, the “gas” of particles of a dissolved substance, expanding and doing work, cools. In this case, of course, the entire solution is cooled.

Osmotic pressure plays a very important role in the life of humans, animals and plants. The membranes of many organs are a semi-permeable partition - for example, the walls of blood vessels and intestines, damage to plant roots, and so on. It is osmotic pressure that determines the volume of substances.

This problem was correctly solved by many readers: V. Sokolov (village of Myshansk, Gomel region, VSSR), S. Araslanova (village of Nizhnevinko, Kumensky district, Kirov region), N. Fedin (Omsk), N. Kolenko (Bondar, Odessa region), N. Egorov (village of Maya, Mechinokangala district, YASSR), M. Kozlovsky (Moscow) region, Ukrainian SSR), M. Maksimenko (Balaklya, Kharkov region), A. Veretsada (Orekhov, Zaporozhye region, Ukrainian SSR), S. Ivakov (Moscow).

F74

Two capacitors are connected in series. The first has a mixture of $C_1$ and is designed for a maximum voltage of $U_1$, the second has a mixture of $C_2$ and is designed for a maximum voltage of $U_2$. To what voltage can this battery capacitors be connected?

Since the capacitors are connected, their charges are equal, and the mixture of the battery is $C = \frac{C_1C_2}{C_1 + C_2}$. If voltage $U$ is applied to a capacitor battery, then its charge will be equal to $q = CU = \frac{C_1C_2}{C_1 + C_2} U$. This charge should not

exceed the maximum charge for which each of the capacitors is designed: $q_1 = C_1 U_1$ and $q_2 = C_2 U_2$.
Posny must satisfy the inequalities

$$\frac{C_1 C_2}{C_1 + C_2} U \leq C_1 U_1$$

And

$$\frac{C_1 C_2}{C_1 + C_2} U \leq C_2 U_2.$$

It is clear that the charge of the capacitors should not exceed the lesser of the possible maximum charges of the capacitors. Let for definition

$$q_1 = C_1 U_1 < q_2 = C_2 U_2.$$

Then

$$\frac{C_1 C_2}{C_1 + C_2} U \leq C_1 U_1$$

And

$$U \leq \frac{C_1 + C_2}{C_2} U_1.$$

If $C_1 U_1 = C_2 U_2$, then we get the natural result:

$$U \leq \frac{C_2 \frac{U_2}{U_1} + C_2}{C_2} U_1 = U_1 + U_2.$$

The correct solution to this problem when merging V. Chupik (Torzhok, Kalinin region), N. Nikolaeva (village Belaya Glina, Krasnodar region), O. Maslok (Novomirgorod, Kirovograd region), S. Ivanov (Moscow), V. Sokolov (village Myshansk, Gomel region, BSSR), N. Fedik (Omsk), N. Kolenko (Bondar Odesskoy region), N. Alekseenko (Kirovsk, Murmansk region), A. Stachenko (village of Petropavlovka, Petropavlovsk district, Dnepropetrovsk region), M. Kozlovsky (Moscow), P. Khmara (Voroshilovgrad), E. Golovko (Stavropol), V. Parkhomenko (village of Pokolbichi, Gomel region), Yu. Batyshen (Moscow), M. Ramazanov (Makhachkala, DASSR), E. Sviriduk (Rovenki, Voroshilovgrad region), T. Selina (village of Saltykovo, Penza region), V. Sytsev (Alga, Aktobe region), A. Kostyurin (village of Krasovo, Moscow region), K. Kirillov (Minusinsk, Krasnoyarsk region).

I. Sh. Slobodetsky

GAME
ON
PLANES

The maximum charge (of the vertices) is drawn on the plane. Some of them are connected by segments. Vsya and Petya play this game: they alternately mark the vertices - Vsya with crosses, and Petya with zeros. After all the vertices are marked, Vsya counts how many “of his own” segments he has, that is, segments in which both vertices are marked with crosses, and Petya counts how many segments he has. The one with the most segments wins. Explore how to play this game. Who can ensure a win or a draw? Consider the following case:

a) when a polygon is drawn;
b) when the drawn figure has exactly $k$ segments at each vertex;
b) when different numbers of segments are coded at different vertices.

L. Limanov

FUNCTIONAL EQUATION

Given the equation:

$$mf \left( \frac{a+x}{c+x} \right) + nf \left( \frac{a-x}{c-x} \right) = kx.$$

$(m \neq n)$.

Define the function $f(x)$.

Ya. Kilimnik

APPLICANT PRACTICUM

In the current academic year, the publication of materials will continue on the pages of Quaenta under the guidance of the “Enrollment Practice.” Letters received by the editor show that this department is of keen interest to many readers of the magazine. We are very grateful to everyone who gave us their comments and suggestions.

Of course, there is no way to cover all the questions included in the program of entrance exams in mathematics and physics. Yes, there is no need for this - these issues are studied at school and discussed in detail in school textbooks. Therefore, the editors have outlined only a few topics that are of fundamental importance and, in one form or another, are most often used when solving competitive problems, and usually cause difficulties for applicants.

For example, we propose to talk about the development of inequalities, about solving systems of non-linear equations, about problems on combinations of spatial bodies, about a graphical method for solving some problems on composing equations, etc. In physics, articles will be published concerning such issues as the kinematic equation of motion, solving problems on the equilibrium of bodies, thermal processes, current laws and others.

These articles will contain, if possible, a detailed presentation of theoretical material, accompanied by an analysis of tasks taken from the practice of entrance exams.

We also propose, for informational purposes, to include examples of options in mathematics and physics that were offered in 1971 at the entrance exams at various universities across the country. These options will be provided only with answers (or instructions) - we hope that the reader will be interested in trying his hand at solving competition problems on his own.

41

M. S. Atamukas

A square triline is a polynomial of the second degree of general form

$$ax^2 + bx + c, \quad a \neq 0.$$

In entrance exams in mathematics, problems directly or indirectly related to the square trilen are offered quite often. Yes, this is not surprising - in the school curriculum, a lot of space and time is removed from studying the quadratic equation

$$ax^2 + bx + c = 0,$$

finding out the properties of a quadratic function

$$y = ax^2 + bx + c.$$

This function is also found in physics; for example, when describing the movement of materials of a point thrown obliquely to the horizon.

Schoolchildren, of course, are well aware of all the formulas related to the square triangle; they can also draw the graph of a quadratic function - parabolic. However, when it comes to solving a specific problem, most tools try to act purely analytically, using only formulas. Meanwhile, very often the use of geometric considerations allows us to obtain an answer easier and faster.

We will show with several examples how important it is to think simultaneously in both algebraic and geometric languages ​​when solving a problem.

In order to check how well you are familiar with the properties of a square trilen, let’s first consider this problem (it was offered in oral exams at Moscow State University).

1. figure 1 shows several graphs of square trilens. Figure approximate, scale not indicated. For each of these parabolas, determine the signs of the corresponding coefficients $a, b$ and $c$.

Take, for example, paraboly 4; it is the graph of some quadratic function

$$f(x) = ax^2 + bx + c.$$

Since the branches of this parabola are directed upward, the coefficient is $a > 0$. (Prove it! Many applicants explained it this way: “The textbook shows

42

It turns out that if $a > 0$, then the parabola is directed upward.” Is such an obligation convincing?). Further, the coefficient $c$ is equal to the value of the function $y$ at the point $x = 0$, that is, $c = f(0)$. From figure it is clear that parabola 4 intersects the positive semi-axis ordinat, and then $c > 0$. The abscissa of the vertex of the parabola is $-\frac{b}{2a}$; from figure it is clear that it is positive. We already know that $a > 0$, and therefore $b < 0$.

Answer: $a > 0$, $b < 0$, $c > 0$.

Readers can figure out the remaining cases presented in figure 1 on their own.

In many problems, it is necessary to find out the behavior of the quadratic function $f(x) = ax^2 + bx + c$ on a given interval $(\alpha, \beta)$, to describe the location of the roots of this triangle relative to a fixed point $x = \gamma$ of the abscissa axis, etc. In these problems, the geometric language makes interpretations simple and straightforward (and, moreover, quite strict!).

2. Based on the abscissa, the interval $\alpha < x < \beta$ is indicated. For what $a$, $b$ and $c$ the equation

$$ax^2 + bx + c = 0, \ a \neq 0, \ b^2 - 4ac > 0$$

has only one root on this interval?

In other words, we need to find out for which $a$, $b$ and $c$ the graph of the function $f(x) = ax^2 + bx + c$ intersects the x-axis on the interval $\alpha < x < \beta$ exactly once.

If one tries to solve this question analytically, using formulas for the roots $x_1$ and $x_2$ of a square triangle, then the conditions under which only one root lies between the numbers $\alpha$ and $\beta$ are written very grammatically. The geometric approach makes it possible to obtain this condition immediately and in a simple form.

Let’s imagine how the graph of a triangle that satisfies the conditions set in the task can behave; Obviously, one of the four positions represented on figure 2 is possible. Note, further, that in each of these cases the values ​​of the function $f(x)$ at points $\alpha$ and $\beta$ have opposite signs, that is

$$f(\alpha)f(\beta) < 0,$$

or, in more detail,

$$(a\alpha^2 + b\alpha + c)(a\beta^2 + b\beta + c) < 0.$$

This condition is necessary and sufficient for the equation $ax^2 + bx + c = 0$ (for $a \neq 0, b^2 - 4ac > 0$) to have exactly one root on the interval $(\alpha, \beta)$.

43

So the graphs helped us find this condition. They also suggest how to prove that this condition is necessary and sufficient.

Necessity. Let the equation have two roots $x_1$ and $x_2$ ($x_1 < x_2$), with one root lying inside the interval $(\alpha, \beta)$ and the other outside it. Then one of the numbers $\alpha, \beta$ lies inside the interval $(x_1, x_2)$, and the other one lies outside it. But the school textbook proves that the values ​​of the function $f(x) = ax^2 + bx + c$ outside the interval $(x_1, x_2)$ between the roots and the values ​​of the function inside this interval have opposite signs. Therefore, $f(\alpha) f(\beta) < 0$.

There is enough to eat. Let the numbers $f(\alpha)$ and $f(\beta)$ have different symbols. Then $b^2 - 4ac > 0$, equation $ax^2 + bx + c = 0$ has two roots $x_1$ and $x_2$, and the values ​​of the function inside and outside the interval $(x_1, x_2)$ have opposite signs. Therefore, only one of the numbers $\alpha, \beta$ lies inside the interval $(x_1, x_2)$. But this means, in turn, that only one of the numbers $x_1, x_2$ lies inside the interval $(\alpha, \beta)$.

It is clear that the square function

$$y = ax^2 + bx + c, \quad a \neq 0, \quad b^2 - 4ac = 0,$$

has only one root $x_0 = -\frac{b}{2a}$, and it lies on the interval $(\alpha, \beta)$ if $\alpha < -\frac{b}{2a} < \beta$.

Let us apply the messages just presented to solve the following problem, taken from the practice of oral exams.

3. Prove that the quadratic trinomial

$$f(x) = (x - a)(x - b) + \lambda (x - c)(x - d)$$

has real roots for any real $\lambda$ if it is known that exactly one of the numbers $c, d$ lies on the interval $(a, b)$ and the other outside it.

Looking at the graphs of various quadratic trinomials, it is easy to notice one important property: the maximum and minimum values ​​of the quadratic trinomial on the segment $[\alpha, \beta]$ can only be achieved at the following three points: at the point $x = -\frac{b}{2a}$, if, of course, it belongs to the segment $[\alpha, \beta]$; at point $\alpha$; at point $\beta$ (at the ends of the segment). Let’s prove it; we will assume that $a > 0$ (the case $a < 0$ is analyzed in a similar way).

Let us extract a complete square from a quadratic trinomial:

$$f(x) = ax^2 + bx + c = a \left( x + \frac{b}{2a} \right)^2 + c - \frac{b^2}{4a}.$$

Using this notation, it is easy to show that for $a > 0$ the function $f(x)$ at the point $x = -\frac{b}{2a}$ has an absolute minimum equal to $c - \frac{b^2}{4a}$; on the interval $(-\infty, -\frac{b}{2a})$ it kills monotically, and on the interval $(-\frac{b}{2a}, \infty)$ it monotically increases.

*) Check for yourself that $f(\alpha) f(\beta) < 0$ implies $b^2 - 4ac > 0$.

44

Because if
$$\alpha \leq -\frac{b}{2a} \leq \beta,$$
then the function $f(x)$ on the segment $[\alpha, \beta]$ takes the smallest value $c - \frac{b^2}{4a}$ at point $-\frac{b}{2a}$ (see graph 1 on figure 3).

The lowest value, due to monotonicity, is achieved at one of the ends of the interval.

Let now the point $-\frac{b}{2a}$ be located outside the segment $[\alpha, \beta]$ (see graphs 2, 3 on figure 3).

Then the segment $[\alpha, \beta]$ takes one of the monotonicity intervals, and, therefore, the minimum value is taken at one of its ends, and the maximum at the other.

Now try to formulate a necessary and sufficient condition under which the quadratic equation $ax^2 + bx + c = 0$ has two roots on the given interval $\alpha < x < \beta$.

The noticed property of the quadratic trinomial will make it easier for us to solve the following problems.

4. (MSU, mechanism, 1970). For what values ​​of the parameter $\alpha$ does the inequality
   $$\sin^6 x + \cos^6 x + \alpha \sin x \cos x \geq 0$$
   holds for all $x$?

Let’s transform the left side of the inequality:
$$(\sin^2 x)^3 + (\cos^2 x)^3 + \alpha \sin x \cos x =$$
$$= (\sin^2 x + \cos^2 x) (\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) + \alpha \sin x \cos x =$$
$$= [(\sin^2 x + \cos^2 x)^2 - 3 \sin^2 x \cos^2 x] + \alpha \sin x \cos x =$$
$$= 1 - \frac{3}{4} \sin^2 2x + \frac{\alpha}{2} \sin 2x.$$

Let us denote $\sin 2x$ by $y$; Obviously, $-1 \leq y \leq 1$, the coefficient of $y^2$ is negative, therefore, on the interval $[-1; 1]$, the minimum of the square trinomial $-\frac{3}{4} y^2 + \frac{\alpha}{2} y + 1$ is achieved at the ends of the interval.

Therefore the inequality
$$-\frac{3}{4} y^2 + \frac{\alpha}{2} y + 1 \geq 0$$
holds on the entire interval $[-1; 1]$ if and only if the trinomial is nonnegative at the ends of the interval:

$$\begin{cases} -\frac{3}{4} (-1)^2 + \frac{\alpha}{2} (-1) + 1 \geq 0, \\ -\frac{3}{4} \cdot 1^2 + \frac{\alpha}{2} 1 + 1 \geq 0 \end{cases}$$

that is, $\alpha \leq \frac{1}{2}$ and $\alpha \geq -\frac{1}{2}$.

Answer: this inequality holds for all $x$ for $-\frac{1}{2} \leq \alpha \leq \frac{1}{2}$.

45

Fig. 4. Minimum at $t = t_0$.

Fig. 5. Minimum at $t = t_0$.

5. (LSU, Mathematics, 1966). The rocket must travel a distance equal to $H$. It starts moving at a constant speed $v$ and at any moment can turn on an additional movement that gives constant acceleration $a$ ($a > 0$) and works until the end of the journey. Fuel consumption on a section of track at a constant speed is proportional to the time of movement, and when additional movement is turned on, it is proportional to the square of time with proportionality coefficients $k_1$ and $k_2$ correspondence ($k_1, k_2 > 0$). How long should the rocket move with the additional movement turned on in order for the promised fuel consumption to be the least?

Let $\tau$ be the time of movement of the rocket at a constant speed, $t$ be the time of movement with additional movement enabled, $A$ be the fuel consumption. From the equation:

$$H = v\tau + vt + \frac{at^2}{2}, \quad A = k_1\tau + k_2t^2$$

we find

$$\tau = \frac{1}{v} \left( H - vt - \frac{at^2}{2} \right), \quad A = \left( k_2 - \frac{ak_1}{2v} \right) t^2 - k_1t + \frac{k_1H}{v}.$$

Note that $0 \leq t \leq t_0$, where $t_0$ is the positive root of the equation

$$H = vt + \frac{at^2}{2}, \quad t_0 = \frac{-v + \sqrt{v^2 + 2aH}}{a}.$$

(This is the flight time if movement is enabled from the very beginning.) We need to find the minimum of the function $A = A(t)$ (*) defined on the segment $[0, t_0]$.

1. $k_2 - \frac{ak_1}{2v} < 0$. The vertex of parabolas $A = A(t)$ has an abscissa

$$t^* = \frac{k_1}{2 \left( k_2 - \frac{ak_1}{2v} \right)} < 0.$$

The parabola is located as indicated at figure 4. Minimum at $t = t_0$.

2. $k_2 - \frac{ak_1}{2v} = 0$; $A = -k_1t + \frac{k_1H}{v}$. Minimum at $t = t_0$ (Fig. 5).

3. $k_2 - \frac{ak_1}{2v} > 0$. Abscissa of the vertex of parabolas

$$t^* = \frac{vk_1}{2vk_2 - ak_1} > 0.$$

46

Fig. 6. Minimum at $t = t^*$.

If $t^* \leq t_0$, then the minimum is at $t = t^*$ (Fig. 6).
If $t^* \geq t_0$, then the minimum is at $t = t_0$ (Fig. 7).
Answer: If $2vk_2 - ak_1 > 0$ and

$$\frac{vk_1}{2vk_2 - ak_1} \leq \frac{-v + \sqrt{v^2 + 2aH}}{a},$$

That

$$t = \frac{vk_1}{2vk_2 - ak_1}.$$

If $2vk_2 - ak_1 \leq 0$ or $2vk_2 - ak_1 > 0$ and

$$\frac{vk_1}{2vk_2 - ak_1} > \frac{-v + \sqrt{v^2 + 2aH}}{a},$$

That

$$t = \frac{-v + \sqrt{v^2 + 2aH}}{a}.$$

Exercises

1. Indicate the conditions under which both roots of the trinomial $f(x) = ax^2 + bx + c$ are less than the given number $\alpha$.
2. Indicate the conditions under which the given number $\alpha$ lies between the roots of the trinomial $f(x) = ax^2 + bx + c$.
3. (MSU, department of structural linearity, 1968). Find all values ​​of the parameter $b$ for which both roots of the quadratic equation $x^2 - 2bx - 1 = 0$ are real and do not exceed modulus 2.
4. (MSU, physicist, 1965). Find all values ​​of $a$ for which the equations $x^2 + x + a = 0$ are valid and greater than $a$.
5. (MSU, Department of Biology, 1969). Find all values ​​of $a$ for which the system

$$\begin{cases} x^2 + y^2 = a, \\ x - y = a \end{cases}$$

has a unique solution.
6. The radius of the base of a right circular cone is $R$ and its height is $H$. Determine the height of the cylinder inscribed in this cone and having the largest lateral surface.
7. (MSU, department of structural linearity, 1969). List all roots of the equation $x^2 + 1 = \cos x$.
8. (MSU, physicist, 1964). Find all values ​​of $m$ for which the expression $x^2 + mx + m^2 + 6m$ is negative for all values ​​of $x$ satisfying the inequalities $1 < x < 2$.
9. (MSU, physicist, 1963). The projectile leaves the ordium at an angle $\alpha$ to the horizon with a speed $v_0$. At the highest point of the trinomial there are two parts of equal mass outside, and the velocities of the parts immediately after the explosion are horizontal and lie in the plane of the trinomial. One half is outside at a distance of $S$ from the shot point. Determine the place where the second half fell, if it is known that it fell further than the first. Assume that the flight of a projectile occurs in airless space.

47

How do you solve kinematics problems? How many formulas do you remember? 5? 10? 15? Very often, solving a problem begins with writing out a large number of formulas - a formula for the path traveled by a body in the $n$ second, a formula for the ratio of the paths traveled by a body over successive periods of time, a formula for the difference in the squares of the body’s velocity, and so on. Then they try to substitute these formulas into each other, divide, multiply… In this way they manage to find the answer. But most often, kinematics problems are an insurmountable obstacle. This is perhaps one of the most difficult sections for applicants. Although there must be one of the simplest. Let’s look into it.

We will first talk about the primolinear motion of the body. If the straight line along which it moves is taken as the coordinate axis $OX$ (point $O$ is the origin), then the position of the body is determined by its coordinate $x$. When moving with constant acceleration, the coordinate of the body changes with time according to the law

$$x = x_0 + v_0 t + \frac{at^2}{2},$$

(*)

where $x_0$ is the initial coordinate of the body (at $t=0$), $v_0$ is its initial velocity and $a$ is the acceleration. Equation () is the kinematic equation of body motion. Of course, equation () is a projection onto the axis $x$ of the vector equation for the radius vector $r$, which determines the position of the center of mass of the body (Fig. 1). Therefore, the projection onto the $OX$ axis of those vector quantities that are directed in the same way as the $OX$ axis is positive, and those that are directed in the opposite direction are negative. This must be taken into account when substituting these values ​​into equation (*).

Let us emphasize again: equation (*) is an equation for the coordinates of the body, and not for the path taken by it. Path is a positive scalar that cannot decrease. At the same time, the body coordinate can behave in any way. For example, if $v_0$ is positive and $a=-a_0$ is negative (velocity vector $v_0$ is directed in the same direction)

49

that is, at the moment of falling to the ground, the speed of the stone is equal to the initial one, but directed in the opposite direction.

Equation (**) makes it easy to find the time $t_\pi$ for lifting the stone. Since at the point of maximum rise the speed of the stone is zero, then

$$0 = v_0 - gt_n.$$

Hence $t_\pi = \frac{v_0}{g}$. The time it takes to lift the stone is half the flight time.

We see that two equations - for coordinates and for speed - allow us to obtain an answer to any question regarding the movement of a stone. These are the ones that should always be remembered.

Let’s solve one more problem.

Problem 2. A freely falling body has passed $1/3$ of its path in the last second. How many seconds ($n$) and from what height ($h$) did the body move?

We take the throwing point as the origin of coordinates, and direct the coordinate axis vertically downwards. Then the coordinate of the body depends on time according to the law

$$x = \frac{gt^2}{2}$$

($v_0 = 0, x_0 = 0, a = g$). If the body lodged $n$ seconds, the fall time is $n \cdot \tau$ ($\tau = 1$ sec), and

$$h = \frac{gn^2\tau^2}{2}.$$

(1)

After time $(n-1)\tau$ after the start of movement, the coordinate of the body was equal to $\frac{2}{3}h$. That’s why

$$\frac{2}{3}h = \frac{g(n-1)^2\tau^2}{2}.$$

(2)

Solving equations (1) and (2) together, we find,

$$n = \frac{\sqrt{3}}{\sqrt{3} - \sqrt{2}} \approx 5,45, \quad h \approx 145 \text{ м}.$$

It was possible to choose another coordinate system, for example, one associated with the earth and an upward coordinate axis. Then the equation of motion of the body would be:

$$x = h - \frac{gt^2}{2} \left( x_0 = h, v_0 = 0, a = -g \right).$$

Here at $t=n\tau$ $x=0$, and at $t=(n-1)\tau$ $x=\frac{1}{3}h$.

You can choose one more coordinate system - with the origin at the throwing point and the axis directed upward. In this system

$$x = -\frac{gt^2}{2};$$

at $t=n\tau$ $x=-h$, at $t=(n-1)\tau$ $x=-\frac{2}{3}h$.

In this problem, the choice of coordinate system is not very important, but often, successfully choosing a coordinate system can significantly simplify the solution.

Let’s solve, for example, the following problem.

Problem 3. With what time interval $\tau$ did two djedish drops come off the roof, if after a time $t_0$ after the second drop began to fall, the distance between the drops was equal to $l$?

This problem can, of course, be solved in a roof-related coordinate system (do it yourself).

But it is more convenient to go to the coordinate system associated with the second drop. Since the drops fall with the same acceleration relative to the ground, their relative acceleration is zero: the drops are uniformly relative to each other. Their relative speed is equal to the speed of the first drop relative to the ground at the moment of separation of the second drop $v_0=g\tau$. The equation of motion of the first drop in the coordinate system associated with the second one looks like this:

$$x = v_0 t + x_0,$$

where $x_0 = \frac{gt^2}{2}$. At $t=t_0$ $x=l$, that is

$$l = g\tau t_0 + \frac{gt^2}{2}.$$

Solving this equation, we get the answer:

$$\tau = \sqrt{t_0^2 + \frac{l}{g}} - t_0.$$

Let’s solve another problem.

Fig. 5.

distance $L$ from the place of throwing the stone will fall to the ground?
In the coordinate system shown at figure 4, the position of the stone at time $t$ is determined by the coordinates

$$x = v_0 t, \quad y = \frac{gt^2}{2}.$$

At the moment $t_0$ falls to the ground $x = L \cos \alpha$, and $y = L \sin \alpha$, that is

$$L \cos \alpha = v_0 t_0,$$
$$L \sin \alpha = \frac{gt^2}{2}.$$

From here

$$L = \frac{2v_0^2 \sin \alpha}{g \cos^2 \alpha}.$$

It would be possible to choose another coordinate system, for example, shown in figure 5. In Nord, this was convenient. In this coordinate system

$$v_{OX} = v_0 \cos \alpha, \quad v_{OY} = v_0 \sin \alpha,$$
$$a_x = g \sin \alpha \quad a_y = -g \cos \alpha.$$

Pozny

$$x = v_0 \cos \alpha t + \frac{g \sin \alpha \cdot t^2}{2},$$
$$y = v_0 \sin \alpha t - \frac{g \cos \alpha \cdot t^2}{2}.$$

Fig. 6.

Substituting the values ​​of $x$ and $y$ at $t = t_0$ ($x = L, y = 0$) into these equations and solving them together, we find the answer.

Exercises

1. The stone falls freely from a height $h$. How long does it take him to cover the last meter of his journey?
2. A stone thrown vertically upward at a height $h$ increased the pressure $T$. At what initial speed was it thrown?
3. A stone is thrown vertically upward with a speed $v_0$ from a height $h$ from the surface of the earth. At the same moment, another stone is thrown vertically from the ground at a speed of $u_0$. How long does it take for the stones to be at the same height? What will this height be?
4. The target, located on the top of the stake, is visible at an angle $\alpha$ to the horizon from the place where the gun is located. How should the gun barrel be directed towards the horizon for the hit to be accurate if the initial velocity of the projectile is $v_0$ and the height of the stake is $h$?
5. A stone is thrown into the river. What will the wave pattern be like? Draw it.
6. Two autonomous vehicles have simultaneously passed through an intersection and are driving along streets making an angle $\alpha$ with each other (Fig. 6). Autonomous speeds are $v_1$ and $v_2$. After what time will the distance between autonomy be equal to $l$?
   Indicated and e. It is convenient to select a coordinate system associated with one of the autonomies. In it, the speed of the second autonomy is constant and equal to

$$v_{OTH} = v_2 + (-v_1).$$

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about a hundred years ago by the famous Russian chess player K. Janisch. It is easy to see that this square is magical, and all the mentioned sums are equal to 260. The given “magic move of the knight” can be classified as especially interesting, since it has not only the indicated property. The very construction of it is true that when the board is rotated 180°, the first one placed (fields 1-32) turns into the second (fields 33-64). It is curious that such a “knight’s move” was brought to N. Rudina’s hypothesis, which claims that chess originates from magic squares *).

Of particular interest is the closed “knight’s move.” Let us show that on a board of size 4×n, where n is any natural number (Fig. 6), there is no closed route (there is a solution, for example, for n=3).

Let’s assume the opposite, that is, that a detour route exists. The knight gets to the outer squares only from the middle ones (we call the outer squares those located on the upper and lower horizontal lines, and the middle ones - the rest). If the knight went around all the squares, observing the conditions of the problem, then 2n moves were made from the outer squares to the middle ones. then he made the remaining 2n moves from the middle fields to the outer ones. With each move, the color of the field on which the knight stands changes, so all the fields on the outermost horizontal lines must be painted one color, and the middle ones - in another. Contradiction.

Concluding the discussion of the problem of the knight’s move, it is a special case of an important mathematical problem involving finding the so-called Hamiltonian path in a symmetric graph**). This, apparently, explains its special attractiveness for mathematicians.

Let’s consider another problem about a horse.

Let there be a white knight and a black kon on the chessboard, and a number of “corresponding” klees are determined

**) You can read more about this hypothesis in the book by its author: N. Rudin and N. From the magic square to chess, M., “Enlightenment”, 1969.

**) See Berge, Graph theory and its applications, IL, M., 1962.

mock (on figure 7 they are filled with ink). You need to get your knight to the square with the king, and then return it to the original one, and along the way you cannot stand either “matching the square” or on already eaten squares.

This problem is known as the Attila’s Horse Problem. “Grass does not grow where my horse has stepped!” the leader of the Huns boasted, hinting that the hordes he led were destroying all living things in their path. This type of problem involves a path leading out of a labyrinth, and is solved in detail in various books on graph theory. In our example, Attila’s horse should move as follows:

Ng4—f6—e8—g7—e6—f8—g6—e7—c6—a5: b3—d2—b1—a3—b5—d6—f7—h6—g4.

The obstruction that the knight changes the color of the field on each move plays an important role, as we already have the opportunity to verify. Here’s another question on this topic.

Can a knight, starting from a1, reach h8 by stepping on each square of the board equal to one?

Of course, the answer is no. On every odd move, the knight lands on a white square, since the initial appearance is black. However, the number 63 (on this move

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the knight must pin on h8) - odd, and the h8 square is black. The problem is very simple, but it is curious that after a game a chess player sometimes comes to solve similar problems.

Look, for example, at figure 8. In this ending, nothing is achieved only by 1. Krs1! Now the white king will move from cl to c2 and back, each time occupying a square of the same color as the knight. Otherwise (1. Krs2) the king takes it lightly. The black knight comes to d3 at the moment when the white king is on c2. After this, the cl square did not reach him, the black king escapes from the turning, and the pawn is queened. The analogy between this chess problem and the above mathematical estimate. In conclusion, let’s look at another, more complex chess example. On figure 9 the basis is acute

Fig. 7.

Fig. 8.

smart sketch (V. Chekhover, 1937. White starts and wins) also lies in the knight’s ability to change the color of the field on each move.

There is only one way to win: destroy the black knight on h8 with the king and queen the pawn, since the rest of White’s pieces are pinned down. However, the white squares for the king are “mined” - the bishop f1 will move away with check and then f2-f1Q will follow. A straightforward attempt gives nothing: 1. Kb2 Kf7 2. Kp3 Kh8 3. Kpd4Kf7 (covering the e5 square) 4. Kp5 Kh8 5. Kpd6 Kg6! (covering the square e5 and e7) or 4. Croesus Kh8 5. Npf4 Kf7! (covering the square e5 and g5). We need to change the color matching for the king and the bishop. First, along the black squares, the king goes to the only safe white square - a8, and only then breaks through to the h8 square.

Next time we will meet again with the task

Fig. 9.

chat about the horse, but for now we offer several tasks for you to solve on your own.

1. There is a 3x3 chessboard. In its upper corners there are two black knights, and in the lower corners there are two white ones (Fig. 10). Prove,

Fig. 10.

that you need to make at least 16 moves to swap the places of white knights with black ones.

2. Prove that a chessboard of size (4k+1)×(4k+1) can be bypassed by moving a knight, visiting each square exactly once.

3. How many ways can white and black knights be placed on a chessboard so that they cannot be each other? Same question for board $m \times n$.

4. On how many different squares of an infinite chessboard can a knight move after making $n$ moves from a given square?

5. Normal knight move $2 \times 1$. Let us now consider the move of the knight $m \times n$. What should the numbers $m$ and $n$ be so that from a given square a knight can get to any square of an infinite chessboard?

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HISTORY OF INVENTION
ELECTRICAL CAPACITOR

A. K. Kikoin

The first half of the 18th century was a time of rapid accumulation of experimental facts about electrical phenomena. It was at this time, for example, that it became clear that there were two electricity sources.

One thing is the electrification of bodies; the nature of electricity remained completely mysterious. It was generally believed that electricity is a special vitality that contains a charged body in every body. And the observed decrease in charge on natural bodies was interpreted as the “evaporation” of this electric liquid. An equally natural idea was to try to prevent such “evaporation” by placing a charged body in … a bottle, using water as the charged body.

This is exactly the experiment that Jurgen von Kleist carried out in one of the cathedrals in Pomerania in 1745 (according to other sources, the experiment was carried out with the aim of obtaining charged water, supposedly beneficial for health). He filled the bottle with water, closed it with a cork, and through the cork a metal rod (simply a bunch) into the water. By connecting the outer end of the rod to an electric machine, which in those days was a rotating glass ball against which the experimenter’s hand rubbed, Kleist imparted a significant electric charge to it. And then the unexpected happened. Taking the bottle with one hand, he had the imprudence to touch the end of the nail protruding from the cork with the other hand, and in doing so he made a strong blow to his arms and shoulders, which caused the mouse to become numb. The necessary case was shimmy, he reported this in a letter to one of his friends.

By coincidence, almost the same experiment and almost at the same time was carried out in the Dutch city of Leiden by university professor Muschenbrock. Only together with the walled bottle did Muschenbrock use a thin-walled glass jar. Having charged the water and taken the jar in one hand, he also brought it with the other hand to the metal rod that served to supply the charge to the water. At the same time, Muschenbrock made such a strong blow to his arms, shoulders and chest that he lost consciousness. and came to his senses for two days. Reporting this “adventure” in a letter to his French correspondent, Muschenbrock adds that he would not agree to repeat the experience, even if he were promised the French kingdom!

At first, the observations of Kleist and Muschenbrock were understood as manifestations of the so-called “living electricity”, since human hands played such an important role in these experiments. But pretty soon it became clear that the hand holding the jar and the charged liquid in it were, as we now say, the plates of a capacitor and that an even more effective device would be obtained if the outer and inner surfaces of the walls of the jar were covered with a layer of metal, for example, tin foil.

This is how the first electric capacitor was born, which the French physicist Jean Nollet called the Leyden jar - a name that has not been forgotten to this day.

Probably, an echo of the then prevailing ideas about electricity and the “bottle” origin of the capacitor remained the word denoting the main characteristic of the capacitor - capacitance.

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HOW THEY SEE THE INVISIBLE

People incredibly quickly get used to the most amazing achievements of science and technology. Is it possible to introduce cognition today with a story about a television, a jet plane, or an artificial Earth satellite? Even the lunar rover - this amazing creation of modern science and technology - has taken a common place in our imagination next to an electronic computer, a nuclear power plant and the mechanism for transmitting hereditary characteristics from generation to generation. We are spoiled by the abundance of new wonderful opportunities that the modern scientific and technological revolution continuously puts at the disposal of man and scientist. Almost nothing can surprise you anymore. But this, in general, is bad. For without surprise there is no interest, and without interest there is no desire to truly learn something, to study it, to explore it yourself.

But let’s cut ourselves off from this “all-importance” for a minute and think about this question. Atoms consist of electrons and atomic nuclei. Atomic nuclei are formed from protons and neutrons. What are the sizes of these particles? The radius of a proton or neutron is of the order of $10^{-13}$ cm. Consequently, any of these particles, roughly speaking, is contained in a volume of the order of $10^{-38}-10^{-39}$ cm$^3$. The dot that comes at the end of the previous phrase has a radius approximately equal to a quarter of a millimeter. A sphere with such a radius will have a volume of the order of $10^{-4}$ cm$^3$. Therefore, it would be possible to measure $10^{35}$ elementary particles in it! How small these particles must be to decrease in such a gigantic quantity inside our point. Neither the eye nor the microscope, even with the highest magnification, can notice them. But you have all seen clear photographs of traces of such particles in textbooks and popular books. They were obtained using cloud chamber signs. Look at them carefully, and a new world will open before you. Here, for example, are photographs of the traces left by electrons. One trace of length, they pass through the chamber, slightly bending in the magnetic field penetrating it. These electrons have high energy. Other traces were short, highly curved, and their remaining electrons had a small amount of energy. Some traces, as it were, will disperse at the end - a collision has occurred with an atom, ending with the removal of one of the electrons from its shell.

Remember how they discovered the invisible man, the hero of Herbert Wallace’s one-shot novel? The pursuer saw traces of bare feet, right before his eyes, appearing one after another in the snow. The man was invisible, but traces of him remained. Also here - the particles are invisible, but the trace of their destructive work is present.

Looking at a photograph of a person unknown to us, we can learn very little about him in essence. Unless we can establish the gender and roughly estimate the age. A photograph will not tell us whether he is pure or not, whether he likes music, what his profession is, whether he has children and, if so, how he feels about them.

Another thing is photographs of traces of elementary particles in a cloud chamber. By studying them, we can accurately determine the nature of the particles, establish their energies, patterns, the nature of interaction with each other, the interaction of other energies, protons or electrons, one cannot help but admire the perfection with which all the details of their short, but full of dramatic life events are recorded. The Wilson chamber has become a helpless assistant in a wide variety of studies. This device is in some ways the highest court of cassation, which

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the experimenter can trust completely. No person endowed with the most brilliant talent of scientific foresight could have predicted all the capabilities of this instrument, possessing such tempting powers and inexhaustible possibilities.

In 1970, the Atomizdat publishing house published an interview with the popular science book by I. M. Bekkerman “The Invisible Leaves a Trace”*). It contains a detailed story about how the Wilson chamber was created and how amazing discoveries were made using this chamber. The undoubted advantage of the book is that I. M. Bekkerman introduces the reader not only to instruments and research, scientific ideas and the way to implement them. The book also tells about the author of these ideas and devices, the remarkable English physicist Charles Thomson Rhys Wilson. Reading the book, we learned interesting details of his life, we see what enormous work and direction the creation of his camera required. Telerb, everything seems surprisingly simple to us. “Condensation of supersaturated vapor on charged parts!” - a modern physicist will say. But in those days when Wilson began the research season, no one could say almost anything about either the mechanism of vapor condensation or the nature of the charged particles. In 1895, Wilson began to study the conditions for the formation of fog, and only in 1911 his “fog chamber” was born.

The book traces in detail the fate of the cloud chamber - how the Soviet physicists P. L. Kapitsa and D. V. Skobeltsyn placed it in a magnetic field (which made it easy to determine the sign of the charge of the part by the characteristics of their paths), how the Japanese physicist Shimzu and the English physicist Blackett automated its work, how it was forced to work much faster, to study minimator lightning - electrical discharges in gases, etc.

The appearance of some of the high energies created by modern accelerators of charged particles has necessitated a radical reconstruction of the cloud chamber. In order to prevent particles from leaving the field of view of the camera, it was necessary to sharply increase the density of the substance in the path of the particles. Cloud chambers with high pressures of gas filling them are created to take into account. But even at a pressure of 200 amps, the particles travel only a small fraction of their path in the chamber. This means that many events happen to them outside the camera and no traces remain. It was necessary to replace the supersaturated steam that fills the general Wilson chambers with a superheated liquid one, in which charged particles form gas bubbles along the path. Just like fog droplets, these bubbles in a boiling liquid form around charged particles. The first such “bubble” chamber was created by the American physicist Donald Glaser. It is sometimes called an “anti-Wilson chamber”, since in Wilson’s chamber the supersaturated steam condenses into droplets of liquid, and in Glezer’s chamber bubbles of steam appear in overheated liquid.

And again, talking about the birth of chamber bubbles, the author traces in detail the train of thought of Donald Glaser, step by step, with logical inevitability, he leads the reader to the final goal.

In the book by I.M. Bekkerman we talk about complex ss

temporary research in the field of particle physics. But the author surprisingly chooses the material and quite clearly presents the essence of the matter, the main thing is in the research he covers.

The events described in the book come close to the present day, to the problems that physicists intend to solve at the most powerful accelerator of charged particles near Moscow, near the city of Serpukhov in the young scientific center of Protvino. A huge live-hydrogen bubble chamber “Mirabelle”, created by French scientists for our accelerator, is mounted there. The chamber diameter is 1.6 m, length 4.5 m. All of it with the necessary supports is over 200 tons. Moreover, there is a magnet weighing 1150 tons, inside of which a camera is placed. What a huge distance separates Mirabelle from the first Wilson camera, which easily fit inside a small suitcase! But this path was covered in a little more time than a vole.

Wilson died this year from the first and only serious illness in his life. Until his last days, he worked hard, showing deep interest in the study of new, unknown natural phenomena.

《Of all the great scientists of our century, he was, perhaps, the most gentle and calm, the most indifferent to honors and glory. He drew his boundless passion for this from his exceptional love for the natural world, in admiration for its beauty.” This is how the outstanding researcher of cosmic rays, the English physicist Blackett, assessed the outstanding English physicist Charles Thomson Rhys Wilson, the creator of the “foggy chamber” signs.

V. A. Leshkovce

“PARTICLES”
AND “ANTIPARTICLES”
IN A GLASS OF NARZAN

If you drink some kind of carbonated water in a glass, for example, narazan, you can see how the carbon dioxide dissolved in the water is released in the form of small bubbles that quickly float up. Such bubbles can be contrasted with liquid droplets. Droplets are “particles” of matter. Gas bubbles in a liquid behave like “antiparticles” with properties opposite to those of the corresponding particles. For example, drops fall down under the influence of gravity, while bubbles rise up. If the vessel in which the drops fall is rotated, the drops will be thrown to the periphery. In a rotating liquid, bubbles tend to the axis of rotation. If there is any heavy body inside the liquid, then the antiparticle bubbles do not hide, but push away from it. At the same time, the two bubbles will be mutually hidden.

The surface of the liquid under the conditions under consideration has interesting properties. Rising from this surface, the bubble collides with the surface in a layer and disappears almost instantly, but not without a trace: from the place where the bubble disappeared, a small drop of water flies up. On the surface of the liquid, “antiparticles” turn into particles. These particles, as if continuing the movement of the rising bubbles, fly up until the force of gravity stops them and makes them fall.

G. I. Pokrovsky

BRAND AND LABEL,
DEDICATED
M. V. OSTROGRADSKY

To mark the 150th anniversary of the birth of the outstanding Russian mathematician M. V. Ostroogradsky, a postage stamp with a portrait of the scientist was issued in December 1951. The miniatura is made according to figure by the famous artist V. Zavyalov in brown and black on pink paper.

In 1967, a label dedicated to M. V. Ostrogradsky appeared in a series of match labels with images of outstanding Russian mathematicians and physicists. In addition to the portrait of the mathematician, it shows the so-called “Ostroogradsky formula”, which is one of the most fundamental results of mathematical analysis. This formula is very often used in mathematical calculations, and is widely used in physics and mechanics - in the study of thermal, magnetic, electrical and hydromechanical phenomena (for example, in deriving the basic law of hydrostatics - Archimedes’ law).

N. N. Kolesnikov

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LOGICAL
CUBES

The game “logical cubes” was released in France. The game set consists of 48 tiles (“cubes”). In their manufacture, three colors were used (red, blue and green), two sizes (large and small dies of the same shape and color), two thicknesses (thick and thin) and four shapes (circle, square, rectangle and triangle). By combining these four features, you can make just 3×2×2×4=48 dies.

With this set you can come up with many games for any age, starting from kindergarten. All games have one thing in common: they teach the ability to alienate, that is, logic, introduce mice to the basic concepts of set theory without confusing them with complicated explanations, and allow them to master basic operations on sets in practice.

Games for the little ones

The simplest one. Place the cubes on the table in some order, let the child remember it and, after he answers, rearrange something. Let him agree on what has changed.

The other is simple. The condition is the same, you just need to agree on which cube is hidden or how many are hidden.

Third. You need to arrange the cubes in an invented order (for example, arrange them by color).

Fourth. Ask the mouse to list how two cubes are different from each other and how they are similar.

Exchange. Two people hold a cube in both hands and exchange those in their right hand. The one with the most similarities in the dice wins; he takes all four dice for himself. All 12 four dice are played this way.

In these games, the concept of an object’s attribute is developed: the child learns that cubes differ in no more than four properties: either color, or size, or thickness, or shape.

Games for average

Dominoes “one difference”. Like in dominoes, players make a chain of cups with one condition: each next cube differs from the previous one by one (exactly one) feature.

Dobino “two differences”. The same, but neighbors in the chain must differ in two characteristics (Fig.1).
Next, you can try playing “tri differences”.

Treasure hunters. Two players each think of one cube (“treasure”) and try to find the treasure hidden by their opponent, pointing alternately at one or another cube. The opponent only reports how many differences they have from what was planned. The one who quickly agrees wins.

Three villages, two villages. Games that develop the concept of intersection of a set. “Two villages”: draw two circles on the floor - “villages” - and write names on them, for example, “red squares” on one and “small fat ones” on the other. You need to place all the red squares in the first circle, and all the small thick cubes (Fig.2) in the second circle.
We need to tell you how to do this. A similar game “tri villages” is even more interesting.

Names of the villages. One player places the cubes into “three villages”, but does not write their names on the circles, and the other must guess them (Fig.3).

Games for seniors

These games can be played by two people; the one who needs fewer questions to answer correctly wins.

Selection rules. One of the players comes up with some rule for selecting cubes, for example: “select only yellow ones,” and the other takes random cubes and receives the answer “yes” when the cube falls under the intended rule, and “no” otherwise. You need to guess the rule. The rules gradually become more complex, creating sums and intersections of features.

Settlement of villages. One player comes up with the name of the villages, and the other tries to populate them with cubes, receiving only “yes” or “no” instructions from the first.

Treasure dumpers (folded version). To guess the “treasure”, 3-4 “test” cubes are immediately laid out on the table and you get an answer as to how many signs of each test differ from the intended one. “Treasure” must then be indicated the first time.

Guessing the law. They come up with a rule according to which you can lay out a goal from the cubes. For example, “every even cube is large” or “after the squares there are thin circles.” The guesser begins to build the goal scientifically, cube by cube, each time receiving the answer “yes” when he places them without breaking the rule, and “no” when the rule is broken. We need to guess the law.

G. I. Kopylov

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4. Use conditions under which both trilene roots
   $$f(x) = x^2 + x + a$$
   large numbers $a$, namely
   $$1 - 4a \geq 0, a^2 + 2a \geq 0, -\frac{1}{2} > a.$$
   Answer: $a < -2$.

5. $a = 0$ and $a = 2$.

6. $$\frac{H}{2}$$.

7. The smallest value of the function $x^2 + 1$ is 1 and is achieved at $x = 0$. Therefore $x^2 + 1 > \cos x$.
   with all $x \neq 0$. The value $x = 0$ is the root of the equation.

8. Use the conditions under which both square roots are trilenic
   $$f(x) = x^2 + mx + m^2 + 6m$$
   lie outside the interval $1 < x < 2$, namely, $f(1) < 0, f(2) < 0$.
   Therefore, we need to find a value of $m$ that satisfies two simultaneous inequalities
   $$m^2 + 7m + 1 < 0, m^2 + 8m + 4 < 0.$$
   Solving these inequalities and selecting their general solution, we find the answer:
   $$\frac{-7 - 3\sqrt{5}}{2} < m < -4 + 2\sqrt{3}.$$
   $$9. \frac{2v_0^2 \sin 2\alpha}{g} - S.$$

To the article
“Chess and mathematics notes”
(see «Kvant» No. 8)

1. It is enough to note that each figure contains either 3 white fields or 3 black ones, and there should be 25 such figures in total.

2. It is easy to verify that for any covering, each trimino contains exactly one field through which the red line passes and exactly one through which the black line passes. Since there are 22 “red” and “black” fields, and 21 triminos, one “red-black” field out of the four marked at figure is not covered. Check that appropriate coverages exist.

3. On each field of the board we will indicate the number of the horizontal line in which this field is located. Check that each trimino covers fields with a sum of numbers that divides by 3, and we need to cover fields with a sum of numbers of 296, which gives a remainder of 2 when divided by 3.

4. a) No. The “white” pawn (a pawn standing on a white square) should become “black” during the secondary arrangement, and the “black” one should become “white”. This means that the number of “white” pawns is equal to the number of “black” ones, which contradicts the odd number of the board.

5. No. Prove that the pawns located on the leftmost vertical remain in place (from a1 to a8 you can go along 6 adjacent squares). Next we consider the fields b1 and b8.

To note
“A few more problems of Vasya Smekalkin”
(see «Kvant» No. 8, 3rd cover page)

1. Let’s make a table of differences between numbers of a given sequence (let’s write down their difference between adjacent numbers). The third line contains a sequence of powers of two. After this it is clear how to continue the table.

| 4 | 3 | 7 | 5 | 12 | 9 | 21 | 17 | 38 | 16 | 33 | 71 | 55 | 126 | 119 | 245 | … |
| :— | :— | :— | :— | :— | :— | :— | :— | :— | :— | :— | :— | :— | :— | :— | :— | :— | … |

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