Kvant 1971, No. 12 · Pages

Popular science physics and mathematics
quantum
12
magazine
Academy
Sciences of the USSR
And
Academy of Pedagogical
Sciences of the USSR

In the room:

The movement of comets and the discovery of the atomic nucleus 1 Ya. A. Smorodinsky
Johannes Kepler 8 A. Einistein
Once again about machine translation 11 V. V. Raskin
Models of molecules 18 A. I. Kitaygorodsky

Math circle
Problem about numbers in table 24 M. L. Gerver

Laboratory “Quenta”
Vortex rings 28 R. Wood

Problem “Quenta”
Problems 31
Problem solutions
M74—M76, f83—f88 33 A. L. Toom,
A. G. Kushirenko,
N. B. Vasiliev,
I. Sh. SLOBODECHY,
L. G. Aslamaov

Applicant’s workshop
Problems for the oral exam in physics 39 M. M. Kozlov,
I. G. Kochnev
at the Leningrad Political Institute
Area method 41 I. D. Novikov
Chess and mathematics notes 47 E. Ya. Gik
Review of reader letters 51 V. N. Berezin,
M. L. Smolyansky

Information
Mathematical optimization at MESI 52 I. G. Venetsky,
Yu. I. Sorkin
XIII International Mathematical
Optimization of schoolchildren 53 I. S. Petrakov,
V. A. Skovortsov
Impressions of an optimida participant 54 M. A. Tsfasman
V International Physical
Optimization for schoolchildren 56 M. D. Karasev
G. S. Tarasyuk
V city scientific conference
schoolchildren of Kyiv 58 A. I. Shapiro

Collector’s Corner
Stamps dedicated to Johannes Kepler
and Tycho Brahe 59 A. V. Altykis
Badges dedicated to international
mathematical optimizations 60 M. L. Smolyansky

Answers, directions, solutions 62
Index of Articles 1971 62

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The last statement, however, needs more precise formulation. It should be added that the formula for potential energy is written in such a way that potential energy tends to zero when the distance $r$ between the planet and the Sun tends to infinity:

$$U(r) = -\gamma \frac{Mm}{r}$$

(γ is the gravitational constant, $M$ and $m$ are the masses of the Sun and the planet).

So, a planet cannot move away from the Sun and rotates in an ellipse if its total energy is negative.

And how will a comet move in the field of the Sun (or a spaceship in the field of the Earth) if its total energy is positive, that is

$$E = \frac{mv^2}{2} - \gamma \frac{Mm}{r} > 0$$

The energy equation (2) in this case formulaically coincides with the energy equation for the elliptical motion of the planet, only with a different sign $E$. The formula for angular momentum remains the same:

$$L = mv\rho$$

(ρ is the impact parameter, see below). Therefore, at $E > 0$ we should obtain the same formulas that we obtained by considering the elliptical trajectories of the planets ($E < 0$), only in some places there will be a “minus” sign along with the “flat” sign. The trajectory of movement in this case will be a hyperbola.

A hyperbola is defined as the locus of points, the difference in distances from which to two given points—foci—is constant. Let us remember that the ellipse is defined in the same way, only together with the “difference” the word “sum” was built.

Figure2 shows how both types of curves are constructed on a grid composed of two families of concentric circles centered at $F_1$ and $F_2$. Find out for yourself why ellipses and hyperbolas pass through the intersection points of circles.

Main theorem

Let us now check that if $E > 0$, then the comet or spaceship really moves along the hyperbola. For our purposes, since we will be talking about the atomic nucleus and Rutherfort’s experiments, it is more convenient to talk not about a comet, but about a loaded particle that moves in the field of a heavy stationary charge.

This can be done because the law of the dependence of the interaction force on the distance is the same in both cases.

Let the heavy part be positively charged and located at the focus $F_1$ (Fig. 2). If the flying light part is negatively charged (its charge is $z_1e$), then it should fly along the trajectory indicated in the upper part of the figure by the sign “—” (we have not yet proven that this is a hyperbola!). The negatively charged part will fly along the same trajectory if the stationary part is negatively charged and is at focus $F_2$. This can be proven by using the mechanics equation - the total energy and momentum are the same in both cases. If the flying part is positively charged, then it will move along the trajectory marked by the “+” sign.

Further we will argue in the same way,

Fig. 2.

as was done in the article “Motion of the Planets”.

Let’s consider two problems at the same time. In one there is a positive charge +$z_1e$ located in $F_1$, in the other there is a negative charge - $z_2e$ located in $F_2$. Let in both cases the energy and angular momentum of a particle that flies along the same trajectory “—” be the same. Let’s create expressions for energy *):

$$E = \frac{mv_1^2}{2} - \frac{z_1z_2e^2}{r_1}$$

(3)

*) Let us not forget that the sign of potential energy is negative in one case and positive in another.

(for a positive particle at focus $F_1$),

$$E = \frac{mv_1^2}{2} + \frac{z_1z_2e^2}{r_2}$$

(4)

(for a negative particle at focus $F_2$).

Let us find from here the kinetic energy and both expressions in variables:

$$\frac{mv_1^2}{2} \cdot \frac{mv_2^2}{2} = \left(E + \frac{z_1z_2e^2}{r_1}\right) \left(E - \frac{z_1z_2e^2}{r_2}\right)$$

(5)

Let us introduce angular momentum (the same in both cases)

$$L = mv_1p_1 = mv_2p_2.$$

(6)

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$p_1$ and $p_2$ are the lengths of the perpendiculars dropped onto the tangents to the trajectory. Let us transform formula (5) to the form

$$\frac{L^4}{4m^2p_1^2p_2^2} - E^2 =$$

$$= \frac{z_1z_2e^2}{r_1r_2} \left[ E(r_2 - r_1) - z_1z_2e^2 \right]$$

(7)

For this equality to hold for any $r_1$ and $r_2$, there must be

$$r_2 - r_1 = \frac{z_1z_2e^2}{E} = 2a,$$

(8)

$$p_1p_2 = \frac{L^2}{2mE} = b^2.$$

(9)

The first equality, by definition, is a hyperbolic equation. It is possible, using the optical property of hyperbola (see below), to prove the theorem that the product of the lengths of two perpendiculars dropped onto the same tangent from the foci of the hyperbola does not depend on the point at which the tangent to the hyperbola is drawn*. From this theorem it follows that the second equality also defines hyperboly.

Hyperbole equation and its properties

On figure 3 two hyperboles are drawn. There are two of them, not four, since the two red curves are two branches of the same hyperbola. The blue curves are two branches of another hyperbole. The blue hyperbola is the conjugate of the red one.

The equation of hyperbolas is written next to them. If we recall the equation of the ellipse (see, for example, the article by I. N. Bronstein “Ellipse” in «Kvant» No. 9, 1970)

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,$$

then we will see that the hyperbolic equation is obtained from the ellipse equation if we change the sign of $y b^2$, that is

*) The proof of this theorem is available, for example, in the book of J. Adamr “Elementary Geometry”.

Fig. 3.

how to make the semi-minor axis mini $((ib)^2 = -b)^2$. Therefore, $a$ in a hyperbola is called the same as in an ellipse, the major semi-axis or real semi-axis, and $b$ is called the mini semi-axis.

For comparison, figure 4 shows two conjugate ellipses and their equations are written. For example, that an ellipse is a locus of points, the sum of the distances from which

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find the angle at which an α-particle will deviate when flying near an exact positive charge.
By measuring the angle of deflection on the test, it would be possible to find out how much the scattering charge is disconnected from the exact one. In practice, the situation is somewhat more complicated.

In reality, of course, it is impossible to determine the trajectory of an individual particle, but one can study the scattering of a beam of particles.

Let us consider a beam of particles in which 1 particle per second passes through each $1 \, cm^2$ of its section. When such a beam falls on a nucleus, particles passing at different distances from the nucleus will be deflected at different angles. From figure 6 it is clear that the number of particles flying past the core (it is located in the center) at distances between $\rho$ and $\rho + \Delta \rho$ is $2\pi \Delta \rho$. If a particle has an impact parameter $\rho$, then it scatters through an angle $\theta$, determined by the formula

$$\text{ctg} \frac{0}{2} = \rho \cdot \frac{mv_0^2}{z_1 z_2 e^2}.$$

$\theta$ is the angle between the asymptotes of the hyperbola.

This formula is derived like this.

Since $\theta$ is the total change in the angle during scattering (the angle between the velocities before scattering and after scattering), it is related to the angle $\alpha$, which is made up of asymptotes with the axis $x$ (Fig. 7):

$$\theta = \pi - 2\alpha.$$

Then, using formulas (8) and (9), we find

$$\text{ctg} \frac{\theta}{2} = \text{tg} \frac{\alpha}{2} = \frac{b}{a} = \frac{L}{a \sqrt{2mE}}.$$

(Since the asymptote is the tangent to the infinitely further hyperbola, then $\rho$ is simply equal to $p_1$ (or $p_2$) from formula (9) for the infinitely further and always $p_1 p_2 = \rho^2$.).

If the speed of a particle at infinity is $v_0$, then $L = mv_0 \rho$, and

$$E = \frac{mv_0^2}{2}.$$

Then

$$\text{ctg} \frac{\theta}{2} = \frac{mv_0 \rho}{\sqrt{2mE}} \cdot \frac{2E}{z_1 z_2 e^2} = \rho \frac{mv_0^2}{z_1 z_2 e^2}.$$

Particles for which $\rho$ lies in the shaded interval will scatter in a certain range of angles from $\theta$ to $\theta + \Delta \theta$. The value $\Delta \theta$ is not difficult to calculate *). Let’s write the formula for the “outer” edge of the ring:

$$\text{ctg} \frac{\theta + \Delta \theta}{2} = (\rho + \Delta \rho) \cdot \frac{mv_0^2}{z_1 z_2 e^2}.$$

*) Those who knew how to differentiate can almost say that next we will simply calculate the derivative $d\rho/d\theta$.

A. Einstein

December 27 marks the 400th anniversary of the birth of Johannes Kepler (1571-1630). Kepler’s name is firmly entrenched in the history of physics and astronomy. First of all, it connects with the crown of the head the well-known laws that formed the basis of celestial mechanics.

Based on the observations of the famous astronomer Tycho Brahe, Kepler was able to find general patterns of planetary motion.

Before Copernicus, science was dominated by the geocentric system of the world, according to which the planets and the Sun revolve around the Earth. In the first half of the 17th century, the great Polish scientist Nicolaus Copernicus proposed a geocentric system of the world, according to which the Sun is at the center of the Universe, and planets move around it in a circle.

From the very beginning, Kepler considered the geocentric system of world structure to be correct. He was sure that the Sun was the reason for the movement of the planets. He was the first to consider further (as we speak now) not only possible, but also necessary. But the concentric circles of the planet’s orbits did not agree with observations. Kepler proposed that the trajectories of the planets were oval, and then came to the conclusion about elliptical orbits.

The laws that laid the foundation for celestial mechanics are Kepler’s most significant, but far from the only work. His works on the prelimination of light rays are widely known. Kepler’s research on prelimination led to the discovery of total internal reflection. Kepler described the production of images on the retina, studied reflection from a mirror, the nature of light and color, and explained the operation of a telescope.

In his research, Kepler proceeded from the conviction that the laws of nature should be simple and beautiful. The task of finding these laws stood before him throughout his life. Despite constant need, lack of any help, religious struggle, family misfortunes, he managed to complete this task. He could say: “What I devoted the most and best part of my life to has now been found.”

Below we place an article by A. Einstein about Kepler, taken from Einstein’s “Collected Scientific Works”,

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Once again about machine translation

We have already written about the difficulties that arise during machine translation *). As a result of their cunning efforts, the machine has, in general, been taught to cope with most of them.

However, there is also the goal of “placing” the city of more significant difficulties, which we discussed in this article. And they are all connected with the mysterious word “semantics”.

Semantics - the science of the meaning of words

The first semantic difficulty is that the same word can mean different things, and in different languages ​​in different ways: the different meanings assigned to words do not coincide. The Russian word “table” most often needs to be translated into English “table”, but sometimes it means “food” (“food”), and in turn the English “table” should sometimes be translated into the Russian word “table”.
In addition, in all languages ​​there are so-called homonyms - words that are spelled the same but have different meanings:
“Give me the bow!” - and for what - for shooting or for soup?
Further, in different languages, words are endowed with different emotional load and have different ability to be used in the enumerative sense. In Russian and in many other European languages, the word “heart” means, among other things, orran, which the popular consciousness considers to be the main thing and gives the ability to feel and experience.
But in the Kabalaka language (West Africa) this function is performed by the word “liver”; in the Jejonob language (a type of Mayan language in Guatemala) the word denoting the abdominal cavity is used in this sense. Imagine that

*) See «Kvant» No. 9, 1971

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the machine translates from the Kabbalah language into Russian and gives you the following phrase: “She has a good liver.”
Next trouble. In different languages, the “amounts of meaning” assigned to a given word are not the same. Let’s take, for example, the concepts of “brother” and “sister” and compare them with the concepts of “elder and “junior”. We get the following table:

| Concept | Hungarian | Russian language | Malay |
| :— | :— | :— | :— |
| older brother | batya | bram | sudara |
| younger brother | occs | | |
| older sister | néne | sesstra | |
| younger sister | húg | | |

The Hungarian language expresses each of the four concepts with a separate word, the Russian language “cleans” them two at a time, and the Malay language covers all concepts with one word. It is easy to translate these words from Hungarian into Malay and generally in the direction of the word to the right according to the table. What if, on the contrary, you need to translate the words “sudara” from Malay into Hungarian? The human translator will look at the context here and try to figure out the gender and age of the person designated by this word, and then select the desired Hungarian equivalent. But how to teach this difficult operation to a machine?

Words that are seemingly “the same” for all languages, such as personal pronouns, can have “unusual” meanings that make translation difficult. You will see this now.

Task 1. In the Hanuno language (one of the languages ​​of the Philippines) there are the following pronouns (and only them):

dah - they tah - me and we
kuh - I tam - we (including me)
mih - we (without me) yah - he, she
muh - we yuh - you

By what criteria should these pronouns be classified according to their meaning? Russian pronouns are classified according to the following criteria: number (this feature has two meanings: singular and plural), gender (three meanings: masculine, feminine and neuter), pitso (three meanings).

The system of these features is not suitable for classifying the pronouns of the Hanuno language, since the given pronouns do not differ from each other (Russian “he” and “she” correspond to the same pronoun “yah”); As for person and number, in the Hanuno language more subtle differences are captured than in Russian (for example, together with one Russian pronoun, the first person plural has three).

In this case, it is possible to find more suitable features for classification that satisfy all the semantic differences of the given pronouns.

Given and e. Select such semantic features (and there should be a minimum number of them, and each with exactly two meanings) so that the given pronouns represent all possible combinations of meanings and so that no two combinations are repeated.

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But that’s not all. Every language has stable phrases that do not mean exactly what, or even not at all, what the words that make them up mean separately. These are the so-called phraseological units: “beat the baklish,” “don’t think twice,” “play the fool,” etc.

Let’s imagine that we have taught the machine to overcome all the difficulties described above. Let’s take an English phrase:

«The President said that the article was red herring».
Happy, we are waiting for the translation. And we get:

“The President said that Stamya is a red herring.”
Hm… Got it? Not good. But ask any translator and they will instantly dispel your confusion, because the phrase actually translates like this:
“The President said that this story is a duck (lie, nonsense).”
Immediately everything fell into place. So, we add one more rule to the machine (how many of them are already written there?!): replace the phrase “red herring” with the word “duck” (or even better, since ducks are not only newspaper, but also fried - “lie” or “nonsense”). But there are a lot of phraseological turns, is it really possible to write for every word?
But a person who knows the language translates so easily and simply!

In order to overcome these and other semantic (and not only semantic) difficulties, in order to correctly select the equivalent word in the target language, that is, in order to perform a translation, the machine must remember the meaning of the words. One of the most prominent specialists in machine translation, American linguist V. Yngve writes:

“Machine translation work has encountered a semantic barrier. We had to realize that we will only get targeted automatic translation when the machine can “understand” what it is translating, and this is a really difficult task.”
And in order for the machine to understand what it is translating, the linguist must provide it with information about the meaning of words, give an accurate, formatted and clear semantic record, which the machine can handle in the same way as with a dictionary and processing rules. Machine translation of the present requires linguistics to record this. But…linguists are not yet able to cry on this score.

One of the interesting attempts to formally study semantic plurals. In this case, the approach is to highlight the so-called semantic plurals, that the meaning of each word is a combination of them (similar to how a natural number is decomposed into simple factors or a vector is specified by its coordinates). This idea is illustrated by the following problem.

Problem 2. Given 4 sets of words, each of which is divided into two subsets a and b:
1a) Beast, worms, ciliates, grass, posture;
b) table, ground, book, plane, coat.
2a) Whale, swallow, perch, tiger, ant;
b) wife, entry-level, hairdresser, husband, cashier.
3a) Sparrow, airplane, Baba Yaga, flour, rocket;
b) book, tunnel, already, well, telephone.
4a) Dog, horse, pigeon, donkey, mule;
b) bear, kangaroo, snake, tiger, perch.

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translator or editor." This definition is given in the book in order to show that machine translation in this sense does not exist at all. The text that the machine produces is so inconvenient and difficult to understand that it has to be subjected to “post-editing,” that is, subsequent editing by a person. 11% of the time during machine translation is spent on inputting material into the machine, 40% on recomposing output phrases, 31% on post-editing, and only 18% of the time remains on the translation itself. Assessing the quality of a translation (survey by an expert) according to the following parameters: accuracy of translation, speed of reading, ease of understanding, giving the following summary assessments (indicators for translation performed by a person are zero; the question of what all these “indicators” are based on and how the experts chose them, fortunately, has nothing to do with the matter):

| | Machine translation | Machine translation | Human translation |
| :— | :— | :— | :— |
| “raw” | “post-edited” | | |
| Accuracy | —16% | —5% | 0 |
| Speed | —21% | —11% | 0 |
| Understanding | —25% | —23% | 0 |

The main difficulty that reduces all three parameters is unnatural syntactic structures and word order in “raw” machine translation, as well as many translation options for one word. Now a little calculation. Let us denote by

$$\mathcal{U}_\text{ч} = \mathcal{U}_\text{м} - N \cdot P \left( B_\text{м} - B_\text{ч} \right).$$

The finished translation is intended for specialists who spend time reading it, and time is money (salary!), so the formula includes the value $P$. Let’s calculate the values ​​of $\mathcal{E}$ and $N$ with average statistical data.

Due to the fact that the “raw” translation takes longer to read, we get an amazing pattern:

| $N$ | 1 | 10 | 15 | 17 | 18 | 20 | 80 | 175 |
| :— | :— | :— | :— | :— | :— | :— | :— | :— |
| $\mathcal{E}$ | 13,17 | 5,97 | 1,97 | 0,37 | —0,43 | —2,03 | —40,13 | —127,03 |

The more people read the machine-translated and edited text, the more wasted and the lower the savings! The savings disappear when this number of readers exceeds 17. But it makes sense to translate exactly what is widely used by specialists!

One more (last) table. Let us explain the last column: according to statistics, exactly 11 million Russian words make up the annual US need for translation from Russian.

| Transfer type | Quality | Cost (in dollars) of translating 11 million Russian words |
| :— | :— | :— |
| Staff written translation | good | 440 000 |
| Non-standard translation | Decent | 350 000 |
| Regular interpretation in (voice recorder) | good | 440 000 |
| Commercial translation company | Average | 240 000 |
| Raw machine translation | Bad | 80 000 |
| Post-edited machine translation | Average | 400 000 |
| Translation: “with the help of machines” | Excellent | 310 000 |
| Specialist who knows the language | — | 0 |

Good translation of roads, “raw” machine translation is cheap, but “angry”. We need to look for a compromise. Our gaze falls on the last line. What is “machine translation” and how is it different from machine translation? This is a smart distribution of responsibilities between man and machine. The machine is entrusted with doing what it can do well and quickly. The man does the rest. In practice this is done as follows. The translator reads the text he must translate and marks the words he does not understand. They are brought into the machine. She gives their translations. All this is simple for her. The remaining operations, which, as we have seen, are not easy to teach a machine, are performed by humans. The translation quality is excellent. The cost is less than not only “good”, but also “decent” translations.

In many countries, more money is spent on research on machine translation, and yet we can say as a souvenir that the bulk of it was not wasted. Firstly, as a result of these studies we know more about the language. Secondly, and this is the main thing, machine translation “gave birth” to modern applied literature.

Machine translation and applied literature

Applied literature is, as its name suggests, the application of literary knowledge in practice. However, one of them is the form of ionic search.

It is estimated that the amount of printed information accumulated throughout the world doubles every year. In 1961 there was less input than now; in 1981 there will be more input than has been accumulated from the beginning of accumulation to the present moment. In other words, during the period

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Molecular models

A. I. KITAIGORODSKY

It is useful to remember that words are made by man.
Words that are used in life are often blurry in nature. Not all concepts are the same - the words cilia and beauty, energy and tension. And “good” weather is different for different people.
In science, this state of affairs does not exist, or at least it should not exist. Particularly intolerant is the imprecise use of words in physics.
The simplest physical concepts of nature were used to describe the properties and behavior of objects and bodies among which our life goes, in short, for “large” bodies or, as we also say, for macro-nutritional bodies.
What concepts from the macrocosm can be applied to molecules? All or some? The truth lies in the middle.
What about geometric and mechanical concepts?
Is it possible to talk about the shape of the molecule, its elasticity, the model of bending and torsion, and finally, the laminarity of molecules? Whether and what sense the concept of intramolecular and mechanical concepts make is not only possible, but also advisable.

This phrase means the following.
In some cases, a ULE molecule can be spoken of as a large body.
A body that can be compared to a given molecule is called a mechanical model of the molecule.
Our task is to explain how this model is constructed and how to use it to solve various physical problems.
The mechanical model of the molecule has recently become widespread due to interest in the huge (compared to an atom) molecules from which synthetic polymers are built - nylon, nylon, polystylene (these names are now known as home), as well as substances that are most important for the life of animals and plants - proteins, nucleic acids, and so on.
Any “image” of a molecule must be composed of a description of the relative position of atomic nuclei and the characteristics of electrons moving around these nuclei.
Chemical experience allows you to establish the atomic structure of a molecule (build its atomic model), that is, indicate which atoms the molecule is made of and how they are connected to each other. Some electrons are required to be associated with certain atoms, while others are “generalized.” Chemists say about these electrons that “they carry out a chemical bond.”

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with a carbon atom. (This group of atoms is called methyl.) Now, please connect a carbon atom of this group with a valence line to the second carbon atom. This second atom, in addition, must be connected to a pair of hydrogen atoms, and draw the fourth line (there are four lines from one atom, which means it is tetravalent) to the oxygen atom. The remaining hydrogen atom should be attached to an oxygen atom.

The physicist will immediately ask a question. And at what distance are the atoms, at what angles do the valence lines go to each other? Answers to such questions can be obtained by physical research. Let us leave aside for now the question of how the geometry of a molecule is established physically by experiments. Extensive data is collected in only reference books. In them you can find information about the distances at which chemically bonded atoms are located and what angles (they are called bond angles) form “rods” between each other, symbolizing chemical valence forces. If you don’t get too caught up in the subtle differences, it turns out that the distances between one-orth atoms are quite universal, although bond angles are more variable. Therefore, predicting the structure of a molecule is not always easy. But we will talk about this later.

Fig. 2.

We can now turn to the problem of mechanical forces.

That forces operate between molecules is evident from the most elementary messages. The vapor of any substance under suitable conditions follows into a drop. If so, then the molecules will undoubtedly be accepted. The substance resists the rampant. This means that being at short distances, molecules repel each other. If there is acceptance at large distances, and repulsion at small distances, then there is an equilibrium distance when these forces are balanced.

Instead of interaction forces, it is much more convenient to talk about interaction energy ). The energy of interaction is easier to measure, and the concept is simpler and clearer than force*). The energy of interaction between molecules (or atoms, or any other particles or bodies) is called the work that must be done in order to move the particles away from each other—so that the interaction stops. The mathematician will say - give it to an infinitely large distance. The closer the particles are, the more work is required to pull them apart. This work reaches its maximum value when the particles are at an equilibrium distance from each other. This work is called binding energy. If the particles collide and repel, that is, if they are at a distance less than the equilibrium one, then the work of discontinuity becomes, of course, less.

A typical interaction energy curve is shown in figure 2. All curves have this character. But for specific purposes, you need to know the parameters of the curve. First of all it is important

*) See the article by G. Ya. Myakishev “Interaction of atoms and molecules”, «Kvant» No. 11, 1971.

**) From the interaction energy curve you can also find the force: sleep is numerically equal to the tangent of the angle of inclination of the tangent to the interaction energy curve.

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hydrogen - hydrogen and carbon - hydrogen.

Where can I get information about these curves? Unfortunately, there are no theoretical ways to derive them. Energy curves have to be determined from experience. Since the selection of the necessary curves is a weight operation typical for many areas of physics, we will describe this interesting work in more detail.

When researching, starting from scratch, one must first of all estimate, at least roughly, the parameters of the curves. From experience we know the close distances at which atoms of neighboring molecules approach each other. Looking carefully at these data, we find out that two hydrogen atoms are not closer to each other than at a distance of 2-3 angstroms, two carbon atoms of different molecules are not closer to each other than 3-4 angstroms, etc. These data are taken from studies of solids. molecules in a solid have several slopes of slimy interactions. Therefore, we think that the equilibrium distances on the corresponding atom-atom interaction curves should be somewhat larger than the above figures. Based on a number of indirect reports, it is clear that this excess should be ten percent. Therefore, to begin with, you can select the abscissa of the minima on the curves, which we are discussing as an example, at points 2.6 and 3.8 angstroms.

Information about the binding energy, that is, about the depth of the well of the energy curve, must be “extracted” from the data on the heats of evaporation. For example, that at low pressures a solid substance evaporates. The heat of this phase transformation can be measured quite accurately. The heat expended on the evaporation of the crystal was used to disperse its molecules over a long distance. With some corrections, which we will not dwell on, the measurement of the heat of evaporation leads us to the value of the interaction energy of all molecules.

Of course, the state of affairs is not very simple - the total value changes, but it is necessary to judge the weaker ones. Hydrocarbon substances, consisting of molecules containing 10-20 atoms, have a heat of evaporation of 10-20 kcal/mol. If we assume that the interaction of connecting molecules plays a major role, then this figure adds up to thousands of atom-atom interactions. This means that the depth of the well of the curve should be of the order of 0.01 kcal/mol.

Thus, one arrives at approximate estimates of the parameters of the energy curves of atom-atom interaction and, with their help, constructs first-approximation curves; Then they calculate (this is where computers are needed) the properties of a large number of substances and compare the calculation results with experience. It becomes very necessary to make certain amendments. Actions are sequential, trial and error (physicists say so - work by trial and error), in the end we come to the optimal atom-atom interaction curves for many pairs of atoms.

In general, the calculation is cumbersome and the aditivity hypothesis has only recently been studied. The result turned out to be positive. It turned out that the assumption is fulfilled with good accuracy.

In advance it is not very clear that the work will be successful. If the hypothesis were unfair, if propipeditivity did not work, if atoms of the same chemical type contained in different molecules did not behave the same way, then the computer would tell us that it cannot select the curves so that they give numbers, at least in ways that coincide with experience. For example, successful work on fitting curves shows that the hypotheses underlying the calculation are valid.

Having proved the validity of the additive representation of the interaction energy of molecules, we provide a basis for the concept of molecular forms.

Having built the “skeleton” of a molecule based on experimental data, we can “dress” it

Fig. 4.

Flesh" using so-called mechanical radii. If on the curve of the atom-atom interaction of hydrogen the minimum lies at 2.6 angstroms, then the mechanical radius of hydrogen should be taken equal to 1.3 angstroms.

Having figured out the radii values, we can proceed to edging the molecule as shown in figure 4. Mechanical radii of molecules can be calculated by following the same steps as described in the previous section. For example, the radii of molecules in the water molecule are approximately 1.3 cm.

Fig. 5.

Answers to crossword puzzle published in No. 11 (1971)

Horizontal: 4. Matter. 7. Confidence. 10. Phenomenon. 11. Regression. 12. Erbium, 13. Copal. 16. Telephone, 17. Fedorov, 18. Physicist. 19. Chemistry, 21. Lebedev.

Vertical: 1. Calorie. 2. Center. 3. Ricochet. 5. Television. 6. Stereophone. 8. Quartet. 9. Remainder. 14. Polymer. 15. Desibel. 20. Ice.

24

Problem about numbers in a table

M. L. Gerver

In «Kvant» No. 7 of 1971, on page 21, the following problem was formed.

In the square table $6 \times 6$ the numbers 1, 2, 3, …, 36 are arbitrarily considered. Prove that there are two average numbers a and b, standing in adjacent cells, and also that a—b $\geq 5$.

Several years ago, a similar problem was proposed at the Moscow Mathematical Olympiad: to prove that in the table $9 \times 9$, filled with polar numbers, there are two adjacent numbers with a difference greater than or equal to 6.

Below we will solve the general problem for the table $n \times n$, and we will obtain an exact (unimprovable) estimate for the maximum difference between neighboring numbers.

Problem formulation

In the table $n \times n$ the numbers 1, 2, …, $n^2$ are arbitrarily considered. Prove that there are two adjacent numbers whose difference is not less than n.

Explanation. Let us clarify the term “neighboring numbers”. We call two numbers adjacent if the cells of the table in which they appear have a common side.

Preliminary remarks

It is easy to consider the numbers 1, …, $n^2$ in the table $n \times n$ so that the difference of any two adjacent ones does not exceed 了她. Examples of such an arrangement at $n=5$ are given in Tables 1 and 2.

From Table 1 it is possible (by rearranging the columns) to obtain another 119 = 5!—1 tables in which the difference between adjacent numbers is not

Table 1.

1 2 3 4 5

6 7 8 9 10

11 12 13 14 15

16 17 18 19 20

21 22 23 24 25

Table 2.

1 3 6 10 15

2 5 9 14 19

4 8 13 18 22

7 12 17 21 24

11 16 20 23 25

ascending 5. In table 2, without increasing the maximum difference between adjacent numbers, you can rearrange the numbers located further from the diagonal going from the lower left to the upper right corner of the table; for example, you can swap 1 and 2 or 23 and 24, etc. Completely similar examples can be constructed for any n.

There are many ways to consider the numbers 1,…, $n^2$ in the table $n \times n$ such that the maximum difference between adjacent numbers is n. He asks, is it possible to consider numbers so that the maximum difference is less than n?

It turns out that this cannot be done - this is exactly what we are going to prove.

25

Solution of the problem for $n=4$

The idea on which the solution is based is especially easy to present with small $n$. Let’s take table $4 \times 4$, select 4 cells in it, which contain the numbers 1, 2, 3 and 4. Let’s put asterisks in those of the remaining cells of the table that are adjacent to the selected ones (Table 3). Note that even if the selected cells are on the edge or corner of the table, there will be no less than four stars (check this yourself). Now, no matter how we arrange the numbers 5, 6, …, 16 in the table, one of the cells with an asterisk will contain the number $a$, greater than or equal to 8 (since there are at least 4 asterisks, and there are only three numbers less than 8 left: 5, 6 and 7). This cell is adjacent to at least one of the cells containing the numbers 1, 2, 3 and 4. So, the number $a \geq 8$ is adjacent to some number $b \leq 4$. Thus $a-b \geq 4$.

We have proven that for any arrangement of the numbers 1, 2, 3, 4, …, 16 in the table, $4 \times 4$ will find two neighboring numbers $a$ and $b$ such that $a-b \geq 4$.

What’s next?

The above can lead to the following final conclusion.

Let us select cells in the table $n \times n$ that contain the numbers 1, 2, …, $n$. Let’s put stars in those of the remaining cells that are adjacent to the selected ones. Let’s give them $n$ some stars. Therefore, no matter how you put the numbers $n+1$, …, $n^2$ in the table, there will be a number $a \geq 2n$ that falls into the cell with an asterisk and turns out to be adjacent to some number $b \leq n$. The difference of the numbers $a-b$ is obviously greater than or equal to $n$.

There is one error in the above reasoning: the statement printed in italics is incorrect. It is already incorrect for $n=5$: if we arrange the numbers 1, …, $n^2$. Then there is a number $m$ ($1 < m < n^2$), which has the following property: if we mark the mouse cells in which the mouse is on the edge or in the corner of the table, then there will be no less than four stars (check this yourself). Now, no matter how we arrange the numbers 5, 6, …, 16 in the table, one of the cells with an asterisk will contain the number $a$, greater than or equal to 8 (since there are at least 4 asterisks, and there are only three numbers less than 8 left: 5, 6 and 7). This cell is adjacent to at least one of the cells containing the numbers 1, 2, 3 and 4. So, the number $a \geq 8$ is adjacent to some number $b \leq 4$. Thus $a-b \geq 4$.

Lemma. Let the numbers 1, …, $n^2$ be arranged in the table $n \times n$. Then there is a number $m$ ($1 < m < n^2$), which has the following property: if we mark the mouse cells in which the mouse is on the edge or in the corner of the table, then there will be no less than four stars (check this yourself). Now, no matter how we arrange the numbers 5, 6, …, 16 in the table, one of the cells with an asterisk will contain the number $a$, greater than or equal to 8 (since there are at least 4 asterisks, and there are only three numbers less than 8 left: 5, 6 and 7). This cell is 相邻 in the table of numbers 1, …, $n^2$, there is a number $a \geq 2n$ that falls into the cell with an asterisk and turns out to be adjacent to some number $b \leq n$. The difference of the numbers $a-b$ is obviously greater than or equal to $n$.

1 case. Each column contains a number not exceeding $k-1$, but there is a row that does not contain any of the numbers 1, …, $k-1$.

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Table 5.

| 1 | 2 | 3 | 4 | 5 |
| :—: | :—: | :—: | :—: | :—: |
| 6 | 7 | 8 | 9 | 10 |
| 11 | 12 | 13 | 14 | 15 |
| 16 | 17 | 81 | 19 | 20 |

Table 6.

| 1 | 2 | 4 | 6 | 8 |
| :—: | :—: | :—: | :—: | :—: |
| 3 | * | * | * | * |
| 5 | * | | | |
| 7 | * | | | |

In this case, we put $m=k-1$ (for example, $m=20$ in Table 1 and $m=8$ in Table 4).

Let us show that $m$ satisfies the requirements of the lemma.

Let’s mark the cells containing the numbers 1, …, $m=k-1$, and put asterisks in those of the remaining cells that are adjacent to the marked ones (Tables 5, 6).

Each column contains check boxes. In addition, each column has at least one unchecked cell (since the entire row does not contain marked cells). Therefore, there is at least one asterisk in each column. All columns $n$; this means that the asterisk is at least $n$.

Psluay. Each line contains a number not exceeding $k-1$, but there is a line that does not contain any of the numbers 1, …, $k-1$.

This case will coincide with the one already discussed: it is enough to reflect the table relative to the main diagonal (this operation is called painting).

at $l \neq i$ in the $l$-precise row the cell $a_{lj}$ is not removed.

So, each line contains both deleted and undisplaced cells, which means it contains at least one asterisk. Consequently, in this case there is a minimum of $n$ stars.

Since cases I, II and III exhaust all possibilities, the lemma is proven.

Solution of the problem for arbitrary $n$

Let the numbers 1, …, $n^2$ be arranged in the table $n \times n$. Let us choose the number $m$ in accordance with the lemma. Let’s mark the cells containing 1, …, $m$, and put asterisks in those of the remaining cells that are adjacent to the removed ones. By the lemma, the asterisks turn out to be no less than $n$. All numbers in cells with asterisks are greater than $m$. This means that most of them are $a \geq m + n$. A cell containing the number $a$ is adjacent to some offset cell containing the number $b \leq m$. Since $a \geq m + n$, $b \leq m$, then $a - b \geq n$. So, we indicated two adjacent numbers $a$ and $b$, the difference between which is greater than or equal to $n$. Thus, our problem is completely solved.

Exercises

1. In a rectangular table $m \times n$ $mn$ are randomly arranged in pairs of distinct integers. Accurately estimate the maximum difference between adjacent numbers.

2. Above, we called two table cells adjacent if they have a common side. Let’s give a new definition of “neighborhood”: we will now call two cells neighboring if they have either a common side or a common vertex (previously, a cell that was not on the edge of the table had four neighbors, now it has eight). How will the estimate of the maximum difference between neighboring numbers change with a new understanding of the term “neighboring”?

3. The last question does not apply to all table problems, but only to the lemma.

In the table, $n \times n$ are offset by $\frac{n(n - 1)}{2}$ cells. Those of the remaining cells that have common sides with the removed ones are given an asterisk. Is it true that there are at least $n$ stars?

(End. See page 8 for beginning.)

But it was still a colossal work, especially considering the state of mathematics at that time. It was necessary to solve the second part of the problem: the orbits are known from observations, now it is necessary to find their laws based on the results of experiments. That is, make a definition of an assumption about the form of orbital curves, and then test it on a huge amount of empirical material! If the results do not coincide, then come up with a new hypothesis and check again! After countless attempts, Kepler came to the following conclusion: the orbit is an ellipse, at one of the foci of which the Sun is located. He also found a law according to which the speed changes over the course of one year: the segment Sun - planet describes equal areas in equal periods of time. Finally, he found that the squares of the revolution time are related to the cubes of the axes of ellipses. Kepler’s entire life was spent solving these problems.

To admiration for this wonderful man is added a feeling of admiration and awe, but this does not relate to the person, but to the mysterious harmony of nature that gave birth to us. Even in ancient times, people came up with curves that correspond to the simplest laws. Along with the straight line and the circle, they included the ellipse and hyperbola. We see the latter implemented in the orbits of celestial bodies, at least with a good approximation.

It seems that the human mind must freely formulate hypotheses before their actual existence is confirmed. The remarkable work of Kepler’s life shows especially clearly that knowledge cannot grow out of the minds of practice. Such a calculation is only possible by comparing what is invented with what is observed.

28

Vortex rings

The questions prepared for the lecture on vortex rings revealed interesting changes.

A typical box for demonstrating good vortices is well known and does not require a detailed description*). Our device is significantly larger than those commonly used. This is a Cubic wooden box about a meter long on the side; one of the walls is made of thin oilcloth, freely suspended, with two diagonals of rubber tubes, firmly attached at the corners. Rubber tubes are needed to ensure that the oilcloth returns to its original position.

Such a box throws out air vortices of great force, and the impact of the ring on the wall of the lecture hall is distinctly loud and similar to the sound of a light blow with a towel. The audience can get an idea of ​​the “hardness” of the rotating air whirlwind by successively releasing invisible pegs into the hall. The blow of a peg in a person’s face is mistaken like a soft push from a down pillow.

In order to make the wells visible, you need to fill the box

*) See the article by N. E. Zhukovsky “Fundamentals of the theory of vortices” in “Kvanit” No. 4, 1971.

I mix ammonia and hydrogen flakes using rubber tubes connected to two flasks in which NH₄OH and HCl are boiling. This method gives good results. Figure 1 was made with photographs of large rings obtained in this way. The side view is interesting: it shows the tail (similar to the tail of a comet), which is formed due to the friction of the outer sections of the ring with the atmosphere as it moves forward.

The power of air rings can be shown in this way. For example, place them on a flat cardboard box, standing at some distance from the installation. In this case, the box immediately turns over or even falls to the floor. By hitting the vortex ring you can extinguish the flame of a gas burner. After some training, you can learn to release two rings quickly in a burst, and so that the second one flies with a few

29

faster than the first. Then the second ring catches up with the first, moves away from it and bounces off; both rings remain intact and turn into vibrating ellipses. This shows that the gas outlet is elastic. All the experiments that I have done with two boxes have produced unsatisfactory results.

Although the large rings produced by the setup described are best suited for lecture demonstration, I find that much more beautiful and symmetrical rings can be obtained by blowing smoke from a paper or glass tube with a diameter of 2.5 cm*). If you look from the side at the blown rings in still air near a lamp or in sunlight, you can see very beautiful spiral streamlines. I managed to form one of these rings as follows. A momentary shutter was mounted on the door of a dark room, and an arc lamp was focused onto its slit using a large concave mirror. The photographic plastic was installed in a dark room so that it was illuminated by a diverging beam of rays coming from the reflection of the arc in the mirror (when the shutter was open). A red lamp was placed in front of the plate, and then the rings were blown out of the tube. As soon as the ring, symmetrical in shape and not moving too quickly, appeared in front of the plate, we held the cord leading to the shutter, and the plate was illuminated by a blinding flash. A clear darkness fell from the ring due to the small size and separation of the light source. Figure 2 taken from the resulting photograph. It can be seen that the ring consists of a layer of smoke and a layer of air, forming a spiral of several terminal turns.

Apparently, the angular velocity of rotation increases as

*) Smoke can be produced, for example, by placing smoldering paper in a closed box.

30

prices with dotted lines. This experiment does not work on a large scale, although I have had some success by spraying ammonia at the upper edge of the hole with a yggago-shaped iron wire heated by a current.

By taking some precautions, it is possible to obtain a smoke half-ring such as in figure 4. This is a brilliant illustration of the fact that the formation of a ring is not dependent on the presence of smoke. The best way to get a half ring is to blow the smoke very lightly into a paper tube, letting it flow down the bottom of the tube until it reaches the end. Then the colcesia is pushed out with a light exhalation. It may be better to conduct the experiment in a large erosive tube with a hole at the bottom, since

Fig. 4.

in this case, one can observe the phenomena occurring inside. It’s quite easy to get a koltsezu, in which the flash part of the smoke is concentrated in the lower half; but obtaining a ring, one half of which is entirely invisible, and such that the edge of the smoke is sharply judged (as shown in figure 4), requires much practice. I tried various schemes to produce these half rings on a large scale, but none of them produced results worth mentioning. It seemed that the use of a hot wire with ammonia was the most promising method, but I did not get a sharply assessed smoke limit, and this is what distinguishes the small rings obtained using a pipe.

Explaining the formation of output rings, the rotational movement is often

Fig. 5.

attributed to friction between the flowing air streams and the edge of the hole. However, most of the formation of outlets is due to the interaction with atmospheric air. To illustrate this point, I have come up with an outlet box in which the trenne with the edge of the hole is absent or, more correctly, is compensated by equalizing it over the entire cross-section of the exiting jet.

Approximately 200 holes with a diameter of 1.7 mm (Fig. 5) are drilled in the bottom of a cylindrical tin box. If the box is filled with smoke and a strong stream of air is released, a red outlet ring is separated from the surface, which is burnt on the sieve. You can simply cover the end of a paper tube with a piece of tightly stretched linen cloth and blow a smoke ring through it.

When experimenting with a box equipped with two round holes, I observe the merging of two rings moving side by side into one large ring. If the rings rotate at a high speed, they will bounce off each other, but if the rings are rotated slowly, they will connect. At the moment of connection, the emerging form is unstable. The connected rings abruptly change their horizontal position to a vertical position so quickly that it is difficult to notice, and then slowly lose the shape of the ring. The same can be demonstrated by using two paper tubes, holding them at different corners of the mouth and almost parallel to each other. In any case, the air in the room should be practically motionless.

Stramy was published in the journal “Nature” in 1901. The publication was prepared by L. A. Savina

31

TASKS

M116. Prove that if you connect the midpoints of successive sides of a convex $n$-gon $M$ (Fig. 1), then the resulting polygon
a) perimeter is not less than half of perimeter $M$ ($n \geqslant 3$);
6) the area is at least half the area $M$ ($n \geqslant 4$).
G. A. Galperin

Fig. 1.

M117. Several people observe the snail for $t$ minutes. Each one is observed for approximately 1 minute and then, during this minute, the snail crawls approximately 1 meter. At no point in time was the snail left unobserved. What kind of path could she have crawled in these $t$ minutes?

Try solving this problem first for small values ​​of $t$, such as $t=2,5$.

N. N. Konstantinov

M118. With four sides of a chessboard of size $n \times n$, a border two fields wide is constructed.

Prove that this border can be bypassed by a chess knight by raising on each square once and only once, if and only if $n-1$ is a multiple of 4.

Y. I. and Y. Y. Sorkinyi

M119. Prove that if you select a point on each face of a convex polygon and draw from this point a vector that is directed perpendicular to the corresponding face to the outside and whose length is equal to the area of ​​this face, then the sum of such vectors will be equal to zero.

N. B. Vasiliev

M120. In some sets, the operation * is introduced, which assigns each two elements $a$ and $b$ of this set an element $a*b$ from this set. It is known that

1. for any three elements $a$, $b$ and $c$
   $a*(b*c) = b*(c*a)$;
2. if $a*b = a*c$, then $b=c$;
3. if $a*c = b*c$, then $a=b$.
   Prove that the operation *
   a) commutative, that is, for any two elements $a$ and $b$
   $a*b = b*a$;
4. associative, that is, for any three elements $a$, $b$ and $c$
   $(a*b)*c = a*(b*c)$.

S. A. Yanovskaya

32

Fig. 2.

F128. A thin homogeneous stick is classified so that it rests on two planes inclined to the horizon at angles α and β; the angle between the planes is 90° (Fig.2). What will happen to the wand? What will be its final position if the friction between the stick and the planes is very small?

B. B. Bukhovtsev

F129. A body located at point A inside a stationary hollow sphere. In what case does it reach the lower point B of the sphere more quickly: if it slides along the surface of the sphere or if it slides along straight AB? Friction in both cases is negligible, the initial velocity of the body is zero and the distance AB is much less than the radius of the sphere.

G. L. Komkin

F130. Inside the smooth sphere there is the smallest loaded ball. What is the size of the contamination?

placed at the bottom point of the sphere so that the ball is held at its top point? The diameter of the sphere is d, the charge of the ball is q, and its mass is m.

G. L. Komkin

F131. A light glass tube of length l and cross-section s, filled entirely with rue and sealed at one end, is located horizontally in a tank with rue near the surface of the rue. What minimum work must be done in order to move the tube to a vertical position in which its open end will touch the surface of the rut?

F132. A cylinder of mass m was untwisted and placed between two closed planes located at an angle of 2α to each other (Fig. 3). Knowing that the coefficient of friction between the cylinder and the planes is equal to k, determine the forces with which the cylinder acts on the walls.

Fig. 3.

34

Note that in more modern algebra textbooks the equality of polynomials is usually understood as the equality of the corresponding coefficients (after reduction to the standard form $a_0 + a_1x + a_2x^2 + \dots + a_nx^n$). If all the equalities in our problem are understood this way, then in the transition from (3) to (4) the word “obviously” must be replaced with the word “by definition”. We will devote separate notes in «Kvant» next year to a detailed discussion of what a polynomial is and how to accurately define various operations on polynomials.

A. L. Toom

M75

a) Prove that in any convex polyhedron the sum of the lengths of all edges is greater than triple the diameter. (The diameter of a polyhedron is called the smallest possible segment with ends at the vertices of the polyhedron).

b) Prove that for any two vertices $A$ and $B$ of a convex polyhedron there are three broken lines, each of which goes along the edges of the polyhedron from $A$ and $B$ and no two go along the same edge.

b) Prove that if two edges are cut in a convex polyhedron, then for any two of its vertices $A$ in $B$ there is a broken line, and a line from $A$ to $B$ along the remaining edges.

d) Prove that in problem 6) it is possible to choose three broken lines that pairwise do not have common vertices, with the exception of the ends $A$ and $B$.

Obviously, problem d) implies problem 6), and problem 6) implies c) and a)*. We will sketch short proofs of a) and b) and solve d).

a) Let $AB$ be the diameter of $M$. Let’s project all edges $M$ onto $AB$. Let’s see how many points are projected into some point $C$ lying inside $AB$. If through $C$ we draw the plane $P$ perpendicular to $AB$, then $P$ will intersect $AB$ along the convex polygon. The vertices of this polygon (and there are at least three of them!) are the intersection points $P$ with the edges $M$ and, therefore, at least three edges $M$ are projected into each internal point of the segment $AB$. From here it is easy to derive the statement of problem a).

When solving problem b) and d) we will use the following intuitively obvious statement: for any two vertices $A$ and $B$ of the convex polyhedron $M$ there is a path from $A$ to $B$ along the edges $M$. We leave the proof to the reader.

*) We also note that using the Ford-aFalerson theorem on the maximum flow and minimum cut (see «Kvant» No. 4, 1970, p. 24, problem 4) it is easy to derive 6); it is enough to apply this theorem to the network formed by the edges of a polyhedron (which edge is counted twice—“there” and “back”).

b) Let the edges $r_1$ and $r_2$ be cut. Obviously, there is a path $L_1$ connecting the ends of $r_1$ and not passing through $r_2$. Indeed, two faces are attached to the edge $r_1$. Edge $r_2$ does not lie on at least one of these faces, so $L_1$ can be taken to be a path along its boundary. Similarly, for the edge $r_2$ there is a path $L_2$. Let us now consider two arbitrary vertices $A$ and $B$. As mentioned above, there is a path $L$, and a path from $A$ to $B$ along the edges $M$. By replacing the edge $r_1$ in the path $L$ with the path $L_1$, and the edge $r_2$ with the path $L_2$ (if, of course, these edges meet in $L$), we obtain the desired path along the remaining edges. Note that this path may have self-computations, which, however, are not difficult to establish.

d) As was said above, for any two vertices $A$, $B$ of a convex polyhedron there is a path from $A$ to $B$, and a path along the edges. Let’s call the length of such a path the number of edges of which it consists, and call it the distance between $A$ and $B$ - let’s denote it by $d(A,B)$ - the length of the shortest (by the number of edges) path. To prove the statement of the problem, we fix the point $A$ and will carry out induction on the distance between the points $A$ and $B$. If $d(A,B) = 1$, then $A$ and $B$ lie on the same edge, and the statement is obvious, since one can go from $A$ to $B$ along the edge $AB$ and the faces of two faces adjacent to $AB$. Let us assume that the statement has been proven for all vertices lying from $A$ at a distance $n-1$. Let $d(A,B) = n-1$ and $d(B',B) = 1$. Let’s consider all faces containing $B'$. The set of all edges belonging to all such edges consists of “radushkas” (edges with endpoint $B'$) and a “ring” (which forms the remaining edges). However, one of the vertices of the ring is the point $B$. By the induction hypothesis, there is a path $l_1, l_2, l_3$ from $A$ to $B'$ that has no pairwise common points except the ends. Obviously, each of these three paths passes through at least one vertex of the ring. Such vertices of the ring are called vertices $A_1$ and $A_2$ (one of them may be vertex $B$). There are two possible cases.

Case 1. Occupy vertices $A_1$ and $A_2$ belong to different paths, say $l_1$ and $l_2$. Then we construct the path from $A$ to $B$ as follows. The first path: from $A$ to $A_1$ to $l_1$, then from $A_1$ to $B$ along the ring (note that by constructing the point $A_1$ in the last section we do not intersect with paths $l_2$ and $l_3$. The second path: from $A$ to $A_2$ to $l_2$, then from $A_2$ to $B$ along the ring. The third path: from $A$ to $B'$ to $l_3$, then along the edge $B'B$. Obviously, the three paths constructed in pairs have no common points except the ends.

Case 2. Occupy vertices $A_1$ and $A_2$ belong to the same path, say $l_1$.

35

Let us then determine which of these vertices will be equal when moving from $A$ to $B^1$ along $l_1$ and take the segment $l'_1$ of the path $l_1$ from $A$ to the first of these vertices. After this, the distance is three paths $l'_1, l_2, l_3$, the distance of the set of vertices of the bulb occupied by path $l'_1, l_2, l_3$ and we will again carry out the previous calculations in relation to these tramuts. If we encounter case 1, then we will construct the required paths in the manner described above, but if we encounter case 2 again, then we will construct the distance of the new triple path $l'_1, l_2, l_3$. If case 2 arises for this triple, then we will distance a new triple, etc. Since we are moving further along the column and there are vertices on the column that belong to the paths $l_2$ and $l_3$, then at some step we will get case 1 and the required paths from $A$ to $B$ will be constructed. Problem d) solution. An even stronger statement is true: for any two vertices $A$ and $B$ one can find three chains of faces that do not have common faces and in each of which the first face contains the vertex $A$, the last one contains the vertex $B$, and neighboring faces have a common edge.

A. G. Kushiirenko

M76

In a company of $n$ people, every two strangers have two mutual acquaintances, and every two acquaintances there are no more mutual acquaintances. Prove that in this company each person knows the same number of people and has a common edge.

For each person, $X$ will be denoted by $m_X$ the number of his acquaintances.

Let $A$ and $B$ be familiar (Fig. 1). Let us denote by $M_A$ the set of acquaintances, and $A, M_B$ is the set.

Fig. 1.

The cruelty of acquaintances $B$. Let’s prove that

(1) $M_A$ and $M_B$ do not have common elements,
(2) each of $M_A$ has exactly one acquaintance in $M_B$ and
(3) each of $M_B$ has exactly one acquaintance in $M_A$.

Statement (1) immediately follows from the condition: $A$ and $B$ cannot have common acquaintances. Let’s prove (2); (3) proves similarly. Let $X \in M_A$ (this notation means “$X$ belongs to the set $M_A$”). Or $X$ is $B$, and then, according to (1), he has only one acquaintance in $M_B$ - $A$ himself. Or $X \neq B$, then, according to (1), $X$ is unfamiliar with $B$, and, therefore, $X$ and $B$ must have, in addition to $A$, exactly one common acquaintance, that is, $X$ has exactly one acquaintance in $M_B$, which is what needs to be proven. From (2) and (3) it follows that a one-to-one correspondence can be established between the sets $M_A$ and $M_B$, therefore $m_A = m_B$.

So, if $A$ and $B$ are familiar, then $m_A = m_B$. If $A$ and $B$ are unfamiliar, then they have a common friend $C$, and, therefore, $m_A = m_B$.

$A$ sent us much clearer solutions of this type. Zinchenko from Arkhangelsk, V. Lozinov and A. Slinkin from Moscow, A. Karimov from Ufa, K. Radyunov from Irkutsk, V. Pronin from Odessa, V. Faimelevy from Armavir, S. Gritsoryan from Baku.

Many readers have solved the inchete problem. For each person $A$ let $M_A$ denote the set of his acquaintances, and $N_A$ the set of those not acquainted with $A$. Then each element from $N_A$ can be associated with a pair of elements from $M_A$ (those with whom he is familiar), and it is not difficult to prove that this correspondence between the set $N_A$ and the set of all possible pairs of elements $M_A$ will be one-to-one. Consequently, if $M_A$ contains $m_A$ elements, then $N_A$ contains $\frac{m_A(m_A-1)}{2}$, and in total there are $n = 1 + m_A + \frac{m_A(m_A-1)}{2}$ people in the company. This equality is true for any person $A$. But the equation

$$n = 1 + x + \frac{x(x-1)}{2}$$

has (at $n > 1$) only one positive root, therefore $m_A$ is the same for all $A$. (A similar solution is analyzed in A. Leman’s book “Collection of Moscow Mathematical Optimization Problems”, problem XXIII, 11, 10, 3).

It is interesting to find out at what $n$ such companies actually exist. We do not know the complete answer to this question. We have seen that $n = \frac{m(m+1)}{2} + 1$, where $m$ is the number of acquaintances of one person, but not for any such $n$ company is essential. For example, it is not difficult to prove

37

Fig. 6.

This answer was added by O. Zaumyslov (Zelenograd, Moscow region), A. Zeleny (Izmanl, Odessa region), V. Rozhkin (Perm), N. Fedin (Omsk), A. Demidov (Moscow), A. Zhukov (Kirovsk, Khariza district, Donetsk region), V. Kovtun (Kyiv), A. Vorobyov (Torzhok Kalininskaya region).

F85

Is it possible to check a closed process (cycle) with an ideal gas so that points A and B (Fig. 7) lie on the same isotherm?

The temperatures $T_1$ and $T_2$, which correspond to the isotherms drawn at figure, and the volume $V_1$ (or $V_3$) are given. Temperatures $T_1$ and $T_2$ define two isotherms:

$$pV = \frac{m}{\mu}RT_1 \quad \text{and} \quad pV = \frac{m}{\mu}RT_2$$

(m is the mass of the gas, μ is the mass of one gram molecule of it, R is the gas constant). Since the volume $V_1$ is given, we know points C and E of the cycle: volume $V_1$ and $V_3$ ($V_3 = V_1 \frac{T_2}{T_1}$), temperatures $T_1$ and $T_2$, and pressure $p_2 = \frac{m}{\mu}RT_1V_1$.

All other points in the cycle can be determined.

Fig. 7.

View only when we specify one of the points - B or A.

We will assume that we have a given volume $V_2$. It defines point B ($P_B = P_2$, $V_B = V_2$, $T_B = T_1 \frac{V_2}{V_1}$) and point D ($V_D = V_2$, $T_D = T_2$, $p_D = p_3 = \frac{m}{\mu} \frac{R T_2}{V_2}$). Point A lies on the isochore AD and on the isobar AC. Therefore $p_A = p_3 = \frac{m}{\mu} \frac{R T_2}{V_2}$, $V_A = V_1$ and $T_A = T_2$, $\frac{V_1}{V_2}$.

The gas temperatures at points A and B are the same if $V_2$ is chosen so that

$$T_2 \frac{V_1}{V_2} = T_1 \frac{V_2}{V_1},$$

that is, if $V_2 = V_1 \sqrt{\frac{T_2}{T_1}}$. The gas temperature at points A and B of the cycle in this case is equal to $T = \frac{p_2V_2\mu}{mR} = \sqrt{T_1T_2}$ (since $p_2V_2 = p_2V_1 \sqrt{\frac{T_2}{T_1}}$ and $p_2V_1 = \frac{m}{\mu} RT_1$).

The correct solution is the addition of N. Fedin (Omsk).

F86

In Millikan’s experiment, the position of drops with mass $10^{-5}$ e is determined with an accuracy of 10 μm. What is the uncertainty in determining its speed?

Due to the uncertainty principle $\Delta p \Delta x \geq h$. Therefore $\Delta v \geq \frac{h}{m \Delta x}$. That is, the speed of the drop cannot be changed with an accuracy greater than $10^{-16}$ cm/sec.

This answer was added by B. Glukhovsky (Sverdlovsk), B. Kovtun (Kyiv) and A. Zlomnik (Zhukovsky, Moscow region).

I. Sh. SLOBODECHY

F87

Source of spherical waves of motion on water surfaces with speed u. Draw pictures of waves on surface water when the wave speed v is greater and less than the speed of the source.

Waves generated on surface water by a stationary source; have the form of concentric circles (Fig. 8, a). If the source is moving, then the waves that it generates at later moments in time are mixed in the direction of movement of the source. Let the speed of the source be less than the speed of the waves. Then the wave front will surround the source, and the wave pattern will be as shown in figure 8, b. However, if the speed of the source is greater than the speed of wave propagation, then the wave front can no longer

38

Fig. 8.

can overtake the source, and the waves will travel inside the angle formed by two rays emanating from the source (Fig. 8, c). The magnitude of the angle is easily found from figure. During the time during which the source travels the distance AO, the front of the wave emitted at point A travels the distance AB, and $\frac{AB}{AO} = \frac{v}{u}$. That’s why

$$\sin \alpha = \frac{v}{u}.$$

The correct solution was sent by A. Zlomnik (Zhukovsky, Moscow region), V. Kovtun (Kyiv), N. Fedin (Omsk), A. Demidov (Moscow), A. Zeleny (Izmail, Odessa region), O. Zamyslova (Zelenograd, Moscow region), G. Zaitsev (Gagra), V. Belov (Vologda), N. Sidelev (Tsiropinsk, Kherson region), O. Trunov (Jalal-Abad Kyrgyzstan, USSR), G. Furman (Chernivtsi, Ukrainian SSR), S. Araslanova (Nikhne-Ivkino, Kumensky district, Kirov region).

L. G. Aslamaov

F88

A vessel filled with sand moves with horizontal acceleration, as a result of which the surface of the sand is inclined to the horizon at a certain angle. The top of the vessel is closed with a lid. How will this angle change if the vessel is filled to the top with water?

Consider a fluid element with mass Δm (Fig. 9). Two forces act on it: the force of gravity Δmg and the force N - the resultant of the pressure forces from the surrounding fluid. Under the influence of these forces, the fluid element moves horizontally with acceleration a. This means that the resultant forces Δmg and N are directed horizontally and equal to Δma. For this to be possible, the force N must make an angle α with the vertical such that tanα = $\frac{a}{g}$.

These considerations do not depend on where exactly the liquid element is isolated. Let’s choose

Fig. 9.

it on the surface. Then it is clear that it is in equilibrium relative to the vessel and water only if the surface of the liquid is perpendicular to the force N: otherwise the component of the force N, parallel to the surface of the liquid, will cause the element of liquid we have identified to move along the surface. So, the surface of the liquid in a moving vessel is inclined to the horizon at an angle α = arctan $\frac{a}{g}$.

Let now a liquid element with mass Δm be inside the liquid. Let’s assume that we placed together a body with the same volume, but with a density less than the density of water. It is clear that in this case our body will float in the direction of the force N. It follows that, unlike a vessel at rest, the lines of constant pressure are not horizontal, but make an angle α with the horizon. Obviously, the constant pressure in the rode that we need in the vessel will be the same. Therefore, it is clear that the angle of inclination of the liquid interface will be the same as the angle of inclination of the surface of the surface before adding water to the vessel: α = arctg $\frac{a}{g}$.

This is natural. The observer, who is on the cart, will say that the liquid behaves as if the vessel would end, and the forces and gravitational acceleration g’ in its coordinate system are directed at an angle α to the vertical. In a vessel at rest, the fluid interface is perpendicular to the force of the fluid, regardless of the shape of the vessel.

The correct solution was sent by A. Aleksandrov (Glazov, UASSR), V. Belov (Vologda), Y. Intin (Rechnitsa, Gomel region, VSSR), V. Kovalev (Kommunarsk, Voroshilgovgrad region), A. Karimov (Ufa), S. Araslova (Nikhne-Ivkino, Kumensky district, Kirov region), A. Demidov (Moscow), N. Fedin (Omsk), E. Dolev (Moscow), L. Bratsinsky (Frunze).

I. Sh. Slobodetsky

39

Problems for the oral exam in physics at the Leningrad Polytechnic Institute

M. M. Kozlov, I. G. Kochnev

The oral exam in physics consists of answering theory and solving problems. We will not touch upon the answers received to theoretical questions. This is an important topic and certainly deserves special consideration.

The task in the entrance exam represents a specific goal - to check how deeply and informally the examinee understands the existence of physical phenomena. A simple example. If an applicant formulates Newton’s laws, but does not have the ability to use these laws when solving a problem, then his theoretical answer, of course, provides. Thus, the task plays a significant role in assessing knowledge.

The applicant must be able to analyze a problem, choose the optimal way to solve it, justify it and correctly perform the technical part of the task.

Below are a number of problems that have appeared in the last 3 years in the entrance exams of the LPI. When solving problems, special attention should be paid to some “difficult” questions and typical mistakes that are typical for applicants.

In mechanics, with the exception of purely kinematic problems, the most challenging problems for applicants are those on the work of a variable force, the friction force, and the calculation of the centripetal force. Many people do not have a clear idea that the centripetal force is either the resultant (that is, equal) of all forces acting on the body - equally

measured motion along a circle, or, in the general case of linear motion, is the projection of the resulting force onto the radius of curvature. Those who consider the centripetal force to be a force of a special nature cannot solve even the simplest problems involving linear motion, since they cannot use Newton’s II law when solving it.

When solving problems on the molecular-kinetic theory, in particular on gas laws, the most difficulties were caused by problems that required some transformations of the unified gas law, and tasks in which it was necessary to move from one unit of measurement to others (for example, from mm rm. cm to amp).

In the “Electrostatics” section, the largest number of errors were associated with the inability to perform numerical calculations. Not all applicants have sufficient skills in applying the concepts and formulas of electrostatics to solve the problem of the movement of individual contaminants (for example, electrons) in an electrostatic field. Quite difficult for applicants are tasks in which they need to find an equal electrostatic force and other forces.

An important place among the exam problems is occupied by problems on constructed electric current. The most difficult tasks for applicants are those in which it is necessary to calculate a section of a circuit with an emf source, for example, with a charging battery. They don’t always do the right thing

40

calculation for a circuit in which electrical energy is converted into other types of energy, in particular mechanical energy.

Difficulties that arise in such problems can be easily overcome based on the law of conservation and transformation of energy (see problem 6).

Analysis of solutions to problems in optics shows that the most difficult application of the lens formula is in the calculations of optical systems consisting of not one, but two lenses. In such problems, compliance with the rules of signs in the lens formula becomes especially important (task 7).

Let us now present several problems from the exam papers.

1. A small ball of mass $m$, carrying a charge $q$, is suspended on a thread of length $l$. The ball moves along a circle lying in a horizontal plane, and a second, stationary ball carrying the same charge is fixed in the center of the circle. Find the angular velocity of the ball.

2. The air pressure inside the bottle, tightly sealed with a stopper, at a temperature of $7^\circ$ C was equal to 1 amm. To what temperature must the air in the bottle be heated in order for the cork to fly out of it, if it is known that in order to remove the cork from a cold bottle at $7^\circ$ C, it is necessary to do 1 J of work. The length of the plug is $l = 4$ amm, the section of the plug is $s = 4$ amm².

3. When manufacturing pressure increased by 0.5 mm, the gas density changed by 0.5 kD/m³. What was the initial density of the gas if the initial pressure was 1 amm?

4. A metal ball with a volume of 1 cm³ has a mass of 8 g and contains 8.6 x 10²² of free electricians. What part of these electrics must be removed from each of two such balls so that the layer of electrostatic repulsion completely compensates for the force of gravitational attraction between them ($\gamma = 6,67·10^{-11} \cdot \text{н} \cdot \text{m}^2/\text{к}^2$)?

5. The winding of the electric motor of the constructed current is made of wire with a total resistance of $R = 0,2$ ohm. When switched on with voltage $U = 120$v, current $I = 30$a flows through the winding of a running motor. Find the mechanical power, develop the electric motor.

6. Change of current in the oscillatory counterretor of the electrinite wave.

$I = 0,3 \sin 15,7 t$.

Find the length of the electrinite wave emitted by the circuit.

7. The rounder consists of two identical thin biconvex lenses having a focal length of 5 cm and located at a distance of 2.5 cm from each other. Find the distance between the focus of the eyepiece and the lens close to it.

8. Find the force acting at the point of attachment of the movable block to the ceiling if a load of mass 60 kg suspended from the block shown at figure is supported by the weight of a person who pulls a vertical downward rope.

9. How much work must be done to place three identical, located at infinity in the rest frame, exact charges of magnitude $q = 5 \cdot 10^{-8}$ k at the vertices of an equilateral triangle on a side of 5 cm ($\varepsilon = 1$)?

10. The potential difference at the battery terminals is 8.5 V, while a current of 3 A flows through the battery. If you change the direction of the current to the opposite and the magnitude of the current is reduced by 1 A, then the absolute value of the potential difference will increase by 2.5 V. Find Ed. With. and the internal resistance of the battery.

11. Two electric heaters, labeled 600 W, 220 W and 400 W, 220 W, are connected in series and connected to a 220 W network. How much electrical power is consumed by each heater?

12. At a distance of 15 cm from the center of the silvered Christmas tree ball there is a Christmas tree tree 1 cm long. The diameter of the ball is 8 cm, and the tree is located perpendicular to the symmetry axis of the ball passing through the middle of the tree. Find at what distance from the center of the ball the image of the tree is located and determine the size of the image.

41

The art of solving geometric problems is somewhat reminiscent of trio or lozonists - sometimes, even knowing the solution to a problem, it is difficult to understand how one could come up with it. Almost every somewhat difficult geometric problem requires an individual approach. But I would like to have a general solution method, which would also allow the solution to be “adventured” to the end without carrying out calculations explicitly.

Unfortunately, there is no such universal method. But there are techniques applied to many problems. We will consider one of them.

Let’s start with a very simple problem, which we will solve in two ways.

Fig. 1. AC = 2, BC = 3, ⇒ ACD = ⇒ DCB = = 45°.

Problem 1. Find the bicsectrix CD of a right triangle ACB with legs AC = 2 and BC = 3 (Fig. 1).

Solution I. By Pythagorean theorem AB = √AC² + BC² = √13. By the property of the bicsectrix of the angle of a triangle inside it

$$\frac{AD}{BD} = \frac{AC}{BC}.$$

From the equality AD + DB = √13 we obtain AD = \frac{2}{5} √13. Next, it’s easy to find the cosine of angle A:

$$\cos A = \frac{AC}{AB} = \frac{2}{\sqrt{13}}.$$

It remains to apply the cosine theorem to determine side CD of triangle ACD:

$$CD^2 = AC^2 + AD^2 - 2AC \cdot AD \cos A = \frac{72}{25}, CD = \frac{6}{5} \sqrt{2}.$$

Solution II. Let’s calculate the area of ​​triangle ACB in different ways. On the one hand, S△ACB = \frac{1}{2} AC \cdot BC = 3. On the other hand,

$$S△ACB = S△ACD + S△DCB = \frac{1}{2} AC \cdot CD \sin 45^\circ + \frac{1}{2} BC \cdot CD \sin 45^\circ = \frac{5 \sqrt{2}}{4} CD.$$

42

Equating the part of the two given expressions for $S_{\Delta ACB}$, we obtain $CD = \frac{6}{5}\sqrt{2}$.

The second solution is clearly shorter and simpler. Moreover, it easily leads to the goal in a more general situation. To verify this, find out for yourself how both of these solutions change if

Task 2. Find the bisector $CD$ of $C$ of triangle $ACB$ if $AC=b$, $BC=a$, $\Rightarrow ACB=\alpha$. (Say: $CD = \frac{2ab}{a+b}\cos \frac{\alpha}{2}$)

However, the main advantage of the second solution is its ideological “originality”. In the first solution, a certain “chain” of triangle elements is constructed: each element of this chain is calculated based on the problem data and other elements already found. The last element of the chain is the desired quantity (in this case, the bisector $CD$). At the same time, it is not immediately clear which elements will be needed when determining the desired value, that is, which elements must be pre-calculated.

In the second solution for the quantity that needs to be found, we obtain an equation by writing out expressions for the area of ​​the triangle in terms of the known and desired elements of the triangle. This example - comparing different expressions for the area of ​​a triangle - turns out to be very fruitful. Why specifically for the area? The fact is that the plane of a triangle is quite simply expressed through various combinations of elements of this triangle.

Before giving the following examples, let us recall the most important of these expressions. First, let’s agree on notation. In triangle $ABC$:

$a, b, c$ - sides, opposite angles $A, B$ and $C$ respectively;

$h_a, h_b, h_c$ — heights, $m_a, m_b, m_c$ — medians, $l_a, l_b, l_c$ — bisectors, respectively;

$r$ — radius of the inscribed circle;

$R$—radius of the circumscribed circle;

$p$ - half-perimeter;

$S$ are planar triangles $ABC$.

Fig. 2. About the center of the inscribed circle.

The formulas that we will use are written in this notation as follows:

$$S = \frac{ah_a}{2},$$

$$S = \frac{ab \sin C}{2},$$

$$S = \frac{a^2 \sin B \sin C}{2 \sin A},$$

$$S = \frac{abc}{4R},$$

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

$$S = pr.$$

All these relationships are known from the school course, so we will not prove them in detail, but will limit ourselves to just a few comments. To derive formulas (3) and (4) from (2), you can use the sine theorem:

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R.$$

Formula (5) (Heron’s formula) is sometimes more convenient to use in a transformed form:

$$S = \frac{1}{4} \times \sqrt{2a^2b^2 + 2a^2c^2 + 2b^2c^2 - a^4 - b^4 - c^4}.$$

Formula (6) is easily obtained from a very (Fig. 2) equality:

$$S_{\Delta ABC} = S_{\Delta AOB} + S_{\Delta BOC} + S_{\Delta COA}.$$

Let’s now consider a few more problems. In each of them, the application of the area method leads to a short and pronounceable solution.

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Task 3. Inside a regular triangle with sides a, take an arbitrary point M. Find the sum of the distances from this point to the sides of the triangle.

Solution. It is necessary to find the sum of the lengths of the perpendiculars MP, MQ and MR (Fig. 3) dropped from the point M to the sides AB, BC and CA, respectively. If we connect the point M with the vertices of the triangle, then it is clear that $S_{\Delta ABC} = S_{\Delta AMB} + S_{\Delta BMC} + S_{\Delta CMA}$. Using the formula for the area of ​​a regular triangle $S_{\Delta ABC} = \frac{a^2 \sqrt{3}}{4}$ and replacing that (by formula (1)).

$$S_{\Delta AMB} = \frac{1}{2} a \cdot MP,$$

$$S_{\Delta BMC} = \frac{1}{2} a \cdot MQ,$$

$$S_{\Delta CMA} = \frac{1}{2} a \cdot MR,$$

we immediately receive the answer:

$$MP + MQ + MR = \frac{a \sqrt{3}}{2}.$$

Task 4. Line AD divides triangle ABC into two. Prove that the radius r of the circle inscribed in triangle ABC is less than the sum of the radii $r_1$ and $r_2$ of the circle inscribed in triangles ABD and ACD, respectively.

Solution. Let CD and CE be the rays that are discussed in the conditions of the task (Fig. 5). Obviously $S_{\Delta ABC} = S_{\Delta ACD} + S_{\Delta DCB} = S_{\Delta ACE} + S_{\Delta ECB}$.

Let us use formula (2), taking into account that $ACD = DCE = ECB = \frac{\alpha}{3}$:

$$\frac{AC \cdot DC \sin \frac{\alpha}{3}}{2} + \frac{DC \cdot BC \sin \frac{2\alpha}{3}}{2} = \frac{AC \cdot EC \sin \frac{2\alpha}{3}}{2} + \frac{EC \cdot BC \sin \frac{\alpha}{3}}{2}.$$

44

Fig. 5.

From here we immediately get

$$\frac{DC}{EC} = \frac{AC \sin \frac{2\alpha}{3} + BC \sin \frac{\alpha}{3}}{AC \sin \frac{\alpha}{3} + BC \sin \frac{2\alpha}{3}}$$

$$= \frac{2 \cos \frac{\alpha}{3} + 3}{6 \cos \frac{\alpha}{3} + 1}$$

Here is an example of a more difficult problem. Although in it you simply need to “solve the triangle”, that is, determine its sides by three known elements, it is not possible to obtain a solution in the usual way - using the theorems of sines and cosines.

Task 6. Name triangle ABC if $h_a=6$, $r=2$, $R=5$.

Solution and e. Let us use formulas (1), (4), (5) and (6) and write the following system of equations for the unknown sides $a, b, c$

$$\begin{cases} S = 3a, \\ S = \frac{abc}{20}, \\ S = \frac{1}{4} \sqrt{(a+b+c)(b+c-a)} \times \sqrt{(a+c-b)(a+b-c)}, \\ S = a+b+c. \end{cases}$$

Equating the rules of the parts of the first and last equations, we obtain that $b+c=2a$. Equating the rules for parts of the first and second equations, we obtain that $bc=60$. Finally, equating the rules for parts of the first and third equations, we obtain (after squaring) that

$$144a^2 = (a+b+c)(b+c-a) \times (a+c-b) \times (a+b-c) = [(b+c)^2-a^2] \times [a^2-(b-c)^2] = [(b+c)^2-a^2] \times [a^2-(b+c)^2+4bc] = (4a^2-a^2) \times (a^2-4a^2+240) = 3a^2 (240-3a^2).$$

Therefore, to determine $a$ we obtain the equation

$$144a^2 = 3a^2 (240-3a^2),$$

whence $a=8$. From the system of equations $b+c=16$, $bc=60$ it is now easy to find two pairs of values ​​$b$ and $c$ (two answers):

$$a = 8, b = 6, c = 10 или a = 8, b = 10, c = 6,$$

which are the geometric definitions of odin triangle (to the point of designating the sides $b$ and $c$).

In this example, the main feature of the area method is especially clearly visible - it “makes” an algebraic problem out of a geometric problem, reducing everything to solving some systems of equations.

Task 7. (MSU, physicist, 1966) Prove that if the lengths of the sides of a triangle form an arithmetic

45

Fig. 6a.

the metische cut, the center of the circle inscribed in this triangle, and the intersection point of its median lie on a straight line parallel to the middle side of the triangle.

Solution. Let the sides of triangle ABC be shaped so that $a \leq b \leq c$. Then according to the conditions of problem $a+c=2b$. Let $O$ be the center of the incircle and $M$ the intersection point of the medians $AA_1$, $BB_1$ and $CC_1$ (Fig. 6a).

First, we derive one useful property of the point $M$ of intersection of the median of triangle $ABC$. On figure 6b, triangle ABC is divided by medians into 6 smaller triangles. The area of ​​each of them is equal to $\frac{1}{6}$ the area of ​​the entire triangle $ABC$. This is very easy to prove: using

Fig. 66.

formula (1), writing the equality
$$S_{\triangle BB_1C} = \frac{1}{2} S_{\triangle ABC}, B_1M = \frac{1}{3} BB_1,$$
$$S_{\triangle MCB_1} = \frac{1}{3} S_{\triangle BB_1C} = \frac{1}{6} S_{\triangle ABC}.$$

The area of ​​the remaining triangles can be found similarly.

Now let’s find the length of the perpendicular $MM'$ dropped from the point $M$ to the side $AC$:
$$\frac{1}{3} S_{\triangle ABC} = S_{\triangle AMC} = \frac{1}{2} MM' \cdot AC =$$
$$= \frac{1}{2} MM' \cdot b, \frac{1}{3} S_{\triangle ABC} = \frac{1}{3} rp =$$
$$= \frac{1}{3} r \left( \frac{a+b+c}{2} \right) = \frac{1}{3} r \left( \frac{2b+b}{2} \right) =$$
$$= \frac{1}{2} rb,$$

whence $MM'=r$, that is $MO \parallel AC$.

The area method often turns out to be convenient at certain stages of solving stereometric problems.

Task 8. (MSU, mechanism, 1969) A pyramid is inscribed in a ball, the side edges of the side are equal to c. Its base is a rectangular plane, the sides of which are connected by the arcs $\alpha$ and $\beta$ radians in the sections of the ball by the planes of the lateral faces. Determine the radius of the circumscribed sphere.

Solution (Fig. 7). Since the inscribed angle changes by half

46

no arc on which it rests,
That
$$\Rightarrow DQC = \frac{\alpha}{2}, \Rightarrow AQD = \frac{\beta}{2}.$$

Pozny $DC = 2c \sin \frac{\alpha}{4}$, $AD = 2c \sin \frac{\beta}{4}$.

Let us consider the section of the ball by the plane $AQC$. Since the center of the ball lies in this plane (prove it!), the radius of the ball is equal to the radius $R$ of the circle circumscribed about the triangle $AQC$. But $AQ = QC = c$, and $AC = 2c \sqrt{\sin^2 \frac{\alpha}{4} + \sin^2 \frac{\beta}{4}}$ (like the diagonal of a rectangle $ABCD$). Therefore, to determine the radius $R$, it is enough to use the expressions for the area of ​​the triangle $AQC$ using formulas (4) and (5).

As a result of obvious transformations, we get the answer:

$$R = \frac{c}{\sqrt{2 \left( \cos \frac{\alpha}{2} + \cos \frac{\beta}{2} \right)}}.$$

The area method is also convenient to use in problems in which it is necessary to find various types of estimates for geometric quantities.

Problem 9. Within what limits can the ratio of the radius of a circle described in a right triangle to the height lowered by one change?

Solution. If $c$ is the hypotenuse of the right triangle $ABC$, then we can write (see formulas (1) and (6)):

$$\frac{ch_c}{2} = pr, \text{откуда} \frac{r}{h_c} = \frac{c}{2p}.$$

Thus, it is necessary to evaluate the relationship

$$\frac{c}{2p} = \frac{c}{a + b + c} = \frac{1}{\frac{a}{c} + \frac{b}{c} + 1} = \frac{1}{\sin A + \sin B + 1}.$$

Since $A + B = 90^\circ$, then

$$\sin A + \sin B + 1 = \sin A + \cos A + 1 = \sqrt{2} \cos (45^\circ - A) + 1.$$

The acute angle $A$ of a right triangle can be any value between 0 and 90^\circ. Therefore, the angle $45^\circ - A$ is smaller in the range from $-45^\circ$ to $+45^\circ$, as can be easily obtained from the properties of the cosine, while $2 < \sqrt{2} \cos (45^\circ - A) + 1 \leqslant 1 + \sqrt{2}$.

Consequently, in any right-angled triangle, an indefinite solution to many problems is always satisfied. Of course, it does not replace all other methods, but it is an approach that should not be forgotten when starting to solve a particular problem.

Exercises

1. Find the median $m_a$ of the triangle, the value of its side $a, b, c$.
2. Prove that in an arbitrary triangle

$$\frac{1}{r} = \frac{1}{h_a} + \frac{1}{h_b} + \frac{1}{h_c}.$$

3. (MSU, Faculty of Economics, 1969). In an acute triangle $ABC$ given: $AB = c$ drawn from the vertex $B$ median $BD = m$. Angle $BDA$ is acute and equal to $B$. Calculate the area of ​​triangle $ABC$.
4. (MSU, Faculty of Psychology, 1969). Given an equilateral right triangle. Prove that the radius of the circle tangent to the hypotenuse and extensions of the legs is equal to the sum of the hypotenuse and the radius of the circle described in the triangle.
5. It is known that in the triangle $ABC$:

$$S_{\triangle ABC} = 16, m_a = 6, m_b = 4.$$

Prove that the medians $m_a$ and $m_b$ are perpendicular.
6. Prove that the sum of the distances from an arbitrary point taken inside an unequally lateral triangle to the side of this triangle is enclosed between the smallest and lowest point.

7. The base $ABCD$ of the straight line $ABCDA_1B_1C_1D_1$ (where $AA_1, BB_1, CC_1, DD_1$ are the side edges) is a rhombus on the side $a$ and an acute angle $φ$, and the height of the line is $h$. Find the distance from the vertex $B_1$ to the diagonal $A_1D$ of the side face $ADD_1A_1$.

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THE UNTOUCHABLE KING

Previously, we looked at a wide variety of mathematical problems on the chessboard. Now we will focus on problems that can also be classified as mathematical, but their chess content is richer than before. In other words, we will be interested in elements of mathematics that occur unrepresented in the chess game itself.

First, let us recall the feature of “chessboard gamemetry” associated with distances. The fields e1 and a5 on figure 1 are at a “diagonal” distance. The king with e1 can reach square a5 in 4 moves. It turns out that the diagonal path is the shortest possible; it cannot be replaced by any other matriroot.

Now consider two fields located on the same straight line:

fields e1 and e8 on vertical e. Moving along it, king e1 can reach square e8 in 7 moves. It is clear that the king’s maturation could have been different. There are exactly 393 (!) ways to reach square e8 in the same 7 moves. The king can move to the e8 square with the most bizarre zig-zag matrirs, as long as they are within the framework of the next figure 1 piece and as long as the king moves from one rank to the next on each move. For proof

Fig. 1.

you can count the number of matriroutes containing 0, 2, 4 or 6 diagonal moves. Using some information from combinatorics (see “Quant” No. 1, 1971), it is easy to find it:

$$1 + A_7^2 + C_7^2 \cdot C_5^2 + C_7^3 \cdot C_4^3 = 393.$$

So, the movement of the king in a straight line, if necessary, can be replaced by movement along a broken line. Here is a practical illustration of this important consideration.

On figure 2 the a7 pawn is defenseless, and Black’s only chance is to respond to the inevitable K:a7 with Krs7 with a draw, since the white king will not be able to get out of captivity. The white king’s path to the pawn will take 5 moves. However, there is only one winning matrice: 1. Npf7-e6! Npb2—c3 2. Kreb—d5!!, and not-

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Fig. 2. I. Maizelis, 1921 White start and win.

the possibility of 2… Kd4 is disastrous for Black. (Does not work, for example, 1. Kb Krs3 2. Kd6 Kd4 3. Ks6 Krs5! 4. Kpb7 Kd6 5. Kd: a7 Ks7).

Of course, in practice there are more subtle cases.

Mathematicians have long despaired of creating the structure of a mathematical theory of the game of chess and are limiting themselves only to particular questions. However, the study of many chess positions is completely mathematical in nature. Let’s give a few examples, all of them relate to the theory of pawn endings.

In position on figure 3 the white king does not take part in the game.

Fig. 3.

re, and it all depends on whether the black king manages to catch up with the h3 pawn. Inexperienced chess players usually reason like this: a pawn goes here, a king there, a pawn here, a king there… that is, what is called “on the fingers.” At the same time, they usually get confused (especially when there are still some pawns on the board) and miscalculate. It turns out that the outcome of the game is very easy to evaluate using the “square rule”: it is enough to find out whether the king, during his move, can enter the pawn’s “square”, in this case depicted at figure 3: c3—c8—h8—h3. More

Fig. 4.

It’s easier to mentally draw just one line - the diagonal of the square (h3-c8). It follows from this that in the position on figure 3 Black makes a draw on his move (goes into the square), and loses on his opponent’s move.

Let us now consider the position on figure 4. Black immediately loses on his move, since the white king moves to b6 and eats the a6 pawn. Thus, White needs to pass the turn of the move to the opponent. How to do this? After 1. Kd5 Ks8 nothing gives 2. Kd6 (2… Kd8 3. c7 + Ks8 4. Ks6 stalemate!), and 2. Ks5 Ks7 leads to the original position. Gaining tempo is achieved using the so-called “triangle method”. For this case, this triangle (c4-d4-d5) is depicted on the same figure 4.

After 1. Krs5—d5 Krs7—c8 2. Kd5—d4 Ks8—b8 3. Kd4—c4! Kb8-c8 4. Kc4-d5 White achieves his goals, since on 4… Kc8-d8 5. Kd5-d6 now wins, and on 4… Kc8-c7-5. Kd5-c5.

Let us now turn to the position at figure 5a. Although White has two extra pawns here, it is extremely difficult to win. To analyze positions with a blocked pawn structure, various methods are used: “Bianchetti critical distances”, “Ebers coordinate system” and others. The general theory of blocked pawn endings is called the theory of corresponding fields. The study of each specific position can be considered as a solution to a subtle mathematical problem, but a general mathematical theory has not yet been created. Of course, we are not able to remain

49

update in our notes on all the details of the theory of the corresponding fields*), we will limit ourselves to only studying the position of Figure 5a, displacing the results of the analysis to figure 56.

White intends to check either through the square d6 or through f4, the black king must have time to prevent both of these plans. Thus, if the white king gets to c5, then black must meet it on the e7 square (with the queen on d7, black does not have time to defend the g4 pawn: Kpc5-d4-e3-f4). In other words, field c5 corresponds to field e7. If the white king is on f4, black must be in time on h5, that is, the f4 square corresponds to the h5 square.

The d4 square is very important for White. It is easy to see that in this case the king

*) You can read more about this, for example, in the book “Chess Endings”, vol. 1, edited by Yu. Averbakh, M., 1956.

must stand on f7 in order to respond to Kpd4-c5 with Kpf7-e7, and to have a response to Kpd4-e3 with Kpf7-g6. From square c4 White can play either Kpc5 or Kpd4. The black king in this case should be located on f8, from where both Kpe7 (with Kpc4-c5) and Kpf7 (with Kpc4-d4) are possible. From square d3 the possible moves are Kpc4, Kpd4, Kpe3. Therefore, field d3 corresponds to field g7. From square c3 the possible moves are Kpd4, Kpc4, Kpd3. In this case, the only square from where the black king can get to f7, f8, g7 is the g8 square.

By traversing sequentially all the most important squares at the disposal of the white king and the moving matching square for his black counterpart, we obtain figure 56. The corresponding squares here are designated by the same number. Note that the found correspondence is not unique, as can be seen in the figure. For example, squares b4 and d4 for the white king correspond to the same square f7 for black.

Now the finding solution is almost automatic. In this case, you should be guided by only one rule: place the white king on a square that corresponds to the current location of the black king, or on a square that corresponds to which the black king cannot get to in one move. Since the field b1 corresponds to the field g7, the field b2—h7, and the field a2—h8, then only 1. Kpa1—a2!! is solved. Aftermath Kpb1? Kpg7! or 1. Kpb2? Kph7! blacks in random will be added to draws.

Since the further course of the game is simple, we will present only the main option (if Black plays differently, then, as you can easily establish for yourself, he loses even faster):

1. Kpa1—a2!! Kph8—h7
2. Kpa2—b2! Kph7—g7
3. Kpb2—b3! Kpg7—g8
4. Kpb3—c3! Kpg8—f8
5. Kpc3—c4! Kpf8—f7
6. Kpc4—d4! In the opposite direction, shown in Figure 5a, the white king has made the “way up” and black is defenseless.

Obviously, with the help of Figure 56 it is easy to estimate the position for other initial positions of the queens, but only for this stove configuration.

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In 1971, the post office delivered twice as many letters to the editorial office «Kvant» as in the previous year. Their total number in individual places reached 350 and did not fall below 250 letters per month. Of course, this is not so much, but we hope that the number of letters will increase.

The authors of the overwhelming number of letters are, as one would expect, schoolchildren, actively interspersing mathematics and physics or both sciences. Not only older classics, but also 7th and 8th grade students squeal. In task «Kvant» for younger schoolchildren, elementary school students answer. There is even a letter from the debtor, written by his father under his dictation (Seryozha Abamov from Kuibyshev).

The activity of readers is especially eloquently evidenced by the special attention they pay to the tasks placed in the magazine under the heading “Problem «Kvant».” If in the phenomenon the number of reader letters with solutions to problems from this section is, then in febrero it was already 40 percent, and in March 50, in April 60, and in subsequent months their number reached 65 percent of the total number of all letters. Let us add to what has been said that most letters contain solutions not to one, but to several problems at once, the degree of difficulty of which significantly exceeds the average. Thus, many schoolchildren using the page identify their solutions and express them clearly and originally. And these readers continue to participate in the work of this section of the magazine even after graduating from high school.

Schoolchildren who submit solutions to problems are placed in the “Problem Book” and are taught to work in science at the level available to them now.

At the reader’s conference, some readers expressed the opinion that it would be desirable to somewhat reduce the scientific level of the articles and problems published in the “Zadchnik”.

However, the majority of science-minded schoolchildren strongly object to reducing the level of difficulty of materials placed in «Kvant». For example, this is what N. Mozharovsky from Kiev wrote to us about this: “I personally really like the magazine. This, in my opinion, is the only magazine in our country that publishes articles at a fairly scientific level, intended for those who have chosen physics and mathematics as their future specialty… I am sure that only schoolchildren who almost never open textbooks and books on physics and mathematics, who solve few problems, can apply for too difficult articles. Indeed, in articles on physics one can rarely find, unfortunately, at least some kind of formula. Where can we become more popular? I want to say that I personally know 20-30 other students who share this with me 等内容…

And here is the opinion of L. Suvorova from the city of Aleksaandrovsk, a representative of other readers «Kvant» - teachers: “I have been working as a physics teacher for the 5th year. I teach elective physics classes at school. This year I am working with tenth graders. We regularly monitor all numbers «Kvant». It is the scientific nature of the journal that attracts us.” A. Kolpakov, a physics teacher from Kursk, creates with it: “With great attention and use, I use the materials from your magazine for extracurricular activities and for preparing the most difficult students for optimization. Your magazine writes a lot about the teachers and students of our school. Students share what they read in «Kvant». Here is what E. Zavyavkin from the city of Kirovo-Chepetsk wrote, for example: “…Your magazine is as good, no, as an excellent teacher and landowner. He interested me in many ways and helped me figure things out. I probably don’t express my gratitude to you in a very literary way, but that’s not the main thing… When I gave it to the kids to read, they and the teachers copied a lot of it this year. Probably, the demand for «Kvant» will grow every year.”

However, in addition to these letters, we also receive letters in which readers, not without reason, wish for an insufficiently clear understanding and the difficulty of individual articles on physics.

“My name is Turbina Galina, I study in the 9th grade of secondary school 22 in Voronezh. I got a 5 in mathematics and physics for 8th grade. But reading your magazine, I don’t understand much. Even, more precisely, in each article the physical meaning of the formulas is not clear. For example, in the 9th issue of 1971, in the article “Conversation on the Uncertainty Principle” by Azbel, it is not clear that he will read “Einstein’s theory of relativity is studied in higher education. It’s hard to believe that everyone, or at least half of the readers-students out of 200 thousand people (circulation 266,135) understood the essence of the indicated formula.”