Kvant 1970, No. 2 · Pages

Popular science physics and mathematics
quantum
2
1970
magazine
Academy
Sciences of the USSR
And
Academy of Pedagogical
Sciences of the USSR

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table the sign $V$ is the sign of the “arithmetic” square root. It is clear that the formula

$$y = f(x) = (\sqrt{x})^2$$

(1)

allows you to calculate the corresponding value $y$ from a given $x$ only if $x$ is non-negative (otherwise the square root is “not extracted”).

For non-negative $x$

$$y = f(x) = x.$$

(2)

Formula (2) is simpler than formula (1), and we would like to consider it a formula that defines our function. But the domain of definition of the function given by formula (2) consists not of only non-negative numbers $x$, but of all numbers $x$. If we want to give a new definition of the very function that is defined by formula (1), we need to write

$$y = f(x) = \begin{cases} x \text{ при } x \geq 0, \\ \text{ не определена при } x < 0. \end{cases}$$

(3)

Similarly, the function $g(x) = \frac{x^3 - 1}{x - 1}$ can be written as follows:

$$g(x) = \begin{cases} x^2 + x + 1 \text{ при } x \neq 1, \\ \text{ не определена при } x = 1. \end{cases}$$

(4)

In school and university exams, complete accuracy is required in such matters.

2. FUNCTION SCHEDULE

Consider the following duty schedule (see No. 1 «Kvant», p. 29):

| | Mon | W | Wed | Thu | Fri | Sat | Sun |
| :— | :— | :— | :— | :— | :— | :— | :— |
| Petya | | | | | | | |
| Kolya | | | | | | | |
| Sasha | | | | | | | |
| Voloda | | | | | | | |

We already know that this is a graph of a function: the name of the attendant can be considered a function of the day of the week. Since there are a network of days of the week, and there are four boys, we drew

$$7 \times 4 = 28$$

cells, but only seven of these cells were painted red. If the boys had decided to alphabetize their names, the table would have been the one shown on page 6. It looks different, but shows the same distribution of duties - the same function.

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In both tables, 28 cells correspond to 28 possible pairs (day of the week, boy).

Of these 28 pairs, seven pairs were identified
(Mon, Petya), (Tue, Kolya), (cp, Sasha), (Thu, Voloda),
(Fri, Petya), (Sat, Kolya), (Sun, Sasha),

i.e. all pairs in which the day of the week is connected to the day of the week for this day:
(day of the week, day for this day),

or abstractly: pairs of the form
$(x, f(x))$.

Only the choice of these pairs is essential for specifying functions.
After this example, you may not think this definition is necessary:
The graph of the function $f$ is the set of all such pairs $*$
$(x, y)$,
that: 1) the first element of the pair $x$ belongs to the domain of definition of the function,
2) second element of the pair
$$y = f(x).$$

| | Mon | W | Wed | Thu | Fri | Sat | Sun |
| :— | :— | :— | :— | :— | :— | :— | :— |
| Voloda | | | | | | | |
| Kolya | | | | | | | |
| Petya | | | | | | | |
| Sasha | | | | | | | |

In our example, the graph of the function $f$:
$$\Gamma_f = \{(пн, Петя), (вт, Коля), (cp, Саша), (чт, Волода), (пт, Петя), (сб, Коля), (вс, Саша)\}.$$
For functions specified by the table

| $x$ | $f_1$ | $f_2$ | $f_3$ | $f_4$ |
| :— | :— | :— | :— | :— |
| $A$ | $A$ | $B$ | $A$ | $B$ |
| $B$ | $A$ | $B$ | $B$ | $A$ |

*) Throughout this article we mean “ordered pairs”. The pair (1,2) is different from the pair (2,1). The first and second elements of a pair may coincide: (1,1) or (2,2) are also pairs.

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in accordance with these definitions we obtain graphs

$$\Gamma_1 = \{(A, A), (B, A)\}, \quad \Gamma_2 = \{(A, B), (B, B)\}, \quad \Gamma_3 = \{(A, A), (B, B)\}, \quad \Gamma_4 = \{(A, B), (B, A)\}.$$

It is clear that for a function with a finite domain of definition, the number of elements of the graph (i.e., the number of pairs included in the graph) is equal to the number of elements of the domain of definition of the function. For a function with an infinite domain of definition, all pairs

$$(x, f(x))$$

cannot be written out. We have to describe these pairs using their properties. For example, for the function

$$y = f(x) = \sqrt{1-x^2}$$

the graph consists of all pairs $(x, y)$ for which two conditions are met

$$x^2 + y^2 = 1 \text{ и } y \geq 0.$$

This definition of the graph of a function can be written as

$$\Gamma_f = \{(x, y) | x^2 + y^2 = 1, \ y \geq 0\}.$$

The very general definition of the graph of the function $f$ can be written in the form of the following formula *):

$$\Gamma_f = \{(x, y) | y = f(x)\}.$$

By defining the graph of a function as a set of pairs, each of which consists of an argument value and a function value corresponding to this argument value, we have freed the concept of a graph from everything random. In this abstract understanding, each function has one single graph.

3. NUMERICAL PLANE

Let us turn to the most general real functions of a real variable in school. At school you learned that the graph of such a function $f$ is the set of those points $P(x, y)$ of the number plane whose coordinates $x$ and $y$ satisfy the equality

$$y = f(x).$$

This formulation and the general definition of a schedule given above in paragraph 2 are similar, but slightly different. Section 2 talks about the set of pairs $(x, y)$, and in the general school definition - about the set of points $P$ with coordinates $x$ and $y$. But is it possible to bring these two formulations into complete agreement?

*) We use the standard notation adopted in set theory.

$$\{x | A(x)\}$$

denotes the set of all objects $x$ satisfying the condition $A(x)$. For example, $\{x | x^2 = 1\}$ is the set of all $x$ for which $x^2 = 1$, i.e. the set of two numbers: $\{1, -1\}$.

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$$P = R^{-\alpha} o(Q).$$

We see that the rotation $R_0^\alpha$ is always an invertible function.

When applied to rotations, people often talk about “mappings.” The inverse of rotation $R_0^\alpha$ is rotation $R_0^{-\alpha}$. Symbolically we can write:

$$R_0^{-\alpha} (R_0^\alpha(P)) = P,$$
$$(R_0^\alpha)^{-1} = R_0^{-\alpha}.$$

A rotation maps a set of plane points onto itself. If we assume that a plane is nothing more than a set of its points (this is what is done in the modern presentation of geometry), then we can say that rotation is a reversible mapping of the plane onto itself.

Invertible mappings of a plane onto itself are called geometric transformations of the plane. You have repeatedly encountered geometric transformations on the pages of our magazine. For now, we will give just one more example of a geometric transformation of a plane. Parallel translation is the mapping of a plane onto itself

$$P \rightarrow Q = T(P),$$

in which all points $P$ move the same distance and in the same direction (Fig. 4).

5. VECTORS

Although it turned out that you have already established from acquaintance with new concepts and unusual interpretations that you already know, we will make another successful one. Let’s try to understand what a parallel transfer graph $P \rightarrow Q = T(P)$ is. By general definition, this is the set of all such pairs of points

$$(A, B),$$

for which

$$B = T(A).$$

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3. What additional condition should be added to the formula?
   $$f(x) = 1,$$
   to get certain functions
   $$f_4(x) = (\sqrt{x})^2 + (\sqrt{1-x})^2?$$

Note e. In problems 1-3, the sign $V$ is understood as an “arithmetic” square root, i.e., a non-negative number.

2. Graph of a function

3. How many functions exist with domain 1, 2, 3 whose graphs are subsets of the set of pairs
   $$\{(1,1), (1,2), (2,1), (2,2), (2,3), (3,3)\}$$

(cm. Fig. 6)?
How many of these functions have an inverse?

5. Show that the graph of the inverse function $f^{-1}$ is given by
   $$\Gamma_{f^{-1}} = \{(x, y) | (y, x) \in \Gamma_f\}.$$
   (Naturally, the function $f$ is assumed to have an inverse.)

6. Number plane

7. Describe the structure of the graph of the Dirichlet function:
   $$D(x) = \begin{cases} 1, если x рационально, \\ 0, если x irrationalно. \end{cases}$$

7*) Number $x$ from the segment [0; 1] decomposes into an infinite track fraction
$$x = 0, x_1 x_2 x_3 \dots (x_n = 0, 1, 2).$$

The value of the function $y = C(x)$ is determined by the double fraction
$$y = 0, y_1 y_2 y_3 \dots (y_n = 0, 1)$$
as follows:
if $x_n = 0$ then $y_n = 0$,
if $x_n = 1$ or $x_n = 2$, then:
$$y_n = 1$$ provided that there were no ones among the digits $x_1, x_2, \dots, x_{n-1}$,
$$y_n = 0$$ provided that among the digits $x_1, x_2, \dots, x_{n-1}$ one unit has already been encountered.

Try to graph this function. Prove that it contains an infinite number of horizontal segments. If you are familiar with understandable function continuity, try to prove that our function is continuous.

Note e. In this problem we do not avoid ternary fractions, in which all signs, starting from some, are twos, and double fractions, in which all signs, starting from some, are ones. For example, we write in the ternary system
$$0,2222 \dots = 1; 0,1222 \dots = 0,200000 \dots$$

and in the dual system we believe that
$$0,1111111 \dots = 1; 0,01011111 \dots = 0,01100000 \dots$$

*) This is a difficult task (see § 15 in the book: S.V. Fomin “Number Systems”, 1968.

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4. Geometric transformations

5. Describe in geometric terms transformations of the number plane that are similar to formulaic problems:

a) $(x, y) \rightarrow (-y, x)$,
b) $(x, y) \rightarrow (x, -y)$,
c) $(x, y) \rightarrow (y, -x)$,
d) $(x, y) \rightarrow (x+1, y)$,
e) $(x, y) \rightarrow (x+a, y+a)$.

9. Prove the formula for rotations around the center $O$

$$R_O^\alpha [R_O^\beta (P)] = R_O^{\alpha+\beta}(P).$$

(1)

10. Prove that for any centers $O_1$ and $O_2$ the transformation

$$F(P) = R_O^\alpha [R_O^{-\alpha}(P)]$$

will be a parallel transfer. How far and in what direction?

5. Vectors

6. Prove the formula

$$T_a [T_b(P)] = T_{a+b}(P).$$

(2)

12. Show that the transformation

$$F(P) = T_a [R_O^\alpha(P)]$$

is a rotation through angle $\alpha$. Around what center?

Comment. Formulas (1) and (2) are written in short

$$R_O^\alpha R_O^\beta = R_O^{\alpha+\beta},$$

$$T_a T_b = T_{a+b}.$$

Taking a function from functions is similar to multiplication in many ways. But this is a special topic, the development of which is not decreasing in this country. We will use this short notation for functions of functions (compositions of mappings) in Problems 13 and 14.

13. Prove that you always

$$T_a T_b = T_b T_a$$

And

$$R_O^\alpha R_O^\beta = R_O^\beta R_O^\alpha$$

when turning around the center. Show with an example that, in general,

$$R_O^\alpha R_O^\beta \neq R_O^\beta R_O^\alpha$$

when turning around centers $O_1, O_2$.

14. Find out fully the question of when

$$R_O^\alpha R_O^\beta = R_O^\beta R_O^\alpha.$$

ON. V. KARLOV AND A. M. PROKHOROV

LASERS

About two thousand two hundred years ago, Roman troops took the Greek city of Syracuse, located in Sicily. This fact would probably have been known only to specialists in the history of the formation of the Roman Empire if Aphimedes, one of the greatest physicists on Earth, had not died during the capture of Syracuse.

Many legends are associated with the personality of Aphimedes, his discoveries, and the circumstances of their birth.

or their use. Most of these legends have a basis in reality. But one, the most beautiful, the legend about the destruction of the Roman fleet by concentrated sunlight, is, unfortunately, pure fantasy.

The fact is that the laws of geometric optics for irradiated light sources indicate the impossibility of concentrating light energy in the form of a beam cord and focusing

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the position of the crests of one wave in relation to the crests of another wave. If this relative position remains unchanged all the time, then the waves are called coherent. The word “coherence” literally means coherence, connectedness. The concept of coherence plays an important role and is widely used in physics.

The overall light wave is not monochromatic. The fact is that the light that we see - sunlight, light from fluorescent lamps or incandescent lamps - is emitted by a large number of individual independent emitters (atoms), each of which emits light without any connection with whether its neighbors emit or do not emit light. As a result of such chaotic emission, the light waves that make up the radiation we observe are not coherent with each other, and the resulting light wave is not monochromatic. It is the non-monochromaticity and incoherence of light that translates into the concentrations of light energy in the beam cord (hence the impossibility of constructing the “hyperbolus of engineer Garin”).

We know that light is emitted by atoms. It is atoms that are those individual independent emitters that we just talked about. At the atomic level, representing light radiation as a continuous wave of light is unfair. Atom can absorb or emit light only in the form of certain portions of light energy - light quanta. The quantum energy value is $\varepsilon = hv$, where $h = 6,6 \cdot 10^{-34} \text{ дж}/\text{сцц}$ is Planck’s constant, and $v$ is the frequency of light.

The energy of visible light quanta will range from $2,5 \cdot 10^{-19} \text{ дж}$ to $5 \cdot 10^{-19} \text{ дж}$. Although this energy is very small, the quantum nature of light is significantly manifested in the interaction of light with matter. At the same time, light has not lost its luminous wave nature, which manifests itself in all those phenomena where physical relationships are important (the addition of waves, the formation of light beams, etc.). Thus, light has its stable wave and corpuscular, that is, properties inherent in particles.

So, micro-receivers of light are atoms, which under normal conditions produce a flow of incoherent radiation quanta. It is obvious that the radiation is emitted by atoms that have some excess energy, the so-called excited atoms. But an excited atom is unstable; it can spontaneously emit radiation quanta without any external reason. The spontaneous emission active occurs by chance. Such spontaneous emission is irregular and chaotic in nature. It is incoherent. But the possibility of emitting light energy is not limited to spontaneous processes. It must be remembered that the internal energy of atoms can only take on certain specific values. As they say, an atom can only be on one energy or another. Transitions of an atom from one level to another are accompanied by a change in its internal energy. When transitioning from a lower energy level to an upper one, an atom absorbs energy; when transitioning from an upper to a lower energy level, an atom releases energy. These transitions are therefore accompanied by either absorption or emission during the transition from energy level $E_1$ to level $E_2$, given by

$$v = \frac{E_2 - E_1}{h}.$$

In addition to spontaneous transitions of atoms from an upper energy level to a lower one, there is an amazing phenomenon of stimulated radiation, which is described by A. Einstein. The process of induced radiation differs from spontaneous emission primarily in that it does not occur on its own, but under the influence of an external

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in relation to atomic radiation. In this case, secondary quanta are indistinguishable from primary ones. They exactly repeat the frequency and direction of propagation of primary quanta. Induced radiation is called induced because the acts of its emission are induced (caused) by radiation external to the atom. As a result, unlike the spontaneous emission of many atoms, stimulated emission is coherent.

Thus, in optics there is a phenomenon suitable for obtaining corent radiation. This is stimulated radiation. But for one atom, the probability of stimulated emission during transitions from upper energy levels to lower ones is equal to the probability of absorption during transitions from lower levels to upper ones. In order for induced emission to prevail over absorption, it is necessary to have more atoms at the upper levels than at the lower ones. But under conditions of thermal equilibrium, the situation is just the opposite: at the lower energy levels there are more atoms than at the upper ones (Fig. 1). When radiation interacts with such a system of atoms, the following will happen. Part of the radiation is absorbed, and some of the atoms will move to the upper energy level (Fig. 2). If the system of atoms is so pre-prepared that at the upper level the energy of the atoms is greater than at the lower level, then during the interaction of radiation

Fig. 1
Fig. 2

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and move randomly and that forces of attraction and repulsion act between them.

But with all this, atoms remained only a certain idea in people’s heads. Not only did no one see them, because they were very small, but no one knew how small they were, what the masses of the atoms were, how many atoms there were in this or that body. It cannot be said that attempts are not made to somehow find out all this.

But these attempts did not lead to convincing results. Some scientists have even become confident that nothing can be learned about atoms and molecules, because they don’t really exist. These scientists believed that atoms are something like meridians and parallels on a geographical map: they are convenient to use, but in reality they do not exist in nature.

Physicists, however, did not lose hope of proving the reality of atoms and molecules, weighing them, counting their number, determining their sizes.

In this excerpt from the book “Atoms, Electrons, Nuclei” by the outstanding Soviet theoretical physicist Matvei Petrovich Bronstein, he talks about how for the first time it was possible to measure the mass of atoms and even count them. This book, published in 1935 in a small edition (only 10,000 copies), has long become a bibliographic rarity.

The publication was prepared by Professor A.K. Kiko and the additions he made are placed in straight brackets.

… And in the end atom really left to weigh. This was helped by one very strange phenomenon, discovered in the first half of the 19th century and to which physicists at one time did not pay due attention. This phenomenon is called Brownian motion.

In 1828, the famous English botanist Robert Brown made one extremely interesting observation. While testing a new, improved microcopper with an achromatic lens that had just been sent to him, Robert Brown decided to use this microcopper to examine an insignificant drop of liquid containing tiny grains of plant pollen. Such a liquid always contains large amounts of microcoptic solid particles. How surprised Brown was when he saw that these particles did not remain in place, but were moving, moving continuously, exactly

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His explanation, brilliant in its simplicity, is as follows: microsoptic particles move because they are influenced by invisible molecules and fluids surrounding them. By observing the movement of Brownian particles, we get some idea of ​​how the invisible molecules of a liquid move, in exactly the same way as we guess about the volition at sea when, standing from the shore, we see the rocking of a boat, thrown by the waves in all directions. Brownian motion is therefore the place that connects the invisible world of atoms and molecules with the world accessible to perception with the help of our senses.

Why can Brownian motion be observed only in the case where the particles are very to the right, it always turns out to be approximately equal to the number only obtained from the left, and an insignificant difference in the number will be completely insufficient to move a large and heavy particle from its place. If the particle has an insignificant mass and insignificant dimensions, then in the chaos of the molecular movements of the liquid it can always be seen that on one side of the particle at a given moment there will be by chance only more than on the other, and therefore the easily mobile particle moves in the direction where its molecules are located. After some very short period of time, the excess molecular will only merge the Brownian particle in a different direction, after another short period of time - in a third, etc.

If this explanation proposed by Carbonell is correct, then the lighter and smaller the particles, the more intense the Brownian motion should be. This is so in reality - Brown was already able to notice this. In addition, we know that the higher the temperature the higher the temperature, the faster the molecules of a liquid move; and in fact, Gouy found that as the temperature increases, the Brownian motion becomes more and more intense. When Zsigmond invented the ultramicrosopt and was able to observe tiny particles of gold in the colude solution (the diameter of the particle is less than a millionth of a centimeter), the Brownian motion of these particles turns out to be so fast that some kind of continuous fineness resulted. Zsigmond describes his first impression as follows: “This is some kind of continuous jumping, floating, galloping, colliding and flying apart, so it’s difficult to understand this confusion.”

We now move on to talk about the classic work done by the French physicist Jean Perrin (1908). In these works, it was finally verified and established that Brownian motion in liquids is caused by the movement of molecules and thereby provided decisive proof of the active existence of molecules and atoms.

Perrin took pieces of gummiguta rubber resin and rubbed them with his hand in water until it turned a bright yellow color. After this, Perren took some of this liquid under the microsopp. Under the microsopp, it turns out with the help of nerastoril in the water, and broke up into many spherical small grains, which were scattered throughout the entire volume of water. These grains are very different in size. And Perrin wanted to have a liquid in which there would be gummigut particles of exactly the same size. To do this, he used a “centrifuge” (centrifugal machine), the same one that is used on large dairy farms to separate cream from milk or in medical laboratories to remove blood globules from the blood, after which a homogeneous liquid remains - blood plasma. Perren’s centrifuge made 2500 revolutions

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etc. Note that in this case the number of grains is considered to be the average of several observations, because the grains move completely chaotically and, therefore, their number in the microscope grain field is either too large or less depending on the case. Therefore, at the same height, Perren counted the grains many times and then calculated the value characteristic of each such height.

Photographs of gum gum emulsion taken by J. Perrin through a microscope.

Let us present the results of one of Perrin’s experiments. The depth of the cuvette was, as we have already said, 100 microns (that is, 0.1 mm). Readings were made at heights of 5, 35, 65 and 95 microns above the level of the bottom of the cuvette. It turned out that the average part number at 35 microns was half that at 5 microns, the part number at 65 microns was half the part number at 35 microns, and the part number at 95 microns was half the part number at 65 microns.

So: if the heights form an arithmetic cut, then the numbers of grains form a geometric cut.

This law of decreasing density of grains with height should have greatly amazed and interested Perrin: after all, according to the same law, density decreases when rising in our atmosphere. Blizzard Pascal, a famous French scientist who lived in the 17th century and first

Results of Perrin’s experiments.

This is how the number of air molecules changes with height.

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The barometer, invented by the Italian Torricelli, applied to the study of the atmosphere, discovered the law according to which the density of atmospheric air decreases with increasing altitude. This law, called the barometric formula, states the same thing: the density of each of the gases that make up the atmosphere kills along with the increase in altitude in geometric projection.

[This law can also be expressed mathematically, in the form of a formula.

Suppose that at some height $h_0$ above the Earth, each cubic centimeter contains $n_0$ molecules of some gas. At some other, higher altitude $h$ there will, of course, be fewer of the same molecules in one cubic centimeter, for example $n$. Then the barometric formula can be written as follows:

$$\lg \frac{n_0}{n} = A mg (h - h_0)^*.$$ (1)

Here $A$ is some constant value, the same for all gases (at a given temperature), and $m$ is

*) If $k$ is the number of “steps” for which we rise to the height $h$, and $\Delta h$ is the height of the “step”, then $h = h_0 + (k - 1) \Delta h$, and $n = n_0 \left( \frac{1}{q} \right)^{k - 1}$, where $\frac{1}{q}$ is the denominator of the progression, which is the density of the gas ($q > 1$). From these two formulas, eliminating $k$, we find that

$$\lg \frac{n_0}{n} = \frac{\lg q}{\Delta h} (h - h_0).$$ (*)

A similar formula can be written for another gas; we provide the quantities relative to it with a prime:

$$\lg \frac{n'}{n'} = \frac{\lg q'}{\Delta h'} (h' - h_0').$$ (**)

It turns out to be very important that if the densities of the first and second gases decreased by the same number of times, that is, $\frac{n_0}{n} = \frac{n'}{n'}$, then the ratio of the height differences at which this will happen, $\frac{h - h_0}{h' - h_0'}$, is exactly equal to the inverse ratio of the mass of the molecules of this gas in question, $g$ - the acceleration of gravity.

It is immediately clear that the left side of the equality is the ratio of the numbers of the part per unit volume (though under the logarithm sign), and the right side is the difference in heights. This means that if the heights form an argumentative progression, then the numbers of the part form a geometric progression.

Let us choose a height difference $h$ - $h_0$ such that the number of part $n$ at height $h$ is half as large as the number of part $n_0$ at height $h_0$. Then formula (1) will take the form

$$\lg 2 = A mg (h - h_0).$$ (2)

In the table of logarithms it is easy to find that $\lg 2 = 0,30103$, so

$$0,30103 = A mg (h - h_0).$$ (3)

If the numerical value of the constant $A$ was known, then by calculating the number of parts per unit volume $n$ and $n_0$ at heights $h$ and $h_0$, it would be easy to calculate the mass of the molecule $m$, that is, weigh it. True, count

*) If $k$ is the number of “steps” after which we rise to the height $h$, and $\Delta h$ is the height of the “step”, then $h = h_0 + (k - 1) \Delta h$, and $n = n_0 \left( \frac{1}{q} \right)^{k - 1}$, where $\frac{1}{q}$ is the denominator of the progression, which is the density of the gas ($q > 1$). From these two formulas, eliminating $k$, we find that

$$\lg \frac{n_0}{n} = \frac{\lg q}{\Delta h} (h - h_0).$$ (*)

A similar formula can be written for another gas; we provide the quantities relative to it with a prime:

$$\lg \frac{n'}{n'} = \frac{\lg q'}{\Delta h'} (h' - h_0').$$ (**)

It turns out to be very important that if the densities of the first and second gases decreased by the same number of times, that is, $\frac{n_0}{n} = \frac{n'}{n'}$, then the ratio of the height differences at which this will happen, $\frac{h - h_0}{h' - h_0'}$, is exactly equal to the inverse ratio of the mass of the molecules of this gas in question, $g$ - the acceleration of gravity.

It is immediately clear that the left side of the equality is the ratio of the part numbers in a unit volume $n$ and $n_0$ at heights $h$ and $h_0$; it was easy to calculate the mass of the molecule $m$, that is, weigh it. True, count

$$\lg \frac{n_0}{n} = A mg (h - h_0).$$

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The coincidence of two numbers that were obtained by completely different methods is the best proof of the correctness of all the assumptions made. This means that molecules and atoms really exist, and are not just a fiction convenient for chemists. Even those who for a long time and stubbornly did not want to admit the existence of atoms were forced to make this conclusion.

… The age-old dispute between the parties and opponents of atoms thus ended with the victory of the supporters of the atomic theory. And now we can confidently say that all things in the world - water, stones, plants, animals, air, iron, etc., etc. - all of this consists of tiny atoms invisible to the eye.

OBJECTIVES FOR THE ARTICLE
“HOW ATOM WAS WEIGHED”

1. Using the data given in the article, find at what altitude the pressure decreases by half. Air is lighter than oxygen in the ratio 28.8:32.

2. $P_0$ - air pressure at sea level, $P_1$ - at altitude $h$. What is the air pressure at altitudes: $2h$, $3h$, $nh$? Assume that the air temperature and the acceleration of gravity do not change with height, and therefore the constant $A$ does not depend on height.

3. Find the height above the Earth’s surface where the air pressure is 0.25 and 0.125 atmospheric. Draw a curve of air pressure versus altitude.

LOGICAL PROBLEM

Let us conventionally assume that if a person does not eat for seven days or sleep for seven days, he will die. Let a person not eat or sleep for a week. What should he do first by the end of the seventh day: go to bed or sleep in order to stay alive. (Despite its humorous nature, the problem has a strict and unique solution.)

QUESTIONS
IN PHYSICS

1. Why is it that when you go swimming on a hot day, you enter the water and the water feels colder than the air, but when you get out, it feels the opposite?

2. Why does the refrigerator come and thaw from time to time?

3. The box containing the mouse is thrown vertically upward. When is the mouse in a state of weightlessness if there is no air resistance?

4. Two identical pans with equal amounts of water were placed on the stove at the same temperature. After some time, boiling water from the kettle should be poured into one pan. In which pan will the water boil first?

4a. From the pan with more water, add so much to the second pan that they again contain equal amounts of water. In which pan will the water boil first?

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CURVES
DRAGON

From this article
you will learn about interesting things
personal family,
one of which
drawn here
and several others -
on pages 42 and 43.

N. B. VASILIEV, V. L. GUTENMACHER

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HOW TO DRAW LONG
BROKEN DRAGONS?

We will call any broken line obtained from a piece of paper folded in half $n$ times, unfolding the folds up to $90^\circ$, a dragon’s broken line $n$. Let’s find out how broken dragons are arranged and how to draw them for large enough $n$.

First way

The broken dragon wound $n$ consists of $2^n$ links and accordingly has $2^n-1$ vertices (not counting the ends). Thus, it has a middle vertex on $(n>0)$ as well, since the number of vertices is odd. On Fig. 2 and 3, the middle vertex of each broken line is offset by a green circle. It can be noted that each of these broken lines consists of two identical pieces, obtained from each other by turning by $90^\circ$. It turns out that this is a general pattern.

Theorem 1. If you continue any broken dragon wound $n$ with end at point $O$ can also be a polyline obtained from this by rotating by $90^\circ$ around the point $O$, then you get a broken dragon wound $n+1$, and back, love breaking the dragon wound $n+1$ comes from some broken wound $n$ this way.

In fact, let’s say we want to fold the strip $n+1$ in half. First, fold it in half once. Then its two halves coincide and, when further folded, will bend in half perfect (Fig. 4). Now let’s expand on $90^\circ$ the last $n$ stibs of the strip. We get two matching broken dragon wounds $n$; All that remains is to separate them by $90^\circ$ - and we will get a broken dragon wound $n+1$ (Fig. 5). Think about how to derive both statements of Theorem 1 from these messages.

Using Theorem 1, it is easy to draw long polygonal dragon lines.

Since the drawing of the dragon follows the lines of a square grid, it is convenient to draw it on checkered paper.

Let’s take any short dragon line (for example, just a segment). Let’s agree that one of its extreme points is the beginning, and the other is the end. Let’s continue it with the same broken line, rotated relative to its end by $90^\circ$ clockwise (or counterclockwise). We continue the resulting new broken line from the end in the same way; We do this as many times as we want and as much as we can. This can be done automatically and quickly if you have tracing paper or, even better, slightly transparent checkered paper at hand. (Think how!)

Obviously, in the same way we can immediately draw both the sides and necks of the dragon: we only need to round off the middle mushroom each time.

The remarkable property of all dragon curves is that they do not intersect themselves, or, what is the same, dragon curves never pass along the same segment twice. Thus, although the dragon’s breaker can pass through the same point (the vertex of the grid) twice, it does not hit the same point more than twice.

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From Theorem 1 it is not immediately obvious how to prove that dragon lines do not pass twice along the same segment - on the contrary, the longer and more intricate broken lines or curves you draw, the more surprising is how well their “protrusions” and “valleys” fit together (see the “locomotive” dragon curve on page 43).

However, it is not difficult to prove this *), using another theorem about the “doubling” of dragon polygons, which, by the way, gives one way to draw long polygons.

Second way

On Fig. 6 the vertices of the black dragon lines are connected by red segments through one another. We see that the red segments again make up the broken lines of the dragon. It turns out that this is also a general pattern.

To formulate it precisely, then also that each red segment is the hypotenuse of an isosceles right triangle, the legs of which are the links of the original broken line (at figure the triangles are slightly shaded), and every two adjacent such triangles are obtained from each other by a rotation of 90° around their common vertex; in other words, if you go along the red broken line of the dragon, then these

*) See problem 9 at the end of the article.

triangles will appear alternately on the right and then on the left.

Theorem 2. If on each link of the dragon’s polyline n, as a zinotenuse, we construct a right-angled isosceles triangle, and in such a way that for two adjacent links these triangles are one of two by rotating by 90° relative to the common vertex, then the stones of the constructed triangles constitute the dragon’s polyline n+1. And vice versa, each broken line n+1 can be obtained in this way from some broken line n.

In fact, let’s follow the last one and folding the strip in half. For convenience, we will refer to the conditional Fig. 7. We want to fold the strip in half n+1 times. Let’s first fold it n times and draw it in profile (red line at Fig. 7). Then fold it in half again and rotate this last fold 90° (black line on the same figure*). The paper goes from mushrooms A to mushrooms B and back not along the hypotenuse, but along the legs of an isosceles right triangle ABC. Let’s unfold all other folds

Fig. 6

the segment is the hypotenuse of an equilateral right triangle, the legs of which are the links of the original broken line (at figure the triangles are slightly shaded), and every two adjacent such triangles are obtained from each other by rotating by 90° around the vertex point; in other words, if you go along the red broken line of the dragon, then these

up to 90° (at Fig. 8 it is shown how a separate fold will unfold). Then the stones of the rectangles form the dragon’s broken line n+1, and the hypotenuse forms n.

Using these messages, it is easy to prove both statements of Theorem 2. You can try to give another proof: derive Theorem 2 from Theorem 1.

*) On our figure the black line is $\sqrt{2}$ times longer than the red one, but this does not matter since we are interested in the shape of the broken line, not its size.

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Using Theorem 2, from each dragon’s broken line of rank $n$ one can obtain two different broken lines of rank $n+1$: it all depends on which side of the first link to complete the triangle.

Please note that when we move from a broken line of rank $n$ to a broken line of rank $n+1$ according to Theorem 1, the broken line becomes twice as long, and the length of each link does not change; if we use Theorem 2 for “doubling,” then the length of the broken line increases by $\sqrt{2}$ times, and the length of the link decreases by $\sqrt{2}$ times.

Now practice drawing polygonal dragon lines using Theorem 2.

WORDS

The property discussed in Theorem 2 can be explained quite simply if we look at dragon lines (or, if you prefer, at ways of folding a piece of paper) from a slightly bluer point of view.

Let the turtle crawl along the dragon’s broken line (Fig. 9). Every time she gets to the top, she has to turn $90^\circ$ left or right. From the turtle’s point of view, its path will be determined by a sequence of turns. For example, for a broken line at Fig. 2, $a$ (starting at the red dot), this sequence will look like this: left, left, right.

We will denote a turn to the right by the letter $R$, a turn to the left by the letter $L^*$). Then the entire broken line will be written in the following “word”:

LLR

(“a word” in many branches of mathematics and logic simply means any sequence of letters). In the same way, you can write any dragon line as a word from the letters $L$ and $R$.

To get the broken line $a$ from the broken line $\delta$ (Fig. 2), you need to fold the piece of paper one more time “to the left” (see Fig. 2, $\delta$ and 10). In this case, between each two vertices of the broken line $a$ there will be one more turn, and at Fig. 10 it is clearly seen that these new turns will be through

*) We chose the first letters of the English words right (right) and left (left), because the Russian letters P and L are too similar to each other (by the way, the German Rechts and links begin with the same letter).

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This curve is obtained if, starting from the segment, the “doubling” process is repeated 12 times, and through turns clockwise and counterclockwise. To better show how this curve goes, we rotate each time not by 90°, but by 95° (black line); if you reduce the angles to 90°, you get the dragon curve, which is depicted on the same figure colored line; it fills an equilateral right triangle with a uniform pattern.

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This rank 12 dragon curve has the special name “Father, Mother and Son”. Find its middle. Can you imagine that the curve develops from this point into two perfectly identical pieces, resulting from each other by turning 90°? To believe this, we will have to faithfully trace the curve with a colored pencil from beginning to middle.

Here, one of the two halves, the curve (“it’s called the “locomotive”), is drawn on a colored background to show how these two curves “stage” each other. One half, after rotating it 90° around the midpoint, will coincide with the other

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to give in; thus we get
$$L L R \rightarrow L L R L R R,$$

i.e., words that are a recording of broken line 6. With one more bend to the left, we would get a broken line characterized by the word

$$L L R L L R R \rightarrow L L R L R L R R,$$

Draw this broken line. This is the main broken line of a rank 4 dragon. If you want, twist the corners of it, as Heiwei suggested.

If you continue the same operation on the last word received again (starting the sequence of interlaced letters again with $L$), you will get words of 31 letters - a record of the Main Dragon Line of rank 5, shown at Fig. 3.

Of course, the sequence of intersecting letters can begin not with $L$, but with $R$ - this will result in other broken lines.

It is easy to see that our method of constructing a word for a broken line of rank $n+1$ from a word for a broken line of rank $n$ exactly corresponds to the geometric method of doubling, which is discussed in Theorem 2 (new letters correspond to enough verbs). In general, the entire “theory” of broken dragon lines could be built not geometrically - using rotations, obtaining triangles, etc., but “algebraic” - using operations on words from two

Fig. 10

letters $L$ and $R$, “records” of broken dragons *).

The reader will be able to partially follow this path and get acquainted with a number of interesting patterns that words have - records of dragon curves, solving the following problems.

TASKS

1. How long do you need to take a strip so that, after folding it in half 30 times, you get a distance between adjacent folds of 1 cm? Are the distances from the Earth to the Moon larger or smaller?

2. How will the dragon’s line change if the strip of paper is placed on the table with the other edge? How will the words that write the broken line change?

3. a) How many different (not similar to each other) broken dragons of rank 4 are there? Draw them all. Write the corresponding words from the letters $L$ and $R$.

6. How many different broken lines of rank $n$ are there?

4. Let’s say that the turtle crawled along the dragon’s broken line and found words from the letters $L$ and $R$. What word will she read if she crawls along this broken line in the opposite direction?

5. Let $s$ be some word from the letters $L$ and $R$. Let us denote by $s$ the words that will be obtained from $s$ if we rearrange the letters in it in the reverse order and then place $L$ on $R$ and $R$ on $L$. (For example, if $s=LLR$, then $s=LRR$).

Let $s$ and $t$ be some words from the letters $L$ and $R$. We will denote by $st$ the words obtained if the words $s$ and $t$ are written side by side. (For example, if $s=LLR$, $t=RL$, then $st=LLRRL$).

Prove that:

a) For any words $s$ and $t$

$$\overline{st} = \overline{t} \overline{s}.$$

6. If $s_{n+1}$ are words corresponding to the dragon line of rank $n+1$, then

$$s_{n+1} = s_n x \overline{s}_n,$$

where $s_n$ are some words corresponding to the dragon’s polyline of rank $n$, and $x$ are words corresponding to one letter.

*) It was this method of presentation that was chosen by the Canadian mathematicians Knuth and Davis, from whose manuscript “Number representations and dragon curves” (Ch. Davis, D. E. Knuth) the authors of the article first became acquainted with curved dragons. A series of figure length dragon curves are also studied from this manuscript.

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c) If $s_n$ is a word corresponding to the dragon’s line, then $s_n$ differs from $s_n$ only by one middle letter.

d) A word that writes a dragon’s broken line of rank $n$ can be constructed in a unique way by arbitrarily specifying $n$ letters that should appear in this word in places with numbers $2^k$, where $k=0,1,2,...,n-1$ (i.e., in the first, second, fourth, eighth,…, $2n^{-1}-m$ places); after that, to find which letter is in the $m$th place, you need to represent $m$ as $m=2^k(2l+1)$, where $k$ and $l$ are integers; if $l$ is even, then the $m$th place has the same letter as the $2^k$th, if $l$ is odd, the opposite letter.

e) Two words $s_n'$ and $s_n''$ define broken dragons of the same shape (similar) if and only if, after crossing out the middle letter, these words either coincide, or one is obtained from the other by replacing $L$ with $R$ and $R$ with $L$. (For example, the words LLRLLRR, LLRRLRR, RRLRRLL define the same broken lines.)

6. Let us have a broken line of a dragon of rank $n$. From it you can obtain two broken dragon lines of rank $n+1$, using either Theorem 1 (taking one or the other end of the broken line as the point $O$) or Theorem 2 (extending the triangle in one direction or the other). Will both cases produce the same two broken dragons of rank $n+1$ or, generally speaking, different ones? How to obtain their words from a word of a given broken line of rank $n$?

7. Two points $A$ and $B$ are given on the plane. Let’s draw a square with the center in the middle of the segment $AB$, one of whose sides is equal to $2AB$ and parallel to the segment $AB$. Prove that any dragon line with ends at points $A$ and $B$ lies inside this square. Prove that for any point $M$ inside this square and any positive number $e$ one can find a dragon line with ends $A$ and $B$ that passes from the point $M$ at a distance less than $e$ (in mathematical language, “the union of all dragon lines with ends at the points $A$ and $B$"). Its first $2^n$ links form a broken line of rank $n$.

Prove that:

a) The end of such a broken line of rank $n$ will be at the point

$$\left( \frac{n}{2} \cos \frac{\pi n}{4}, \frac{n}{2} \sin \frac{\pi n}{4} \right)$$

(red dots on figure on page 38).

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---

running_head:
folio:

6. The word that is the recording of this broken line can be quickly written like this. First, we write the letters $L$ and $R$ through the numbers, leaving spaces between them for the missing letters:

L_R_L_R_L_R_L_R_L_R...

and then we proceed as follows. With the finger of our left hand we point to the first letter, and with our right hand we write it in the first gap, then with our left hand we point to the second letter, and with our right hand we insert it into the second gap, with our left into the third… and so on sequentially through all the letters (without missing new ones), and with our right hand we insert each indicated letter into the first free gap.

$$LLRLLRLLLRRLRLLLRL$$

c) If, as in Problem 10, we release four main broken lines of rank ∞ from one point, then along each segment of the grid exactly one link of one of these broken lines will pass (Fig. 11).

(This is a difficult theorem; it was first proved by D. E. Knuth.)

Fig. 11.

LITERATURE

M. Gardner, Mathematical games, Scientific American 216, 1967, March (124-125), Appel (118-120), July (115).

The day before
olympids

For those who take part in the final section of the All-Russian Olympiad for schoolchildren, April turns out to be “hot”. It is in this place that the event will take place: young mathematicians - in Simferopol (from 23 to 29), physicists - Sverdlovsk, chemists - Voronezh (from 8 to 15). Four representatives from each of the region, regional and resource Olympids, eight Muscovites, six Leningraders, guys who took first and second places at the previous Olympiad, and four from special physics and mathematics interns will come here.

The very next day after the grand opening of the Olympiad, its business part begins - all participants will be charged for written work in their “specialty”. They will have to solve 4-5 problems each. Physicists and chemists, in addition, will have an experimental tour on April 12th.

The most pleasant and easiest day is the closing of the Olympiad and the presentation of awards.

Thirteen graduating class students - eight mathematicians and five physicists, who successfully performed at the All-Union Olympiad, will receive tickets to the next international Olympiad this summer.

The magazine «Kvant» will constantly inform its readers about everything that happens on these modern ones.

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In this section, continuing from issue to issue, we place several tasks, usually quite difficult. The best solutions to these problems, submitted by readers, will be published in the journal in a few months.

In mathematics

M6. There is a clock in front of you. How many hand positions are there that cannot be used to determine the time unless you know which hand is the hour hand and which hand is the minute hand?

(It is believed that the position of each of the arrows can be determined accurately, but following how the arrows move cannot.)

M7. a, b, c are the sides of the triangle. Prove that

$$\frac{a}{b+c-a} + \frac{b}{c+a-b} + \frac{c}{a+b-c} \geq 3.$$

S. Berkolaiko

M8. Two people play this game. From a pile of 25 matches, everyone takes one, two or three matches in turn. The winner is the one who, at the end of the game - after all the matches have been sorted out - turns out to have only matches.

Who wins when played correctly - the starter or his partner? How should he play to win? How will the answer change if we assume that the one who takes an odd number of matches wins?

Explore this game in the general case when there are 2n+1 matches and you are allowed to take any number of matches from 1 to m.

M9. Consider the following properties of a tetrahedron (we call an arbitrary triangular paramid a tetrahedron):

a) all faces are equal in size;
b) each edge is equal to the opposite one;
b) all faces are equal;
d) the centers of the circumscribed and inscribed sphere coincide;
d) the sums of plane angles at each vertex of the tetrahedron are equal.

Prove that all these properties are equivalent.

Try to find other equivalent properties of the tetrahedron.

M10. Four circles, the centers of which are the vertices of a convex quadrilateral, completely cover this quadrilateral. Prove that from them it is possible to choose three circles that cover a triangle with vertices at the centers of these circles.

G. Galperin

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F7. A horizontal rod $O_1A$ of length $l$ rotates around a vertical axis $O_1$ (see Fig. 1). A wheel of radius $r$ is mounted on the axis attached to the end of the rod $O_1A$. The wheel axis is horizontal and makes an angle $\alpha$ with the rod $O_1A$. The wheel rotates on an axle without friction and rolls along the ground. There is more friction between the wheel and the soil. How many revolutions will the wheel make when the rod $O_1A$ makes one revolution around the vertical axis?

G. Komkin

F8. A long rod $AB$ with a threaded hole at the end is screwed onto a vertical screw (see Fig. 2). The rod is released. The friction between the screw and the rod is negligible. How will the rod move after it comes off the screw?

Physics and Mathematics Olympiad MIEM

Fig. 1

Fig. 2

F9. On the horizontal table there is a weight attached to the table using a long spring. At first the spring was not stretched. The weight is then moved 20 cm from its equilibrium position and released. The weight began to oscillate along the spring. Due to friction, the amplitude of its oscillations per period decreases by 7%. How many oscillations will the weight make before stopping? At what distance from the equilibrium position will it stop?

Weighty correspondence physics and mathematics Olympiad 1967

F10. How to construct a heater of the highest possible power from four thin wire spirals with resistances of 10 ohms, 20 ohms, 30 ohms and 40 ohms, designed to produce a power of no more than 2 W each, if there is a current source with e. d.c. 20 V and internal resistance 20 ohms? B. Bukhovtsev

F11. Three open barrels are filled with water and installed at different heights (see Fig. 3). From each barrel, rough pieces are carried upward and connected together. The weights are also filled with water. Where will the water flow through the pipes if taps $K_1$, $K_2$ and $K_3$ are opened at the same time?

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Many boys and girls are concerned about issues related to entrance exams, especially in mathematics and physics. After all, these exams are held in all technical types, in many (including the humanities!) faculties of universities and pedagogical institutes. What awaits applicants during the entrance exams? How are they carried out? How best to prepare for them? All this undoubtedly interests readers «Kvant».

On the pages of the magazine we will regularly publish “Practicum for applicants” - information about the materials of entrance exams in mathematics and physics, provide consultations on certain topics that cause educational difficulties, and talk about books specifically addressed to university applicants.

This issue will contain information about written examinations in mathematics. Typically, these exams primarily test the ability to solve problems using techniques learned in school. The tasks proposed in the exams do not require knowledge beyond the scope of the “Program of entrance examinations for applicants to higher educational institutions of the USSR.”

Naturally, higher requirements are imposed on applicants to mathematics, physics, and some other university faculties than to applicants to most universities. This, however, does not mean that, for example, future physicists are required to have some additional knowledge that goes beyond the scope of the program. They just need to be more fluent in the material, demonstrate a certain independence of thought, and be able to solve more complex problems.

To learn how to confidently solve the problems of entrance exams, you need to carefully repeat the theoretical school course and get enough practice in solving similar problems, use one of the common collections of problems in mathematics, compiled specifically for university applicants.

When you start preparing for entrance exams, don’t put it off, especially if you are going to take the exams in six months!

Editorial.

50

N. S. BAKHVALOV, N. N. KUZNETSOV

The written exam in mathematics is the main exam for applicants to the Faculty of Mechanics and Mathematics of Moscow State University. Often there comes a case about its excessive difficulty, that only graduates of Moscow mathematical schools can get even a “C” on this exam.

These cases are completely unfounded *). The tasks (the so-called “variants”, consisting of 4 problems) offered in the written exam are composed so that in each variant there is one problem

*) In 1969, out of 922 Moskovey applicants, 469 (about 50%) received a positive mark (three or higher) on the written exam, and out of 2,356 nonresidents - 912 people (about 40%).

are quite accessible to those with average mathematical training. Solving such problems already guarantees a three. Therefore, to successfully pass the written exam, you only need a solid knowledge of the school curriculum in mathematics, and only those applicants who have not shown preparation enough to successfully study at the faculty receive an unsatisfactory grade.

Of course, I will break with easy problems; the version also includes tasks whose solutions require sufficient mathematical culture (but not additional knowledge that goes beyond the scope of the program). Therefore, it is impossible to get a high mark on this exam.

Option 1

1. Points A and B are on roads intersecting at an angle ACB=60°. You can get from point A to B by bus - first along one road to intersection C, then along the other - spending 11 minutes. If you go from A to B in the right direction, it will take 1 hour. 10 minutes, and if you first take the shortest route to the road on which point B is located, and then take the bus, then even more time, even if there is a bus right away.

What is the distance from point A to the intersection if the pedestrian speed is 3 km/h and the bus speed is 30 km/h? (Roads are considered straight.)

2. In triangle ABC, bissertrices AD and BE are drawn, intersecting at point O. It is known that segment OE has length 1, and vertex C lies on the circle passing through points E, D, O. Find the sides and angles of triangle EDO.

3. Find all values ​​of x that simultaneously satisfy the following conditions:

$$\cos 13x = \cos x,$$
$$\cos 2x + \sin 5x = 1,$$
$$|x| < 3.$$

4. The primordial processes of a plane quadrilateral into two mutually perpendicular planes are squares with sides of 2. Find the perimeter of the quadrilateral, knowing that one of its sides is equal to $\sqrt{5}$.

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This requires special preparation and the use of appropriate literature *).

Here we will consider, as an example, one typical set of problems in the version presented in the written exam for MEXMATH in 1969 (varnant 1 on p. 50). He was given 4 hours to decide. To receive a grade of “3” it was enough to solve at least one problem; a score of “5” was given if there was a complete solution to any three problems, and four were in the main solution.

We will analyze these problems in detail and note the characteristic mistakes made by those who arrived when solving them, and in conclusion we will present the readers of Kvanat with another option for independent solution.

TASK ANALYSIS

Task 1

Let’s make a drawing (Fig. 1). $D$ is the point of line $BC$ close to $A$. Note that the location of point $B$ relative to $D$ is unknown in advance (the drawing shows two location options). Let $x=AC$, $y=CB$ (these distances are measured in kilometers). Writing down the conditions of the problem, we obtain a system of two equations:

$$\begin{cases} x+y=5,5 \\ x^2+y^2-xy=12,25 \end{cases}$$

and inequality:

$$10x\sqrt{3}+|2y-x|>70$$

*) As suitable problem books, we point out, for example, the books by E. B. Vakhovsky and A. A. Ryvkin “Tasks in elementary mathematics of increased difficulty” (about it, see the article “How to solve a problem” on page 60 of this issue of “Kvanat”), N. Kh. Rozov, M. K. Potapov, G. V. Dorofeev, “A manual on mathematics for university applicants”, V. B. Lidsky, P. V. Ovsyannikova, A. N. Tulaikova and M. I. Shabunna “Tasks in elementary mathematics.”

System of equation (1) has two solutions: a) $x=4$, $y=1,5$; 6) $x=1,5$, $y=4$, and we must choose from them the one that satisfies the additional condition (2). The second solution to this condition does not satisfy, or $15\sqrt{3}+6,5<30+6,5<70$. The first decision satisfies him. To verify this, you need to check that $40\sqrt{3}+1>70$, or $40\sqrt{3}>69$. By squaring the last inequality, we obtain $4800>4781$ and thereby verify that it is true. So, the solution to the problem is $x=4$ km.

Let us note the typical mistakes that applicants made when solving this problem.

Most of them managed to compile and solve the system of equation (1); however, choosing the desired solution using the additional condition (2) turned out to be an insurmountable obstacle for many. Please, the most convenient way to understand the order is the incorrect visual representation of the relative position of the point $B$ and $D$, as a result of which, instead of inequality (2), the incorrect inequality was checked

$$10x\sqrt{3}+2y-x>70.$$

At the same time, quite often the examiner is limited to checking only one of the solutions found. For example, having established that the second solution of system (1) ($x=1,5$; $y=4$) does not satisfy inequality (3), they automatically did on this basis

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the conclusion is that the first solution is the one we are looking for (although it is easy to see that it also does not satisfy inequality (3). Let us note one more mistake that a considerable part of those constructed made: when checking condition (2), the value $\sqrt{3}$ was replaced by an approximate rational value, for example, 1.7, and the equality was checked incorrectly

$$17x + |2y - x| > 70$$

(it is easy to check that both solutions of the system of equations (1) also do not satisfy it).

Problem 2

Let’s connect the points $C$ and $O$ (see Fig. 2). Line $CO$ is the bisector of angle $C$, or it passes through the intersection point of bisectors $AD$ and $BE$. That’s why

$$\angle ECO = \angle DCO$$

(1)

Angles inscribed in a circle $OECD$

Fig. 2.

$DEO$ and $DCO$ are equal, as if they rest on the same arc. The same is true for angles $EDO$ and $ECO$. Based on (1) we conclude that

$$\angle DEO = \angle EDO,$$

i.e., the triangle $EDO$ is equilateral.

The inscribed angles $ECD$ and $EOD$ are supported by additional arcs; therefore their sum is $180^\circ$. On the other hand, we have an integer equality:

$$\angle EOD = \angle AOB =$$

$$= 180^\circ - (\angle OAB + \angle OBA) =$$

$$= 180^\circ - \frac{1}{2} (\angle A + \angle B) =$$

$$= 180^\circ - \frac{1}{2} (180^\circ - \angle C) =$$

$$= 90^\circ + \frac{1}{2} (\angle C).$$

Because

$$180^\circ - (\angle C) = 90^\circ + \frac{1}{2} (\angle C),$$

where

$$\angle C = 60^\circ.$$

From the reasoning given above it follows that

$$\angle DEO = \angle EDO = 30^\circ,$$

$$\angle EOD = 120^\circ,$$

so the sides of triangle $EDO$ are 1, 1 and $\sqrt{3}$.

Nibaulier’s typical mistake in this problem was to “prove” that $\triangle ABC$ is equilateral. Some of the examinees, based on a geometric illusion, assumed that the center of the circle $OECD$ lies on the straight line $CO$. It follows that the angles $OEC$ and $ODC$ are right angles, so the bisectors $AD$ and $BE$ are simultaneously altitudes, which is only possible in an equilateral triangle.

Another part of those who prove the equilaterality of the triangle $ABC$, based on the following reasoning: let us drop from the point $O$ the perpendiculars $OE'$ and $OD'$ to the sides $AC$ and $BC$, respectively. The distance is two options for the location of the point $D'$ and $E'$ relative to the point $D$ and $E$ (see Fig. 3). In case a) the angles $CDO$ and $CEO$ are obtuse, and in case b) they are both acute. Therefore, the vector of points $C$, $D$, $O$, $E$ can describe a circle only when the point $E'$ coincides with $E$, and $D'$ with $D$.

This leads to the conclusion that $AD \perp BC$ and $BE \perp AC$.

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We want to leave it as an exercise for readers to find the error in the above reasoning.

It should be noted that the number of examinees who correctly solved this problem (and similar problems of other options) was relatively low: approximately 1 out of 9 people solved it.

Task 3

Solution of the first equation of this system:

$$\cos x - \cos 13x = 0$$

or

$$2\sin 7x\sin 6x = 0$$

gives two series of values

a) $x = \frac{1}{7}n\pi$ ($n = 0, \pm 1, \pm 2, \ldots$),

6. $x = \frac{1}{6}m\pi$ ($m = 0, \pm 1, \pm 2, \ldots$).

Selecting from these values ​​those that satisfy the inequality $|x| < 3$, we find:

a) $x = \frac{1}{7}n\pi$, $|n| \leq 6$,

6. $x = \frac{1}{6}m\pi$, $|m| \leq 5$.

Now we must choose those values ​​of $x$ that satisfy the second equation of the system. This equation can be represented as:

$$1 = \sin 5x + \cos 2x = \sin 5x + \sin \left(\frac{\pi}{2} + 2x\right) = 2\sin \left(\frac{7x}{2} + \frac{\pi}{4}\right)\cos \left(\frac{3x}{2} - \frac{\pi}{4}\right).$$

Let us first substitute series (a) into it. Since $7x = n\pi$ ($|n| \leq 6$), the equality must be satisfied:

$$\sin \left(\frac{n\pi}{2} + \frac{\pi}{4}\right)\cos \left(\frac{3n\pi}{14} - \frac{\pi}{4}\right) = \frac{1}{2}$$

Since $\sin \left(\frac{n\pi}{2} + \frac{\pi}{4}\right) = \pm \frac{V^2}{2}$, then $\cos \left(\frac{3n\pi}{14} - \frac{\pi}{4}\right) = \pm \frac{V^2}{2}$ and, therefore, $\frac{3n\pi}{14} - \frac{\pi}{4}$ must differ from $\frac{\pi}{4}$ by an initial multiple of $\frac{\pi}{2}$. However, $\frac{3n\pi}{14} = \frac{3n}{7}\cdot \frac{\pi}{2}$, so $\frac{3n}{7}$ must be an integer. Under the constraint $|n| \leq 6$ this occurs only at $n = 0$. Substituting the value $n = 0$ into equation (1), we make sure that it satisfies it. So from series (a) only $x = 0$ is a solution.

Let’s now check series (6). Presenting the second equation of this system in the form:

$$\sin 5x = 1 - \cos 2x = 2\sin^2x,$$

we note that $\sin 5x$ must be non-negative, that is, the angle $5x = \frac{5}{6}m\pi$ ($|m| \leq 5$) is contained in the first two quadruples. Substituting possible integer values ​​$m$ from $-5$ to $+5$ into the expression $\frac{5}{6}m\pi$, we establish that only six are subject to verification

54

of which: $m=0, 1, 3, 5; -2, -4$. Calculating the left and right sides of equation (2) with these values, we obtain the table:

$$\begin{array}{c c c c c c c} m & 0 & 1 & 3 & 5 & -2 & -4 \\ \sin \frac{5m\pi}{6} & 0 & \frac{1}{2} & 1 & \frac{1}{2} & \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} \\ 2\sin^2 \frac{m\pi}{6} & 0 & \frac{1}{2} & 2 & \frac{1}{2} & \frac{3}{2} & \frac{3}{2} \end{array}$$

from which we obtain the required solutions: $m=0, 1, 5, t. e.$

$$x=0, \frac{\pi}{6}, \frac{5\pi}{6}.$$

Three values ​​of (3) give a complete solution to the problem, since the solution of series (a) gave the value $x=0$, already included in (3).

Note that in the same way, by brute force, we can check series (a).

When solving this problem, a typical mistake was the following: many forgot that the solution must satisfy all the conditions of the problem at the same time, and often simply did not understand this. Some limited themselves to solving one of the equations and selecting a solution based on inequality, others combined equations and ultimately solved only one equation that was not equivalent to the proposed system.

Task 4

Let $ABCD$ be the given quadrilateral, $AB=\sqrt{5}$. The primangular processes of any polygon onto parallel planes are equal to each other; Therefore, we will draw planes $P_1$ and $P_2$ parallel to the original planes through the vertex $A$ and in the future we will talk about the processes of the quadrilateral precisely on these planes.

For each point $M$ of space we will denote by $M_1,M_2$ and

55

Since $AB_1=2$, then from the right triangle $ABB_1$ there is $BB_1=\sqrt{5}-4=1$. Since in a right-angled triangle $AB_2B_3$ the hypotenuse $AB_2$ is equal to 2, and the leg $B_2B_3=BB_1=1$, then $\angle B_2AB_3=30^\circ$; therefore, the acute angle $AD_2D_3$, the sides of which are perpendicular to the sides of the angle $B_2AB_3$ ($AB_2\perp AD_2$, $D_2D_3\perp AB_3$), is also equal to $30^\circ$; therefore from the right triangle $AD_2D_3$ it is found that

$$D_2D_3=D_2A\cdot\frac{\sqrt{3}}{2}=\sqrt{3},$$

where

$$DD_1=D_2D_3=\sqrt{3}$$

And

$$AD=\sqrt{4}+3=\sqrt{7}.$$

Receptions $AB$ and $DC_2$ are the lines of intersection of parallel planes $ABB_1$ and $DD_1C_1C$ with the plane of a given quadrilateral $ABCD$; therefore $AB||DC$. We similarly conclude that $BC||AD$. Therefore, $ABCD$ is a parallelogram, and therefore its perimeter is equal to

$$2(\sqrt{5}+\sqrt{7}).$$

Please note that we wrote down the solution in such a way that we never had to refer to the drawing to justify the explanations (although in order to understand the solution, it is useful to make figure for yourself like 4 a, 6, c). Often, when solving such problems, they rely heavily on the drawing, forgetting to analyze all possible cases of one of them.

This problem has been solved very little; Therefore, it is impossible to talk about typical mistakes in connection with this task.

In conclusion, we offer the reader another version of the written task for the exercise.

Option 2

1. At points $A$ and $B$ two channels lead away from the river, crossing each other at right angles. The boat travels from $A$ to $B$ along the river for 1 hour 15 minutes, and along the canals for 1 hour 10 minutes. From point $C$ on the river, located between $A$ and $B$, one and a half times closer to $A$ than to $B$, you can get to the point of intersection of the canals using an automatic route in less than 13 minutes.

How long does it take the boat to travel from $A$ to the point of intersection of the channels, if the speed of the boat along the river is one third less, and the speed of the automatic vehicle is twice as high as the speed of the boat along the canal? (The river and canals are considered straight.)

2. In the isosceles triangle $KLM(KL=KM)$ the bissertrices $KN$, $LP$, $MQ$ are drawn. It is known that the vertex $K$ lies on the circle drawn through the points $N$, $P$, $Q$. Find $KM$ if $KP=2m$.

3. Find all values ​​of $x$ satisfying the simultaneous conditions:

$$2\cos^2 9x=1+\cos 10x,$$

$$\cos 9x+\sin 5x=1,$$

$$|x|<5.$$

4. The primugal processes of triangle $ABC$ onto two mutually perpendicular planes are regular triangles with sides equal to 1. The median $AD$ of triangle $ABC$ is equal to $\sqrt{\frac{9}{8}}$. Find $BC$.

56

WHAT CAN BE ADDED TO THE TITLE
OF OUR MAGAZINE?

(REPLICA MATHEMATICS)

From the article “The History of Quantum,” written in the first issue of the magazine, the reader learned about the important concept of modern physics and the origin of the term “quantum.”
It is interesting to note that another term, this time a mathematical one, owes its origin to this gelatinous root. So the answer to the question posed in the title is not difficult for a mathematician.

Fig. 1

The term “quantifier” belongs to one of the most important branches of modern mathematics—mathematical logic—and entered scientific circulation at approximately the same time as “quantum,” i.e., at the beginning of our century.

Let us try to briefly explain the meaning of this term and the designations associated with it. Consider, for example, the following equality:

$$x^2 = 2^x.$$

(1)

and we will assign to the variable $x$ all possible natural numbers as values: 1, 2, 3, 4, 5, etc. It will turn out that for some natural values ​​$x$ equality (1) will be correct (true), and for others it will be incorrect (false). Thus, the equality $1^2 = 2^1$ is false, and the equality $4^2 = 2^4$ is true.

In connection with our initial equality, it is appropriate to consider two statements: 1) these equality are satisfied by this and (any) natural values ​​$x$, 2) there are natural values ​​$x$ that satisfy this equality.

Statement 1) is denoted in mathematical logic as follows:

$$\forall x (x^2 = 2^x).$$

The symbol $\forall$ is readable for everyone and is called a general quantifier. And this symbol comes from the English word All (which means everything), in which only the first letter was left (mathematicians are very economical), and then also turned over (mathematicians, scales, are inventive).

57

---

running_head:
folio:

For statement 2) the following notation is accepted:
$$\exists x (x^2 = 2^x).$$

The symbol $\exists$ is read by an existent and is called an existential quantifier. The origin of this symbol is similar: the English word Exist (which means to exist) was treated in the same way as the word All.

Telery, in the new abbreviated notation, we get
$$\forall x (x^2 = 2^x) = JI \quad (\text{ложь}),$$
$$\exists x (x^2 = 2^x) = H \quad (\text{истина}).$$

Let’s look at another example, in which the initial set will be the set of all adult people - thus, in the statements below, we resolve together $x$ “substitute” any adult. In this case, it is not without embarrassment to admit that the statement $\forall x (x - \text{честный человек})$ false, but this is somewhat reassuring about the truth of the statement $\exists x (x - \text{честный человек}).$

That’s all that can be said in the first case.

The origin of the term quantifier (the Latin word quatim means “how much” (connected to the fact that in traditional logic statements are classified according to “quantitative content” - according to how many objects they relate to. For example, “Petrov is an honest man” is a single narrowing. \exists x, x - \text{честный человек}} — частное сужение. «\forall x, x - \text{an honest man}} is a universal narrowing.

F. Leonidov

REFLECTIONS ON THE WAY TO SCHOOL

1. On foot

The sum of house numbers on a block (a section of street between two intersections) is 33. What numbers are these?

It is easy to report the following:

1. The houses are located on the odd side, since the sum of four numbers is always.

2. The number of houses on a block is odd because the sum of four odd numbers is also even.

3. If the number of houses is odd, then there is an average house. Let his number be $l$. Neighboring houses have numbers $l+2$ and $l-2$, neighboring ones $l+4$ and $l-4$, etc. Thus, the sum of the numbers equidistant from the average is 2$l$ and, therefore, the double sum of house numbers on the block is 2$lm$, where $m$ is the number of houses. This means that the sum of the house numbers on the block is $lm$ and is not a prime number.

In our case, the number 33 can be decomposed into two copartners in two ways. 1) 33=3×11, 2) 33=1×33 It is clear that $m<l$ since negative numbers are not assigned to houses. This means that either $l=11$ and the block consists of houses numbered 9, 11, 13, or $l=33$ and the whole block consists of one house 33.

Solve the same problem if the sum of the house numbers is 333.

A. Chulsky

(Continued on page 59)

59

which are oriented in a given direction are called polarized.) The more ordered the directions of the magnetic moments of radioactive atomic nuclei, that is, the higher the degree of their polarization, the more clearly the difference in the intensity of the gamma ray emitted by such nuclei in different directions is manifested. But gold is not a ferromagnetic material, and the magnetic field of even very strong magnets cannot cause a noticeable orientation of its atomic nuclei.

However, B. N. Samoilov and V. V. Sklyarevsky showed that if radioactive gold is included in a ferromagnetic substance, such as iron, the resulting alloy is cooled to an ultra-low temperature (below 0.3 ° K) and placed in a strong magnetic field, then the production of gamma radiation disappears. Gold nuclei “more willingly” emit gamma rays in some specific directions. Later it turned out that the atoms of any element, once in a ferromagnetic material, are influenced by the polarized atoms of the ferromagnet. They are affected by magnetic fields of up to a million oersteds, causing ordering in the orientation of any atoms. Thanks to this, it will be possible to carry out studies with the nuclei of non-magnetic elements on a number of physical effects (the so-called superfine splittings), which were previously carried out only on the ordered orientation of the nuclei of magnetic elements.

This discovery makes it possible to polarize the nuclei of many atoms and obtain unique information about the atomic nuclei themselves, the electron shells of atoms and the properties of solids. Using it, one can already determine many values ​​of the magnetic moments of excited atomic nuclei.

The State Prize was also awarded to two famous Leningrad mathematicians O. A. Ladyzhenskaya and N. N. Uraltseva “for a series of works on boundary value problems for linear and quasalinear parabolic equations”, published in 1962-1967.

Parabolic equations are one of the types of mathematical physics equations that explore the most important problems in the natural and technical. They describe a large group of physical phenomena associated with the spread of bodies, diffusion of various substances, changes in their concentration and other processes in which the values ​​of physical quantities (temperature, concentration, etc.) are equalized.

The works of O. A. Ladyzhenskaya and N. N. Uraltseva considered the main problems of solving various equations of parabolic type and obtained important results that make a significant contribution to the general theory of differential equations.

V. A. Leykovtsev

REFLECTIONS
ON THE WAY TO SCHOOL

2. Vavtobus

The Indian bus without a conductor included 15 independent people. None of them have copper children, but only coins in denominations of 10, 15 and 20 kopecks. (A ticket costs 5 kopecks. You can throw any coin into the ticket office and tear off the corresponding number of tickets.) Nevertheless, the passengers who entered were able to pay for the fare by taking change from each other. Think about how this could be?

For those who have solved this problem, we suggest solving three more difficult ones:

1. Prove that these passengers had at least 20 silver coins (otherwise they would not have been able to pay).
2. Prove that they had at least eight 15-point coins.
3. Prove that all passengers together had at least 2 rubles. 50 k.

Of course, it is possible to come up with an example where N passengers pay only with coins in denominations of 10, 15, 20 kopecks for any N≥2. Think about what the exact and estimates are for the total number of coins, the total number of coins and the total number of coins in this case? N.V.

60

HOW TO SOLVE A PROBLEM!

What is better - to solve the problem yourself or read its solution? For a math lover, there can be no two answers to this question; the most interesting and useful thing is to find the solution yourself. Well, what if you spend a long time working on a problem, but you stubbornly fail to achieve a solution? It seems that both this and that have been tried, but still it is not possible to find the path leading to the goal.

However, advice on what to do if a problem “cannot be solved” is usually not given to mathematical problems. You have to open the ready-made solution placed at the end of the book and read it. At the same time, I involuntarily felt a feeling of dissatisfaction, annoyance with myself - after all, everything was clear, even simple, some small sense was missing. Now, if only someone could tell me where to start, what additional construction to make, then I could complete the decision myself! It’s even worse if, along with the solution, only the answer is included in the book: then you remain stuck with a treacherous task without any help…

Numerous observations (and, in particular, the experience of applied university exams) show that many students do not imagine how to attend to the solution of a problem if it is not an exercise on a “known type”, but is presented in a rather unusual way, if its formulation differs from the canons accepted in textbooks. Great difficulties are presented by tasks in which the center of gravity lies not in formal calculations, but in consistent logical reasoning, requiring the complex use of various sections of the school course.

Yungeyel will find such problems in sufficient quantities in the new problem book by E. B. Vakhovsky and A. A. Ryvkin on elementary mathematics*). There are 477 problems in total. Almost each of them requires an individual approach and a fresh approach, i.e., it provides an opportunity to test a small independent study, and not just apply some standard solution algorithm.

But what’s interesting about the book by E. B. Vakhovsky and A. A. Ryvkin is that the author did not just select various problems and provide them with solutions. To the traditional formula “problem condition - solution” an intermediate link is added - “instruction”, which has become its goal to give the necessary “additional impetus” to those readers who experience difficulties in solving a particular problem.

The authors, as it were, put themselves in the place of the person solving the problem, try to see and understand the source of its occurrence, direct his efforts in the most natural direction, dividing the solution into several stages (many tasks are given two instructions, and for some even three!), each of which the student can carry out himself. By skillfully and naturally helping the reader, leaving him a reasonable share of work and therefore maximally stimulating his independence, Youngheim allows amateur mathematicians to develop their “mathematical understanding”, to accumulate that experience that will later help to find approaches to new problems.

There is no doubt that the new problem book, both in its content and thanks to the “semi-programmed” system for presenting materials, will be very useful to students, especially those who live in rural areas, towns and small towns where there are no physics and mathematics schools.

The book by E. B. Vakhovsky and A. A. Ryvkin is no less

61

and certain shortcomings. Not all instructions can be considered successful; Some of them are not advice on how to look for solutions, but a condensed plan of the entire solution, and if the plan is clear, then all that remains is to carry out very extensive calculations. In many cases, it would be more appropriate to ask the reader a leading question rather than immediately recommending “consider this” or “do this.”
In the preface, the authors write that the theoretical introductions that are presented to the chapters of the problem book “must either be read from beginning to end, or not read at all.” Unfortunately, in many cases it is better to recommend the second second to another reader. The introduction to some chapters (for example, 7, 9, 23) is completely unsuccessful, or it contains fragmentary information, often debatable and not corresponding to the material of stable textbooks.
A number of other defects can be noted. The solutions to some problems are not the best. For example, read problem 6.10 “head-on”, following the rules for multiplying a multi-digit number by a single-digit number; the solution to Problem 13.13 is greatly reduced if we analyze its conditions and note that for all admissible x the inequality cos x>0 must hold. There are inaccuracies and typos in the book. So, the answer to problem 18.13 should be: 120+90√2 km; in the driving solution of problem 2.9, it is essential that the points A and C around which the rotation is made are the vertices of the acute angles of the triangle (we strongly recommend that the reader figure out where in the solution this is implicitly used!), the condition of problem 12.7 is not precisely formulated - it is necessary to additionally suggest that aB+bA≠0.

However, it hardly makes sense to give here a complete list of individual shortcomings. The young reader, we hope, understands that creating a book is a very complex matter, and it is not always possible, despite the successful authors, to do everything perfectly in the first edition.

The overwhelming majority of the problems included in the book were taken by the authors from versions of written works offered in different years of the advanced exams in mathematics at Moscow University, as well as in a number of other universities in Moscow. It is very good that these problems, equipped with instructions and solutions, become accessible to the broad masses of mathematics lovers and are included in school practice. But at the same time, it would be appropriate to accurately indicate the source of knowledge, since these tasks are the fruit of the collective creativity of the examination commissions. Anyone who has ever been involved in composing a problem knows how much work and how long it takes to create a new original problem. By the way, these links would also be useful to the reader - he could get an idea from them about the level of requirements for applicants to various universities).

The famous American mathematician D. Polya, in his book “How to Solve a Problem,” dedicated to the general methodology for solving mathematical problems, wrote: “If he (the mathematics teacher) awakens the curiosity of students, offering them problems that challenge their knowledge, and with his leading questions helps them solve these problems, then he will be able to instill in them a taste for independent thinking and develop the necessary abilities for this.” It is a pleasure to present and recommend to the reader a book of just this nature, a book that has its own purpose not only to show how to solve problems, but also to suggest (where necessary) how to solve them.

N. Kh. Rozov

RELEASED BY THE MSU PUBLISHING HOUSE, the book “Introduction to Electronics” by candidate of physical and mathematical sciences A. M. Khazen is addressed to schoolchildren who are interested in physics, mathematics and especially electronics and radio engineering.

The book is intended for the general reader and at the same time written on a good physical level. The clarity of the presentation and the original examples given are combined with sufficient rigor. It can be said with certainty that for a schoolchild who reads this book carefully, many of the understandings of electronics will become clear.

Young readers will be interested to know that a fundamental vacuum diode can be used to create an electric current generator operating from an external source of the body, how a linear electron accelerator and a traveling wave tube are similar and different, what an explosive magnetic generator is, etc.

The book highlights key issues for various classes of electronic devices. Despite the large number of specific devices described in the book, it is not overloaded with examples.

The books can be recommended as a guide for schoolchildren who have chosen technical disciplines or physics as their future specialization.

62

ANSWERS, INSTRUCTIONS, SOLUTIONS

By the way
“What is function graphics”

1. 1. $x \neq 0$; 2) $x \leq 1$; 3) $x \neq +1$.
2. $0 \leq x \leq 1$.
3. 6, of which 2 are reversed.
4. All points of the graph lie on two segments $D=0$, $0 \leq x \leq 1$, and $D=1$, $0 \leq x \leq 1$, and they are located on each of these segments “densely everywhere” (i.e., for any point $M$ segments and for any number $\varepsilon > 0$ there is a graph point at a distance less than $\varepsilon$ from the point $M$).
5. See Fig. 1.
6. a) $R^{90^\circ}_{(00)}$ (rotation relative to the origin counterclockwise by angle $90^\circ$);
7. a) $R^{90^\circ}_{(00)}$ (rotation relative to the origin counterclockwise by angle $90^\circ$);

Fig. 1

6. transformation symmetrical about the $Ox$ axis;
   e) parallel translation to a distance $[a]$ in the direction of the bisector of the angle $Oxy$, if $a > 0$, and in the opposite direction, if $a < 0$.

10. Parallel transfer to a vector with start at point $O_2$ and end at point $R^{\alpha}_{O_1}(O_2)$ (or, what is the same, to a vector with start $R^{-\alpha}_{O_2}(O_1)$ and end at $O_1$).
11. Flip the point $B$ symmetrically about the bisector.
12. Prove that if in the triangle $ABC$ there is imputation and $AB > BC$, then $ABM < MBC$, derive from the condition $AH = BM$ the equality $<MBC = 30^\circ$.

By the way
“As if an atom is weighed”

1. 5.55 km.
2. $P_2 = P_1 \left( \frac{P_1}{P_0} \right)$,
   $P_3 = P_2 \left( \frac{P_1}{P_0} \right) P_1 \left( \frac{P_1}{P_0} \right)^2$,
   $P_n = P_1 \left( \frac{P_1}{P_0} \right)^{n-1}$.
3. 11.1 km; 16.65 km. It is convenient to use the result of problem 1. Pressure is proportional to air density.
   To the “Logical task”
   A person cannot sleep and eat at the same time. Therefore, the period of seven days after sleep and after eating occurs at different times. A person must do what he did a week ago: sleep or eat.

Answers to questions about physics

1. Due to its high specific heat, water warms up more slowly than air. Therefore it is colder than air. When you get out of the water, the droplets of water remaining on your body will evaporate. Absorbing a lot of heat, they take it away not only from the surrounding air, but also from the body. The body cools and the air begins to seem colder than the water.

2. Most products contain water. As it evaporates, it freezes on the coldest part of the refrigerator - the evaporator. Therefore, the evaporator is covered with a thick snow coat, which has low thermal conductivity. This leads to a decrease in heat flow from the chamber, and the temperature in the refrigerator does not drop enough.