Kvant 1972, No. 2 · Pages

KEANT
2
FEBRUARY
1972

POPULAR SCIENTIFIC PHYSICS AND MATHEMATICS JOURNAL
USSR ACADEMY OF SCIENCES and USSR ACADEMY OF PEDAGOGICAL SCIENCES

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

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

On the first page of the cover there is a photograph for the article by A. S. Varpakhovsky and A. N. Vilenkin “Snow Crystals” (photo by A. N. Vilenkin). The article is located on page 32.

On the fourth page of the cover there are graphs of exponential and exponential functions.

Editorial Head L. V. Chernova Editor-in-Chief A. N. Klimanov
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The manual will not be returned

FOUNDED
IN
1970

2
FEBRUARY
1972

POPULAR SCIENTIFIC PHYSICS AND MATHEMATICS JOURNAL
USSR ACADEMY OF SCIENCES AND USSR ACADEMY OF PEDAGOGICAL SCIENCES

PUBLISHING HOUSE • SCIENCE•
MAIN EDITORIAL OFFICE OF PHYSICAL AND MATHEMATICAL LITERATURE

IN THE ROOM:

2 Relation of equivalence and partitioning of a set
M. M. Glukhov
10 Water on the Moon
M. A. Ginzburg
16 Switching circuits
P. B. Freivald
20 Measuring the speed of light
B. Vinetsky
26 Irrational sum of radicals
J. H. Kamnev
28 Radioactive memory
V. I. Kuznetsov

LABORATORY “QUANTA”
32 Snow Crystals
A. S. Varpakhovsky,
A. N. Vilenkin

MATHEMATICAL CIRCLE
36 Areas of polygons
P. P. Kantor,
J. M. Rabb

“QUANTUM PROBLEM”
42 Problems
44 Problem solutions: M84—M88; F92—F99

APPLICANT PRACTICUM
54 Methods for calculating equivalent resistances
A. Hatzem
60 Pyramid and sphere
Yu. V. Sidorov
65 Options for entrance exams in mathematics
1971.

REVIEWS, BIBLIOGRAPHY
67 About Polya’s book “Mathematical Discovery”
M. L. Smolyansky,
Yu. A. Gastev

COLLECTOR’S CORNER
70 Stamps associated with international geophysical scientific programs
A. B. Altykis

71 ANSWERS, GUIDELINES, SOLUTIONS
“QUANTUM” FOR JUNIOR SCHOOL CHILDREN (3rd page of the region) A. P. Savin,
MIXTURE
25 Problems about straight lines
B. M. Rosenthuler
31 Arithmetic and algebraic problems
41 Symposium of Young Mathematicians (Gorky)
53 Symposium of Young Mathematicians (Tbilisi)
Graphs of exponential functions (4th page area)

2

M. M. Shukhol

Misunderstood equivalence
and partitioning of the set

Try to imagine all the variety of semantic relationships of the word “class”: “working class”, “student of the 10th grade”, “kαιota of the 1st grade”, “class of mammals”… No matter how different these relationships may seem, the word “class” in all cases has the same origin - it is always associated with the classification, or partitioning, of any set into disjoint classes. The concept of “class” and its numerous subdivisions (family, type, category, species, variety, genus, category, and others) is widely used in literally all areas of human activity. For example, when we need to study the objects of a set in connection with some of their properties, we strive to divide this set into classes so that all elements of one class behave in the same way with respect to interrelated properties, and then it becomes possible to study a whole class of objects based on one of its representatives.

Thus, biology does not study each plant individually, but divides all plants into classes, families, genera and species, so that when studying, say, the logical family, one flower is constructed as a representative of an entire family (or species). When studying a class by its representative, one usually tries to choose the most simple one from the point of view of other properties (not of interest to us at the moment). Each of us has used this opportunity more than once, for example, when solving an equation, replacing one equation with another, simpler, but having the same roots, that is, contained in the class of equations equivalent to the given one.

To partition a set, that is, to distribute its elements into classes in such a way that each element falls into one and only one class, one usually uses various characteristics inherent in the elements of a given set. For example, the objects depicted in Fig. 1.

Fig. 2, a

Figure 1, can be divided into classes by color (Fig. 2, a), by shape (Fig. 2, b) or by some other characteristic. However, to divide a set into classes, we need not only the properties inherent in each element separately, but rules indicating whether two arbitrarily selected elements of a given set should be classified into the same or different classes. For example, all straight planes can be divided into classes if we agree to classify two straight lines into one class if and only if they are parallel. In the resulting partition, any two (not necessarily different *)) lines of the same class are parallel, and any two lines from different classes are not parallel. In this example, we do not use a property that each individual line has or does not have, but a property that refers to pairs of lines (whether a and b are parallel or not parallel). About the construction of a partition of lines they say that it is induced by the parallel relation

flattery. We can also say about the partitions of objects figure 1 considered above that they are induced by certain relations. Thus, partitioning by color is induced by the relation “having the same color.”

At the same time, not every relationship between elements of a set induces a partition of this set into classes. For example, if we tried to distribute all people into non-overlapping groups so that any two people from one group respected each other, but two people from different groups did not respect or did not know each other, then, obviously, nothing would come of this idea.

Check whether the perpendicularity relation will induce a partition of all straight lines of the plane.

What is attitude?

Let’s start by remembering some familiar relationships. So, in school mathematics you often

Fig. 2, b

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there were relations of equality (of numbers, figures, functions), relations of “more than” and “less than” for numbers, relations of divisibility for integers, relations of parallelism and perpendicularity for lines and polygons. In everyday life, relationships between people were widespread: relationships of seniority, kinship, neighborhood, partnership, belonging to the same nation, profession.

Now look more closely at one relation, for example, the relation of equality of numbers. Numerical equalities are correct (true) and incorrect (false). For example, the equality 5 = 5 is true, but 5 = 2 is false. If $x_1$ and $x_2$ are unknown numbers, then the equality $x_1 = x_2$ (the statement “$x_1$ is equal to $x_2$”) does not have a specific meaning: it depends on the “unknowns” $x_1, x_2$, is their function: for each specific numerical value $x_1, x_2$ it will be true or false.

In the general case, we will call a biary (or binary *) relation on the set $M$ a statement about the elements of the set $M$, which depends on two variables and turns into a true or false statement when substituting into it the variables of any specific elements from $M$. In general, the relation is written as $x_1\Theta x_2$ ($x_1 = x_2, x_1 < x_2, x_1 \parallel x_2, x_1 \perp x_2$, etc.). Often together “attitude $x_1\Theta x_2$” is simply said “attitude $\Theta$”. For example, together “relation $x_1 = x_2$” we say “equality relation”. The term “order relation”, “neighborhood relation”, etc. has a similar meaning. If the relation $a\Theta b$ is true, then it is also said that $a$ and $b$ are related by the relation $\Theta$ or are in the relation $\Theta$. If the relation $\Theta$ has the property that the truth of $a\Theta b$ always implies the truth of $b\Theta a$, then it is called “symmetrine”. (In what follows, we will sometimes call symmetrically “property 1.”) For example, the relations of parallelism and perpendicularity of straight lines are symmetrical, but the relation < on the set of real numbers is not symmetrical (5 < 8 is true, and 8 < 5 is false).

The relation $\Theta$ on the set $M$ is called reflexin (“property 2”). if $a\Theta a$ is true for any element $a$ of $M$, and transimitive (“property 3”) if the truths of $a\Theta b$ and $b\Theta c$ entail the truth of $a\Theta c$. For example, the relation of equality of numbers is reflexive and transitive, the relation of perpendicularity of lines is not reflexive and not transitive, and the relation “less than” for numbers is transitive, but not reflexive.

Problem. Find out which of the properties 1-3 have the divisibility relation for integers, the “intersect” relation (that is, would have one common point) for plane lines, and the neighborhood relation for people.

Sickening equivalence

Let us now return to the question that interests us: what relations, defined on a certain set $M$, induce partitioning of this set into classes? In other words, what properties should relations $\Theta$ have, according to which the set $M$ can be divided into disjoint classes such that

1. any two elements (different or the same) from the same class are in the relation $\Theta$,
2. no two elements from different classes are in the relation $\Theta$?

The answer to this question is:

Theorem. The relation $\Theta$ on the set $M$ can and only can induce a partition of the set $M$ when it is reflexive, symmetrical and transimitive.

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Proof. Let the relation $\Theta$ induce a partition of the set $M$, that is, the sets $M$ can be partitioned into classes satisfying conditions 1) and 2). Whatever element $a$ we take, it belongs to some class of our partition. According to 1), $a$ is in relation to $\Theta$ with itself: $a\Theta a$ is true, so $\Theta$ is reflexive. If, further, $a\Theta b$ is true, then, as follows from 1)–2), the elements $a$ and $b$ belong to the same partition class. Then from 1) it follows that $b\Theta a$ is also true. This means that for any $a, b \in M^*$) from $a\Theta b$ follows $b\Theta a$, that is, the relation $\Theta$ is symmetrical.

Do not read the data until you are sure that the relation $\Theta$ is transitive!

Thus, if the relation $\Theta$ induces a partition of the set $M$, then $\Theta$ is reflexive, symmetric and transitive: properties 1–3 are necessary for $\Theta$ to induce a partition of the set $M$.

Now it is clear why we could not divide a person into groups of people respecting each other and why the relation of perpendicularity did not induce the division of straight planes. The relation of “reflection” and perpendicularity is not transitive (and not reflexive).

Let us now prove sufficient properties 1–3. Let $\Theta$ have properties 1–3, and $a$ be an arbitrary element of the set $M$. Let us denote by $M_a$ the class (set) of all such elements $x$ from $M$ for which $a\Theta x$ is true:

$$M_a = \{x \mid a\Theta x \text{ — истинно}\}.$$

In this way we will obtain a certain set of classes. Since by property 1 $a\Theta a$ is true, then $a \in M_a$ (by definition of the class $M_a$) and, consequently, each element of the set $M$ belongs to at least one of the resulting classes. Let us show that any two classes $M_a, M_b$ either coincide or do not intersect. Indeed, if $M_a$ and $M_b$ have at least one common element $c$, then $a\Theta c$ and $b\Theta c^*$. By property 2, from $b\Theta c$ it follows $c\Theta b$, and from $a\Theta c$ and $c\Theta b$, by property 3 we obtain $a\Theta b$. So $a\Theta b$ is true. Let us now take an arbitrary element $d$ from the class $M_b$. By the definition of the class $M_b$ true $b\Theta d$, and by property 3 from $a\Theta b$ and $b\Theta d$ it follows $a\Theta d$; means $d \in M_a$. So, if $d \in M_b$, then $d \in M_a$. It is similarly proven that from $d \in M_a$ it follows $d \in M_b$ (carry out the corresponding reasoning yourself). This proves that $M_a = M_b$, that is, classes that have at least one common element, coincide. Thus, each element of the set $M$ is contained in some class and different classes do not intersect. This means that we have a partition of the set $M$. Let us show that this partition is real and induces the relation $\Theta$, that is, conditions 1)—2) are satisfied. Let the elements $a$ and $b$ be contained in the same class. Since $a \in M_a$, this means that $b \in M_a$ (in the previous form both to be committed are equal!), that is, $a\Theta b$. So, any two elements from the same class are in relation $\Theta$: condition 1) is satisfied. Now let $a$ and $b$ be two arbitrary elements from different classes, so that $M_a \neq M_b$. If there were $a\Theta b$, then this means that $b \in M_a$, that is, the classes $M_a$ and $M_b$ having at least one element $b$. But then, according to what was proven above, it would be $M_a = M_b$, which contradicts the conditions. This means that $a\Theta b$ is false, that is, elements from different classes are not in the relation $\Theta$ (condition 2 is satisfied). The theorem has been proven.

When fixing the set $M$, elements containing in one class are usually called an equivalence relation, and the classes of the resulting partition themselves are called equivalence classes.

*) The sign ∈ is read “belongs”: $u \in M$ means that $a$ is an element from the set $M$.

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Now the theorem proved above can be formulated more briefly: the relation $\Theta$ on the set $M$ can and only can induce a partition of the set $M$ when it is an equivalence relation.

Thus, if we want to find out whether some relation $\Theta$ induces a partition of some set $M$, then we need to check whether the relation $\Theta$ is an equivalence relation on $M$.

For example, the set of all circles of a plane falls into classes of concentric circles, since the relation of concentricity of circles is reflexive, symmetrical and transitive. (At the same time, we naturally believe that

Fig. 3.

every circle is concentric to itself.) It is also very simply verified that the similarity relation of triangles is reflexive, symmetrical and transitive. Consequently, the set of all triangles is divided into classes of similar triangles.

Let’s take one less obvious example. Let $Q$ be the set of all fractions of the form $\frac{a}{b}$, where $a, b$ are integers and $b \neq 0$. For each fraction $\frac{a}{b}$ we determine the class $Q_{a,b}$, assigning the fraction $\frac{c}{d}$ to this class if and only if $ad = bc$. Will the result be a division of all fractions into disjoint classes?

We have defined some relation $\Theta$ on the set $Q$:

$$\frac{a}{b} \Theta \frac{c}{d}$$ if and only if $ad = bc.$

Let us find out whether $\Theta$ is an equivalence relation.

1. Reflect. For any fraction $\frac{a}{b} \frac{a}{b} \Theta \frac{a}{b}$, since $ab = ba$.

2. Symmetry. If $\frac{a}{b} \Theta \frac{c}{d}$, then $ad = bc$, or $cb = da$. The last equality means that $\frac{c}{d} \Theta \frac{a}{b}$.

3. Transitivity. Let $\frac{a}{b} \Theta \frac{c}{d}$ and $\frac{c}{d} \Theta \frac{e}{f}$, that is, $ad = bc$ and $cf = de$. Multiplying the last equality by $f$ and $b$, respectively, we get:

$$adf = bcf, \ bcf = bde.$$

Hence we have $adf = bde$ and $af = be$ (can be shortened to $d$, since $d \neq 0$!). But $af = be$ just means that $\frac{a}{f} \Theta \frac{e}{f}$.

Thus, $\Theta$ is an equivalence relation and therefore the set of all fractions falls into disjoint classes of equivalent (or, as they say in school, equal in magnitude) fractions. So, having an equivalence relation on a certain set, we can divide this set into classes. But you can also do the opposite: first divide the set into classes, and then determine the equivalence relation, considering by definition two elements to be equivalent if and only if they belong to the same class of the partition in question.

For example, this is how the attitude of “being classmates” appears: first, children new to school are divided (almost arbitrarily) into classes, and then children who find themselves in the same class are called classmates.

Isn’t it possible to save money?

If we often have to check whether a given relation is reflexive, symmetrical and transitive, then naturally

The question naturally arises: is not any of the indicated properties 1-3 logical a consequence of the other two (in this case we would not have to check all three properties)?

It turns out that this is not so: the properties of reflectivity, symmetry and transnitivity are independent. To prove that a certain property does not depend on the other two, it is enough to give an example of a relation that has the last two properties, but does not have the one in question.

For example, the relation ≤ to the set of real numbers is reflective and transnitive, but not symmetric. This means that symmetry is not a consequence of reflectivity and transnitivity. The property of transnitivity is also not derived from the other two. For example, for lines in a plane, the relation “line $x_1$ coincides with line $x_2$ or perpendicular to $x_2$” is reflective and symmetrical, but not transnitive. (Check!)

An example showing the independence of reflectivity from the other two properties can be given yourself.

Disappointment of equivalence in mathematics

We have already talked about the significance of partitions of a set into classes (or, as we can now say, equivalence relations) in various areas of the person of action. The equivalence relation plays a very important role in mathematics. In addition to those mentioned earlier, you can give many other examples of relative equivalence that you have encountered in school mathematics. Such, for example, are the relations “to have the same absolute value” for numbers, the relations of equality of two algebraic expressions, the relations of equivalence of polygons and polygons.

Let’s take a closer look at two examples showing how equivalence relations are used when processing the most important concepts in mathematics.

Natural numbers. The concept of a natural number was developed by man in the process of long historical development. Before a person learned to abstractly imagine the number 5, he learned to open 5 pebbles, make 5 notches, tie 5 knots, etc. and compare these “standard” sets with other counter ones by establishing a one-to-one correspondence between them. This process of abstraction “at an accelerated pace” for each of us goes through in childhood, learning to first see two eyes, two apples, two cars, etc., and only then imagine the number 2. Now that this abstraction of each of us, presumably, has long been completed, try to understand the existing concept of a natural number. For this purpose, we define the following relations:

Between finite sets: we will assume that two finite sets $M$ and $N$ are in the relation $\Theta$ if and only if there is a one-to-one mapping from the set $M$ to $N$. (You can read about the concept of a set mapping, for example, in the article by A. N. Kolmogov “What is a function” in No. 1 of “Quantum” for 1970.) It is easy to verify that the relation $\Theta$ introduced in this way is reflective, symmetric and transnitive. (Check!) Therefore, it is an equivalence relation, and therefore all finite sets are divided into classes in accordance with it. Any two sets from the same class of the resulting partition are called equivalent. However, in order to distinguish the resulting partition from other partitions inspired by arbitrary reflective, symmetrical and transnitative relations, any relation is an equivalence relation; An example would be the relations identified, on the one hand, by all four-numbered numbers, and on the other, by all odd-numbered ones)—sets equivalent in this sense are called equal-numbered. For example, a class containing a set of letters {a, b, c, d, d} will contain a set of fingers on a person’s hand,

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a set of cosmonauts on the Voskhod and Voskhod-2 spaceships, a set of “parts of the world” (without Antarctica!), etc. Already from this small list it is clear that one class includes sets of very different natures and the only thing common to all of them is that they are all of equal power, that is, any one of them can be uniquely mapped onto any other. Now we will form the definition of a natural number: by a natural number we call any class of equally powerful finite sets.

At first, such a definition can be quite surprising: such a seemingly simple thing as a natural number is defined through something cumbersome and hard to see - through an entire class of sets! However, if you really think about this definition and get used to it, you will agree that it really reflects the “quantitative essence” of the concept of a natural number. This definition reveals the so-called abstraction of identification: first we divide all finite sets into classes of equal sets, then we “identify” all sets of one class and think of the result of this identification as a certain natural number.

Since each class of equipower sets is completely determined by any of its representatives, we can speak of a natural number determined by any set of a given class. The number defined by the set $M$ is denoted by $|M|$ and is called the order, or cardinality, of the set $M$. For natural numbers, you can introduce a notation system that is not associated with specific sets. Of course, this can be done in different ways. Historically, this was the case: different peoples had different notation systems and named natural numbers. (This is discussed in more detail in the article by I.M. Yaglom “Systems of Numbers”, see «Kvant», No. 6, 1970). Adhering to the currently generally accepted notation system and named, we can say that the number 1 (one) is the class of a set equal to the set of capitals of the Soviet Union, the number 2 (two) is the class of a set equal to the set of human eyes, and so on. By adding one more element to any finite set, we get a new set that is not equal to the original one. Carrying out this process, we will obtain an infinite sequence of sets that are not equal to each other and the natural numbers determined by it:

1, 2, 3, …

Rashchina says that a rational number is a fraction of the form $\frac{a}{b}$, where $a, b$ are integers and $b \neq 0$. However, such a definition is not complete, since it does not immediately follow from it whether, for example, the numbers $1/2$ and $3/6$ are equal or not. Therefore, it comes immediately to add: the rational numbers $\frac{a}{b}$ and $\frac{c}{d}$ are equal if and only if $ad = bc$. As a result, it turns out that a rational number is not just a fraction, but a whole class of fractions. Thus, the concept of a rational number is introduced more carefully as follows.

Consider the set $Q$ of all fractions and define on $Q$ the relation $\Theta$ by putting $\frac{a}{b} \Theta \frac{c}{d}$ if and only if $ad = bc$. We have already shown above that $\Theta$ is an equivalence relation and, therefore, the set $Q$ will develop into disjoint classes. We will call any two fractions of the same class of the specified partition accumulative. Now we can formulate a definition: a rational number is any class of equivalent fractions. After everything said earlier about the definition of a natural number, we define a rational number as a class of fractions in a very simple way.

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We do not know all the resources of the Moon, but what is known is enough to assume that lunar stations and colonies will appear on the Moon in the future. One of the main difficulties in organizing a colony is providing people with water and oxygen. A person needs approximately 0.9 ke of anhydrous food, 0.9 ke of oxygen and 2.7 ke of water per day. No traces of water were found in lunar images delivered by Soviet and American spacecraft.

However, there is reason to believe that there was water on the Moon.

Nowadays, the theory of evolution has become widespread, according to which the Moon, like the Earth, was formed from a cloud of small solid particles. These particles on the Moon contained approximately the same amount of water (the so-called juvenile, that is, original water) as the substance of the Earth contained it: the “building material” of the Earth and the Moon was approximately the same. The decay of radioactive uranium and radioactive potassium led over time to the heating of the deep layers of the Moon to temperatures of 1000-2000°C. The water contained in the Moon’s substance evaporated and penetrated outward in the form of steam, to the surface. The fate of this steam depended on the conditions at the surface of the Moon and its temperature.

The surface temperature of the Moon reaches +120°C during the day, and drops to −160°C at night (in the polar regions to −200°C). The materials that make up the surface of the Moon (small fragments of various minerals - the so-called “regolith”) do not conduct heat well. Therefore, daily temperature fluctuations penetrate very shallowly, and already at a depth of less than a meter below the surface, the temperature does not depend on the time of the lunar day. In low latitudes, near the equator, it is approximately −30°C. At such a low temperature below the surface, the water vapor released from the depths of the Moon should have frozen, forming a layer of ice.

Let’s estimate the boundaries of the ice layer. Temperature increases with depth. On Earth, the temperature increases by 28°C for every kilometer traveled into the interior. Let us assume that the radioactive processes of heat generation on the Moon are the same as on Earth. Then the temperature on the Moon rises by 30°C to the ice melting point (0°C) at a depth of about 1 km. This is the lower boundary of the lunar permafrost layer (Fig. 1), and below it is a layer of water.

At the upper boundary of the ice layer, molecules continuously break off from its surface, and the ice evaporates. Rocks on the Moon, how are they established?

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updated during the study of lunar samples, they always contain pores. The steam molecules flow through these pores and emerge onto the surface of the Moon.

We know the age of the Moon’s rocks from the relative abundance of radioactive isotopes and their decay products in lunar samples. This age is about 4.5 billion years. The calculation shows that with $t = -30^\circ\text{C}$ and the physical properties and structure of lunar rocks known to us, not all of the led from the lunar crust evaporated over these 4.5 billion years.

The hypothesis about permafrost and a deep layer of water on the Moon was envisioned in 1961 by the American astrophysicist Gold. However, his conclusion about the existence of ice and water deep in the lunar crust is based on some assumptions.

Fig. 1.

Firstly, it is believed that the total amount of juvenile water in 1 kg of matter of the Earth and the Moon was the same, and this total amount of juvenile water is estimated by the average depth of the world’s oceans (layers of water obtained by dividing the volume of water of the oceans, seas, ice of the Arctic and Antarctic, ice of mountainous regions on the surface of the Earth). On Earth this depth is ~1 km.

Secondly, it is believed that a significant part of the juvenile water comes to the surface after the lunar crust has cooled and therefore froze in the led, which evaporated slowly - through the pores.

If vapors of juvenile water managed to emerge on the surface of the lunar crust, in order to reliably calculate the process of evaporation of ice from it.

There are also subsurface ices on Earth. In those areas where the average pressure temperature is below 0°C, a layer of permafrost soil remains under the Earth’s surface, which only partially thaws in the summer. Permafrost areas make up 15% of the Earth’s land mass (and another 10% of the landmass is occupied by glaciers). In our country, permafrost covers the common areas of Siberia and the Far East.

Transformations led—water—steam on the Moon and on Earth *)

Let us now rise from the depths to the surface of the Moon. How water, LED and steam will behave during the day ($t = +120^\circ\text{C}$)

*) This recommendation from the reader who popularized the book, Academician L. D. Landu and Prof. A.I. Kitaygorodsky “Physics for everyone”, M., “Nayka”, 1963. Chapter 12 of this book will serve as a guide on all physical issues raised in our country.

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Fig. 2.

and night ($t = -160^\circ\text{C}$) on the surface itself, without a protected layer of the lunar crust? The answer to these questions is given by the so-called state diagram of water (Fig. 2), which we will now discuss.

We will plot the pressure under which ice, water or steam is located along the vertical axis, and their temperature along the horizontal axis. Each point on the plane bounded by the coordinate axes $P$ and $t^\circ$ corresponds to a certain state of water (ice, water or steam).

Let’s start with ice. The blue curve shows the dependence of the pressure under which ice melts on temperature. The higher the pressure, the lower the melting point. Let’s move away from the blue curve a little to the right: at the same pressure we will increase the temperature. The ice will completely melt - the points of the plane to the right of the blue curve correspond to water. If you move to the left from the blue curve and reduce the temperature, the water will freeze and turn into ice. At the blue points of the curve (and only at these pressures and temperatures) ice and water are in equilibrium with each other - how much ice melts, how much water turns into ice. Therefore, the blue curve is called the equilibrium curve of ice and water.

If we increase the temperature of water at a constant pressure, we will find ourselves in a region of temperatures at which only steam can exist. This region of temperature and pressure is to the right of the red curve - the equilibrium curve of water and steam, or, in other words, the curve of the dependence of the boiling point of water on pressure. This curve, as can be seen from figure, shows that the lower the pressure under which water is, the lower its boiling point. We know this well - for example, high in the mountains, where the air pressure is low, it is difficult for a climber to boil an egg in boiling water, so we have to fry the eggs. At the top of Elbrus, where the pressure is equal to the atmospheric atmosphere, water boils at $82^\circ\text{C}$.

Let’s reduce the pressure to 0.006 am. Experience shows that the boiling point drops to zero (more precisely, to $0,0075^\circ\text{C}$) - “boiling water” has the freezing temperature of water. What will happen if the pressure drops further? Boiling temperature drops below zero, “boiling water”

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equal to or greater than atmospheric if the ice is covered with a layer of water (Fig. 3).

For the water-steam equilibrium curve, everything is circumstantial. All points to the right of the red curve correspond to the steady state - steam, to the left - to the steady state - water. If the water vapor pressure is less than the pressure on the red curve, then the water at a given temperature will completely evaporate (the steam above it is unsaturated). In atmospheric exposure, the vapor pressure of water is usually 0.013 am, that is, 60% of the saturated vapor pressure. Therefore, the water absorbed in the dish will always evaporate over time. When air is cooled, the pressure of 0.013 am becomes greater than the saturated vapor pressure at the new temperature and water vapor condenses (for example, morning dew).

But evaporation from the surface of a liquid is a very slow process. There is another process of transition of liquid into vapor, much more quickly - evaporation during boiling. In this case, vapor bubbles form throughout the entire volume of the liquid. External pressure (total pressure 1 am) is transmitted from the surface of the liquid to all points in its depth, so the resulting internal pressure is equal to (or greater than) the total pressure on the liquid, regardless of whether the cause of this external pressure is water vapor or atmospheric air pressure.

On the Moon, water always boils on the open surface, but freezes below. But depending on the time of day (that is, on heating by the Sun), there can be a wide variety of effects.

Whether steam freezes or not depends not only on its temperature, but also on pressure. On the Moon, at low pressures, water vapor could escape on its surface and not freeze. On figure 2 we see that under lunar conditions water vapor can exist at temperatures close to -200°C. It all depends on conditions (dependence of pressure on depth, temperature, porosity of the Moon’s substance), which over the 4.5 billion years of the Moon’s history could be very different.

Thus, both main points of the theory of subsurface ice - the release of a sufficient amount of water vapor from the depths of the Moon and its freezing at the surface - are based on certain assumptions and need to be tested experimentally. We are working on developing experimental methods for searching for ice and using them to examine the Moon over its entire surface, as well as in depth.

The temperature range on the lunar surface is very wide (from -160 to +120°C). It can be suggested that the properties of ice in such a wide temperature range are also not constant. The search for ice on the Moon (still theoretical) prompted new research into the properties of ice in laboratories on Earth.

In 1969, the process of freezing water was studied under conditions of the lunar shadow, that is, at a pressure less than ten millionths of the atmosphere, and a temperature of (-170°C) - (-180°C). At the same time, a hitherto unknown form of ice was discovered - amorphous ice. It is twice as dense as water: when water freezes at elevated temperatures, it does not expand (remember that general ice floats on the surface of water), but collapses.

The experimental design is very simple. Distilled water was introduced into a vacuum chamber and frozen into a thin layer of ice on a metal cone. The volume of ice was determined by comparing photographs of the cone before and after ice freezing, the mass was determined by weighing. From here comes the density. Ice frozen on the cone at a temperature below -175°C had a density of 2.3 e/cm³ and was sticky on the glass. Using two-mechanical methods of measuring the temperature of water, on the same scale, the temperature of water is measured using a thermometer.

Are rivers possible on the Moon?

At first glance, no. If water is in a solid or gaseous state, then how can it flow? However, upon closer analysis of the state diagram (Fig. 2), such a simple answer turns out to be incorrect.

Let’s imagine that there is a stream of water flowing across the surface of the Moon. Water will evaporate intensely from the surface. Evaporation wastes a lot of heat. This removes heat from the still uncorrected water and it begins to freeze. The ice that forms is lighter than water. It will float on the surface and creates additional pressure on the surface of the water with its weight. When the pressure of the ice layer $P = \rho gh$ exceeds the pressure at the triple point $P_0 = 0,006 \text{ am}$, the water will stop evaporating.

On the Moon $g = 1,16 \text{ m/cm}^2$, the density of ice is $0,9 \text{ e/cm}^3$, that is, the minimum thickness of the ice layer that protects water from boiling at $0^\circ \text{ C}$ is equal

$$h = \frac{P_0}{\rho g} = \frac{0,006 \cdot 10^5 \text{ m/m}^2}{0,9 \cdot 10^3 \text{ kg/m}^3 \cdot 1,6 \text{ m/cm}^2} \approx 0,4 \text{ m}.$$

Are there grooves on the surface of the Moon similar to the beds of earthly rivers? It turns out yes. The winding grooves were replaced on the Moon as early as 1788. They usually originate in craters, on high ground, and end in lowlands. Recently, American scientists Lingenfelter, Peel and Schubert proposed that the winding grooves are the former beds of lunar water rivers flowing under a protective layer of ice. Indeed, in their structure and size, the winding furrows are similar in appearance to the beds of earthly rivers. The subglacial layer can serve as a source of water. A large meteor is able to break through the lunar crust to the water layer and open a path for water on the surface.

But there are other theories of the origin of winding furrows, for example, from lava flows. Whether this or that theory holds true must be shown by further research. In “earthly” laboratories,

Attempts have been made to model “lunar rivers.”

It was placed in a vacuum chamber under the nose with dimensions $30 \times 42,5 \text{ cm}$, up to which it was filled with a layer of crushed rock with particles from 100 microns to 4 mm (in most experiments, the diameter of the particles of this “rheolite” is 100-500 microns). The thickness of the porda layer varied from 0 to 5 cm. The tray was placed obliquely. Water was introduced through a tube at the end of the tray, raised upward, at a speed of up to 800 milliliters per minute. The pressure in the chamber is maintained at about 0.0013 am, that is, five times lower than the pressure at the triple point.

The results of the experiment were as follows. At atmospheric pressure, the flow of water actually broke through a system of channels in the “rheolitta”. True, the meanders characteristic of terrestrial rivers were not observed, but these channels also branched out and changed direction.

During experiments in vacuum, a completely different picture was observed. The flow of water intensively closed along its entire length, dispersing particles of “lunar dust” and forming ice to the sides. The boiling continued for several seconds until the water source was covered with an ice crust. The water continued to flow under the sheet of ice, but did not necessarily flow downwards, in the direction of the slope of the tray with soil. It seeped through the “rheolite,” breaking through the surface in one place or another. This process reveals the character of an explosion on the surface—a jet of steam carried away particles of the rheolate, and then an ice crust formed. At the end of each, a layer of ice covered the entire tray of soil. Then he evaporated. No channel formations were observed, and the soil surface became cold.

So the hypothesis about “lunar rivers” was not found experimentally confirmed.

It is more likely that water is released from the depths of the Moon by volcanoes. Volcanic activity (the release of gases from the depths of the Moon) was discovered by the Leningrad astronomer Processor Kozyrev. If there is still water in the depths, then when gases erupt, water vapor should also be released.

From the editor

Recently there was a report in our press that a group of American scientists who were conducting a study of the Moon using instrument readings left on its surface by the crews of the Apollo 12 and Apollo 14 spacecraft discovered a “lunar geyser” spewing water vapor. This message needs further verification.

16

SWITCHING DIAGRAMS

R. V. Freivald

At the V All-Union Mathematical Optimization*) the following problem was proposed:

A switch (Fig.1) with two inputs and two outputs can be in two different states.

figure 2 shows a telephone communication circuit with three inputs and three outputs, which has the following property of “universality”:

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

by changing the state of the switch, you can make any of the six connections of three inputs with three different outputs, that is

(Check this out. Note that the total number of different states of this circuit is 2³=8, since each of the switches can be in two states.)

a) Construct a circuit with four inputs and four outputs that would be “universal”, that is, any of the 24 possible connections of inputs and outputs would be implemented.

b) What is the minimum number of switches needed for such a circuit?

b) We call a circuit with n inputs and n outputs n-universal if

*) For more information about random optimization, see «Kvant» No. 11, 1971, pp. 35-42.

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Divide into blocks $A$ and $B$ so that 4 connections pass through $A$ and 4 through $B$. In doing so, we must ensure that two connections do not pass through the same input or the same output of blocks $A$ and $B$.

Let’s consider the first and fifth inputs of our circuit. Each of them can be connected either to the first input of block $A$ or to the first input of block $B$. Therefore, one of these inputs must be connected to block $A$, the other to block $B$, otherwise two connections will be connected to the same input of some block. The following pairs of inputs must also be connected to different blocks: 2nd and 6th, 3rd and 7th, 4th and 8th. It is easy to see that such a condition for distributing connections across blocks is not only necessary, but also sufficient to ensure that different connections in the circuit do not pass through the same input of blocks $A$ and $B$. To form the same condition for the inputs to blocks $A$ and $B$, it is useful to pay attention to the symmetry of the arrangement of the inputs and outputs of the circuit. We conclude that the following pairs of circuit outputs must be connected to different blocks: 1st and 5th, 2nd and 6th, 3rd and 7th, 4th and 8th.

Let us depict on figure these relationships between different connections of the same compound. Each individual connection will be represented by a rectangle, inside of which it is written which input this connection connects to which output. If two connections must pass through different blocks, then the corresponding rectangles are connected by a linear line. In this case, we will draw a linear line in red if there are 4 differences with inputs, and in blue if there are 4 differences with outputs.

For example, connection

$$\begin{array}{c c c c c c c c} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \downarrow & \downarrow & \downarrow & \downarrow & \downarrow & \downarrow & \downarrow & \downarrow \\ 3 & 8 & 1 & 6 & 2 & 7 & 5 & 4 \end{array}$$

depicted in this way on figure 7.

Whatever the connection, exactly one blue and exactly one red line emerges from each rectangle. Therefore figure always consists of one or more “rings”. Moreover, in each “ring” red and blue lines alternate, otherwise two lines of the same color would come out of a certain rectangle. Consequently, each “ring” contains a number of primogols.

Using this figure, it is easy to distribute connections among blocks so that all the above restrictions are met. To do this, you need to ensure that the connections that correspond to rectangles connected by a linear line pass through the blocks. Therefore, in each “ring” figure you need to randomly select one slug and the connection corresponding to this slug should be made through block $A$, then the connection corresponding to the corresponding slug should be made through block $B$, then the connection corresponding

19

shue to the next rectangular, through $A$ and so on until the “ring” is closed. The connections corresponding to the first and last rectangles will be made through different blocks, since the number of rectangles in the “ring” is even. The 8-universal scheme (see Fig. 3) has been proven. We only note that the distribution of communication across blocks can always be carried out so that one of the switches (for example, the lower left) is in some predetermined state. (In other words, this switch can be removed from the circuit, but the circuit cannot be changed to be universal.) Then, in the “ring” containing the connection passing through the selected switch, the distribution of the connection across blocks must begin with this connection.

To estimate from above the number of switches in a minimal $n$-universal circuit, we will construct $n$-universal circuits with as few switches as possible. The above proof of the 8-universal circuit (see Fig.3) can be easily generalized to the case of any number of inputs. This method makes it possible, having an 8-universal circuit, to build a fairly economical 16-universal circuit, then a 32-universal circuit, and so on. It can be calculated that if $n=2^k$ and the circuit (Fig. 4) is used as blocks of a 4-universal circuit, then such a circuit has $2k \cdot 2^k - \frac{11}{4} \cdot 2^k = 2n \cdot \log_2 n - \frac{11}{4} \cdot n$ switches. If $n$ is not equal to a whole power of two, then you can use a circuit that is universal for a large number of inputs equal to a full power of two. Since the closest complete power of two exceeds the given number $n$ by no more than twice, it turns out that for any $n$ there is a $n$-universal circuit containing at most $4n \cdot \log_2 n$ switches.

(For ending, see page 27.)

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Even 300 years BC, Greek scientists already knew that light travels in a straight line, has the property of being reflected from the eye surfaces, and that the angle of incidence is equal to the angle of reflection. Ancient knowledge is also about the refraction of light. Claudius Ptolemy, the great astronomer and mathematician of antiquity, compiled tables of his measurements of the angles of incidence and refraction. But there was no consensus on the issue of the speed of light in ancient science. Most scientists of that time, including Aristotle, believed that the speed of light was infinitely high - after all, it could not be measured even approximately by any of the instruments and methods that existed at that time. However, 900 years ago, the Arab scientist and physician Avicenna proposed that the speed of light, although very high, is finite. His contemporary Algazen also agreed with him. But, of course, none of them could support their opinion with facts or experiments.

One way or another, all generations over the centuries have passed or interest in the nature of light, as evidenced by the myths, legends, and traditions that have reached us. This inexhaustible interest in natural phenomena was at first the only motive that prompted scientists to accept the experimental measurement of the speed of light. However, from the largest experimental physicists

tator Albert Michelson, who devoted almost his entire life to measuring the speed of light, when asked by correspondents: “Why is this needed?” replied: “This is terribly interesting!” But even before Michelson’s work, determining the speed of light and its dependence on various conditions acquired the character of not only satisfying the curiosity of scientists. The increasingly exciting idea of ​​the “luminiferous ether”, the question of the corpuscular or wave nature of light, and then the relative theory required an accurate knowledge of the speed of light. In our time, the precise determination of this quantity has purely practical significance. It is necessary to know it when calculating the energy of nuclear reactions, and for measuring distances between space objects in connection with the expansion of space research, and in many other equally important problems.

The first person to try to experimentally determine the speed of light was Galileo Galilei. His method was extremely simple, but, unfortunately, it did not produce results. This is how Galileo himself describes it:

“The experience I came up with is as follows. Two persons each hold a fire, enclosed in a lantern or the like, which can be opened or closed by the movement of the hand in plain sight.

22

Fig. 2.

Will spread in the telescope tube ($\tau = \frac{l}{c}$, where $c$ is the speed of light, $l$ is the length of the tube), the Earth will travel along the orbit a path equal to $v \cdot \tau$, and the image of the star will move away from the center of the field of view to the same distance (Fig. 2, a). To return the image of the star to its place, you need, very much, to tilt the telescope in the direction of the Earth’s movement at such an angle $\alpha$ that $tg \alpha = \frac{v \tau}{l}$ (Fig. 2, b). Since $l = c \tau$, then $tg \alpha = \frac{v \tau}{c \tau} = \frac{v}{c}$. Hence $c = \frac{v}{tg \alpha}$. The same will happen in pro-
opposite point of the earth’s orbit, but the telephone will be tilted in the opposite direction. Thus, directions to the same star can differ by a maximum angle of $2\alpha$. The changes showed that $2\alpha = 41''$. During astronomical observations, the telescope, of course, is not tilted. Thanks to the change in the direction of the Earth’s speed, the image of the star will be at different points in the telescope’s field of view on different days and will form an ellipse throughout the year. The major axis of this ellipse will be visible from the center of the lens at an angle of $2\alpha$. Knowing the angle $\alpha$, Bradley found the speed of light and obtained a value for it of 303,000 km/sec. This was in 1728.

However, scientists were not satisfied; they would like to change the insufficient early light on Earth, without leaving the boundaries of their home planet, so that all conditions of the experiment were under control. The solution to this problem had to wait one hundred and twenty years.

In 1849, the French physicist Armand Hippolyte Louis Fizeau came up with a fairly simple method for this. Its scheme is as follows (Fig. 3). A ray of light from a source $S$ falls on a semi-open mirror plate $\Pi$, reflected from it, directed through collecting lines to a stationary

Fig. 3

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The length of the base is $r$, then let the light travel $3_13_2$ and back in time $\Delta t = \frac{2r}{c}$. During this time, the mirror $3_1$, rotating with a frequency $n$ (rpm), will rotate through an angle $\alpha = 2\pi n \cdot \frac{2r}{c}$. According to the law of reflection, the angle between rays 1 and 2 (see Fig. 4) on the path $3_1\Pi$ will be equal to $2\alpha = 8\pi n \cdot \frac{r}{c}$.

Thus, the image $S'_1$, which is seen by the observer with a rotating mirror $3_1$, turns out to be mixed from the image $S'$, obtained with a stationary mirror $3_1$, by an angle equal to $8\pi n \cdot \frac{r}{c}$.

Using this method, Foucault obtained the value $c = 298000$ km/sec for the speed of light in air.

In accordance with Arago’s original plan, Foucault used his installation to change the speed of light in water. To do this, he slightly changed the property by equipping it with two symmetrically located spherical mirrors and placing water between one of them and the rotating mirror. This allows him to compare the speed of light in water and air and find that the speed of light in water is $3/4$ of its value in air. This result was the final blow to Newton’s corpuscular theory of light.

The methods of physics and Foucault made it possible to change the speed of light with an accuracy of $\pm 500$ km/sec.

In 1873, Alfred Cornu, who repeated the experiment of physics, changed the speed of light with an accuracy of $\pm 200$ km/sec.

In 1877, the American physicist Albert Michelson, the conditions of light with an accuracy of $\pm 50$ km/sec.

But Michelson was not satisfied with the result. From 1877 until the end of his life, he repeatedly returned to changing the speed of light, continuously improving his installation. The value $c = 299796 \pm 4$ km/sec obtained by Michels was considered for a long time to be the most accurate.

In the problem of changing the speed of light, Michelson’s experiments completed a stage that can be called classical. But even before the end of Michelson’s work, new experiments were carried out, already on a new technical and ideological basis, provided by the rapid development of physics. These experiments continue to the present day; their goal is to establish the most accurate value for the speed of light.

Modern methods of changing the speed of light can be divided into

26

Recall that a number is called rational if it is equal to $\frac{m}{n}$, where $n$ and $m$ are integers, and rational otherwise. Everyone knows that the number $\sqrt{2}$ is irrational. It is easy to prove a more general statement: if the natural number $k$ is not a perfect square, then the number $\sqrt{k}$ is irrational. Note that $k$ can be represented as $b^2a$, where $a$ and $b$ are natural numbers, and $a$ is not divisible by the square of an integer (as they say, “free from squares”), that is, $a$ is equal to a prime number or the product of different prime numbers: $a=p_1p_2...p_r$; $r \ge 1$. The statement that the number $\sqrt{k}=b\sqrt{a}$ is irrational properties the following: there are no non-zero integers $b_1$ and $b_2$ such that $b_1\sqrt{a}+b_2=0$. Indeed, if such $b_1$ and $b_2$ exist, then $b_1^2a=b_2^2$. But this equality contradicts the theorem on the uniqueness of the decomposition of a natural number into prime factors: $p_1$ enters the right side to an even degree, and the left side to an odd degree.

In this note we will prove that if $b_1, b_2,..., b_n$ are non-zero integers, $a_1, a_2,..., a_n$ are square-free natural numbers, then $b_1\sqrt{a_1}+b_2\sqrt{a_2}+\dots+b_n\sqrt{a_n} \neq 0$.

Up to $n=4$ this is easy to prove by contradiction by raising both parts (grouped approaches in a manner) into a square several times. In the general case, we will carry out a proof of induction on the number of distinct prime numbers $p_1,..., p_N$ that make at least one of the numbers $a_i$. As is often the case, it is more convenient to prove by induction a stronger statement (interesting in itself):

there are integers $c_1, d_1, c_2, d_2,..., c_e, d_e$ such that $d_i \neq 0$, $c_i \ge 1$, all prime divisors of the numbers $c_i$ are contained among the numbers $p_1,..., p_N$ and the product

$$\left(d_1\sqrt{c_1}+\dots+d_e\sqrt{c_e}\right)\times\left(b_1\sqrt{a_1}+\dots+b_n\sqrt{a_n}\right)$$

is a non-zero integer.

Let us denote $b_1\sqrt{a_1}+\dots+b_n\sqrt{a_n}$ by $S$, and eliminate the sum $d_1\sqrt{c_1}+\dots+d_e\sqrt{c_e}$ by $S'$.

If $N=1$, then s has the form $b_1\sqrt{p_1}+b_2\sqrt{p_2}-b_2$: we proved above that $SS'=b_1^2p_1-b_2^2 \neq 0$. Let us now assume that $N \ge 2$ and that our statement is valid for all smaller values ​​of $N$.

We will denote by letters $S_1, S_2,..., S_8$ sums of the form $\beta_1\sqrt{\alpha_1}+\dots+\beta_m\sqrt{\alpha_m}$, where $\beta_i$ are integers, $\alpha_i$ are square-free natural numbers, all of whose prime divisors are contained among the numbers $p_1,..., p_{N-1}; S_1, S_2,..., S_8$, unless otherwise stated, can be equal to zero.

The sum $S$ can be written as $S=S_1+S_2\sqrt{p_N}$, where $S_2 \neq 0$. By the induction hypothesis, there is a sum $S_3$ such that $f=S_3S_2$ is a nonzero integer. Produced

28

V. I. Kuznetsov

One day, builders are paving a road and find an ancient cave filled with volcanic rocks. It contained a warehouse of shoes - three hundred pairs of sandals made of woven grass. When, in what building was this amazing warehouse founded? There was no answer. It was obtained only a few years later, when a method was developed for determining the age of a product made of wood, bone and other materials containing carbon.

To understand how the time of creation of such products is determined, let us turn to the nuclear processes occurring in the upper layers of the atmosphere (at an altitude of about 15 kilometers).

The earth’s atmosphere is continuously bombarded by cosmic rays. The bulk of cosmic radiation comes from fast-moving protons—the nuclei of hydrogen atoms. Their energy is enormous - it reaches billions of electron volts. Passing through the atmosphere, protons collide with the nuclei of atoms that make up the air. When atomic nuclei collide, they release neutrons and other particles. The resulting neutrons, in turn, interact with atmospheric matter, most often with nitrogen. In some cases, neutrons knock a proton out of the nitrogen nucleus (consisting of 7 protons and 7 neutrons) and take its place. As a result, nitrogen turns into an isotope of carbon, the nucleus of which contains 6 protons and 8 neutrons. The nuclear reaction that occurs can be written as follows:

$$\frac{14}{7}N + \frac{1}{0}n \rightarrow \frac{14}{6}C + \frac{1}{1}p.$$

The new atom has two more neutrons than the nucleus of the most common isotope of carbon, $^{12}$C. Heavy carbon $^{14}$C is called radiocarbon. Radiocarbon nuclei emit electrons (beta particles) and turn into nitrogen: $^{14}$C $\rightarrow^{14}$N. The rate of this process is characterized by the time during which half of the first initial number of atoms $^{14}$C is converted into nitrogen. This time is called the half-life and is usually designated $T_{\frac{1}{2}}$. For the isotope $^{14}$C $T_{\frac{1}{2}} = 5600$ years. If another $T_{\frac{1}{2}}$ years pass, then of the remaining half of the atoms, half will decay again and one quarter remains - $(1/2)^2$, and in a time of $t$, equal to $n$ half-lives,

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keep only the $(1/2)^n$ part of the $^{14}\text{C}$ atoms:

$$N = N_0 \left(\frac{1}{2}\right)^n = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}.$$

(1)

Here $N_0$ is the number of radiocarbon atoms at the initial moment, $N$ is their number at time $t$.

The half-life of radiocarbon is a value of the same order as the age of a person in a culture. History has reliable data only within 5000 years. Knowing the number of radiocarbon decays per unit volume or weight per unit time in the atmosphere, water and living materials, it is not difficult to calculate how much of it decays per year and how much of it is contained on Earth. Calculations have shown that there are only about 60 tons of radiocarbon on Earth.

For every thousand billion atoms of earth’s carbon there is one atom $^{14}\text{C}$. Radiocarbon atoms, together with common carbon atoms, are produced by plants, and from there they enter the organisms of animals and humans. In any animal organism, radiocarbon is continuously decayed and replenished. In every gram of carbon in a living organism, approximately 14 radiocarbon atoms disintegrate per minute. The concentration of isotone $^{14}\text{C}$, as a rule, is the same in all carbon that is part of living organisms.

As soon as the organism dies, it falls out of the chain of radioactive carbon cycle going on on Earth. In such an organism a new radio-

Fig. 2. This rope was used by the ancient inhabitants of Peru about 3000 years ago.

Fig. 3. Scraps of canvas in which one of the handwritten scrolls found near the Dead Sea was wrapped. Their age is about 2000 years.

carbon does not enter, and previously accumulated carbon disintegrates. The amount of radiocarbon in a dead body will be halved after 5,600 years. If an object is made from a tree cut down approximately 5,600 years ago, then in a gram of carbon released from it, the radioactive radiation counter will register not 14, but about 7 decays per minute. Based on the number of decays, the age of objects made of wood, as well as animal bones and plant remains can be calculated. To do this, you need to use formula (1). From this formula it follows that

$$\frac{N}{N_0} = \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}};$$

$$\lg \frac{N}{N_0} = \frac{t}{T_{1/2}} \lg \left(\frac{1}{2}\right);$$

$$t = T_{1/2} \frac{\lg(N/N_0)}{\lg(1/2)} = T_{1/2} \frac{\lg(N_0/N)}{\lg 2}.$$

The number of decays per minute $A$ is proportional to the number of atoms $N$. So, at the initial moment of time $A_0 = kN_0 = 14$, and at the moment of time $t$, when the number of atoms is $N$, $A = kN$. Hence,

$$N = A/k, N_0 = A_0/k = 14/k,$$

And

$$N_0/N = 14/A$$

$$t = T_{1/2} \frac{\lg(14/A)}{\lg 2} = 18400 \lg(14/A).$$

Here $t$ is age in years.

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Fig. 4. Reed mat from Nevada [USA]. She is about 3000 years old.

When carbon from sandals found in the cave was placed into a beta particle counter, the value of A was found to be 4.5 times per minute. It follows that the age of the find is approximately 9,000 years.

Measuring the number of $^{14}$ C decays is a difficult task. For this purpose, carbon is converted into one of the gaseous compounds - carbon dioxide ($\text{CO}_2$) or methane ($\text{CH}_4$). The radioactive radiation counter is filled with gas. It is he who registers every decay of radiocarbon. But the beta particle counter is also sensitive to other particles: mezons, gamma quanta and some other particles formed by cosmic rays give impulses that are no different from the impulses caused by beta particles. Therefore, coming to record the counter only with a layer of iron. However, it is very difficult to obtain iron completely purified from radioactive impurities. Therefore, a cylinder filled with rue is placed between the counter and the iron record (it is put on the counter). The double “coat” reliably covers the counter from almost all radiation that can distort the measurement results, but not from mezons. Mezons weakly interact with matter and penetrate even through such protection. To eliminate mezonic pulses, the main counter, filled with the gas under study, is surrounded on all sides by auxiliary counters. Before getting into the central counter, the mezon is located in one of the surrounding counters. In this case, the electronic equipment will register two simultaneous pulses in the central and side counters. These double pulses are excluded during counting.

Thanks to such sophisticated equipment, it is possible to change very small radiocarbon concentrations and reliably determine the age of objects that have been buried underground or on the ocean floor for up to 50,000 years.

With what accuracy can the age of radiocarbon be determined by the method? The amount of radiocarbon in the atmosphere is proportional to the flux of primary protons falling on the Earth. The flux of protons varies depending on conditions on the Sun. The lower the solar activity, the greater it is. The fact is that during periods of high solar activity, the magnetic field of the Sun increases and captures some of the cosmic protons that would have fallen to Earth during the years of the quiet Sun.

I had to check the accuracy of the “carbon clock”. They checked on objects whose age was known—wooden products from Ancient Egypt, mutiae of the pharaohs. An even more accurate check was carried out using tree rings. There are trees on Earth that are up to 3,000 years old. This is how long American redwood trees live. Having cut a long-lived tree trunk, you can count the growth rings and take the substance, for example, from the layer corresponding to a three-layer ring (if you count from the edge to the core). Carbon from the isolated substance tree conditionally increases in radiocarbon atoms $^{14}$ C—value $N$. From formula (2) we have:

$$N_0 (3000 лет назад) = \frac{N}{(1/2)^{T/1/2}} = N \cdot 2^{\frac{t}{T/1/2}}.$$

It turned out that carbon concentrations changed little over the years. However, great difficulties will arise in determining the age of buildings. The house can be built from wood that has already been used in another structure. In addition, the age of the tree itself in some cases plays a big role. Somehow scientists decided on the “age” of roadside grass. The counters registered a very small proportion of radiocarbon in the substance of roadside plants. The calculations led to an appreciating result - the “age” of the grass turned out to be equal to several years! A quick explanation was found. Exhaust gases automatically passing along the road contain carbon without the heavy isotope $^{14}$ C. The age of the oil from which gasoline is obtained is measured in millions of years, and naturally, all the radiocarbon in it has decayed.

Determining the age of a shoe warehouse of acceptable times is one of the first archaeological problems solved by physics using the radiocarbon method. This method helped determine the timing of ancient Vykanichic eruptions and establish the dates of ice ages.

Advancing on the continent, the ice destroyed plant life, people went south, marking their path with the coals of cooled ochars. The trunks of ancient trees and human sites have been preserved to this day under a layer of glacial sediment. But only after the advent of the radiocarbon method was a “schedule” of ice movement established.

Measurements have shown that over the past 40 thousand years there have been three ice ages on Earth. The latest is about 11,400 years ago. Since then, the Earth has been relatively warm.

Arithmetic problems

1. Find all numbers $x$ such that $x^3=7p+1$, where $p$ is a prime number.

2. Odd numbers are arranged in ascending order 1, 3, 5, 7, …; after which they are divided into groups containing 1, 2, 3, 4, … numbers, that is, like this: (1), (3,5), (7, 9, 11), (13, 15, 17, 19), … Prove that the sum of the numbers belonging to a group of $r$ numbers is equal to $r^3$.

Next, let the numbers be divided into groups consisting of 1, 1, 4, 3, 9, 6, …, $r^2$, $\frac{r(r+1)}{2}$, $(r+1)^2$, … numbers.

Prove that the sum of the numbers in a group of $r^2$ numbers is equal to $r^5$.

3. Prove that there exists an infinite quadruple of natural numbers $a, b, c, d$ such that $(a^3+b^3)(c^3+d^3)$ can be represented as $M^3+N^3$, where $M, N$ are natural numbers.

4. How many numbers divisible by 37 can be obtained from the number 3333777 by rearranging its digits?

5. Find rational numbers $x$ and $y$ such that $0<x<3$, $y>4$, and $x^3+y^2=3^3+4^3$.

… and algebraic

6. Write an equation whose roots are the ratios of the roots of the equation $y^3+py+q=0$.

7. Let us denote by $t_k$ the arithmetic mean of the $k$ powers of positive numbers $a_1, ..., a_n$. Prove that the sequence $t_1, t_2^{1/2}, ..., t_k^{1/k}, ...$ does not kill.

32

LABORATORY “QUANTA”

SNOW
CRYSTALS

A. S. Varpakhovsky, A. N. Vilenkin

Photographs and figure seeds can be found in many physics textbooks in the chapters that talk about symmetry. But until recently, this was the limit of scientists’ interest in sooty crystals. Serious study of the nucleation, growth and structure of soot crystals began not so long ago. Interest in sozh crystals was mainly associated with the study of the formation of jedya and the phenomena that originated in the area. It turned out that most of the djedeva drops begin their life as soot crystals, taloshes, before they fall to the ground. However, only cold, high-altitude cirrus clouds are composed of ice crystals. Basically, clouds are collections of small water droplets that are suspended in the air in the same way as smoke particles. For many years, it remained a mystery how these droplets grew to the size required for them to fall to the ground. It also remained a mystery that often these droplets “refused” to freeze, although the temperature of the cloud was much lower than the normal freezing point of water, that is, below 0 ° C.

We now know that a supercooled cloud remains stable until it contains at least a small number of small ice crystals that originate on particles of earthly dust. Water molecules that fall on an ice crystal form a strong bond with it, which is quite difficult to break. Water molecules that condense on a drop are relatively easy to tear off - the heat of evaporation is less than the energy required to separate a water molecule from an ice crystal. Therefore, the cloud consists of water droplets and ice crystals, and ice crystals grow much faster than drops. Moreover, due to the growth of ice crystals, the humidity of the surrounding air decreases. This causes the water droplets to gradually evaporate and disappear. At the same time, the ice crystals grow to a size large enough for them to fall to the ground. When falling, several crystals can unite, forming a seedling.

Although soot crystals are varied, they can be classified according to their basic shapes; hexagonal recognition columns

34

fluctuates from - 5 ° C to 0 ° C, you can find crystals in the form of hexagonal plates, short prisms and strikingly beautiful stars with a diameter of about several millimeters. These stars are the basis of snowflakes. At temperatures several degrees below zero, the crystals fuse together to form snowflakes.

All this suggests that the shape of crystals is determined mainly by the temperature at which they grow. This was confirmed by experiments on growing ice crystals in the laboratory. Ice crystals are grown in a special chamber in which the temperature and amount of water vapor are strictly controlled. A thin thread was used as a seed. The temperature in the chamber was different in different areas along the thread. figure 3 shows a photograph of a section of thread overgrown with crystals. The size of this area is about 6 cm. On the right at figure the temperature in the chamber is indicated. In the upper part of the thread, mostly tiny hexagonal plates were grown. Below are hexagonal plates, branched stars and again hexagonal plates. At temperatures below - 25° C, hexagonal features are formed again.

Experience has shown that it is the temperature that determines the shape of the crystal. The amount of water vapor affects the growth rate. However, the exact nature of snow crystal growth still remains unclear.

It is very interesting to study the growth of snow crystals on earth. Often in winter, when there is a sharp drop in temperature, tree branches and house walls become covered with frost. The clouds where snowflakes originate are difficult to reach. Frost is easily achieved and can be observed during its formation. Frost usually appears on objects that have a high heat capacity and low thermal conductivity. When res-

When warming occurs, the temperature of these objects turns out to be lower than the temperature of the surrounding air, and water vapor in the air condenses on them. If there is little vapor in the air, then red fluff flakes are obtained. When the air humidity is high, cold objects are covered with a crust of ice. Water simply condenses on cold objects and then freezes.

Particularly interesting are the patterns that cover the windows of carts, buses and trams in winter. In a sudden position, the temperature of the windows becomes lower than the air temperature in the room. Vapor molecules in the humid air in the room settle on them, forming red patterns. It is also important that the air in the room is not very humid. Otherwise, the steam first condenses on the glass and then freezes, forming a layer of ice. Patterns do not appear on the window if the window is open. In this case, the air temperature in the room near the glass decreases, becoming the same as the temperature of the glass itself. In the ice patterns that grow on the glass, you can see the large shapes that snow crystals can take.

You can watch the growth of snow crystals on your windows at home. However, unfortunately, patterns on windows are short-lived, and it is difficult to change the conditions of their growth. But you can “grow” patterns that are very similar to ice ones, using not water, but dissolving hyposulfite (it is sold in stores in the photo departments) or Carlsbad cols (you can buy it at the pharmacy). These patterns should not differ from ice ones and, by knighting them, you can change the growth conditions - the rate of heat supply or removal, the solution concentration (a change in the solution concentration corresponds to a change in air humidity), etc. The growth of crystals forming the patterns can be monitored using a microcoppa.

You can get the pattern like this. First, prepare a saturated mixture in six batches from which you will work. Wipe the glass with this brush and place it under the Fan. After about 5 minutes, the water will evaporate and a pattern will appear on the glass. The hardest thing here is to wet the glass well. The fact is that water generally does not wet the surface of the glass and does not spread over it, but forms droplets. Then, together with the patterns, you will get spots in which just dried sediment remains.

If you do not blow the glass or apply a lot of abrasive on it, then together with the pattern you will get pieces of crystals, they will grow “from below” (from the surface of the glass) and will appear above the glass. We want the crystals to grow small and onto all glass surfaces at once. In order for the solution to wet the glass, the surface of the glass must be thoroughly cleaned with gasoline or alcohol (you can also use cologne). It’s even better to use cols alcohol solution rather than water. Of course, you may not get the pattern right the first time - don’t despair. Remember that any physical experiment may require several attempts.

The artificial patterns you receive can be shaped. Two methods of formation are convenient.

The first is basic: patterns are photographed on film and then printed on photographic paper.

The second one is not entirely ordinary. The patterns are grown on a narrow plane of glass. This strip is placed in a photo enlarger along with the film, and is printed directly onto the paper. The crystals that form the patterns are non-transparent and look black when transmitted by light (on the print after development - white). Places where there are no crystals allow light to pass through and appear black on the print. This is how the pattern placed on the cover of our magazine was obtained. It was grown from the Carlsbad cols.

36

Here are some problems related to the concept of area of ​​a polygon. We give an axiomatic definition of the area of ​​a polygon, listing those properties that we need.

A1. The area of ​​a polygon is a positive number.
A2. The areas of equal (congruent) polygons are equal.
A3. If a polygon is cut into several parts, then its area is equal to the sum of the areas of these parts.
A4. The area of ​​a triangle is equal to half the product of its base and height. (We consider the unit of scale to be given; it is easy to show that this product does not depend on which side and the corresponding height we take.)
The letter A was chosen because the word “Axiom” begins with it. From these four properties of the axioms you can derive all the theorems about the areas of polygons that you study in school.

It is much more difficult to prove that, indeed, it is possible to associate with each polygon $M$ in the plane the area number $S(M)$, its area, so that properties A1–A4 are satisfied. This is true, but it is usually not proven in a school course and is considered very dynamite. Instead of A4, we can consider the main property from which A4 is derived:

A4’. The area of ​​a square with side 1 is 1.

For example:

a) The area of ​​a parallelogram is equal to the product of its base and its height;
b) The area of ​​a trapezoid is equal to the product of half the sum of the bases and the height.
Think about how to prove these statements (that is, derive from axioms A1-A4).
We’ll start by proving another formula for the area of ​​a trapezoid.

1. The area of ​​a trapezoid is equal to the product of one of the sides and the perpendicular dropped onto it from the middle of the other side.

Proof. Let $ABCD$ be the given trapezoid $(AD \parallel BC)$, $K$ the midpoint of side $CD$, $KH$ the perpendicular dropped from the point $K$ to the line $AB$. Let us draw through the point $K$ a line parallel to the line $AB$. Let $M$ and $P$ be its points

38

7. Prove that the area of ​​the right one is the largest and the smallest of its diagonals.

Note. Take a look at figure4.

8. Prove that the area of ​​a polygon circumscribed about a circle is equal to the product of the radius of this circle and the perimeter of the polygon.

In these problems, you can bypass the “cut and fold” technique. Note that, in principle, this technique can always be used when you need to prove the equality of the areas of two polygons; The following theorem is true: if two polygons are equal in size, then one of them can be cut into parts from which the other polygon is composed.

This theorem is not difficult to prove *). But practically, in order to establish equality of areas or find the ratio of areas, it is not at all necessary to “cut and add”. For example, you can very quickly compare the areas of two triangles using the formula we called A4. It immediately follows, in particular, also correspondences for areas:

if the vertex of a triangle is moved along a straight line parallel to its base, then the area does not change (Fig. 5; triangles ABC

*) See the book by V. G. Boltyansky “Equal-sized and equally spaced figures”, series “Popular lectures on mathematics”, vol. 22, Gostekhidat, M., 1956.

and ADB have a common base AB and equal heights);

if one of the sides of a triangle adjacent to a given angle is increased by k times, then its area will also increase by k times (see Fig. 6; triangles ABC and ABD have a common height AH, therefore the ratio of their areas is equal to the ratio of the bases:

$$S_{\triangle ABC} / S_{\triangle ABD} = BC / BD.$$

This implies a more general correspondence:

If two triangles have a common angle, then their area is proportional to the product of the sides enclosing the angle.

(see Fig. 7; $$S_{\triangle ABC} / S_{\triangle ABD} = BC / BD,$$
$$S_{\triangle ABD} / S_{\triangle EBD} = AB / EB,$$
$$S_{\triangle ABC} / S_{\triangle EBD} = (AB \cdot BC) / (EB \cdot BD)).$$

In particular, if two triangles are similar and the sides of one are k times larger than the corresponding sides, then the area of ​​the first is $k^2$ times the area of ​​the other.

In the following problems, the solution is based precisely on these considerations: on comparing the areas of the triangle

39

nicknames that either have the same bases, or the same heights, or the same angles. Try to find the simplest and most beautiful solution!

9. O is the intersection point of the segments AC and BD (Fig. 8). In order for the areas of triangles AOB and DOC to be equal, it is necessary and sufficient that lines BC and AD are parallel. Prove it!

To solve this problem, you need to prove two statements: (1) if the areas of triangles AOB and DOC are equal, then lines BC and AD are parallel; (2) if lines BC and AD are parallel, then the areas of triangles AOB and DOC are equal.

10. Prove that a convex quadrilateral is a moed and polko moed parallel when each of its diagonals bisects its area adb.

And here, as in the previous problem, you need to prove two theorems: direct and inverse.

11. In triangle ABC, the line passing through vertex A and dividing the median BM in the ratio 1:2, counting from the vertex, we intersect side BC at point K. Find the ratio of the areas of triangles ABK and ABC.

Note: Draw a line parallel to AK through point M and use it to find the ratio of segments BK and BC.

12. On the continuation of the convex side BC of the quadrilateral ABCD, find a point O such that the area adb of the quadrilateral ABCD is equal to the area of ​​the triangle ABO.

Note: Draw a line through point D parallel to diagonal AC.

Does problem 12 always have a solution? Is it always the only one?

This problem allows you to turn any convex polygon into an equal-sized one with fewer sides.

13. Two straight lines are drawn through the middle of the altitude of an isosceles triangle, connecting it with the vertices of the base (Fig. 9). What part of the area of ​​the triangle is made up of the parts on which the triangle is still cut by straight lines?

14. Given triangle ABC. We extend its side AB beyond vertex B with the segment BP=AB, side AC beyond vertex A with the segment AM=CA, and side BC beyond vertex C with the segment KC=BC. How many times is the area adb of triangle PKM greater than the area of ​​triangle ABC?

Note: Contain points M and B, P and C, A and K.

15. Form an analocine problem for a quadrilateral and solve it. Derive the result of problem 6 from it.

16. On the sides of the convex quadrilateral ABCD, points M, P, K, H mak are taken, such that AM : MB = 3 : 5; BP: PC=1: 3; CK: KD=4: 5; DH : HA = 1 : 8. Find the ratio of the area of ​​the quadrilateral MBPKDH to the area of ​​the quadrilateral ABCD. Think, given any relationship between AM and MB, BP and PC and mak, can this problem be solved?

Note: Draw a diagonal and use A3.

17. In a convex quadrilateral, the midpoints of adjacent hundred-

41

Square SVM. But $S_{\triangle ABN} = S_{\triangle CBN}$, because the point $N$ lies on the median. Therefore, $S_{\triangle ABM} < S_{\triangle CBM}$. And we assumed that $S_{\triangle ABM} = S_{\triangle CBM}$. We got a contradiction. The problem is completely solved.

Think about how to use it to solve the following problem, switching with problem 20.

24.* Prove that if the interior of a quadrilateral ABCD there is a maxim O or the segments OA, OB, OC, OD divide it into four

Fig. 14.

parts of equal size, at least one of its diagonals bisects the other diagonal. Form and prove the converse theorem.

25. On the sides of angle A there are points B and C. A straight line is drawn through the midpoint K of segment BC, intersecting the sides of the angle at points D and E. Prove that the area of ​​triangle ADE is greater than the area of ​​triangle ABC.

The solution to this problem allows us to draw a straight line through a given point inside a given angle from the corner of a triangle of the smallest possible area. Think about how to do this.

26. In this page ABCD (BC || AD) draws the diagonal AC (Fig. 14). At what height should the boundary of a line parallel to the bases be crossed so that the sum of the areas of triangles AKL and LMC is the smallest (K, L and M are the points of intersection of the line with the corresponding segments AB, AC and CD)?

27*. The industrial sides of the convex hexagon ABCDEF are parallel. Prove that the area of ​​triangle ACE is at least half the area of ​​the hexagon.

SYMPOSIA
YOUNG MATHEMATICS

The idea of ​​a gathering of students from specialized physics and mathematics schools in different cities originated at school No. 40 in Gorky. In March 1971, during the spring school holidays, at the invitation of this school, representatives of mathematical schools in Moscow, Riga, Rostov, Volgograd, Kishnev, Alma-Ata, Tbilisi, Kostroma, Leningrad and Kyiv gathered in Gorky.

The symposium consisted of several sections, at each of which schoolchildren made reports. The work of the sections was led by scientists from Gorky State University.

The mathematics section was headed by Candidate of Physical and Mathematical Sciences, Associate Professor D. A. Gudkov. At this section, reports were heard on “Non-standard sums”, “Modeling of reflexes”, “Matematika and linguistics”, “Calculating the area of ​​a shadow”, “Self-reproduction of numbers” and others.

Reports were given by: Y. Shneiderman (10th grade, no. 40, Gorky), A. Knafel (8th grade, no. 40, Gorky), N. Pomapkina (10th grade, no. 40, Gorky), O. Malinovsky (9th grade, no. 13, Riga), A. Borisenko (10th grade, no. 40, Gorky), A. Borisenko (10th grade, no. 40, Gorky), 145, Kyiv), I. Rauhvaer (10th grade, no. 34, Kishnev), M. Vishik (10th grade, no. 444, Moscow), V. Kaplinsky (8th grade, no. 444, Moscow) and others.

The physics section was headed by Candidate of Physical and Mathematical Sciences, Associate Professor M. S. Kovner. At this section, reports were heard “Peaceful atom - for the service of human water supply”, “On analogies in physics”, “On the “peaceful mass” method and its application in solving physical problems”, “Production and study of thin crystalline films”, “Sweep generator and oscilloscope” and others.

Reports were given by: A. Mirukholova (10th grade, no. 42, Tbilisi), A. Ezersky (10th grade, no. 40, Gorky), Y. Khandrus (10th grade, no. 145, Kyiv), I. Molan (10th grade, no. 34, Kishnev), N. Chaika (10th grade, no. 42, Tbilisi), M. Gershtein (10th grade, no. 40, Gorky), M. Gomayunov (10th grade, no. 239, Leningrad) and others.

There were also sections of astronomy, chemistry and regional studies.

During the symposium, students not only read and discussed reports. They also managed to get acquainted with the work of school No. 40, with the sights of the city, visited museums, and met with scientists.

(Continued on page 53)

42

Tasks

Solutions to problems from the Kwan Problem Book can be sent no later than a month and a half after the corresponding issue of the journal is published. On the envelope, after the journal address, indicate the solutions to which problems you are sending (for example, “Kwan’s Problem Book” M126, F128). At the beginning of the letter, indicate your last name, first name, patronymic, six-digit postal code and address (as well as the class and school where you are a teacher). The solution to each problem must be written on a separate sheet(s); Please include your last name at the end of each decision. A star mark for a task of increased difficulty.

M126. A polygon circumscribed around a circle of radius $r$ is somehow cut into triangles. Prove that the sum of the radii of the inscribed circles of these triangles is greater than $r$.

I. D. Novikov

M127. For each natural number $n$ we denote by $s(n)$ the sum of its digits (in decimal notation). Let’s call a natural number $m$ special if it cannot be represented as $m=n+s(n)$, where $n$ is some natural number. (For example, the number 117 is not special because $117=108+s(108)=108+9$, but the number 121, as you can easily see, is special.) Is it true that there are only a finite number of special numbers?

M128. Find the ratio of the sides of a triangle, one of whose medians is divided by the incircle into three equal parts.

M129. a) 12 $n$ milk was poured into a bucket. How, using only jugs 5 and 7 $n$, to divide the milk into two equal parts?

6. • Solve the general problem: for which $a$ and $b$ can we divide $(a+b)n$ milk in half and use only the jugs in $a$ $n$, $b$ $n$ and $(a+b)n$ ?*.

V. V. Ushchakov

M130. What is the maximum number of points that can be placed?
a) on a plane;
b) * in space, so that none of the triangles with vertices at these points are obtuse?

G. A. Galperin

F138. An astronaut with a mass of 100 kg is outside a spacecraft whose mass is 5 $m$, on a halyard 64 $m$ long. Find the tension of the halyard if the ship is located between the astronaut and the Earth on the line connecting their centers of gravity.

When calculating, assume that the ship is moving in a circular orbit, the height of the crust $h$ from the Earth’s surface is negligible compared to the radius of the Earth ($R=6400$ km).

*) In one transfusion from one vessel to another, you can pour out everything that is there, or fill the second vessel to the top.

F139. The football player hit the ball, giving it speed $v$ at an angle $\alpha$ to the horizon, and hit the near bottom corner. If a football player hit the ball in the same place on the football field and the ball flew at the same angle to the horizon, but at a speed 5% greater than the speed $v$, then it would hit the top of the goal post. Find the speed with which the ball begins to move if the height of the turn is $h=2\text{ m}$ and the angle is $\alpha=30^\circ$.

I. Sh. Slobodechkiy

F140. Will other barometers give correct readings of quality if part of its tube (below the level of mercury in it) is made of very soft rubber?

F141. Three wires are removed from the “black box” containing an unknown electrical circuit. Two of them are connected to ground and then to ground, plotting graphs for the three possible options for switching on the circuit. These graphs are shown in figure 1. The current is considered positive if it goes to the circuit located in the box and negative otherwise. Come up with the simplest diagrams of the contents of the “black box” and determine their parameters.

B. B. Bukhovtsev

F142. A steel ball suspended on a thread of length $l$ was deflected so that the thread assumed a horizontal position and was released. At the moment when the thread made an angle $\alpha-30^\circ$ with the vertical, the ball hit a stationary steel plate (Fig. 2). To what height does the ball rise after hitting the plate, if the impact can be considered absolutely elastic?

G. L. Komkin

45

Fig. 1.

straight line $l_1$ so that it coincides with $l_2$ ($0^\circ < l_1 < l_2 < 180^\circ$). If the lines $l_1$ and $l_2$ are parallel or coincident, then, by definition, $\Rightarrow (l_1, l_2) = 0$. Verify that for any two given points $U$ and $V$ and a given angle $\alpha$ ($0^\circ < \alpha < 180^\circ$), the set of points $X$ for which $\Rightarrow (UX, VX) = \alpha$ will be rounded (strictly speaking, excluding the points $U$ and $V$ themselves); in other words, what points $U, V, X_1$ and $X_2$ lie on the same circle moeda and $\Rightarrow (UX_1, VX_1) = \Rightarrow (UX_2, VX_2) \neq 0$ (Fig. 3, a; comparison figure 3, a and 3, b explains the advantages of the new method of determining the angle).

Let’s return to our task. Now the statement that five points $C, N, P', S, R'$ lie on the same circle (hence the following)

Fig. 2.

blows that $R'$ coincides with $S$) follows from the equalities

$\Rightarrow (CS, NS) = \Rightarrow (Q'Q, NQ) =$
$= \Rightarrow (Q'P', NP') = \Rightarrow (CP', NP')$;
$\Rightarrow (CR', P'R') = \Rightarrow (MM', P'M') =$
$= \Rightarrow (MN, P'N) = \Rightarrow (CN, P'N)$.

These equalities are easy to check formally, without referring to the drawing - you just need to remember which points lie on the same straight line and which on the same circle; In addition to the theorem in italics above, it is used that $Q'Q \parallel MM' \parallel l$. (Of course, we still need to analyze the “expressed” cases when any two points defining a line in these equalities coincide; they are all quite visible and we will not)

Fig. 3. Blue lines show the set of points for which: a) $(UX, VX) = \alpha$ (left figure) and b) $UXV = \alpha$ (right figure).

Fig. 4. Another formulation of problem M84: as a result of the three mappings shown in figure, from the point $X$ we obtain $Y = f(X)$; if you do the same for the point $Y$, it turns out that $f[Y] = X$.

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Let us dwell on them.) Now the problem is completely solved.

Let us give another, more natural formulation of this problem (think why this is the same problem!).

Let $P$ and $N$ be two distinct points of a circle with center $O$; $A$ is the base of the perpendicular dropped from the point $O$ to the given line $l$, and $f$ is the mapping that assigns to each point $X$ the line $l$ the point $f(X)$ of the same line according to the following rule: if $X'$ is the point of intersection of the line $PX$ with a circle, $X''$ is the point of intersection of the straight line $NX'$ with $l$, then $f(X)$ is a point symmetrical to $X''$ relative to $A$. then the mapping $f$ coincides with its inverse, that is, $f(f(X))=X$ for all $X$ (Fig. 4).

For those who are familiar with the basic concepts of project geometry, we suggest trying to find a simple “projective” proof of this theorem (note that even in its formulation it is necessary to consider $l$ an extended line, that is, add an infinite remote point to $l$ - then the mappings in question will be one-to-one). For a particular case, the solution is given in I.M. Yaglom’s book “Geometric Transformations” (GITTL, 1956; part II, problem 177), but it only applies if the straight line $l$ intersects the circle, and uses the formula of “double relations”. A blueprint for another solution, “without computing,” appears on page 71 of this issue of the journal.

N. B. Vasiliev

M86

The bottom of the rectangular box was made with tiles of sizes $2 \times 2$ and $1 \times 4$. The tiles spilled out of the box and in the process lost one tile $2 \times 2$. Instead, remove the tile $1 \times 4$. Prove that now it is not possible to lay out tiles up to the box.

Let’s count up to a box into squares of size $1 \times 1$ and mark those squares that are in odd places in odd boars (see Fig. 5). Note that each tile $1 \times 4$ covering four squares at the bottom of the box will be

Fig. 5

Fig. 6.

Fig. 7.

It is necessary to cover an even number of squares we have displaced (two or none), and the $2 \times 2$ tile always covers exactly one displaced square. Thus, if the tiles are completely covered to the box, then the number of tiles $2 \times 2$ has the same parity as the total number of offset squares at the bottom of the box. Therefore, if the number of tiles $2 \times 2$ has changed by one (or even by an odd number), then filling the box again will not be possible.

Think about whether the statement of the problem will remain true if, instead of tiles $1 \times 4$ and $2 \times 2$ (Fig. 6), tiles of three squares are considered: rectangular $1 \times 3$ and “corners” (Fig. 7). You will find the solution to this problem at the end of the magazine, on page 64, but don’t rush to write it down!

The correct solutions to our problem were sent by A. Gritsoryan and E. Karasik from Baku, E. Turkevich from Chernovtsy, A. Chernyak from Minsk and A. Shamaev from Moscow.

L. G. Limanov

M87

Prove that if three circles of the same radius pass through one point, then three other points of pairwise intersection of these circles lie on a circle of the same radius.

Let all three circles of radius $r$ with centers $O_1, O_2$ and $O_3$ intersect at point $A$, in addition, intersect in pairs at points $B_1$ ($O_2$ and $O_3$), $B_2$ ($O_3$ and $O_1$) and $B_3$ ($O_1$ and $O_2$). It is obvious that the midpoints $C_1, C_2, C_3$ of the segments $AB_1, AB_2, AB_3$ are at the same time the midpoints of the segments $O_2O_3, O_3O_1, O_1O_2$ of the corresponding (Fig. 8). Therefore, each of the triangle $B_1B_2B_3$ and $O_1O_2O_3$ has sides twice as long as the corresponding side of the triangle $C_1C_2C_3$ (and parallel to them). Therefore, $\triangle B_1B_2B_3 = \triangle O_1O_2O_3$. But it is clear that the circumradius of the triangle $O_1O_2O_3$ is equal to $r$ (indeed,

47

Fig. 8.

Fig. 9.

$$AO_1 = AO_2 = AO_3 = r$$, therefore the radius of the circumscribed circle of triangle $B_1B_2B_3$ is also equal to $r$, which is what needed to be proved.

Many readers have proven that triangles $B_1B_2B_3$ and $O_1O_2O_3$ are congruent in a more complex way, in which we had to separately consider various cases of the relative positions of these triangles (we deliberately did not refer to figure in our solution, where all the segments mentioned there are drawn, in order to emphasize that it does not depend on their location). A solution similar to ours can be obtained by considering the triangle together

$$C_1C_2C_3$$ triangle $D_1D_2D_3$, where $D_1, D_2, D_3$ are points diametrically opposite points $A$ in the circle $O_1, O_2$ and $O_3$, respectively; prove that $B_1, B_2$ and $B_3$ are the midpoints of the triangle $D_1D_2D_3$. In the configuration that we received (Fig. 9), it is also interesting that point $A$ is the center of the circumcircle of triangles $O_1O_2O_3$ and $D_1D_2D_3$ and at the same time the point of intersection of the altitudes of triangles $B_1B_2B_3$ and $C_1C_2C_3$; the latter, however, is no longer used in our solution.

It is curious that D. Polya (and some readers who sent us letters) had a completely different idea for solving this problem. We publish an excerpt from the book “Mathematical Discovery”, which talks about the origin of this idea, on pp. 67-69.

V. N. Berezin

M88

What condition must the coefficients of the equation $x^3 + ax^2 + bx + c = 0$ satisfy in order for its three roots to form an arithmetic progression?

Let’s give the answer right away. The desired match is $c = \frac{ab}{3} - \frac{2}{27} a^3$ (or, what is the same, one of the roots must equal $-\frac{a}{3}$). Indeed, let $x_1, x_2$ and $x_3$ be the roots of our equation. Then, according to Vieta’s theorem,

$$x_1 + x_2 + x_3 = -a.$$

On the other hand, in order for $x_1, x_2, x_3$ to form an arithmetic progression, it is necessary and sufficient that

$$x_1 + x_3 = 2x_2.$$

Pose $x_1 + x_2 + x_3 = 3x_2$, opening $x_2 = -\frac{a}{3}$. Try to derive the conditions under which equations of the fourth degree form an arithmetic progression.

L. I. Moscoviome

In this issue we publish problems F92—F99

F92

The mass hoop $m$ stands on the mass board $M$ (Fig. 10). The friction coefficient between the board and the hoop is $k$. The board lies on a smooth table. With what acceleration will the board move if the hoop is pulled with a force of $F$?

Let’s write down the equation of motion of the hoops and boards (equation II of Newton’s law).

In the horizontal direction, two forces act on the hoop - the force $F$ and the tread force $F_{TP}$ between the board and the hoop (see Fig. 10). The center of gravity of the body—in this case, the center of the hoop—moves as if all the forces acting on the body were applied to it. Therefore we can write that

$$ma_1 = F - F_{TP}$$

($a_1$ is the acceleration of the center of the hoop).

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Fig. 10.

The hoop not only moves horizontally, but also rotates around the center. The acceleration of the rotational movement is imparted to it by the same forces $F$ and $F_{TP}$, but in this case they act “in concert”, spinning the hoop in one direction. The acceleration of the rotational motion of the hoop in the coordinate system associated with its center*) is denoted by $a_2$. Then we can write that

$$ma_2 = F + F_{TP} \quad (2)$$

Now let’s write down the equation of motion of the dock. In the horizontal direction, only one force acts on it - the force of friction of the hoop on the dock. Therefore, if the acceleration of dock is $a_3$, then

$$Ma_3 = F_{TP}. \quad (3)$$

So, we have three equations involving four unknown quantities: $a_1$, $a_2$, $a_3$ and $F_{TP}$. To find these unknowns, you need to create another equation.

When moving the hoop and docks, two cases arise: 1) the hoop does not slip relative to the dock and 2) the hoop slips relative to the dock.

Let’s consider the first case: the hoop does not slip relative to the dock. In this case, the speed $v_1$ of the translational movement of the center of the hoop, the linear speed $v_2$

*) In fact, the coordinate system associated with the center of the hoop is not inertial. But, when considering the hoop equation, this can be ignored, since the force of inertia is applied to the center of gravity of the hoop, and its moment relative to the center of the hoop is zero.

**) Equation II of Newton’s law for a rotating body is written as follows:

$$I \varepsilon = (F + F_{TP}) r, \quad (*)$$

where $I$ is the moment of inertia, $\varepsilon$ is the angular acceleration and $r$ is the radius of the hoop (see the article by A.K. Kion “Rotational motion”, «Kvant» No. 1, 1971).

For hoop $I = mr^2$, $\varepsilon = \frac{a_2}{r}$. If we put these values ​​into equation (*), we get equation (2).

rotation of the hoop around the center and staple $v_3$ docks are connected to staple $v_1$ (Fig. 11). Since the dock moves relative to the Earth to the right with a bracket of $v_3$, then the bracket of the dock in the coordinate system associated with the center of the hoop is equal to $v_1 - v_3$. The lower point of the hoop touching the dock also moves with the same rigidity. This is the linear bending of the hoop around the center.

Since the forces acting on the hoop and dock are constant, the accelerations $a_1$, $a_2$ and $a_3$ are also constant. Let the velocities $v_1$, $v_2$ and $v_3$ of the body arrive after a time $t$ after the start of their motion. Then $v_1 = a_1 t$, $v_2 = a_2 t$ and $v_3 = a_3 t$. Hence,

$$a_2 t = a_1 t - a_3 t.$$

From here

$$a_2 = a_1 - a_3.$$

We have obtained an equation relating the accelerations of bodies. Now solving equations (1), (2), (3) and (4) together, we find

$$a_1 = a_2 = \frac{F}{m}$$

And

$$a_3 = 0.$$

We found that the acceleration of the dock is zero. From equation (3) it is clear that the friction force is also zero. This is due to the fact that if a force $F$ is applied to a hoop standing on an absolutely smooth dock, as shown in figure 10, then the acceleration of the translational motion of the hoop will coincide with the acceleration of its rotational motion. In this case, the lower point of the hoop rolls along the dock without slipping. This happens because all the mass of the hoop is concentrated in its thin rim and the force $F$ imparts to the hoop equal accelerations of translational and rotational motion. But this

49

means that the friction force does not depend on the force $F$ and is always equal to zero. Therefore, the hoop cannot slip relative to the point (the friction force cannot be equal to $kN$), and we can not consider the second case.

F93

In the side wall of the vessel directed to liquid with a refractive index $n$, a hole of small radius $r$ continues. A thin beam of light is directed horizontally from the vessel along the axis of the hole. To what level $h$ must there be liquid above the hole so that a ray of light emerges from the jet without ever experiencing internal reflection?

Neglect the change in the cross section of the jet; the refractive index of the liquid is assumed to be sufficiently large.

At the greatest angle to the surface of the water, a ray of light falls at the point $A$ scab (Fig. 12). The angle of incidence $\alpha$, at which the beam at this point does not experience total internal reflection, must be such that

$$\sin \alpha = \frac{1}{n}.$$

This follows from the law of light refraction ($\frac{\sin \alpha}{\sin \beta} = \frac{1}{n}$) and the condition that the angle is $\beta = 90^\circ$. The tangent to the jet at point $A$ makes an angle $\gamma = 90^\circ - \alpha$ with the horizontal. Thus, a ray of light will emerge from the scab without ever experiencing internal reflection if the angle formed by the tangent to the jet at point $A$ makes an angle $\gamma$ with the horizon such that cos $\gamma = \frac{1}{n}$.

But the tangent to the surface of the liquid is the tangent to the trajectory of the particle emerging at the upper edge of the hole. Therefore the direction of the tangent

$$\cos \gamma = \frac{v_2}{v_1} = \frac{\sqrt{4gr}}{\sqrt{2gh}} = \sqrt{\frac{r}{h}}.$$

Now expressing cos $\gamma$ in terms of tg $\gamma$ —

$$\cos \gamma = \frac{1}{\sqrt{1 + tg^2 \gamma}}$$

and taking into account that cos $\gamma = \frac{1}{n}$, we get

$$\frac{1}{n} = \frac{1}{\sqrt{1 + 2\frac{r}{h}}}.$$

From here

$$h = 2r/(n^2 - 1).$$

F94

A cycle (closed process) is performed over one gram-molecule of an ideal gas, consisting of two outputs and two isobars (Fig. 13). The gas temperatures at points 1 and 3 are $T_1$ and $T_3$, respectively. Determine the work done by the gas per cycle if it is known that points 2 and 4 lie on the same isotherm.

Expanding along isobar 2-3, the gas does work

$$A_1 = P_2 (V_2 - V_1) = P_2 V_2 - P_2 V_1 = RT_3 - RT_2 = R (T_3 - T_2).$$

When the gas slopes along the isobar 4-1 work

*) This also follows from the law of conservation of energy. Indeed, when a mass of water $\Delta m$ flows out of a vessel with liquid, the potential energy of water decreases by amounts $\Delta mgh$ (we can simply assume that the mass of water $\Delta m$ moved with the surface of the liquid to the hole).

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Fig. 13.

already occurs above the gas. It is equal
$$A_2 = P_1 (V_2 - V_1) = R (T_4 - T_1).$$
The total work done by the gas, very low, is equal to
$$A = A_1 - A_2 = (P_2 - P_1) (V_2 - V_1) = R (T_3 + T_1 - T_2 - T_4).$$

Since points 2 and 4 lie on the same isotherm, the gas temperature in them is the same. Let’s denote it $T = T_2 = T_4$. then $A = R (T_3 + T_1 - T_2 - T_4)$. Points 3 and 4 lie on the same isotherm, so $\frac{P_2}{P_1} = \frac{T_3}{T}$. Points 2 and 1 also lie on the same isotherm and $\frac{P_2}{P_1} = \frac{T}{T_1}$. From these two equations we find that
$$T = \sqrt{T_1 T_3}.$$
Hence,
$$A = R (T_1 + T_3 - 2 \sqrt{T_1 T_3}).$$

F95

The surface of the river forms an inclined plane. Is it possible for a word to swim down a river at a speed exceeding the speed of the current?

If the body does not move relative to the water, then the friction force acting on it from the side of the water is zero. Therefore, the component of the gravity force, parallel to the surface of the water (Fig. 14), will straighten the body until it is balanced by the friction force on the water (which at low speeds is proportional to the speed of the body). In this case, it will flow at a speed greater than the speed of the current.

Now imagine that we, having left a thin, rigid, weightless shell, removed the body and poured water into the shell so that its volume was equal to the volume of the immersed part of the body (Fig. 15). Since the body was floating, the mass of the poured water is equal to the mass of the body. Therefore, the shell with water is equivalent to the body that we removed. But the water poured into the shell is no different from the water in the river. The friction in a fluid is related only to the thickness of the fluid itself and does not depend on the surface of the body. This is explained by the fact that a thin layer of water molecules “floats” to the surface of the body and the formation is determined by the interaction of water molecules with each other. This means that you can completely remove the shell, the friction force will not change. It turns out that the body must flow at a speed equal to the speed of the current.

What’s really going on?

Let’s first understand what “flow speed” means. Due to friction, the speed of the river flow varies at different points in the river. It increases from day to surface and from the shore to the middle of the river. Therefore, the speed of the flow in the place where the plush is located along the body river is also different. This water in the shell differs from the surrounding water, because the speed of the water in the shell is the same at all points.

Let us assume that the speed of the body is equal to the speed of the current at the edge of the body, which is close to the shore. Then the speed of the flow of water passing through the body at other points is greater than the speed of the body. Because of this, the water should carry the body, increasing its speed. On the other hand, when the speed of the body increases, it should slow down due to the fact that the speed of its point, located close to the shore, is greater than the speed of the current and the friction force is converted into movement of the body. As a result, the body must flow at a speed greater than the minimum, but less than the maximum of the flow rates throughout the body.

F96

Two identical balls connected by a thread. Find the height of rise of this system if one of the balls is thrown upward with a speed of $v$.

The momentum of the entire system is $mv$, and its mass is $2m$. Therefore, at the initial moment