Kvant 1971, No. 5 · Pages

Popular science physics and mathematics
quantum
5
1971
magazine
Academy
Sciences of the USSR
and the Academy of Pedagogical
Sciences of the USSR

3

In the room:

P. L. Chebyshev (1821—1894)
1
About Bertrand’s postulate
4
Potential Energy Graphs
8
Hippocrates’ moons
17
Math circle
Physics helps geometry
22
Laboratory “Quenta”
Impossible objects
26
Photographing impossible objects
27
Problem “Quenta”
Tasks
30
Solutions to problems M31-M33;
F43, F48-F52
32
Applicant’s workshop
What is here - are we losing or is it here!
40
Entrance written exam
in mathematics at the faculty
computational mathematics
and cybernetics of Moscow State University in 1970
42
Entrance written exam
in mathematics at MIPT in 1970
46
Problems using Newton’s laws
50
Information
Moscow Mathematical…
58
Answers, directions, solutions
60
Collector’s Corner
3rd cover page
N. N. Kolesnikov
M. I. Bashmakov
R. G. Mints
V. N. Berezin
B. Yu. Kodan
L. Penrose,
R. Penrose
V. I. Bakhmin
I. N. Bernitsa,
A. B. Zelevinsky,
A. G. Kushinrenko,
I. Sh. Slobodetsky
V. A. Petrov
Sh. A. Alimov,
F. P. Vasiliev,
V. P. Modenov
A. P. Savin,
B.V. Fedosov
Yu. V. Zaychikov
A. G. Terushkin

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The scientific interests of P. L. Chebyshev are distinguished by their great diversity and breadth. He left behind extensive research in the field of mathematical analysis, especially in the theory of approximation of functions by polynomials, in intercalculus, number theory, probability theory, geometry, ballistics, theory of mechanisms and other fields of knowledge.

In each of these areas of science, Pafnuty Lvovich obtained functional results, developed new ideas and methods that determined the development of these branches of mathematics and mechanics for many years and retained their significance to this day.

At the same time, it gives rise to Chebyshev’s ability to obtain excellent scientific results using simple, elementary means.

Another important feature of P. L. Chebyshev’s scientific activity is his constant interest in practical issues, the desire to connect theoretical problems of mathematics with the needs of natural and technical, practical activities of people. In the light of modern trends in the development of science, the programmatic setting of scientific activity by P. L. Chebyshev seems extremely practical: “The practical activity of man is extremely diverse, and in order to satisfy all its requirements, there are not enough sciences and many different methods. But of these, especially important are those that are necessary to solve various modifications of the same problem, common to all practical human activity: how to allocate one’s means to achieve the greatest possible benefit” (P. L. Chebyshev, Socinnia, vol. II, St. Petersburg, 1907, p. 239).

It should be noted that for Pafnutya Lvovich himself, his interest in practice turned out to be extremely fruitful, since many of his mathematical discoveries were made while solving applied problems. For example, the study of the hinge mechanism, known as the “Watt parallelogram,” led him to the creation of the basic theory of the best approximation of functions by polynomials, which has now turned into a widely developed mathematical field of great applied importance. In terms of the richness of mathematical ideas, the significance of the results obtained and the study of their acquisition, the work of P. L. Chebyshev “The Theory of Mechanisms Known as a Parallelogram” (1854) is outstanding. In this method, Chebyshev introduces and solves the famous problem: “Given a continuous function f(x); among all polynomials P(x) of degree n, find one such that in a given interval the maximum value of the difference |f(x) - P(x)| it was possible to do less." If the function itself is a polynomial of degree n+1, then the problem posed is equivalent to finding a polynomial of the same degree, with the same coefficient for the leading term, which deviates less from zero on the interval under consideration. This is how the famous “Chebyshev complete nomes” arose - polynomials that “deviate least away from zero” in the interval [−1, +1]: T_n(x) = \frac{1}{2^{n-1}} \cos(n \arccos x). (These expressions are polynomials in x of degree n. The coefficients of the highest degree of these polynomials are equal to one.) Chebyshev also studied the approximation of functions after a variety of functions and trionometric polynomials. (A rational function is the ratio of two polynomials, and a trionometric polynomial is a polynomial of sin x and cos’x.)

In probability theory, Chebyshev managed to obtain a number of important results using extremely simple means. Many results and conclusions were only named, not completed, but all of Chebyshev’s work in this area was the basis on which the Russian school of probability theory developed. Rigorous proofs of polynomials, the least

Chebyshev, and their further development was carried out by his students, academicians A. M. Lyapunov and A. A. Markov.

P. L. Chebyshev’s research in number theory was of outstanding importance for science. For the first time since Euclid, with surprisingly witty and surprisingly elementary reasoning, he obtained the most important results in the problem of the distribution of prime numbers in the works “On Determining the Number of Prime Numbers Not Exceeding a Given Value” and “On Prime Numbers.” P. L. Chebyshev proved that the function $x(x) - x$ of simply numbers not exceeding them $x$ satisfies the inequality

$$0,92 < \frac{\pi(x)\ln x}{x} < 1,06^*$$

for sufficiently large $x$. This inequality was then somewhat studied by the English mathematician Sylvester, and the French mathematician Hadamard strictly proved that

$$\lim_{x \to \infty} \left[ \pi(x): \frac{x}{\ln x} \right] = 1$$

or asymptotically $\pi(x) \sim x/\ln x$.

Here Chebyshev proved the statement (the so-called Bertrand postulate), which is that between $x$ and $2x(x \geq 2)$ there is always at least one prime number **).

Classic results were also obtained by Chebyshev in the field of mathematical analysis.

One of the sciences that Pafnuty Lvovich was interested in throughout his life was the theory of mechanisms and machines, and Chebyshev was engaged not only in theoretical research in this area, but also paid more attention to the direct design of mechanisms.

Long before the Soviet Lunokhod 1 continued the first route

$$\ln x - \log_a x \text{ по основанию } e,$$

where $e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$.

**) For the method by which Chebyshev obtained his remarkable results, see pp. 4-7.

Sou on the lunar surface, science fiction and scientists considered various options for machines that would exist for some walking mechanism. P. L. Chebyshev developed a version of a plantigrade machine that imitates the movement of an animal when walking.

The enormous influence of P. L. Chebyshev on the development of mathematics in our country is not limited to his personal achievements. His works, exceptionally rich in new ideas and methods, gave a powerful impetus to the development of diverse branches of mathematics and mechanics; in addition, he personally posed important tasks and problems for young scientists. On his direct advice, A. M. Lyapunov began research on the theory of the figuration of rotating life, where he obtained classical results that are of paramount importance for mechanics and cosmogony.

The great mathematician and mechanic P. L. Chebyshev was a leading man of his time. So, for example, together with two other academician mathematicians - V. G. Imshenetsky and V. Ya. Bunyakovsky - he proposed to the physics and mathematics department of the St. Petersburg Academy to elect a wonderful Russian living spirit - Sofya Vasilievna Kovalevskaya **) as a corresponding member of the Academy.

Chebyshev paid a lot of attention to issues of public education, taking an active part in the Academic Committee of the Ministry of Education.

The scientist’s works, his scientific, pedagogical and educational activities, and the famous St. Petersburg mathematical school he founded played an exceptionally large role in the development of accounting mathematics and mechanics. In 1944, the USSR Academy of Sciences established the P. L. Chebyshev Prize for the best research in the field of mathematics and the theory of mechanisms and machines.

**) S.V. Kovalevskaya was elected on November 7, 1889.

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ABOUT BERTRAND’S POSTULATE

M. I. Bashmakov

Among the 69 works of Pafnuty Lvovich Chebyshev, two works played a particularly important role in the development of number theory. These works: “On determining the number of prime numbers not exceeding a given value” and “On prime numbers,” written by him in his youth, are related to the most difficult question in number theory: the distribution of prime numbers.

In the second of these works, inequalities are indicated that make it possible to fairly accurately estimate the number of prime numbers in a given segment of the natural series and, in particular, the following fact is proven: between $x$ and $2x$ (where $x>1$) there is always at least one prime number. This statement was made by the famous French mathematician Bertrand, who tested it within the existing tables of prime numbers, but was unable to prove it and accepted it as a construct.

We will trace Chebyshev’s main ideas that led him to these estimates, and then propose a series of problems whose solutions allow readers (familiar enough with the properties of logarithms) to reconstruct Ukazyv’s proof of Bertrand’s construction and obtain inequalities for the number of prime numbers not exceeding a given value.

The reader may wonder what does logarithm have to do with prime numbers? This is the merit of Chebyshev, that he pointed out how to use the general functions of analysis - roots, logarithms - in the study of problems about integers. If we want to find out whether there is at least one prime number on the interval from $n$ to $2n$, then we can proceed as follows:

1. come up with some function $y=f(x)$ of the real argument $x$ that would not change its value while $x$ runs through the segment between neighboring prime numbers, and would take on a larger value when passing through a prime number;

2. prove inequalities for the relative function $f$, which would help to estimate the growth rate quite accurately.

For a function chosen in this way, the difference $f(2n) - f(n)$ will be a positive number if and only if the function $f$ returns when moving from $n$ to $2n$, that is, if there is at least one prime number between $n$ and $2n$.

The first condition is easy to satisfy. We can consider the following function: let us denote by $\pi(x)$ the number of prime numbers not exceeding $x$.

For example, $\pi(1) = 0$ (one is not considered a prime number), $\pi(3) = 2$, $\pi(9) = 4$, $\pi(9,5) = 4$, $\pi(10,9) = 4$, $\pi(11) = 5$, etc.

Of course, if we could prove that

$$\pi(2x) - \pi(x) > 0,$$

we would have proved Bertrand to act, since

$$\pi(2x) - \pi(x)$$

just denotes the number of prime numbers contained between $x$ and $2x$.

8

GRAPHICS
POTENTIAL
ENERGY

R. G. MINZ

How to find out whether body movement is limited and to what extent? Do equilibrium positions exist, and if so, are they stable or unstable?

In this article we will show that such questions are easy to answer if the potential energy of the body is known. We’ll discuss when it makes sense to talk about potential energy and how to calculate its magnitude at each point.

If we talk about potential energy, it means that we want to study the motion of a body, more precisely, of material points *), based on the law of conservation of mechanical energy.

This law states that for a closed system of material points (bodies) the quantities

$$E = T + U.$$

(1)

Sum

$$T = \frac{m_1 v_1^2}{2} + \frac{m_2 v_2^2}{2} + \dots + \frac{m_n v_n^2}{2}$$

is called the kinetic energy of the system, and $U$ is the potential energy of the interaction of the body. Kinetic energy in formula (1) reflects the body’s own motion, and potential energy, which depends on the distances between bodies, their mutual influence on each other.

In what follows, for simplicity, we will consider the movement only

*) As usual, by a material point we mean a body whose own dimensions in this problem are negligible.

one of the material points *), assuming that all the others are at rest. In such cases, we usually talk about the motion of a particle in an external strong field, and formula (1) takes the form

$$E = \frac{mv^2}{2} + U.$$

(2)

Speaking about the movement of a particle in an external field, we will consider the influence of stationary parts on moving ones, and all systems of parts as a whole are closed. There is, however, a subtlety to this. Let us trace it visually using the example of two materials of points with masses $m$ and $M$. Let a particle of mass $M$ dig around at some moment. Let us find out under what conditions it continues to remain motionless, that is, it will be possible to talk about the movement of a particle of mass $m$ in an external field. Let’s use the law of conservation of momentum

*) The problem of the movement of a large number of telecommunications interacting with each other is extremely complex. Until now, they had only been able to solve the problem of two bodies in a general form.

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nia (after all, our system of bodies is closed): $M\vec{v} + m\vec{v}_1 = m\vec{v}_0$. (Here $\vec{v}_0$ is the initial speed of a particle with mass $m$, $v_1$ is its speed after some time and $v$ is the speed acquired by particles with mass $M$.)

Hence $\vec{v} = \frac{m}{M} (\vec{v}_0 - \vec{v}_1)$. If $\vec{v}_1 \neq \vec{v}_0$, then $\vec{v} = 0$ only when $M \to \infty$ Physical this means that $m \ll M$. Thus, the external field is the result of the action of heavy stationary particles on light ones. Let us recall two typical examples of an external field: a satellite in the Earth’s gravitational field, an electron in the field of a nucleus. In each of these examples, of course, condition $m \ll M$ is satisfied.

If the total energy $E$ of the particle is given, then correspondence (2) imposes restrictions on in which region of space the particle can be located. Indeed, after all, the kinetic energy $E_{\text{kin}} = \frac{mv^2}{2}$ allows on the basis of $\sqrt{\frac{2}{m}} E_{\text{kin}}$ to be minimal!), and from the law of conservation of energy it follows that

$$\frac{mv^2}{2} = E - U.$$

Thus, $E \geq U$; we have obtained a criterion for determining the region in which movement with total energy $E$ is possible. At points where $E = U$, kinetic energy, and therefore speed, vanishes.

Let us consider as an example the movement of a body along a straight line (axis $x$). Then the potential energy depends only on the position on this line: $U = U(x)$. In this case, our criterion is very convenient to apply graphically. Let us draw a straight line $U(x) = E$ on the horizontal potential energy graph. Wherever the potential energy graph passes below the straight line, $E > U$ and movement is possible. At the points of intersection, speed

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Fig. 2. The body, moving to the left, approaches the point $x_1$, where its speed is zero. One cannot accumulate at this point, since a force acts on it (see the sled rule on page 12). Therefore, it will continue to move to the right, increasing speed.

Knowing the potential energy, it is easy to determine the speed of the particle at each point. Since $\frac{mv^2}{2} = E - U$, then $v = \pm \sqrt{\frac{2}{m} [E - U]}$.

Two signs of speed correspond to two possible directions of movement along the $x$ axis. figure 2 graphically shows the dependence of the particle velocity on its position for the case when the energy and potential energy correspond to figure 1a. The direction of movement is indicated on the graph by an arrow.

So, we see that with the help of potential energy graphs we can simply and clearly find out where the possibility of movement is, what type of trajectories it has and how the speed of a body depends on its position. When clarifying these questions, we needed to know the behavior of potential energy as a function of coordinates “in general,” that is, how it behaves along the entire trajectory.

Let’s now see what is associated with the rate of change of potential energy, that is, the quantity

$$\frac{\Delta U}{\Delta x} = \frac{U(x_2) - U(x_1)}{x_2 - x_1},$$

where $\Delta x = x_2 - x_1$ is very small

According to the law of conservation of mechanical energy

$$\Delta U = U(x_2) - U(x_1) = \frac{mv_1^2}{2} - \frac{mv_2^2}{2} = \frac{m}{2} (v_1 + v_2) (v_1 - v_2)$$

where $v_1$ is the speed of the body at point $x_1$, and $v_2$ is its speed at point $x_2$. In a small area $\Delta x$ the force acting on the body can be considered constant. Movement under the influence of a constant force, which means $F$, equal to acceleration, and acceleration $a = F/m$. In this case, $v_2 - v_1 = a \Delta t = \frac{F}{m} \Delta t$ ($\Delta t$ is the time of movement), and $\frac{v_1 + v_2}{2}$ is equal to the average speed $\frac{v_1 + v_2}{2} = v_{cp} = \frac{\Delta x}{\Delta t}$. Let’s substitute these expressions into formula (4):

$$\Delta U = m \frac{\Delta x}{\Delta t} \left( -\frac{F}{m} \Delta t \right) = -F \Delta x.$$

From here

$$\frac{\Delta U}{\Delta x} = -F.$$

Formula (5) can also be obtained otherwise, taking into account that with uniform motion only the potential energy of the body changes and its change is equal to the work of the force acting on the body.

Let us find out the geometric meaning of formula (5). Let us be given a potential energy graph. It is clear that the smaller the value of $x_2 - x_1$, the more accurately the rate of change of potential energy coincides with the tangent of the tangent to the potential energy curve (Fig. 3). Action

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Indeed, $tg\alpha_1 = \frac{a}{b} = \frac{U(x_2) - U(x_1)}{x_2 - x_1}$, and the smaller $x_2 - x_1$, the more accurately the angle $\alpha_1$ is equal to the angle $\alpha$. Thus, we come to the conclusion that the tangent of the tangent to the potential energy graph, taken with the opposite sign, coincides with the force acting on the body. This means that the more the potential energy increases, the greater the magnitude of the force and vice versa.

After formula (5) has acquired a clear geometric meaning, it is easy, for example, to find out whether equilibrium positions exist and what they are. We call the equilibrium position the point at which a body can remain for as long as possible, provided, of course, that it initially has zero speed. When is this possible? A change in speed is characterized by acceleration: if it is zero, then the speed of the body is unchanged. Acceleration is proportional to the force acting on the body. This means that equilibrium positions are those points where the force vanishes.

Let’s look again at the potential energy graph. As we have already seen, the force acting on a body is equal to the tangent of the tangent to the potential energy curve taken with the opposite sign. If this angle is zero, then the force is zero. Thus, on the potential energy graph, the equilibrium positions correspond to those points at which
Fig. 4

Fig. 5a,6

whose tangent is parallel to the abscissa axis. It is obvious, for example, that equilibrium positions are maxima and minima of potential energy (Fig. 4).

Equilibrium positions can be stable and unstable. What do you mean? Let us slightly deviate the body from its equilibrium position. If at the same time the force acting on the body returns it back, then the equilibrium is called stable; if not, then it is unstable.

Let us examine for stability the equilibrium position corresponding to the minimum on the potential energy curve (point $A$ on Fig. 4). Let the body move from point $A$ to the right. In the new position $A_1$ (Fig. 5 a), the tangent of the tangent angle is positive, and the force is negative, that is, directed in the direction opposite to the displacement and returns the body to

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point $A$. If you should merge to the left of point $A$, then in this position ($A_2$ on Fig. 5 a) the tangent of the tangent angle is negative, the force is positive, that is, directed to point $A$ and forces the body to return to its original position. Thus, the equilibrium position at the minimum potential energy is stable. Similarly, we can consider the stability of the equilibrium position if the potential energy is maximum (Fig. 5 b).

So, we see that from the potential energy curve we can find the magnitude of the force acting on the body, and it is easy to determine where the equilibrium positions are and what they are. In this sense, the potential energy graph can be compared to an ice slide from where it descends to the Sankh. In those places where the hill is round, the sled accelerates quickly; on a gentle descent, it accelerates more slowly. The top of the slide and the depressions in it are the equilibrium positions. After all, if you put a sled there, they will stay there. In this case, the top is the position of unstable equilibrium, and the valley is stable.

Now, perhaps, let’s talk about how to calculate the value of potential energy. To do this, let us return again to formula (5) and rewrite it in the form

$$U(x_2) - U(x_1) = -F \cdot (x_2 - x_1).$$

Recall that the length $\Delta x = x_2 - x_1$

Fig. 6

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The energy of the body is equal to the area taken with the opposite sign *) under the graph of the dependence of the action of force on the body on the coordinates **). Then its graph is a straight line parallel to the x-axis. The area under it is equal to $F \cdot (x - x_0)$, from where we obtain for the difference in potential energies

$$U(x) - U(x_0) = -F \cdot (x - x_0)$$

(Fig. 7a, b).

Using correspondence (8) we obtain

$$\frac{mv^2}{2} - \frac{mv_0^2}{2} = F \cdot (x - x_0).$$

$$\frac{mv^2}{2} - Fx = \frac{mv_0^2}{2} - Fx_0,$$

that is, potential energy $U = -Fx$. However, this is not the only possibility; with the same reason we could write such an equality

$$\frac{mv^2}{2} - Fx + C = \frac{mv_0^2}{2} - Fx_0 + C,$$

where $C$ is a constant number. But then the potential energy is $U = -Fx + C$. Thus, we see that the potential energy is determined exactly up to an arbitrary constant. This can be explained by the very law of conservation of mechanical energy. After all, if we say that the quantities $E = \frac{mv^2}{2} + U$ are conserved, then $E' = \frac{mv^2}{2} + U + C$ is also conserved, where $C$ is some constant. But in this case $U' = U + C$. This means that potential energy can only be determined to an accuracy of an arbitrary constant, which is always chosen from messages of convenience *). Of course, such uncertainty of potential energy does not in any way affect the conclusions we have drawn about the characteristics of motion in a given field, the presence of equilibrium positions in it, and the dependence of the body’s speed on the coordinate. You can easily verify this; you only need to simultaneously change the potential energy by values ​​$C$ and change the total energy of the body by the same amount. The fact is that to fully describe the movement of a body, it is necessary to specify its position and speed at some moment

*) It should be noted that if the force is negative, then the corresponding participation should be assigned a minus sign.

*) A field in which the force acting on the body is constant everywhere is called homogeneous. Homogeneous, for example, is the gravitational field near the Earth’s surface $F = mg$ and the electric field inside a flat capacitor.

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Fig. 8

time, thereby we will know
$\frac{mv_0^2}{2}$ and $U(x_0)$ Polyana energy
$E = \frac{mv_0^2}{2} + U(x_0)$ By adding some constant value to $U$, the energy of $E$ will increase by exactly the same amount.

Let us now consider the motion of the cube outside in the absence of friction (Fig. 8) and calculate what the potential energy is in this case. According to Hooke’s law, the force acting on the cube is equal to $F = -kx$, where $k$ is the stiffness of the spring $x$ is its deformation. In this case, the graph of the force is straight, the tangent of the angle of inclination of which to the axis $x$ equals $k$ (Fig. 9a)

The potential energy difference $U(x_1) - U(x_0)$ is equal to the area of ​​the trapezoid $ABCD$ protected on Fig. 9a, the height of which is $BC = x_1 - x_0$. and we get the main $\frac{AB + CD}{2} = \frac{kx_0 + kx_1}{2}$. Using the formula for the area of ​​a trapezoid, we find that $U(x_1) - U(x_0) = \frac{(x_1 - x_0)(kx_0 + kx_1)}{2} = \frac{k(x_1^2 - x_0^2)}{2} = \frac{kx_1^2}{2} - \frac{kx_0^2}{2}$.

As a result, for $U(x)$ we obtain the following expression: $U(x) = \frac{kx^2}{2}$

*) Similar to the case of a homogeneous field with constant $C$, we chose here so that $U(0) = 0$.

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cases ($F = \text{const.} \quad F = -kx$) it is impossible to calculate this area without knowledge of integral calculus. If we find it, then it turns out that for a force changing according to the law $F = \frac{\alpha}{r^2}$, the potential energy

$$U = \frac{\alpha}{r}$$

Please note that in formula (10) we have chosen the constant $C$ so that the potential energy vanishes at infinity. Why is this convenient and when? Let us remember that potential energy characterizes the action of bodies on each other—interaction; if at infinity the force becomes zero, then the further the bodies are separated from each other, the weaker their mutual influence. This is precisely why the choice of constant in our and similar ($F(\infty) = 0$) cases is accordingly.

From formula (10) it follows that the type of potential energy significantly depends on the sign of $\alpha$ (see Fig. 106). If $\alpha > 0$, then the force is $F = \frac{\alpha}{r^2} > 0$, and we are dealing with repulsion (like charges, $\alpha = q_1 q_2 > 0$). If $\alpha < 0$, then the force is $F = -\frac{|\alpha|}{r^2} < 0$, which corresponds to attraction (opposite charges, $\alpha = -q_1 |q_2|$). Here, please, it is worth returning once again to the analogy with a sled going down an ice slide. Notice that if you put a sled on a slide that has a potential energy interwash profile, they point in the direction of the force. Such a set of rules makes it possible to immediately determine whether attraction or repulsion is taking place. Indeed, if the sank is shown to the point $r = 0$ (for example, the lower curve at Fig. 106), then attraction takes place, if on the contrary (for example, the upper curve), then repulsion occurs.

The same rule allows us to qualitatively construct the type of potential

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Stifling. At large distances, attraction quickly kills, and recoil when approaching, starting from a certain distance, returns very sharply. The described picture of the interaction of atoms corresponds to the potential energy graph. Shown-

16

nyy on figure 11 Make sure of this, use the analogy with an ice slide. Point $r=r_0$ corresponds to the position of stable equilibrium, and therefore $r_0$ is equal to the distance between atoms in the molecule.

In conclusion, let us summarize some results. We have seen that from the potential energy graph we can find out a number of essential characteristics of movement (the number of different trajectories depending on the energy, whether the movement is limited or not, whether there are equilibrium positions and whether they are stable or unstable). Each of them, as it turned out, has a simple geometric meaning. It is also easy to determine the magnitude and direction of the force acting on the body (“the sled rule”). We have shown that the area under the force graph, taken with the opposite sign, is equal to the potential energy of the body. This makes it possible in some cases to directly calculate potential energy. This makes solving various problems much easier.

And finally, we have seen how the qualitative form of potential energy allows us to understand what exists and give an answer to it.

Now try to solve the following problems yourself.

Exercises

1. Find out the area of ​​possible movement when
   $E<U_0$ and $E>U_0$.
   if the potential energy has the form shown in Fig. 12.

2. Find equilibrium positions along the potential energy curve (Fig. 13) and find out their stability.

Draw a graph of force versus coordinate.

3. Show geometrically that when the potential energy changes by a constant value $C$, the force acting on the body at each point does not change.

4. A cube with side $a$ of density $\rho_1$ floats on the surface of a liquid whose density is $\rho_2$. Draw graphs of the force acting on the cube and the potential energy depending on the depth of the piercing.

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HOLES
HIPPOCRATES

In the fifth century BC, that is, almost two and a half thousand years ago, a scientist lived in Greece, named Hippocrates of Chios. It was Hippocrates who made the first known attempt to create the “Elements” of geometry. Unfortunately, this book has not reached us.

According to one legend, the unlucky merchant Hippocrates was limited to his ancestors and went to Athens to desire them. Here he met sages who spent their time mainly solving geometric problems. Having not found justice for the robbers, Hippocrates consoled himself with the fact that he surpassed the most skilled sages in geometry. . .

Hippocrates tried to square the circle, in other words, to construct a circle equal in size to a given square using a compass and ruler. And although he, naturally, did not succeed, he managed to construct several figures, which were pieces of a plane, limited by two circles, and nevertheless equal in size to a square.

In the history of geometry, the name of Hippocrates is associated with figures of a special type, the so-called lune. A lunula (in other words, a crescent) can be defined as follows: on a chord of an arbitrary circle, a semicircle is constructed externally, the ends of which coincide with the ends of the chord. Figure bounded by a smaller arc

the circumference of a given circle and the drawn semicircle is the hole.

On figure 1, four holes are also painted in the desired color. Hippocrates noted that their total area is equal to the area of ​​a square, colored blue here. Indeed, the sum of the areas of the semicircles built on the sides of this square is equal to the area of ​​the circle in which the square is inscribed. If you remove the purple-colored segments from the semicircles, then you will be left with four holes; if you remove them from the large circle, then the status remains a square.

Fig. 1.
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18

Fig. 2.

Figure 2 illustrates another theorem of Hippocrates. It turns out that the area of ​​a trapezoid inscribed in a circle with a center at point O is equal to the sum of the areas of the orange holes and the orange semicircle (all three holes are equal in construction, the semicircle is equal to the semicircles from which the holes are formed). This theorem is proven in the same way as the previous statement.

At figure 3 there is a figure also compiled by Hippocrates. The two sides and the smaller base of the trapezoid are equal to 1, the larger base is equal to $\sqrt{3}$. The lower arc of the figure, protected on figure by black lines and which is naturally called a generalized lone arc, touches the diagonal of the trapezoid. It turns out that the area of ​​the inscribed trapezoid and the lune are equal (prove this by the areas of the lune. The proof of this fact is easy to obtain if you use the Pythagorean theorem. By the way, this figure has another amazing property. The lune has the same width. More precisely, the diameters of the circles most inscribed in them are each equal to half the difference between the sum of the legs and the hypotenuse of the triangle shown in the same figure.

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the sides of the triangle $T$ still serve as the metric.

Since the symmetric segments are equal in size, it is now sufficient to prove (see Fig. 6 f) that the sum of the areas of the quadrilaterals $R_1, R_2$ and $R_3$ without the area of ​​the triangle $L$ is equal to twice the area of ​​the triangle $T$.

Triangles $L$ and $L'$ are symmetrical about side $c$. Let us now reflect the triangles $P$ and $Q$ first relative to the side $c$. We get triangles $P'$ and $Q'$. Then reflect triangle $P'$ relative to side $b$, and triangle $Q'$ relative to side $a$. A figure is formed from two triangles $T_1$ and $T_2$.

Both of these triangles are symmetrical

Fig. 62.

$$\frac{S}{S'} = \pi$$

Fig. 7

Fig. 8a, b, c.

22

Physics
helps
geometry

B. Yu. Kogan

When studying physics, of course, mathematics is used. But it turns out that the other way around is also true: physics benefits mathematics. Here are some examples of how the concept of center of gravity can be used in geometry.

1. Theorem on medians of a triangle. Let us place material points of equal mass at the vertices of the triangle (Fig. 1). The center of gravity of the system of points A and B is located at point K - the middle of the segment AB. This means that the center of gravity of the system of points A, B and C lies on the median of the SC. Reasoning in a similar way, we will come to the conclusion that this center of gravity lies on the median AL and on the median BM. But the system of points A, B and C has only one center of gravity. Therefore, the medians of a triangle intersect at one point.

Further, since point C has mass m, and point K “represents” a system of points A and B, having mass 2m, then KS is half the size of SC. Therefore, the point of intersection of the medians divides each of them in a ratio of 2:1.

2. Theorem on the midlines of a quadrilateral. Let us place material points of the same mass (Fig. 2) at the vertices of the quadrilateral ABCD. Since the center of gravity of the system of points A

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or about points A and D, we will come to the conclusion that the center of gravity of the system of points A, B, C and D lies at the intersection of the mean lines KM and LN. Further, since the system of points A and B has the same mass as the system of points C and D, then KS = SM. Reasoning in a similar way, we find that LS=SN. Consequently, the midlines of the quadrilateral, intersecting, are divided in half.

Note that our reasoning does not use the fact that points A, B, C and D lie in the same plane. This means that the proven theorem is also true for a spatial quadrilateral. (It follows from this that the midlines of such a quadrilateral necessarily intersect.)

3. Generalization of the previous theorem. Consider an arbitrary quadrilateral ABCD (Fig. 3). Let us place material points at its vertices: at vertex A - masses m, at vertex B - masses αm, at vertex D - masses βm and at vertex C - masses αβm, where α and β are arbitrary positive numbers. Then the center of gravity of the system of points A, B, C and D will lie at the intersection of straight lines KM and LN, where K is the center of gravity of the system of points A and B, M is the center of gravity of the system of points C and D, L is the center of gravity of the system of points B and C, and N is the center of gravity of the system of points A and D. Further, since lengthwise arbitrary correspond to weights, then

$$\frac{AK}{KB} = \frac{\alpha m}{m} = \alpha, \quad \frac{DM}{MC} = \frac{\alpha \beta m}{\beta m} = \alpha,$$

$$\frac{NS}{SL} = \frac{\alpha m + \alpha \beta m}{m + \beta m} = \alpha, \quad \frac{BL}{LC} = \frac{\alpha \beta m}{\alpha m} = \beta,$$

$$\frac{AN}{ND} = \frac{\beta m}{m} = \beta,$$

$$\frac{KS}{SM} = \frac{\alpha \beta m + \beta m}{\alpha m + m} = \beta.$$

Therefore, if the sides of this quadrilateral are divided so that

$$\frac{AK}{KB} = \frac{DM}{MC} = \alpha, \quad \frac{BL}{LC} = \frac{AN}{ND} = \beta,$$

then the segments NL and KM will separate in that

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For example, on figure 4 each “vertical” segment is divided into three identical sections, and each “horizontal” segment is divided into four identical sections *).

4. Finding matches. Using the concept of the center of gravity, problems of the following type can be easily solved. On the sides of the triangle $ABC$ there are points $K$ and $L$ (Fig. 5). Given that $\frac{AK}{KB} = \frac{1}{2}$ and $\frac{BL}{LC} = \frac{3}{5}$, find the ratios of $\frac{AS}{SL}$ and $\frac{CS}{SK}$.

Solution. Let us place material points at the vertices $A, B, C$ so that the correspondence holds
$$\frac{m_B}{m_A} = \frac{AB}{KB} = \frac{1}{2}, \quad \frac{m_C}{m_B} = \frac{BL}{LC} = \frac{3}{5},$$
for example, let’s put
$$m_A = 10m, \quad m_B = 5m, \quad m_C = 3m.$$
Then the center of gravity of the system of points $A$ and $B$ will lie at point $K$, and the center of gravity of the system of points $B$ and $C$ will lie at point $L$. Consequently, the center of gravity of the system of points $A, B$ and $C$ lies on the straight line $CK$ and at the same time on the straight line $AL$, that is, it is located at the intersection point $S$. From here it is easy to find that
$$\frac{AS}{SL} = \frac{m_B + m_C}{m_A} = \frac{5m + 3m}{10m} = \frac{4}{5},$$
$$\frac{CS}{SK} = \frac{m_A + m_B}{m_C} = \frac{10m + 5m}{3m} = \frac{5}{1}.$$
Exercise

1. Each vertex of the tetrahedron is connected to the intersection point of the medians of the opposite face. Prove that the four resulting segments intersect at one point.
2. In triangle $ABC$ the median $AK$ is drawn. The line connecting the vertex $C$ with the midpoint of the median intersects the side $AB$ at the point $L$. Find the ratio $AL:LB$.
3. In the triangle $ABC$ shown at figure 6,
   $$\frac{AK}{AB} = \frac{BL}{BC} = \frac{CM}{CA} = \frac{1}{3}.$$
   What part of the area of ​​triangle $ABC$ is the sum of the areas of the protected triangles?
   *) This theorem is also true for a non-planar quadrilateral.

26

Impossible objects

L. Penrose, R. Penrose

The impression of three-dimensional objects can be conveyed using flat images. This can be used to create a contradictory impression.

figure non-existent structures are described here. Each part is perceived as an image of a three-dimensional three-dimensional object, but the entire figure, due to the incorrect connection of these parts, appears to be a ghost object.

An example of such a construction is shown in figure 1. Each part is a block of rectangular cross-section, but the lines in figure are connected so that the object depicted does not exist in reality. As the eye follows the edge of the block, the apparent distance from the object to the object being observed changes.

figure 2 shows a structure, each part of which is a series of steps. However, the parts of this are connected to each other in such a way that the picture as a whole is incompatible: the steps are constantly rising.

The illusion of non-existent structures can also be caused by unknown objects. To do this, they must be designed in a special way and photographed at a certain angle. An example of such a photographer is shown in figure 3. Here the action of an existing model gives the same impression as the image of an unknown staircase (Fig. 2). This is because the rightmost step when photographing is close to the camera and higher than the step that appears to be located directly above it.

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Fig. 3. Impossible cube model.

open-loop three-dimensional model (Fig. 2). If you look at such a model with one eye and find the desired point, you can see how the two disconnected sides of the model will fit together. By photographing the model from this position, you will get a photograph of the impossible Penrose trigon.

This example is also used to obtain an image of an impossible cube. If you make a cage like the one shown in figure 3, then, positioning it so that the back edges of the cage cover the slits in the front ones, you will see an impossible cube: the back edges of the cage will seem to be pushed forward. (A photograph can also be obtained from this position.) The physiologist Gibson, using this idea, came up with a number of interesting illusions in which he used rectangular cardboard with slits in some of them, so that the most distant of the rectangular ones would appear to be located in the foreground. On figure4 the two sets of squares, although seemingly identical, are arranged differently in space, and in the third row the jack is actually close to the six, and the cards are therefore the same size. Such a photograph can be obtained by making appropriate slits in a nearby card and in the middle rectangular column, through which you can see the more distant six (Fig. 5).

Another series of the same kind of optical illusions can be obtained using the so-called artificial Ames room, the geometry of which is shown in figure 6. This room is made somewhat unusually. The floor in it is inclined and rises to the right of the observer, the rear wall moves away as one moves to the left, the windows on this wall are of different sizes and are shaped like a trapezoid. However, if you look into such a room from a certain point, it looks completely ordinary: the floor of each

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Fig. 5. A device that allowed you to observe the illusion that you see on figure 4.

square, the back wall is perpendicular to the direction of view, and the windows are rectangular and identical. Ames made one of these rooms large enough for people to stay in it. The result was an amazing effect: when the viewer looks at people walking along this room, he observes unusual changes in their sizes. Depending on where a person goes, he seems to be either growing or faltering.

A small model of such a room allows you to see a person’s hands or face through the windows of the back wall. As in a large room, hands or faces appear large or small depending on whether they appear in the windows to the right or to the left. We are so accustomed to rooms of a general shape that in the Ames room we perceive even highly distorted objects as real. Perception of an image is a psychological process. In this case, the attitude of the abuser to the subject of abuse and even his condition at the moment is important. It is interesting that familiar faces are perceived as completely normal in Ames’s room, while the geometry of the room always becomes noticeable to the observer.

Of course, building a large Ames room yourself is quite difficult. But it is apparently possible to make a small model. By placing interior items, you can get some pretty effective photos if you want. Why not try it?

Fig. 6. Geometry of the twisted Ames room.

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Tasks

M81. An arbitrary point $P$ is taken inside the square $A_1 A_2 A_3 A_4$. From vertex $A_1$ a perpendicular is dropped to line $A_2 P$, from vertex $A_2$ - to $A_3 P$, from $A_3$ - to $A_4 P$, from $A_4$ - to $A_1 P$. Prove that all four perpendiculars (or their extensions) intersect at one point.

A. N. Vilenkin

M82. There are several identical automatic vehicles on the ring road. It is known that if all the gasoline available in these automatic vehicles were poured into one, then this car would be able to drive along the entire ring road and return to its original place. Prove that at least one of the cars standing on the road can compress all the stakes, taking gasoline from the rest of the cars along the way.

M83. Prove that the numbers 1, 2, …, $n$ for any $n$ cannot be divided into two groups so that the product of numbers in one group is equal to the product of numbers in the other.

M84. Let $A$ be the base of the perpendicular dropped from the center of the given circle onto the given straight line $l$.

On this line two more points $B$ and $C$ are taken so that

$$AB = AC.$$

Two arbitrary secants are drawn through the points $B$ and $C$, one of which intersects the circle at points $P$ and $Q$, the second - at points $M$ and $N$. Let lines $PM$ and $QN$ intersect line $l$ at points $R$ and $S$. Prove that

$$AR = AS.$$

This problem (or some of its variants) is sometimes called the “butterfly problem”; the origin of this name is clear from figure 1.

M85. Prove that if $a_1, a_2, ..., a_m$ are pairwise distinct natural numbers, none of which is divisible by the square of an integer greater than one, and $b_1, b_2, ..., b_m$ are integers other than zero, then

$$b_1 \sqrt{a_1} + b_2 \sqrt{a_2} + \ldots + b_m \sqrt{a_m} \neq 0.$$

L. N. Vaserstein

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F93. In the side wall of a vessel filled with a liquid with a refractive index $n$, a hole of small radius $r$ continues. A horizontal thin beam of light is directed along the axis of the hole from the vessel. To what level $h$ above the hole must the liquid flow out so that the light ray emerges from the jet without ever experiencing internal reflection?

Neglect the change in the cross section of the jet, and assume the refractive index of the liquid to be sufficiently large.

I. F. Ginzburg

F94. A cycle is performed over one gram-molecule of an ideal gas (a closed process consisting of two isochores and two isobars (Fig. 2)). The 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.

F95. The surface of the river forms an inclined plane. Can a body float freely along a river at a speed exceeding the speed of the current or the maximum speed of the current?

F96. Two identical balls connected by thread. Find the rise height of this system if one of the balls is thrown upward with a speed of $v$.

A. Drozdov

F97. A coil of insulated wire is bent in the shape of a figure eight, the bulb of which has a radius of $r_1 = 1$ cm and $r_2 = 3$ cm. The coil is in a magnetic field with induction $B = 10^4$ ec, perpendicular to the plane of the coil.

The wire insulation is designed for a voltage of 10 e. Will insulation breakdown occur if the magnetic field is suddenly turned off? Field switch-off time $10^{-3}$ sec.

L. G. Aslamaov

Fig. 2.

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Solutions

In this issue we publish solutions to problems M31–M33

M31

A square sheet of paper is cut straight into two parts. One of the resulting parts is again cut into two parts, and this is done many times. What is the smallest number of cuts that need to be made so that the resulting parts turn out to be exactly twenty-decagons?

With each cut, the total number of pieces of paper increases by 1 (since one piece disappears and two new ones appear), so after $n$ cuts there will be $(n+1)$ pieces of paper.

Let us now calculate what the total number of vertices in all pieces together can be after $n$ cuts. With each cut, the total number of vertices increases either by 2 (if you cut through the vertices), or by 3 (if you cut through the vertex and side), or by 4 (if you cut through 2 sides). Since at first there were 4 vertices, then after $n$ cuts there will be no more than $4n+4$ vertices in all pieces together.

Let’s assume that after $N$ cuts there are 100 dacagons. Since in this case the total number of pieces obtained will be $N+1$, then, in addition to these decadagons, there will be another $N+1-100$ pieces. Each of these pieces will have at least three vertices, so the total number of vertices in all pieces will be at least $100\cdot 20+(N-99)\cdot 3$. As was proven earlier, this number is no more than $4N+4$. Means,

$$4N+4 \geq 100\cdot 20+(N-99)\cdot 3 = 3N+1703,$$

whence $N \geq 1699$.

So, we have proven that it is impossible to get 100 bidecagons by making fewer than 1699 cuts. This is the main and most difficult part of the proof.

Let us now show how you can get 100 bidecagons by making 1699 cuts. Here is one way: cut the square into 100 rectangles (99 cuts) and turn each rectangle into a 20-sided triangle in 16 cuts, cutting off triangles from the corners (1600 cuts). There will be 1699 cuts in total.

Answer. Can get 100 decanagons by making 1699 cuts, and by making fewer cuts, 100 dedecagons.

In the same way, you can find out what is the smallest number of cuts you need to make to get exactly $k$ l-gons.

The exact solution to this problem was sent by D. Grigoriev from Leningrad and V. Livits from Zaporozhye.

I. N. Bernistey

M32

In all cells of the table $100 \times 100$ is a plane. It is allowed to simultaneously change the signs in all cells of one row or one column. Is it possible, by continuing the operations several times, to obtain a table with exactly 1970 minutes?

Let us change the sign $x_i$ times in the $i$th row, and $y_k$ times in the $k$th column. Then in the cell located at the intersection of the $i$th row and $k$th column, the sign will change $x_i+y_k$ times. Consequently, this cell will contain a minutec if and only if $x_i+y_k$ is odd. Thus, about

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The exact number of minuses in the resulting table depends only on the parity of the numbers $x_i$ and $y_k$; on figure 1, ​​zeros are placed together with even numbers, and ones are placed together with odd numbers. Let $x$ be the number of odd numbers among $x_i$; $y$ is the number of odd numbers $y_k$ (on figure $x = 5, y = 4$). Then, as is easy to calculate, the total number of minuses in the table will be equal to

$$x(100-y)+(100-x)y = 100x+100y-2xy,$$

where $x, y$ are integers.

Now let’s return to our problem and prove that the answer is negative, i.e. it is impossible to get 1970 minuses. Let’s assume that we managed to get 1970 minuses. Then $1970 = 100x+100y-2xy$, from where

$$xy - 50x - 50y + 2500 = 1515.$$

Factoring the left-hand side, we get

$$(x-50)(y-50) = 1515 = 15 \cdot 101.$$

We have

$$-50 \leq x-50 \leq 50, -50 \leq y-50 \leq 50.$$

Since 101 is a prime number, then either $x-50$ or $y-50$ is divisible by 101 (after all, their product is divisible by 101). But from the previous inequalities it then follows that either $x-50$ or $y-50$ is equal to 0. So, we have a contradiction that proves that exactly 1970 minutes of obtaining the impossible.

The correct solution was sent by D. Grigoriev.

A. V. Zelevinsky

M33

There is a natural number $n > 1000$. Let’s take the remainders from dividing the number $2^n$ by the numbers 1, 2, 3,…, $n$ and find the sum of these remainders. Prove that this sum is greater than $2^n$.

Let us denote the sum of remainders from dividing $2^n$ by the numbers 1, 2,…, $n$ by $S_n$. We will succeed in the statement of the problem and prove that for $n > 1000$ $S_n > 3,5n$. Along the way, we will prove the same method as

$$S_n > \frac{n}{2} (\log_2 n - 4)$$

for any $n$.

We follow mainly the decision of D. Grigoriev from Leningrad. The solution is based on a very simple consideration: $2^n$ is not divisible by an odd integer other than 1, which means that the remainder of division of $2^n$ by such a number is not less than 1. From here it is easy to deduce that for any odd $k > 1$ the remainder of division of $2^n$ by $2^l k$ is not less than $2^l$. Indeed, let the remainder of the division of $2^{n-l}$ by $k$ be equal to $r$, that is, $2^{n-l} = mk + r$, where $0 \leq r < k$, then $2^n = m(2^l k) + 2^l r$, where $0 \leq 2^l r < 2^l k$, that is, the remainder of the division of $2^n$ by $2^l k$ is equal to $2^l r$, and since $r \neq 0$, then this remainder is not less than $2^l$.

It follows that in any case

$$S_n \geq 1 \cdot (\text{количество не превышших } n \text{ чисел вида } 2k + 1, k > 1) + 2 \cdot (\text{количество не превышших } n \text{ чисел вида } 2(2k + 1), k > 1) + 2^2 \cdot (\text{количество не превышших } n \text{ чисел вида } 2^2(2k + 1), k > 1) + \dots$$

• $2^l \cdot (\text{количество не превышших чисел вида } 2^l(2k + 1), k > 1), (*)$

34

where $l$ is determined by the condition $3 \cdot 2^l \leq n$, $3 \cdot 2^{l+1} > n$ (there is no point in taking more than $l$, since then the expression in the bracket will be equal to zero).

What can you say about the right side of this inequality?

Let’s consider the $(i+1)$-th term: $2^i(\ldots)$. The number in the bracket is obviously equal to the number of non-exceeding $n$ terms of the arithmetic progression $3 \cdot 2^i$, $5 \cdot 2^i$, $7 \cdot 2^i$, …, that is, a progression with the first term $3 \cdot 2^i$ and the difference $2 \cdot 2^i$. The number of such terms is no less than $\frac{n-3 \cdot 2^i}{2^{i+1}}$ and, therefore, the $i$ term on the right side of the inequality is no less than $\frac{n}{2} - \frac{3}{2} \cdot 2^i$. Basically, the solution to the problem is over. Everything else is various transformations of the right side () using the obtained inequalities. Let us replace, for example, the sum on the right side () when washed with eight terms. Then we will get

$$S_n > 8 \cdot \frac{n}{2} - \frac{3}{2} \sum_{i=0}^{7} 2^i >$$

$$> 8 \cdot \frac{n}{2} - \frac{3}{2} \cdot 2^8 = 3,5n + \left(\frac{n}{2} - 384\right).$$

For $n > 1000$ the expression in the bracket is positive and, therefore, we have proven that $S_n > 3,5n$ for $n > 1000$. Thus problem M33 is completely solved.

However, a more precise inequality can be obtained. Let’s sum up all the terms on the right side (*); we get:

$$S_n > (l+1) \frac{n}{2} - \frac{3}{2} \sum_{i=0}^{l} 2^i >$$

$$> (l+1) \frac{n}{2} - \frac{3 \cdot 2^{l+1}}{2}.$$

Using the inequalities $3 \cdot 2^l \leq n$, $3 \cdot 2^{l+1} > n$, we find that

$$S_n > \frac{n}{2} \log_2 \frac{n}{3} - n =$$

$$= \frac{n}{2} \left(\log_2 \frac{n}{3} - 2\right) > \frac{n}{2} \left(\log_2 n - 4\right).$$

This formula shows, for example, that if $n > 2^{1004}$ then $S_n > 500 n$.

Task. The sum of the remainders from dividing the numbers $2^{l+1}$, $2^{l+L}$ by the odd number $m > 1$ is greater than $\frac{m}{2} \left(\frac{L}{\log_2 2m} - 1\right)$.

Using this problem, the solution of which will be given below, one can prove that for an infinite number of values ​​of $n$ the sum of $S_n$ is actually larger than $\frac{n}{2} \left(\log_2 n - 4\right)$, that is, one can prove that the estimate of $S_n > \frac{n}{2} \left(\log_2 n - 4\right)$ is very larger for an infinite number of values ​​of $n$. This is not surprising, since our method of proof does not take into account multi-wanted numbers other than zero, while these remainders are often large. At the same time, the value of $S_n$ cannot grow too quickly with the growth of $n$, since there are too many of those of $S_n \leq 1 + 2 + \ldots + (n-1) = \frac{n(n-1)}{2}$.

We believe that the following hypothesis is true:

Hypothesis 1. For sufficiently large $n$ the inequality $S_n \leq \frac{n^2}{4}$ is true;

6. there is a positive number $c > 0$ such that $cn^2 \leq S_n$.

We can neither prove nor verify these statements. Moreover, we do not know whether the weaker one is fair.

Conjecture 2. There is a positive number $c > 0$ such that the inequality $cn^2 \leq S_n$ holds for an infinite number of values ​​of $n$.

If you managed to prove or test one of these hypotheses, send us your reasoning. If they turn out to be interesting, we will publish them in one of the issues of the magazine. As a weak test of Hypothesis 2, the following can be proven.

Theorem. For an infinite number of values ​​of $n$ the following inequality holds:

$$S_n > \frac{n^2}{132 \log_2 n}.$$

The theorem is derived from the problem as follows. Consider the sum $\Sigma_k$ of the remainders from dividing the numbers $2^{k+1}$, …, $2^n$ by the number $k$. According to the problem, if $k$ is odd, the sum $\Sigma_k$ is residually large, therefore the number $\Sigma_1 + \Sigma_2 + \ldots + \Sigma_n$ is also large. But how easy it is for*

*) The number $\frac{n^2}{132 \log_2 n}$ for large $n$ cities is greater than the number $\frac{n}{2} \left(\log_2 n - 4\right)$. It is not difficult to prove, for example, that for a sufficiently large $n$ the ratio of these numbers is greater than $\sqrt{n}$.

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tag, $S_1 + S_2 + \dots + S_n = \Sigma_1 + \dots + \Sigma_n$,
poznym, the amount worth the left side is also
is large and, therefore, one of the terms, say $S_l$, is also large. We leave the careful execution of the proof to the reader. Let us now move on to solving the problem.

Solving the problem. Let’s write down the remainders from dividing the numbers $2^{l+1}, \dots, 2^l + L$ by $m$ and put a “tick” between two consecutive remainders if the first is greater than the second. Then, firstly, the remainder before the checkmark is greater than $\frac{m}{2}$, and secondly, the number of numbers between two consecutive checkmarks is less than $\log_2 2m$, since the remainders between two checkmarks form a geometric progression with value 2 (indeed, $2\log_2 2m - 1 = m > m - 1$, and every remainder is not greater than $m - 1$). Thus, the number of spaces between the checkmarks is greater than $\frac{L}{\log_2 2m} - 1$ and since at the end of each gap there is a number greater than $\frac{m}{2}$, then everything is proven.

A. G. Kushiirenko

In this issue we publish solutions to problems F43, F48—F51

F43

Find the pressure at the center of a liquid planet of radius $R$ if the liquid is incompressible and has a density $\rho$.

Divide the volume of the planet into thin spherical elephants with thickness $\Delta r$. It is easy to show that the equilibrium of the granitization forces acting from the side of the layer on the particle inside this layer is equal to zero. Indeed, for this purpose, consider a cone with a small angle at the apex, in which part of the mass $m$ is placed. The cone cuts out areas of areas $s_1$ and $s_2$ (Fig. 1) from the layer. If the mass of the substance per unit surface of the layer is equal to $\mu$, then the granitonic forces acting on the mass $m$ from sections $s_1$ and $s_2$ are equal

$$F_1 = \gamma \frac{m\mu s_1}{r_1^2}, \quad F_2 = \gamma \frac{m\mu s_2}{r_2^2};$$

But

$$\frac{s_1}{r_1^2} \cos \alpha_1 = \frac{s_2}{r_2^2} \cos \alpha_2 = \Omega$$

(Ω is the solid angle at the vertex of the cone), and $\alpha_1 = \alpha_2$; protected triangles are similar, that is, $\angle OA_1 M_1 = \angle OA_2 M_2, \angle OA_1 B_1 = \angle OA_2 B_2$ (they rest on the same arc) and $\alpha_1 = \angle OA_1 B_1 - \angle OA_1 M_1 a\alpha_2 = \angle OA_2 B_2 - \angle OA_2 M_2$.

Pozny $\frac{s_1}{r_1^2} = \frac{s_2}{r_2^2}$. Thanks to this

$$F_1 = F_2,$$ that is, these forces mutually balance each other. Carrying out a similar consideration for other parts of the layer, we will prove the statement made.

The force with which the layer element $\Delta s\Delta r$ bends to the center of the planet is equal to

$$F = \gamma \frac{4}{3} \pi r^3 \rho \Delta s \Delta r\rho$$,

where $r$ is the distance from this element to the center of the planet. From here we find that the increase in pressure in a section of thickness $\Delta s$ is equal to

$$\Delta p = \frac{F}{\Delta s} = \frac{4}{3} \pi \gamma \rho^2 r \Delta r.$$

Posny pressure at a distance $r_0$ from the center of the planet will be equal to

$$p = p_0 + \frac{4}{3} \pi \gamma \rho^2 \Sigma r \Delta r.$$

Since the sum $r\Delta r$ is equal to the area

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The figure bounded by the graph $y=r$ and the axis $r$ (Fig. 2), then

$$\sum r \Delta r = \frac{(R+r_0)(R-r_0)}{2} = \frac{R^2-r_0^2}{2}.$$

Because

$$p = p_0 + \frac{2}{3} \pi \rho^2 (R^2-r_0^2);$$

$p_0$ is the pressure on the surface of the planet, which can be taken equal to zero. Then

$$p = \frac{2}{3} \pi \rho^2 (R^2-r_0^2).$$

At the center of the planet ($r_0=0$) the pressure will be equal to

$$p_\mu = \frac{2}{3} \pi \rho^2 R^2.$$

F48

A rope is wound around the armature shaft of the dynamo, to which a weight is attached. As the weight descends, it rotates the anchor. When the armature was sufficiently untwisted, a load resistance was attached to the adhesives of the dynamo. Draw on a graph the dependence of the speed of rotation of the anchor on the time from the moment the load begins to move.

Since before the load is connected, the moment of external forces acting on the armature is constant, the armature rotates with constant angular acceleration and its speed increases in proportion to time.

After connecting the load, current will flow through the armature winding. Due to this, a moment of force will act on the armature in the magnetic field of the stator, proportional (like the force acting on a conductor with current in a magnetic field) to the magnitude of the current. This moment of force slows down the rotation of the armature. Since the current flowing through the armature winding is proportional to the em.d.c. induction occurring in the winding, and ed.d.c. induction, in turn, is proportional to the rate of change of the magnetic flux propagating the winding, that is, the angular speed of rotation of the armature, then at a certain angular speed $\omega_0$ the moment of force acting on the armature from the side of the load will be equal to the moment of the braking force acting on the armature in the magnetic field. $\omega_0$ is the speed of steady rotation of the armature.

The graph of the dependence of the armature rotation speed on time is shown in Fig. 4. The cool line is for the case when the armature rotation speed at the moment the load was connected was less than $\omega_0$, and the blue line—when it was more than $\omega_0$.

F49

The U-shaped tube is filled with water (Fig. 4). Air has been removed from one elbow; the air pressure in the other leg at temperature $T_0=20^\circ C$ is equal to atmospheric pressure. Both ends of the tube are full. What will be the difference in water levels in the elbows if the tube is heated to $T=100^\circ C$?

37

Fig. 4.

The pressure in the left knee is equal to the saturated vapor pressure. The right elbow contains both air and water vapor, and the pressure is equal to the sum of the partial pressures of air and steam. Moreover, the steam in the right vessel is also saturated, and its partial pressure is equal to the steam pressure in the left knee. Therefore, when considering the equilibrium of water, we can not take into account the vapor pressure in the left and right elbows.

Let us write down the equilibrium condition for water in the tube at $T_0=20^\circ\text{C}$:

$$\rho gh_0 = P_0, \quad P_0 = 1 \text{ amm.}$$

From here

$$h_0 = \frac{P_0}{\rho g} \approx 10 \text{ m.}$$

At $100^\circ\text{C}$, the air pressure in the right knee becomes $P$, and the difference in water levels in the knees becomes $h$. At the same time

$$\rho gh = P. \quad (*)$$

$P$ is related to $P_0$ by the combined gas law:

$$\frac{P_0 V_0}{T_0} = \frac{PV}{T},$$

and $l - l_0 = \frac{1}{2} (h - h_0)$. Since $V_0 = l_0 s$, $V = ls$ (s is the cross-sectional area of ​​the tube), then from here we find

$$P = P_0 \frac{T}{T_0} \frac{l_0}{l} =$$

$$= P_0 \frac{T}{T_0} \frac{l_0}{l_0 + \frac{1}{2} (h - h_0)}.$$

Substituting this expression into equation (*), we get

$$h = \frac{P_0}{\rho g} \frac{T}{T_0} - \frac{l_0}{l_0 + \frac{1}{2} \Delta h},$$

or

$$h = h_0 \frac{T}{T_0} \frac{l_0}{l_0 + \frac{1}{2} \Delta h}.$$

Assuming that $1/2 \Delta h \ll l_0$, we get

$$h = h_0 \frac{T}{T_0} \approx 13 \text{ m};$$

$\Delta h \approx 3 \text{ m}, \frac{1}{2} \Delta h \approx 1,5 \text{ m} \ll l_0 = 15 \text{ m}$. This means that, comparing $1/2 \Delta h$ with $l_0$, we got a result that is not very different from the exact one.

F50

On a smooth horizontal table lie two identical cubes with masses $m$, connected by a spring of stiffness $k$ and length $l_0$ into a non-growth system. A force $F$, constant in magnitude and direction, will begin to act on the left cube (Fig. 5). Find the minimum and maximum distance between the cups as the system moves further.

Let the distance $l$ between the cups be minimum or maximum (we will say “extreme”). Then both cubes move with the same speed $v$, and the kinetic energy of the system is $2 \frac{mv^2}{2} = mv^2$. At the same time

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Time, the potential energy of the system is equal to the potential energy of the sloped spring, that is, $kx^2/2$, where $x$ is the change in the length of the spring ($x$ can be either positive - then the length of the spring $l = l_0 - x$ is less than $l_0$, or negative - then the length of the spring is greater than $l_0$). The total energy of the system is equal to the sum of the kinetic and potential energy, that is, $mv^2 + kx^2/2$. Since the system acquired this energy thanks to the force $F$, we can write that

$$FL = mv^2 + \frac{kx^2}{2}.$$

(*)

$L$ is the distance traveled by the left cube at the moment when the length of the spring became extreme (Fig. 6): $L = S + \frac{l_0}{2} - \frac{(l_0 - x)}{2} = S + \frac{x}{2}$ ($S$ is the distance traveled by the center of mass of the system).

The speed of the $v$ cups is equal to the speed of the center of mass of the system. A constant external force acts on the system, and it moves with uniform acceleration $a = F/2m$. Therefore, if we denote by $t$ the time from now

Fig. 7.

movement began until the moment when the length of the spring became equal to $L$, then $v = at$ and $L = \frac{at^2}{2} + \frac{x}{2}$. Substituting these expressions for $v$ and $L$ into equation (*), we obtain

$$F \left( \frac{at^2}{2} + \frac{x}{2} \right) = ma^2 t^2 + \frac{kx^2}{2}.$$

Hence $\frac{F^2 t^2}{4m} + \frac{x}{2} F = \frac{F^2 t^2}{4m} + \frac{kx^2}{2}$ or $Fx = kx^2$. Therefore, $x_1 = 0$, $x_2 = F/k$.

Thus, the minimum distance between a group is $l_{\min} = l_0 - x_2 = l_0 - \frac{F}{k}$, and the maximum is $l_{\max} = l_0 - x_1 = l_0$.

F51

Photographing a tiger from a distance of 20 meters is dangerous. What size should a 1mm pinhole camera be to make the tiger in the photo look striped? The distance between the stripes on a tiger’s skin is 20 cm.

A camera obscura is a box of responsibility. The photographic film placed in the camera receives a wide beam of rays from every point of the photographed object (Fig. 7). In order for a tiger to appear striped in a photograph, it is necessary that the areas on which the rays falling from the nearest points of neighboring stripes on the tiger’s pelt fall do not intersect. To do this, it is necessary that the film be located at a distance $x$ from the answer, more than $d/tg\alpha$.

Because

$$tg\alpha = \frac{l/2}{L + x},$$ then $x > \frac{2d(L + x)}{l}$.

Hence $x > \frac{2dL}{l\left(1 - \frac{2d}{l}\right)}$. $\frac{2d}{l} \ll 1$, therefore, neglecting $2d/l$ compared to 1 in the denominator of the formula, we get $x > \frac{2dL}{l} = 20$ cm. The chamber size must be greater than 20 cm.

I. Sh. Slovodechkiy

39

In one of the squares in Krasnodar there is a round flower bed located in the center of the square square ABCD (Fig. 1). Two friends - Vasya and Petya - go to school right through this square. Moreover, sometimes they choose the AMC path (red line), on other days - the ADC path (blue line). Still, it is true that the first path is shorter. Petya believes that the second path is shorter, and here’s why: the circle they have to go around has a rather large radius. “Of course,” says Petya, “if the radius of the circle were not large or if there were no flower beds at all, then the path along the diagonal AC would be clearly shorter than the broken line ADC (Fig. 2). But the size of the circle is considerable, and if we assume that the circle were inscribed in a square, then, Petya argues, we would have a picture like the one in figure 3.” Now Vasya is convinced that the blue line is shorter than the red one: along the “blue” road you have to go “straight” all the time, making only one turn, while along the “red” road you have to make several turns, and sometimes it even seems to go back.

Two friends decided to dispel their doubts in this way: from point A they took different paths: Vasya through the park on the red line, Petya - around the blue line. They approached point C almost simultaneously. The dispute remained unresolved. Then they went back (each their own way) to point A. Now Petya arrived before Vasya. And the dispute was finally resolved by Petya’s crawl. True, it seemed to Vasya that Petya was faster than Vasya.

At home, Vasya decided to do mathematical calculations, denoting the radius of the circle by R, and the side of the square by a.

“It’s strange,” thought Vasya, “the answer to the question of the problem does not depend on the size of the radius of the circle.” However,” he continued his thought, “it was possible to do without calculations: after all, every boundary line, as is known, is longer than any boundary line that has the same ends.

What calculations did Vasya perform and what result did he get when using SEL?

40

WHAT DO WE LOSE OR FIND HERE?

V. A. Petrov

Almost every written exam paper in mathematics includes some kind of equation. When solving it, the examinee must not only accurately check the calculations necessary to obtain the answer, but also be required to make sure that these calculations are correct. You need to be able to analyze each of the proposed transformations: does it lead to an equivalent equation, does it threaten the loss or acquisition of a solution.

Issues of tactics of “fighting” with built roots and with the loss of roots were considered in the article by A. G. Mordkovich “Someone loses, someone finds” («Kvant» No. 5, 1970) *). Now we invite you to check whether you are confident enough in this tactic. First, solutions to the equations are given. You need to analyze these decisions and answer the following question. Is the answer correct? If not, then what is here - are the roots lost or the structure found? Why? What is the correct answer? Compare your thoughts with ours on page 62.

1. $2\sin 4x = \sqrt{2(1 - \cos 4x)}$.

Solution. Using the formula
$$2\sin^2\alpha = 1 - \cos 2\alpha,$$
we get
$$2\sin 4x = \sqrt{4\sin^2 2x}$$
or
$$\sin 4x = \sin 2x,$$
which is already very easy to solve.

Answer: $x = n\pi$ and $x = \frac{\pi}{6} + n\frac{\pi}{3}$, where $n = 0, \pm 1, \pm 2, \dots$

*) This “tactics” is covered in more detail in the book by G.V. Dorofeev, M.K. Potanov, N.H. Rozov, very useful for applicants, “Mathematics Manual for Those Building Universities”, M., “Nauka”, 1970.

2. $8\cos x\cos 2x\cos 4x = 1$.

Solution. Multiplying both sides of the equation by sin $x$ and using the double angle sine formula three times, we get the simple equation
$$\sin 8x = \sin x.$$
Answer: $x = (2n + 1)\frac{\pi}{9}$ and $x = 2n\frac{\pi}{7}$, where $n$ is any integer.

3. $\sin \left(\frac{1}{5}\arccos x\right) = 1$.

Solution. From this equation we find
$$\arccos x = \frac{5\pi}{2} + 10n\pi$$
($n = 0, \pm 1, \pm 2, \dots$).

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From here, by the definition of arccos $x$, we obtain $x = \cos \left( \frac{5\pi}{2} + 10n\pi \right) = 0$.

Answer: $x = 0$.

4. $\log_{x-1} (x^2 - 5x + 7) = 1$.

Solution. Let’s find the ODZ:

$$x^2 - 5x + 7 > 0, \quad x - 1 > 0,$$

whence $x > 1$. According to the definition of a logarithmic function, we have $x^2 - 5x + 7 = x - 1; \quad x_1 = 2, \quad x_2 = 4$.

Since both values ​​are included in the ODZ, they are solutions to the original equation.

5. $\sqrt[3]{2x - 1} + \sqrt[3]{x - 1} = 1$.

Solution. Let’s cube both sides of the equation:

$$3x - 2 + 3\sqrt[3]{(2x - 1)(x - 1)} \times (\sqrt[3]{2x - 1} + \sqrt[3]{x - 1}) = 1.$$

The expression in parentheses represents the left side of the original equation - it can be replaced with the right side. After this, the equation is reduced to the form

$$(2x - 1) (x - 1) = (1 - x)^3.$$

Answer: $x_1 = 0, \quad x_2 = 1$.

6. Can the equation

$$\sqrt{x + a\sqrt{x + b}} + \sqrt{x} = c$$

have infinitely many solutions?

Solution. Move $\sqrt{x}$ to the right and square both sides:

$$x + a\sqrt{x + b} = c^2 - 2c\sqrt{x} + x.$$

After bringing similar terms we get: $(a + 2c)\sqrt{x} = c^2 - b$. For $a + 2c = 0$ and $c^2 - b = 0$ (and only in this case), the equation becomes an identity that is valid for all non-negative $x$.

Answer: The equation has infinitely many solutions at $a = -2c$ and $b = c^2$. The solutions in this case are $x \ge 0$.

7. $$\cos \frac{x}{2} \sqrt{x^2 + (7-\pi)x - 7\pi} = 1.$$

Solution. If, when raised to the power of some positive