Popular science physics and mathematics
quantum
7
1971
magazine
Academy
Sciences of the USSR
And
Academy of Pedagogical
Sciences of the USSR

In the room:

Great Law
1
V. I. Kuznetsov

Archimedes and the squaring of parabolsOLS
7
A. D. Bendukidze

Paradoxes of jet propulsion
11
M. S. Livits

Math circle

Pricili Dirichlet
17
A. I. Orlov

Problem «Kvant»

Tasks
22

Solutions to problems M39—M49; F45, F59—F65
24
L. G. Limanov,
V. N. Berezin,
N. B. Vasiliev,
I. Sh. Slobodetsky

Applicant’s workshop

About problems in photometry
41
V. E. Belonuchkin

Entrance exams in mathematics at the Moscow Textile Institute in 1970
45
T. I. Berkhina,
E. M. Kahn

Written examination in mathematics at the humanities faculties of Moscow State University in 1970
48
A. G. Kushitsrenko,
S. V. Fomin

Information

About mathematical preparation of applicants
57
M. L. Smolyansky,
G. N. Dyachenko,
V. V. Goldber

Collector’s Corner
Stamps dedicated to Marie and Pierre Curie
60
A. V. Almykis

Reviews. Bibliography
Useful book
61
M. B. Balk

Answers, directions, solutions
63

«Kvant» for younger students

Our guest is Vasya Smekalkin
64
V. M. Rosenutuler

Problem about places
4th cover page

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Fig. 1.

expressed by the Moon will be equal to $\frac{v^2}{R_\pi} = 0,0027 \, m/cck^2$. But the apple is attracted by the Earth with much greater force. The earth’s acceleration force is also 9.81 $m/cck^2$.

Why does the Earth attract the Moon thousands of times weaker than an apple? Robert Hooke, using the spring balance he invented, tried to change the change in the force of gravity. But at the top of the mountain the load stretched the spring in the same way as on the seashore. Yet Newton suggested that gravity is weakened by stretching. The height of the mountains is small compared to the size of the Earth, and even more so compared to the stretch from the Earth to the Moon.

After studying the laws of planetary motion established by Johannes Kepler *), Newoton managed to find a simple rule for the decrease in gravitational force with stretching. This rule states: if the centers of two balls, the mass of which is uniformly distributed over their volumes, are located at a distance $r$, then the force of attraction $F$ between the balls is directed along the line connecting the centers of the balls, is proportional to the product of their masses $m_1, m_2$ and inversely proportional to the square of the distance $r$:

$$F = \gamma \frac{m_1 m_2}{r^2};$$

γ in this formula is a constant value; it can only be determined experimentally.

Let’s return again to the attraction of the Moon. The Moon is located at a distance of 60 Earth radii, and the apple is only one Earth radius from the center of the Earth. Therefore, the force, and therefore the acceleration, in lunar orbit is $60^2$ times less than at the Earth’s surface:

$$a = \gamma \frac{M_3}{(60R)^2},$$

$$g = \gamma \frac{M_3}{R^2}.$$

Let’s divide the first of these expressions by the second:

$$\frac{a}{g} = \frac{1}{3600},$$ or $a = 0,0027 \, m/cck^*)$.

The obtained value surprisingly exactly coincides with the previously calculated value of the centripetal acceleration of the Moon.

This is how Newton managed to test the law of universal gravitation. True, there was not such good agreement in his calculations: work in Wulsthorpe, Newton did not have the exact value of the earth’s radius. Only a few years later this value was changed with sufficient accuracy.

Despite the obtained coincidence of the magnitude of the accelerations, Newoton was not satisfied with the verification. It was still necessary to solve the problem of the attraction of an apple lying on the Earth. Is it possible to consider that the globe attracts the apple as if all the mass of the Earth were concentrated in the center?

*) The gravitational constant γ cancels out - you don’t need to know it at all to be convinced of the correctness of the law of gravitation discovered by Newoton.

The parts close to the apple are more pleasing to it than those far away (Fig. 2). The resultant force can only be calculated using integral calculus, but at that time it had not yet been discovered. The study of the Moon’s motion was delayed for several years. Newton took up optics: he polished lenses, built a wonderful reflective telescope, and enthusiastically studied optical spectra… But thoughts about thiogeny did not leave him. It took years to create the integral calculus, and on its basis Newton proved a remarkable theorem: a shell of a spherical shape with a mass uniformly distributed over it fits the body, the shell located in its mass was in the center of the sphere (Fig. 3, a). The earth can be represented as a set of spherical shells (Fig. 3, b). The action of each shell does not depend on the others, and the force of tyogeny is not absorbed by the substance of the Earth. If we accept these two positions, of which the second is one of the most amazing properties of gravitation, then naturally we conclude that the apple is accepted by the Earth exactly as Newton assumed.

So, the fall of an apple to the Earth and the movement of the Moon in its orbit are caused by one reason - the Earth’s gravity. What forces control the movement of the planets? The planets move around the Sun, which means that the force of teogeny must be distorted from the Sun. Further generalization - any two bodies agree according to the same law.

At the beginning of the 17th century, Johannes Kepler discovered the laws of planetary motion. Keper’s first law stated that the planets move in ellipses shaped like the Sun. The meaning of Keper’s second law is that planets move faster in their orbits when they approach the Sun, and when moving away, they begin to move more slowly: the line connecting the planet and the Sun “sweeps out” equal areas in equal times.

Fig. 2.

The squares of the formation times of the planets are proportional to the cubes of their average distances from the Sun - this is the content of the third law.

Kepler would derive these laws based on many years of records of astronomical observations. Newton, relying on his laws of motion and the law of thoogeny, was able to theoretically obtain Keper’s laws.

Once upon a time, members of the English Royal Society met in a London coffee shop - the famous architect Christopher Wren, Robert

Fig. 3. F=γ mΣmₙ/r².

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Hooke and Edmond Halley. Halley said that he was able to obtain Kepler’s third law by combining the law of gravity and Newton’s laws of motion, but only for the case of circular orbits. Movement in a circular orbit can occur if an external force acts on the planet, directed towards the center of the circle and equal to $Mv^2/R$. This force is created by the attraction between the planet and the Sun. Pozny $\gamma \frac{M_C M}{R^2} = \frac{Mv^2}{R}$. If we remember that the speed of the planet’s motion in orbit is $v = \frac{2\pi R}{T}$, then, substituting this expression into the previous formula, we obtain Kepler’s law:

$$\gamma \frac{M_C}{R} = \frac{4\pi^2 R^2}{T^2},$$

or $\frac{R^2}{T^2} = \text{const.}$

When Halley finished his story, Wren offered a prize to anyone who could prove that the law of gravity decreasing in inverse proportion to the square of the distance is also true for motion in elliptical orbits.

Halley turned to Newton. Newton replied that he had already done this, and in November 1684 the manuscript with the calculations was in the hands of Halley. It took several years to prepare the manuscript and print it. Finally, in September 1687, Newton’s book was published. It was called “Mathematical Principles of Natural Philosophy.” The laws applied to the study of the planets of the solar system have provided important insights. Thus, Newton “weighed” the Sun, expressing its mass in units of earth masses:

$$\gamma \frac{M_C M_3}{R^2} = M_3 \frac{v^2}{R_3} = M_3 \frac{4\pi^2 R^2}{R_3 T_3^2},$$

$$M_C = \frac{4\pi^2 R_3^3}{\gamma T_3^2},$$

$$\gamma \frac{M_3 M_\pi}{R^2} = M_\pi \frac{v^2}{R_\pi} = M_\pi \frac{4\pi^2 R_\pi^2}{R_\pi T_\pi^2},$$

$$M_3 = \frac{4\pi^2 R_\pi^3}{\gamma T_\pi^2},$$

where $R_3$ and $R_\pi$ are the radii of the Earth and Moon orbits, respectively.

The result was very interesting. The Moon is a satellite of the Earth, and the Earth is a satellite of the Sun. The mass of the satellite decreases each time. In the first equation, when the law of gravity is applied to the Sun and Earth, the mass of the Earth is reduced, and in the second, the mass of the Moon is reduced. Thus, Newton expressed the masses of the Earth and the Sun in terms of quantities easily accessible by astronomical changes - orbital radii and periods of revolution. The value $\gamma$ remained unknown to him. But if we take the ratio of the solar and terrestrial masses, then the constant $\gamma$ is reduced and the formula is obtained

$$\frac{M_C}{M_3} = \frac{R_3^3 T_2^2}{R_3^3 T_3^2} = \left[ \frac{\text{расстояние Земля — Солнце}}{\text{расстояние Луна — Земля}} \right]^3 \times \left[ \frac{1 \text{ месцц}}{1 \text{ год}} \right]^2.$$

Thus, Newton was able to express the ratio of the masses of the Sun and the Earth in terms of quantities known to astronomers.

To calculate the mass of the Sun in more familiar units, say kilograms, it was necessary to estimate the mass of the Earth. Newton knew only its volume. The average density of the Earth’s substance in his time was not known. The only thing that was certain was that the density of the continents was greater than the density of water. But how many times does the average density of the Earth exceed the density of ocean water? Newton concluded that the average density of the Earth lies between 5 $e/cm^3$ and 6 $e/cm^3$. Only ninety years later, Sir Henry Cavendish “weighed” the Earth, changing the values ​​$\gamma$ using elegant round scales. The Earth’s density turned out to be 5.5 $e/cm^3$.

How to determine the mass of the Moon? After all, it is abbreviated in all formulas. Nevertheless, Newton found a way to estimate this value.

For thousands of years, the cause of tides in the seas and oceans remained a mystery to people. The Romans tied

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the wave moves relative to the continents at a speed of 1800 km/h and is slightly affected by the movement of the Moon.

The attraction of the Sun is many times stronger than the lunar one, but the heterogeneity of the solar attraction is not great. Therefore, the solar tidal force $F_C = 2\gamma \frac{M_C}{r^3_{C3}} R$ is less than the lunar one. The mass of the Sun is 27 million times greater than the mass of the Moon, this Sun-Earth distance $r_{C3}$ is 389 times greater than $r$ - the Earth-Moon distance. So $F_C = 2\gamma \frac{27.10^6 M_\pi}{(389 r)^3} R \approx 0.45 F_\pi$, that is, “solar” tides are weaker than lunar tides. Of course, we do not distinguish between these tides. Twice a month the Moon and the Sun are in line with the Earth, the solar and lunar tidal forces add up - a large tide is observed. If the Earth, Sun and Moon are located so that the Earth-Sun direction leaves an angle of 90° with the Earth-Moon direction, then the tidal forces of the Sun and Moon weaken each other and a small tide is observed. On islands lost in the ocean, where tidal phenomena are not distorted by the influence of the shores, large tides rise by 1.30 m, and small tides by only 0.65 m. If you make calculations, you can estimate the ratio of the solar and lunar masses by the magnitude of the large and small tides.

Thus, the untrained “satellite” of the Moon, the ocean tide, helped Newoton calculate the mass of the Moon. This calculation was very inaccurate. The task was complicated by the friction of water masses against the ocean and other circumstances that were difficult to take into account. The exact mass value was found after the launch of the lunar satellites. The measured periods of revolution and characteristics of the trajectory of the Lunars make it possible to determine with great accuracy the mass calculated by Newoton only approximately. Based on the movement of the artificial satellites of the Moon, scientists have determined that the lunar mass is distributed unevenly throughout its volume - Kepler’s laws are satisfied for the approaching Moon satellites.

Before Newton’s actronomers believed that comets passed through the solar system only once. However, Newton showed that comets can move in closed elliptical orbits. Especially their orbits are very elongated. Therefore, comets travel great distances from the Sun and their convergence period is very long. Edmond Halley managed to calculate the time of return of one of the most significant comets, records of the appearance of which have been preserved since ancient times. Calculations confirmed that the comet is returning within the timeframe predicted by Halley. It can be observed for 76 years. In 1986, this comet will again be visible from Earth. Only one actor, Johann Halle, who lived for almost a hundred years, observed Halley’s comet twice. The return of comets is another confirmation of the law of universal gravitation.

Solutions to many remarkable problems in which the law of universal gravitation is applied were contained in the “Mathematical Principles”: this includes the calculation of the precession of the earth’s axis, and the proof that any stone thrown on the earth moves in an ellipse, one of the foci of which is the center of the earth, and, finally, a study of the influence of the mutual gravitation of the planets on their motion. The influence of the planets is small compared to solar gravity, but a planet like Jupiter changes the nature of the orbit of neighboring Saturn, or, as actronomers say, disturbs its movement.

For the third century now, Newton’s laws have been used in calculations of machine tools and artillery pieces, trajectories of Earth satellites and packages. Mathematical formulas written more than two centuries ago are contained in calculating machines of the latest design, irrefutably confirming the properties of these laws.

ARCHIMEDES
AND
QUADRATURE
PARABOLA

Even in ancient times, scientists were engaged in calculating areas. There was no general method yet. (This method gives us integral calculus.) Therefore, when calculating the area of ​​a figure, its specific properties were used. The article describes how Archimedes, using the property of a parabola, calculated the area of ​​a parabolic segment.

Who doesn’t know the name Archimedes? This great thinker left an indelible mark on human history. Researchers of his work speak with admiration of him as a great engineer, an astronomer without equal, a brilliant mathematician, a man who knows all the secrets of nature…

The mathematical works of Archimedes captivate the reader with clarity of thought, elegance, brought to modern technical calculations. The famous Greek historian Plutarch writes:

“In all geometry it is impossible to find more difficult and profound problems that would be solved as simply and clearly as those that Archimedes studied.”

One of the first mathematical works of Archimedes, “Quadrature of the Parabola,” is no exception.

This work is devoted to calculating the area of ​​a parabolic segment, that is, a figure bounded by a parabola and a line, and presents, in essence, a proof of the following theorem:

The area of ​​a parabolic segment is equal to four times the area of ​​a triangle that has a common base and height with the segment.

The work is naturally divided into two parts. In the first part, Archimedes shows how this theorem was discovered by him with the help of mechanics, namely with the help of the “law of the lever” he established, and in the second part he gives a geometric proof.

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Fig. 1.

We will introduce the reader to the contents of the second part of the work.

1. Let us formulate one remarkable property of the parabola *), on which the entire subsequent presentation is based.

If the line $AB$ touches the parabola at the point $P$ and the chords $A_1B_1, A_2B_2, A_3B_3$, … are parallel to this tangent, then the midpoints of the chords - points $C_1, C_2, C_3$, … - lie on the same straight line; this line is parallel to the axis of the parabola and passes through the point $P$ (Fig. 1).

Besides,

$$\frac{A_1C_1^2}{PC_1} = \frac{A_2C_2^2}{PC_2} = \frac{A_3C_3^2}{PC_3} = \dots$$ (1)

Try to prove these properties of a parabola yourself.

2. Let us now consider a parabolic segment bounded by another parabola $APB$ and a chord $AB$ (Fig. 2).

The point of the parabola $P$, the tangent at which is parallel to the chord $AB$, is called the vertex of the segment, and the sum of the chord $AB$ is the base of the segment. Middle of the base. According to property (1), the straight line $PC$ is parallel to the axis of the parabola.

*) Recall that a parabola is the locus of points on a plane that are equidistant from a certain point and do not pass through a line. In accordance with the definition of a parabola, a point on the parabola is a point where the distance from the vertex to the focus is equal to the distance from the focus to the directrix.

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Let us prove one remarkable property of this sequence, namely, we will prove that for any $n$ the following equalities hold:

$$s_1 + s_2 + \dots + s_{n-1} + s_n + \frac{4}{3} s_n = \frac{4}{3} s_1.$$

(6)

In fact, since $4(s_1 + s_2 + \dots + s_{n-1} + s_n) = 4s_1 + (4s_2 + 4s_3 + \dots + 4s_{n-1} + 4s_n)$, it follows from correspondences (5) that $4(s_1 + s_2 + \dots + s_{n-1}) + 4s_n = 4s_1 + (s_1 + s_2 + \dots + s_{n-1})$, that is, $3(s_1 + s_2 + \dots + s_{n-1}) + 4s_n = 4s_1$.

Equality (6) is obtained from here by dividing by 3.

6. Now we can prove that the area of ​​the parabolic segment is $S = \frac{4}{3} s_1$.

Archimedes proves this equality by the so-called “method of exhaustion.” This method, which belonged to one of the greatest Greek mathematicians, Eudoxus (IV century BC), replaced the method of passage to the limit during the time of Archimedes.

Let’s assume $S > \frac{4}{3} s_1$ or $S < \frac{4}{3} s_1$.

Let us prove that this is impossible.

Let us first assume that $S > \frac{4}{3} s_1$.

We have already noted that by continuing the process of inscription, it is possible to ensure that the sum of the areas of the remaining segments is as small as possible. This means that by selecting $n$ appropriately, the difference $S - (s_1 + s_2 + \dots + s_{n-1} + s_n)$ can be made less than any predetermined positive number. Let us select $n$ so that the equality $S - (s_1 + s_2 + \dots + s_{n-1} + s_n) < S - \frac{4}{3} s_1$ holds. Then $s_1 +$

$s_2 + \dots + s_{n-1} + s_n > \frac{4}{3} s_1$, and this, according to equality (6), is impossible *. Thus, inequality (7) is false.

Let us now assume that

$$S < \frac{4}{3} s_1.$$

(8)

Since the terms of sequence (4) tend to zero, $n$ can be chosen so that the following equality holds:

$$\frac{4}{3} s_n < \frac{4}{3} s_1 - S.$$

From this inequality, taking into account equality (6), we obtain

$$S < s_1 + s_2 + \dots + s_{n-1} + s_n$$

which, of course, is impossible.

So, inequality (8) is also false. This proves that $S = \frac{4}{3} s_1$.

7. In conclusion, let us make the following remark.

As is known, the ancient Greeks understood by squaring a figure the construction of a square equal in size to this figure. The problem of quadrature in the “Greek sense” does not always have a solution. It has been proven, for example, that for a circle it is unsolvable. That is, it is impossible to construct (using a compass and ruler) a square equal in size to a given circle.

The result obtained by Archimedes is also remarkable in the sense that it shows the possibility of a positive solution to the problem of quadrature of a parabolic segment: knowing the base and height of a segment, it is always possible to construct a square equal in size to this segment.

*) We assumed that $S > \frac{4}{3} s_1$, therefore $S - \frac{4}{3} s_1$ is a positive number.

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Paradoxes of jet propulsion
M.S. Livshits

Rocket propulsion power

During the operation of jet propulsion, the chemical energy of the fuel is converted into kinetic energy with a certain efficiency: part of this chemical energy is spent on various thermal losses (heating the walls of the engine chamber, heat carried away by escaping gases, etc.). The ratio of the resulting kinetic energy to the chemical energy of the burned fuel is called the internal (or effective) efficiency $\eta_{\alpha\phi}$. Obviously they come in the order of $0,3 \div 0,5$.

While it was deliberately not specified - the kinetic energy of what?

Let’s consider the coordinate system associated with the rocket. In a very short time $\Delta t$, a mass of gas $\Delta m$ will be ejected from the rocket nozzle. If $\Delta t$ is small enough, then the speed of the rocket practically does not change and the coordinate system associated with it can be considered inertial. In this coordinate system, the rocket is motionless, which means that the chemical energy of the fuel is converted into the kinetic energy of gases.

Let $u$ be the relative velocity of gases flowing out of the rocket nozzle; then the kinetic energy of the gas mass $\Delta m$ escaping from the rocket nozzle is

$$E_{\Delta m} = \frac{\Delta mu^2}{2}.$$

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Since power is equal to the work done per unit time, we will obtain an expression for power if, instead of the value of the ejected mass of gases, we substitute the second mass consumption μ into our expression.

Therefore, engine power calculations

$$P_{\text{дв}} = \frac{\mu u^2}{2}.$$

The same expression could be obtained in another way, considering the movement of the rocket and gases in a moving coordinate system. In this case, the kinetic energy $E$ obtained during the operation of the rocket engine is spent on changing the kinetic energy of the emitted gases $\Delta E_{\text{газа}}$ and changing the kinetic energy of the rocket:

$$\Delta E = \Delta E_{\text{газа}} + \Delta E_{\text{ракеты}}.$$

Let us find separately each of the terms of this sum.

1. Let the speed of the rocket before the release of fuel $\Delta m$ be equal to $v$. Then

$$\Delta E_{\text{газа}} = \frac{\Delta m (u - v)^2}{2} - \frac{\Delta m v^2}{2},$$

where $\frac{\Delta m v^2}{2}$ is the kinetic energy possessed by the discarded mass of gases before its expiration (when it is in the rocket and moved with it at a speed of $v$), $\frac{\Delta m (u - v)^2}{2}$ is the kinetic energy possessed by the discarded mass of gases after its expiration from the rocket nozzle with a relative speed $u$ (since in a fixed coordinate system its speed will be $u - v$).

2. Now let’s deal with the second term - $\Delta E_{\text{ракеты}}$. Since during time $\Delta t$ a mass of gas $\Delta m$ is ejected from the nozzle, carrying away the momentum $\Delta mu$, then this mass of gas, in accordance with Newton’s II law, is acted upon by a force

$$F = \frac{\Delta mu}{\Delta t} = \mu u.$$

According to Newton’s III law, forces of the same magnitude, but directed in the opposite direction, act on the rocket - this is the traction force of the engine. If during the time $\Delta t$ the rocket traveled a distance $\Delta S$, then the thrust force of the engine made a perfect movement outside with the coordinate associated with the rocket.

So, the power of the rocket engine $P_{\text{дв}} = \frac{\mu u^2}{2}$ even during stationary operation of the engine remains constant. Let’s remember this.

Net mechanical power of the engine (power developed by traction forces)

The main task of the engine is to increase the kinetic energy of the rocket. We have seen that the thrust force is equal to $F_{\tau} = \mu u$, and the change in the kinetic energy of the rocket during time $\Delta t$ is equal to $\Delta E_{\text{ракеты}} = \mu u \cdot \Delta S$. Therefore, the useful thrust power of the engine is equal to

$$W_{\pi} = \frac{\Delta E_{\text{ракеты}}}{\Delta t} = \mu u \frac{\Delta S}{\Delta t} = \mu u v.$$

During stationary operation of the engine, that is, with a constant second mass consumption and a constant gas flow rate, the useful power of the engine increases in proportion to the speed of the rocket.

At first glance, this conclusion seems paradoxical. In fact, since the formula that we obtained is valid at any rocket speed, it will be correct

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vedlin and at such a rocket speed at which the values ​​​​are obtained, the useful power will become greater than the power developed by the rocket engine due to fuel energy!

From a comparison of the expressions for engine power and useful power, it is clear that at $v = \frac{u}{2}$ the useful power becomes equal to the power of the rocket engine. Consequently, with a further increase in the rocket speed, it will be greater than the engine power. This is the quantitative limit where this paradox appears.

In order to understand this paradox, we will have to consider what the external (flight) efficiency is. However, at the same time, we will find out why it is advisable to characterize a rocket engine not by power, as is customary for other engines, but by the thrust force it develops.

The external efficiency cannot fully characterize the efficiency of a rocket engine, since in the general case, as we saw above, the fuel energy is distributed between the rocket and the emitted gases. Therefore, in theory, the so-called external (or flight) efficiency is also introduced, and in the literature one can find two different definitions and corresponding expressions for external efficiency. Each of these definitions of efficiency From different points, energy characterizes the mechanics of rocket flight. Let us consider these definitions, pointing at the same time to their logical basis.

First definition of external efficiency

A flight with a running engine is a flight of a body of variable mass. Therefore, in addition to the power of the rocket engine, the energy

compliance must also take into account the exchange of energy between the rocket and the gases leaving it, that is, all the power that we have. A change in the energy of the rocket can occur both due to the chemical energy of the fuel (the work of the traction force) and due to the energy that the ejected mass of gases possessed before it left the rocket. It can also be transferred to a rocket.

So, external (flight) efficiency can be defined as the ratio of the work done by the thrust force to the sum of the mechanical energy resulting from the combustion of fuel and the energy possessed by the mass of gases before it left the rocket. This relation can be replaced by the relation of the corresponding powers (the result will not change).

$$\eta' = \frac{\mu u v}{\mu^2 / 2 + \mu v^2 / 2} = 2 \frac{v}{u} / 1 + \frac{v^2}{u^2}.$$

Denoting the relation $\frac{v}{u} = R$, we obtain the final

$$\eta' = \frac{2R}{1 + R^2}.$$

Let us now examine the formula. The value of $\eta'$ is maximum when $\frac{1}{\eta'}$ is minimum, that is, when the expression $\frac{1 + R^2}{R} = \frac{1}{R} + R$ is minimum. But $\frac{1}{R} + R \geq 2 \sqrt{\frac{1}{R} \cdot R}$ (the argument mean of two numbers $a$ and $b$ is greater than the geometric mean or equal to it, if $a = b$). This means that $\frac{1}{R} + R$ is minimal for $\frac{1}{R} = R$ (or, what is the same, for $\frac{1}{R} + R = 2$). This means that the maximum $\eta'$ is achieved at $R = 1$ and at the same time $\eta' = 1$.

Physically, this is completely clear: at $v = u$ the expelled mass of gas has a kinetic energy equal to zero, that is, it completely gives up

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The kinetic energy she had was equal to the package.

The graph of $\eta'$ looks like that shown in Fig. 1.

Reduced flight efficiency at $v > u$ is explained by the fact that the energy carried away by the gas stream appears again and begins to increase.

The second definition of external efficiency is

In the second determination of external efficiency. it is approached in the following way.

The only source of increment in the kinetic energy of the “rocket + discarded gases” system is the operation of the package engine, that is, the energy obtained due to the transition of the chemical energy of the fuel into the mechanical energy of the fuel and packages. And by useful work, you need to remember the work that goes into increasing the kinetic energy of the package itself.

At the same time, the external efficiency is defined as the ratio of useful work to the total mechanical energy obtained in a package engine due to the chemical energy of the fuel.

Let us determine the magnitude of useful work in the understanding accepted here.

The traction force produces over the packet work, which increases the kinetic energy of the packets. But, on the other hand, the mass of gas shown to the package reduces the kinetic energy of the package by the amount that this mass of gas possessed (having, so to speak, its contribution to the total energy of the package) before its expiration. Thus, the net power is $\mu uv - \frac{\mu v^2}{2}$.

Therefore, by the second definition

$$\eta'' = \frac{\mu uv - \frac{\mu v^2}{2}}{\mu u^2} = 2R - R^2$$

Let’s draw a graph of the dependence of $\eta'' = 2R - R^2 = R (2 - R)$ on $R$. It is a parabola intersecting the $R$ axis at $R = 0$ and $R = 2$, with a maximum at $R = 1$ (Fig. 2). That is, the maximum $\eta''$ is achieved at the same time as the maximum $\eta'$ - at $R = 1$ (in this case, all the mechanical energy of the package engine is consumed to increase the kinetic energy of the packet moving forward, or the ejected mass of gas has zero speed and does not carry away any energy). The graph shows that at $R > 2 (v > 2u) \eta'' < 0$.

Here, when considering the second definition of efficiency. packet engine, and reveals another important feature of the physics of the packet engine, considered from the point of view of changes in kinetic energy

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moving packets after reaching the value of the flight speed packets $v=2u$, becomes negative! Another paradox…

What is the solution to these paradoxes?

1. The first “paradox”, that is, the fact that the power developed by the traction force continuously increases with increasing packet speed and at $v > \frac{u}{2}$ becomes greater than the power of packet motion, is explained, as can be seen from the first definition of external efficiency, by the presence of an exchange part, the power of which we will consider. It also increases with increasing packet speed. Thus, with an increase in the speed of the packets, the gases leaving the packet become, so to speak, more and more “rich” and will remove to the packet more and more values ​​(absolute, not relative) of the energy $\frac{\mu v^2}{2}$ that they had before leaving the packet. The packages of gases escaping from the nozzle will give back to the package that part of the energy that was stored earlier. This leads to the fact that the power developed by the traction force continuously grows and at $v > \frac{u}{2}$ becomes greater than the value obtained due to the chemical energy of the fuel.

2. The second “paradox” is also explained quite simply. When the speed of the packet becomes greater than the double speed of gas outflow ($v > 2u$), the energy carried away by the gas increases so much that the flowing stream of gas carries away not only all the energy created at a given moment by the packet motion, but also part of the energy that the “rocket-gas” system possessed up to this moment (accumulated over the previous “history” of the flight). Or, if we approach the appearance of the physical meaning of this “paradox” a little differently: the decrease in the kinetic energy of the package due to a decrease in its mass (due to “losing weight”) is greater than the decrease in kinetic energy due to the decrease in the speed of the package. So, it is advisable to characterize a packet propulsion engine by its traction force. This is sharply different from the usual situation, say, for auto-

16

a mobile or aircraft engine used to provide movement in a resistive environment.

In the latter case, it is the power of the engine that is its most important characteristic, since the more powerful the engine, the higher the speed of movement of a given body it can maintain, predetermining compliance with the increasing resistance of the medium, that is, compensating for energy losses to overcome the resistance of the medium.

During the flight of a rocket with a running engine, the relative kinetic energy (kinetic energy per unit mass) of the remaining part of the rocket continuously increases along with the speed (the effectiveness of this process is characterized by flight efficiency in the first definition), although the kinetic energy of the rocket with the new (remaining) fuel supply at speed $v > 2u$ is less than the kinetic energy of the rocket with fuel before the ejected mass of gas leaves the rocket.

FORUM OF ASTRONOMERS AND GEODESISTS

Six November days (XI. 23-28, 1970) in the assembly hall of Kazan State University. V. I. Ulyanov (Lenin) held the next V Congress of the All-Union Astronomical and Geodetic Society (VAGO) at the USSR Academy of Sciences.

VAGO is a voluntary scientific and public organization engaged in work in the field of astronomy, geodesy and cartography. It was created in 1932. The Society includes 55 branches, uniting 5,640 detective members, 197 collective members (enterprises, institutions, etc.) and 1,040 members of the youth section.

More than 600 people from all over the country took part in the work of the V VAGO Congress. At the congress, organizational issues were considered (report of the Central Council, elections of the governing bodies of the Society and decision-making). At the plenary sessions and sections, scientific reports and messages from the country’s most prominent scientists and amateur astronomers were heard.

At the youth section of the congress, messages from S. S. Voikov, V. V. Martynenko, B. G. Pshenicher, V. V. Chistyakov, S. S. Novikov and others spoke in detail about the work carried out by the members of this section. According to the VAGO Charter, members of the youth section have the right:

“a) participate with a full voice in congresses, conferences, SSSYAH, meetings and meetings of the Society;

b) use the services, laboratories, library, archives and collections of the Society;

b) wear the badge of a member of the Society,”*)

During the period between the IV and V WAGO congresses, members of the youth section, who worked at the 31st branch of the Society, took an active part in amateur scientific work on the study of variable stars, observations of the Sun, service clouds, meteors, artificial Earth satellites, planets, in atmospheric sounding, maintenance of crater-like structures, in the study of crater-like structures - a set of atmospheric phenomena that limit the effectiveness of large telexcopes. In the youth sections of the Moscow, Novosibirsk and Crimean branches of the Society and in the Uglich branch of the Yaroslavl branch, electron meters, photographic cameras, and electron meters were designed and manufactured. In the Crimean and Azerbaijan sections, young astronomers are very successful in amateur telex construction and produce mirrors for amateur astronomers in other cities.

Much attention is paid to organizing various astronomical excursions for schoolchildren. Meteorological expeditions organized in Crimea are regularly held

*) Admission to membership in youth sections will be carried out by branches of the society or their filpals on the basis of a written application from an incoming member of VAGO or an educational institution, enterprise or Komsomol organization.

(For ending see page 21)

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Fig. 1.

In cell 0 we will “put” everyone who has not made a single mistake, in cell 1 - those who have one mistake, in cell 2 - two, … and so on until cell 13, where only Sasha Ivanov ended up.

We apply the Dirichlet principle (note, this is a very important point).

Let us prove the statement of the problem by contradiction. Let’s assume that no three students made the same number of mistakes, that is, less than three students got into each of the cells 0, 1, …, 12. Then in each of them there are two people or less, and in total there are no more than 26 people in these 13 cells. Having added Sasha Ivanov, we still won’t get 30 guys. Contradiction.

Can we say that exactly three people made an equal number of mistakes? Of course not. Perhaps all the guys, except Sasha, wrote the dictation without a single mistake, that is, everyone made 0 mistakes. Can we assume that at least four are in the same “cage”? No, you can’t. A class in which 3 people each made 0, 1, 2 errors, 2 people each made 3, 4, …, 12 errors, and one person made 13 errors satisfies the conditions of the problem.

2. Let there be 41 people in the class, not 30. Prove that there are four who made the same number of mistakes. (The remaining conditions are as in task 1.)

3. There are about 7.1 million residents of Moscow, each with no more than 100,000 hairs on their heads. Prove that in Moscow there are at least 70 people with the same number of hairs on their heads.

Acquaintance

Let us assume that acquaintance is a “symmetrical” relationship between people: if Komarov is familiar with Zhukov, then Zhukov is familiar with Komarov.

4. Let’s choose 5 people in any way. Prove that at least of them have the same number of acquaintances among those selected.

20

Fig. 3.

14. Several arcs of a circle are painted black. The sum of the lengths of the surrounded arcs is less than half the length of the circle. Prove that there exists a diameter whose both ends are not surrounded.

Let’s paint in blue the arcs that are symmetrical black relative to the center of the circle. Since the sum of the lengths of the blue arcs is equal to the sum of the lengths of the black ones, the total length of the surrounded arcs is less than the length of the circle. This means (the Dirichlet principle!), there is an uncolored point. The diameter passing through it will be the desired one.

“Consider the segment AB. Let there be several black segments with a total length of 1.5AB on it. then some point AB belongs to at least two black segments.” This reasoning helps in the next problem.

15. In a square ABCD on a side of 1 cm there are several circles whose sum of radii is 0.6 cm. (The circles can intersect or coincide.) Prove that there is a line parallel to AB that has common points with at least two circles.

Sooner or later it will happen again

16. Given a series
    1,1,2,3,5,8,13,21,34,55,89,…,
    in which each number, starting from the third, is equal to the sum of the previous two. Is there a number among the 100,000,001 first terms of this series that turns out to be zeros?

Let’s consider a series of remainders from dividing these numbers by 10,000. The remainder in the nth place is denoted by the symbol $a_n$. $a_1=1$, $a_2=1$, $a_6=8$, $a_{10}=55$ and so on. There are 10,000 different residues, $10 000^2=100 000 000$ different pairs of residues. Let’s consider the pairs $a_1a_2; a_2a_3; a_3a_4;...; a_{100 000 001}a_{100 000 002}$. According to the Dirichlet principle, of these 100,000,001 pairs of residues, at least two coincide: $a_k=a_m$, $a_{k+1}=a_{m+1}$, where k and m are less than 100,000,002, k is less than m. Based on the remainder of the sum and one of the terms, the remainder of the second term is uniquely determined (check), therefore $a_{k-1}=a_{m-1}$, $a_{k-2}=a_{m-2}$ and so on up to $a_2=1=a_{m-k+2}$, $a_1=1=a_{m-k+1}$. It is now clear (why?) that $a_{m-k}=0$.

17. Prove that for any natural number n in the series from the condition

21

of the previous problem there is a number ending in $n$ zeros.

18. Prove that there is a power of 29 ending in 00001.

19. Let the number $A$ not be divisible by 2 and 5. Prove that for any $n$ there is a power of the number $A$, which when divided by $10^n$ gives a remainder of 1.

20. Prove that there is a multiple of 1970 whose decimal notation contains only ones and zeros.

Table

The table contains $n \times n$ cells containing integers from 1 to $n^2$. You will be interested in the following statement:

“For any arrangement, there are two cells that have a common side such that the difference between the numbers in these cells is greater than 5.”

21. Prove this statement
    a) for $n=10$;
    b) for all $n$, greater than 10;
    c) for $n=9$.

22. Roll it over for $n=5$.

The author does not know whether the statement under consideration is true for $n=6, 7, 8$.

Exercise

23. There are 123 residents in the house; together they are 3813 years old. Is it possible to choose 100 of them that together are at least 3100 years old?

24. There are 7 lines on the data plane, no two of which are parallel. Prove that there are two of them such that the angle between them is less than $26^\circ$. Is the analgetic statement true if $26^\circ$ is replaced by $25^\circ$?

25. Prove that any tree can be cut off $8/15$ of its leaves, leaving at least $7/15$ of the shadow provided by the tree (neglect the shadow from the trunk and branches; the number of leaves can be considered divisible by 15).

26. Twenty-one boys were given 200 nuts. Prove that, no matter how they divide them, there will be two boys who get an equal number of nuts (maybe 0 nuts each).

27. 7 segments are given. The length of each is more than 10 cm and less than 1 m. Prove that any three can form a triangle.

28. From the series 1, 2, …, 200, the 101st number was somehow selected. Prove that one of the chosen numbers is divisible by another.

(End. See page 16 for beginning.)

for young astronomers coming from different places of the USSR. Many members of the junior section were active participants in observing the 1968 solar eclipse.

In recent years, the network of youth astronomical societies and various clubs and groups, uniting about 3,000 schoolchildren, has noticeably expanded. Assistance to these organizations is provided from five regional methodological centers located in the astronomy departments of the Moscow and Baku Pioneer Palaces, in the general school of the Alma-Ata Pioneer Palace, the Crimean region youth general school (Simferopol) and the children’s general school of the Young Technicians Club of the Siberian Branch of the USSR Academy of Sciences, which direct the work of young astronomers to the development of observations using modern methods and technical funds for the design and manufacture of various astronomical instruments and the organization of public schools and astronomical laboratories.

Great attention to the work of young astronomers from the Ministry of Education of the USSR and the RSFSR, the Central Committee of the Komsomol, the Central Station of Young Technicians, VDNKh and the Board of the All-Union Society “Znanie” should be able to develop the work of youth sections of VAGO. All this allowed the First All-Union Conference of Young Astronomy Amateurs to be held in Baku in 1969, which contributed to the exchange and dissemination of the best work of the youth sections of VAGO.

Vice-President of VAGO Professor L. S. Khrenov

22

TASKS

M91. Two people play tic-tac-toe on an endless sheet of checkered paper. The beginner puts a cross in any cell. Each next move, he must place a cross in any free cell; any cell adjacent to one is considered to be any cell that has a common side or a common vertex with it. The second player, with each turn, can place three polykkas at once in any three free cells (not necessarily next to each other). Prove that, no matter how the first one plays, the second one can “lock him up”: make sure that the first one has nowhere else to put a cross.

Explore similar games in which the second player is allowed to place not three, but only two or only one polyk in one turn. What will be the result here if the partners play correctly: did the players succeed in

Fig. 1. Here is one of the positions that can arise after the third move (crosses and crosses placed on the first move are highlighted in black, red - on the second, blue - on the third).

crosses (and what is the maximum number of moves that the crosses can last) or can the game continue indefinitely?

Try to study other variants of this game: when only cells that have a common side with it are considered adjacent to a given one; when the plane is divided not into squares, but into regular hexagons; when the first one is allowed to put ru crosses at once, and the second one - q polikk.

A. P. Savin

M92. Petya plans to spend the entire 90 days of his vacation in the village and at the same time strictly adheres to the following schedule: every second day (that is, every other day) he goes swimming in the lake, every third day he goes to the store for groceries, and every fifth day he solves problems in mathematics. (On the first day, Petya did both, and the third, and was very tired.) How many “pleasant” days will Petya have, when he needs to swim, but does not need to go to the store and solve problems? How many “boring” ones are there when there is nothing to do at all?

M93. Each of the numbers $x_1, \dots, x_n$ is equal to the plane or minus one. It is known that

$$x_1x_2+x_2x_3+\dots+x_{n-1}x_n+x_nx_1=0.$$

Prove that $n$ is divisible by even.

A. M. Leontovi

M94. Prove that there is no polyhedron with at least four edges attached to each vertex and each face.

L. G. Limanov

24

SOLUTIONS

In this issue we publish solutions to problems M39—M49.

M39

Prove that non-negative integers $x, y$ satisfy the equation $x^2 - mxy + y^2 = 1$ (where $m$ is the given integer greater than one) if and only if $x$ and $y$ are adjacent terms of the sequence $\varphi_0 = 0, \varphi_1 = 1, \varphi_2 = m, \varphi_3 = m^2 - 1, \varphi_4 = m^3 - 2m, \varphi_5 = m^4 - 3m^2 + 1$, …, in which $\varphi_{k+1} = m\varphi_k - \varphi_{k-1}$ for all $k \ge 1$.

For example, all solutions to the equation

$$x^2 - 3xy + y^2 = 1$$

(1)

in integers $(x, y)$, $0 \leq x < y$, are pairs

$$(0, 1), (1, 3), (3, 8), (8, 21), \dots$$

(2)

neighboring terms of the sequence $0, 1, 3, 8, 21, \dots$ defined by the conditions $\varphi_0 = 0, \varphi_1 = 1, \varphi_{k+1} = 3\varphi_k - \varphi_{k-1}$ for all $k \ge 1$.

Let us solve this problem for $m = 3$ (for other $m > 1$ the solution is similar.) We must prove that, firstly, all pairs of numbers (2) satisfy equation (1) and, secondly, no other pairs of non-negative numbers satisfy this equation. (Note that this second and more difficult part of the solution was largely overlooked by the readers who sent letters.)

In sequence (2), each pair $(x, y)$ is followed by a pair $(x_1, y_1)$, which can be obtained from $(x, y)$ using the following transformation:

$$x_1 = y, y_1 = 3y - x.$$

(3)

We denote this transformation $(x, y) \rightarrow (y, 3y - x)$ by the letter $T$. Not difficult

25

check that if the pair $(x, y)$ satisfies equation (1), then the pair $(x_1, y_1) = T(x, y)$ also satisfies it:

$$x_1^2 - 3x_1y_1 + y_1^2 = y^2 - 3y(3y - x) + (3y - x)^2 = y^2 - 3xy + x^2.$$

(4)

From here it is immediately clear that all pairs (2) satisfy equation (1): after all, the pair (0, 1) satisfies it, and all the others are obtained from it using a finite transformation number $T$.

Let us now prove that there are no other solutions $(x, y)$ in integers subject to the conditions $0 \leq x < y$ of equation (1). Let $(x_1, y_1)$ be one of such solutions. Let us find a pair $(x, y)$ from which these $x_1$ and $y_1$ are obtained using formulas (3) - for this it is enough to express $x$ and $y$ in terms of $x_1$ and $y_1$:

$$x = 3x_1 - y_1, y = x_1.$$

(5)

We will denote such a transition from $(x_1, y_1)$ to $(x, y)$—the inverse transformation *) to $T$—by $T^{-1}$. From equalities (4) it follows that if the pair $(x_1, y_1)$ satisfies equation (1), then the pair $(x, y) = T(x_1, y_1)$ also satisfies it. Let us show that if, in addition, $0 < x_1 < y_1$, then

$$0 \leq x < y.$$

(6)

Let us represent the equality $x_1^2 - 3x_1y_1 + y_1^2 = 1$ of the following in two ways:

$$x_1^2 + y_1(y_1 - 3x_1) = 1,$$

$$x_1(x_1 - y_1) + y_1(y_1 - 2x_1) = 1.$$

(7)

From equality (7) it immediately follows ($x_1$ and $y_1$ are positive!) that $-x = y_1 - 3x_1 \leq 0$, and if $y_1 - 3x_1 = 0$, then $x_1 = 1$ and $y_1 = 3$. From equality (8), in turn, it follows that $y - x = y_1 - 2x_1 > 0$ (after all, $x_1 - y_1$ is negative). Inequalities (6) have been proven. Note that they can be rewritten like this:

$$0 \leq x < x_1,$$

(9)

since $x_1 = y$. So, any solution $(x_1, y_1)$, $0 < x_1 < y_1$, after the transformation $T^{-1}$ goes into the solution $(x, y)$, for which $0 \leq x < y$. If at the same time $x > 0$, then we can apply $T^{-1}$ one or more times, obtaining new (smaller in magnitude) solutions. Indeed, as (9) shows, the magnitude of the integer $x$ decreases with each transformation. Therefore, after a certain finite number of steps we must stop, and we can stop only when we obtain the solution $(x, y)$, which has $x = 0$ and, therefore, $y = 1$. Thus, we have proven that from each solution $(x_1, y_1)$ for a finite number transformed $T^{-1}$ a solution (0, 1) is obtained. It follows from this that each solution from (0, 1) is transformed by a finite number $T$ (that is, that there are no other solutions different from (2). The problem is completely solved.

This solution can be presented more clearly as follows. We will represent the pair of numbers $(x, y)$ as usual, as a point on the plane. Then the equation $x^2 - 3xy + y^2 = 1$ defines a curve on the plane called a “sperbola”. (On the cover there is a whole family of hyperbolas given by the equations $x^2 - 3xy + y^2 = c$ with different values ​​of the parameter $c$, and the one we are interested in is yellow.) Solving this equation in integers means indicating all the points with integer coordinates that lie on the hyperbola. However, such a point is easy to indicate: it is a black point with coordinates (0, 1). All the rest (within the angle $0 \leq x < y$), as we have proven, are obtained from it by successive application of the transformation $T : (x, y) \rightarrow (y, 3y - x)$. This transformation of the $T$ plane has the following properties:

1. Transforms every straight line in the plane back into a straight line (such transformations are called linear).

*) The concept of an inverse transformation was discussed in the article by A. N. Kolmogorov “What is a function?” («Kvant» No. 1,1970).

28

Fig. 2. Among the $m+k+1$ elements, first select one (such that there are at least $k$ elements to the left of it and to the right of it), draw a line along it, and then select $k$ elements to the left of the line and to the right of the line. On ours figure $m=9, k=3, m+k+1=13$.

la certain $k+m$ elements, and then another $k$ elements from the remaining $n-m$ elements. What has been said can be found like this. Let’s place the data of $n+k$ elements “in a line” and draw a vertical line separating $k+m$ elements on the left from $n-m$ elements on the right. Let’s somehow select $k$ elements to the left of the bar and $k$ elements to the right of the bar, and then form a connection of each of these “right” selections. There will obviously be $C_{k+m}^k \cdot C_{n-m}^k$ such connections.

Now it is not difficult to prove that

$$C_k^k \cdot C_n^k + C_{k+1}^k \cdot C_{n-1}^k + C_{k+2}^k \cdot C_{n-2}^k + \dots + C_n^k \cdot C_k^k = C_{n+k+1}^k \cdot C_{n-k+1}^k.$$

To do this, it is enough to establish a one-to-one correspondence between samples of $n+k+1$ elements of $2k+1$ and samples of $n+k$ elements of the following form: “$k$ elements to the left of the line; trait; $k$ elements to the right of the line,” the total number of which is equal to the sum on the left side of equality (2). So, the required amount is

$$(k!)^2 \cdot C_{n+k+1}^2 = \frac{(n+k+1)!(k!)^2}{(2k+1)!(n-k)!}.$$

In particular, for the first task M40 we have $k=1$ and therefore the sum will be $C_{n+2}^3 = \frac{(n+2)(n+1)n}{6}$.

If the above reasoning seems difficult to you, and even more so if it seems interesting to you, we recommend that you look into N. Ya. Vilenkin’s book “Combinatorics” (M., “Nauka”, 1969). There, in particular, on page 57, formula (24), similar to formula (2), is proven.

V. N. Berezin

M41

Given a circle, its diameter $AB$ and a point $C$ on this diameter. Construct two points $X$ and $Y$ on the circle, symmetrical with respect to the diameter $AB$, for which the line $YC$ is perpendicular to the line $XA$.

Let $K$ be the intersection point of the segments $XY$ and $AB$; by condition $XY \perp AB$ and $XK = KY$. For any point $X$ on circle $\Rightarrow AXB = 90^\circ$. Therefore, requirement $YC \perp AX$ is equivalent to each of the following conditions:

$$YC \parallel BX, \Rightarrow YCB = \Rightarrow CBX,$$

$$\triangle CYK = \triangle KBX, \quad CK = KB.$$

Thus, if we use the perpendicular to the line $AB$ in the middle of the segment $CB$, we will obtain two required points at its intersection with the circle (any of them can be taken as $X$, then the other will play the role of $Y$).

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M42

The digits of some seventeen-digit numbers are written in reverse order. The resulting number is added to the original. Prove that at least one of the digits of their sum is even.

Let us prove indica by $n$ that this statement is true for any $(4n+1)$-digit number. At $n=0$ it is obvious. Suppose that for some $n$ it is false, that is, there is a number $a_1 a_2 a_3 \dots a_{4n-1} a_{4n} a_{4n+1}$ such that in total

$$\frac{a_1}{a_{4n+1}} + \frac{a_2}{a_{4n}} + \dots + \frac{a_{4n-1}}{a_{4n}} + \frac{a_{4n+1}}{a_{4n}}$$

all numbers are odd. Then the sum of the digits $a_1 + a_{4n+1}$ is odd and in the places indicated by arrows will not carry over the unit to the next digit. Therefore, in the sum of $(4n-3)$-digit numbers

$$\frac{a_3 \dots a_{4n-1}}{a_{4n-1} \dots a_3}$$

Also, all digits are odd, therefore, our statement is not true for $(n-1)$. The proof is complete (think about why the usual “indication step” - “if true for $n$, then it is true for $(n+1)$” - can be replaced by a step in the opposite direction: “if false for $n$, then false for $(n-1)$”!).

Note that the proven statement cannot be simplified: for numbers in which the number of digits is even or has the form $(4n+3)$, the statement of the problem may not be true:

$$11 \ 112 \ 222 + 22 \ 221 \ 111 = 33 \ 333 \ 333, \ 72 \ 727 \ 262 \ 626 + 62 \ 626 \ 272 \ 727 = 135 \ 353 \ 535 \ 353.$$

M43

Which side of a regular triangle is divided into $n$ equal parts. Straight lines parallel to the sides are drawn through the division points. As a result, the triangle was broken into $n^2$ small triangles. Known as a “chain” is a sequence of triangles in which none appears twice and each subsequent one has a common side with the previous one. What is the maximum possible number of triangles in the chain?

Answer: $n^2 - n + 1$. To prove that they are often found on odnomas, it is enough to take one unexpected step: color the triangles in a checkerboard pattern, as was done on figure. The rest is quite simple.

In the entire triangle, the red triangles are $n$ larger than the yellow ones (in each horizontal row, the red ones are one larger), and in the chain the color must be black, so there can only be one red ones larger than the yellow ones.

One of the chains of the maximum possible length is shown at figure. (From what was said above, it is clear that in order for a chain to have length $n^2 - n + 1$, it is necessary and sufficient that it begins and ends in red triangles and passes through all yellow triangles without exception.)

30

Prove that for every natural number $K$ there are an infinite number of natural numbers $T$ that do not contain zeros in decimal notation and such that $T$ and $KT$ have the same sum of digits.

Let there be $k$ digits in the number $K$ (in decimal notation). Then in the role of $T$ you can take any number 999…99, written $n$ by nines, where $n \geqslant k$; then both the sum of the digits $T$ and the sum of the digits $KT$ are equal to 9$n$ (in the number $KT = K(10^n-1)$, digits spaced $n$ digits apart add up to 9; for example, $743\ 552 \cdot 99\ 999\ 999 = 743\ 552 \cdot (10^8-1) =$
$= 74\ 355\ 200\ 000\ 000-743\ 552 = 74\ 355\ 199\ 256\ 448$;

We leave the almost obvious proof for the general case to the reader).

M45

Prove that from any 200 integers you can choose 100 whose sum is divisible by 100.

Try to generalize this problem: prove that from any $2n-1$ integers we can choose $n$ whose sum is divisible by $n$, where $n \geqslant 2$.

Lemma. If the statement of the problem is true for $n = a$ and for $n = b$, then it is also true for $n = ab$.

Note that the property “the sum $n$ of integers $x_1, \ldots, x_n$ is divisible by $n$” can also be formulated as follows: “the arithmetic mean $\frac{x_1 + \ldots + x_n}{n}$ is an integer.”

So, let arbitrary $2ab-1$ integers be given. Since the statement is true for $n = b$ and $2ab-1>2b-1$, from the given $2ab-1$ numbers we can choose $b$ whose sum is divisible by $b$. Then from the remaining ones (if there are at least $2b-1$) we choose another $b$ numbers that have this property, and so on.

Since
$$2ab-1 = (2a-1)b + (b-1),$$
then this operation can be repeated $2a-1$ times and obtain $2a-1$ sets of $b$ numbers, in each of which the arithmetic mean of $b$ numbers is an integer. Since the statement is true for $n = a$, from these $2a-1$ arithmetic means we can choose $a$, the sum of which is divisible by $a$. It is clear that then the $ab$ numbers that make up the corresponding $a$ sets of $b$ numbers each have the desired property: their sum is divisible by $ab$. The lemma is proven (this lemma, as well as the problem, was invented by Yu. I. Nonin).

So, in order to prove the statement of the problem for arbitrariness $n = a_1a_2\ldots a_k$, it is enough to prove it for individual factors $n = a_1$, $n = a_2$, …, $n = a_k$ (and apply the lemma several times). Therefore, it is enough to learn how to prove this statement for prime factors of the number $n$. For example, to prove for $n = 100$, it is enough to prove two statements: “from any 3 integers one can choose 2 whose sum is divisible by 2” and “from any 9 integers one can choose 5 whose sum is divisible by 5.” The first is obvious, and the second can be proven with a not very long search. We give the reader the opportunity to verify this, and also try to come up with a proof that would be suitable immediately for any simple $n$. However, we will present such proof in the next issue of the journal.

31

M46

How many sides can a convex polygon have that are equal to the largest diagonal?

Answer: two, one or none.

It is not difficult to construct examples for each of these cases. A polygon in which three or more sides are equal to the largest diagonal does not exist, since two sides equal to the largest diagonal must necessarily have a common vertex (Fig. 5), and three sides, each two of which have a common vertex, can only be in a triangle, and it has no diagonals.

M47

From the digits 1 and 2, five $n$-digit numbers were made so that each two numbers coincided in $m$ digits, but not all five numbers coincided in any one digit. Prove that $\frac{2}{5} \leq \frac{m}{n} \leq \frac{3}{5}$.

Let’s write the numbers one below the other. Each digit (“column”) has 5 digits. Of these, you can make 10 (unordered) pairs. For all $n$ columns there will be 10 $n$ such pairs. Let’s find out how many of them there are that consist of identical numbers, that is, pairs (1, 1) and (2, 2). There are either 4 or 6 such pairs in each column. Therefore, the total number of such pairs is not less than 4 $n$ and not more than 6 $n$. On the other hand, the number of such pairs is equal to 10 $m$, since when comparing any two numbers across all $n$ digits, according to the condition, the result is equal to $m$ such pairs, and all possible pairs of 5 numbers are 10.

So, $4n \leq 10m \leq 6n$, which is what needed to be proved (see Fig. 6).

M48

In an acute triangle $ABC$ bicsectrix $AD$, median $BM$ and altitude $CH$ intersect at one point. Prove that angle $BAC$ is greater than 45°.

It is not difficult to prove that for a given $\Rightarrow BAC = \alpha < \frac{\pi}{2}$ there is exactly one (up to subdivision) triangle $ABC$ in which the height is $CH$, bicsectrix $AD$ and median $BM$

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Diana $BM$ intersect at one point. Indeed, having distinguished the (arbitrary) segment $AC$ on one side of the angle $\Rightarrow A = \alpha$, dropping the perpendicular $CH$ to the other side of the angle and designating its point of intersection with the bisector of the angle $A$ through $O$, we will find the point $B$ of intersection of the second side of the angle and the line $OM$ (where $M$ is the midpoint of the segment $AC$) and thereby construct the required triangle $ABC$; it is obvious that given $\Rightarrow A$ and $AC$ it is defined in a unique way. Note that in this case the angle $B$ is acute (since the point $H$ - the base of the height - lies on the side $AB$).

Let us prove that if $\Rightarrow A < \frac{\pi}{4}$, then $\Rightarrow ACB > \frac{\pi}{2}$. Let us plot the segment $HB_1 = AH$ on the straight line $AB$. Then the bisector $B_1F$ of the angle $B_1$ of the isosceles triangle $AB_1C$ will pass through the point $O$. Since $\frac{AF}{FC} = \frac{AB_1}{B_1C} > 1$, the point $M$ lies between $A$ and $F$, so $B_1$ lies between $A$ and $B$ and $\Rightarrow BCA > \Rightarrow B_1CA > \frac{\pi}{2}$.

In this problem, you can get a more accurate result: find an exact estimate for the angle $\alpha$, provided that $\Rightarrow ACB = \gamma < \frac{\pi}{2}$. To do this, we express $\gamma$ as a function of $\alpha$. Let’s draw $HE \parallel BM$ (E is the point on $AC$). Then (Fig.7)

$$\frac{EM}{MC} = \frac{HO}{OC} = \frac{AH}{AC} = \cos \alpha, \quad \frac{HB}{AH} = \frac{EM}{AE} = \frac{\cos \alpha}{1 - \cos \alpha}, \quad \frac{HB}{HC} = \frac{\cos^2 \alpha}{\sin \alpha (1 - \cos \alpha)},$$

$$\gamma = \gamma(\alpha) = \frac{\pi}{2} - \alpha + \arctg \frac{\cos^2 \alpha}{\sin \alpha (1 - \cos \alpha)}.$$

From here it is clear that the function $\gamma = \gamma(\alpha)$ on the segment $0 < \alpha < \frac{\pi}{2}$ monotonically decreases (the more $\alpha$, the greater the sin $\alpha$ and the less cos $\alpha$, and therefore the fraction under the arc sign is always smaller). Let us now find $\alpha$ for which $\gamma(\alpha) = \frac{\pi}{2}$. This property equation is: $0 < \alpha < \frac{\pi}{2}, \sin^2 \alpha (1 - \cos \alpha) = \cos^3 \alpha$, whence cos $\alpha = \frac{\sqrt{5} - 1}{2}$. So, if $\gamma < \frac{\pi}{2}$, then $\alpha > \alpha_0 = \arccos \frac{\sqrt{5} - 1}{2} \left( \alpha_0 > \frac{\pi}{4} \right)$, since $\frac{\sqrt{5} - 1}{2} < \frac{\sqrt{2}}{2}$; as you can see from the tables, the angle corresponding to $\alpha_0$ in degrees is approximately $51^\circ 50'$).

M49

The cards contain numbers from 11,111 to 99,999 inclusive. Then these cards are placed one whole at a time in random order. Prove that the resulting 444,445-digit number cannot be a power of two.

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It is easy to prove that this number is divisible by 11,111, resulting in the fact that the numbers $10^5A$ and $A$ for any integer $A$ give the same remainders when divided by 11,111, and the sum $11 111+11 112+\dots+99 998+99 999$ is divisible by 11,111. But if the number is divisible by 11,111, then it, of course, cannot be a power of two. (This immediately follows, for example, from the theorem on the uniqueness of factorization.)

Nearly more successful solutions to the problems were sent to us by E. Turkevich from Chernivtsi (he found a precise estimate for the angle $\alpha$ in problem M48, solved problem M46 and M49) and R. Akhmetshin from Osh (M42, M44, M47).

N. B. Vasiliev

In this issue we publish solutions to problems F45, F59—F65

Let’s return once again to problem F45. She has a pretty simple and red solution.

F45

A spherical capacitor filled with a dielectric and charged to a certain potential difference is discharged through its dielectric. What will be the magnetic field of the discharge currents in the space between the spheres?

Since at any point of the dielectric all directions in the plane perpendicular to the radius are equal, then the vector $\vec{B}$ inductance of the magnetic field cannot have a component lying in this plane. Therefore, the vector $\vec{B}$ can only be directed along the radius. But all points of the insulator are equal. This means that the vectors $\vec{B}$ and the magnetic field lines either converge or exit from the center of the sphere. But the magnetic field lines must be closed. This means it is impossible, and $\vec{B} = 0$. The magnetic field of the capacitor discharge currents is zero.

F59

Which edge of the wire cube should be removed so that the resistance between points $A$ and $B$ (Fig. 8) changes as much as possible? The resistances of all edges of the cube are the same.

Number all the edges of the cube and its vertices. The current flowing along edge 1 branches at vertex 1 into two flows, which then follow along edges 7 and 8.

Similarly, the current flowing along edge 2 branches into two equal flows at vertex 2 and so on. Through edge 4 the current flows through edges 7 and 12… Edge 4 is equivalent to edges 5 and 6. This means that through edges 1, 2, 3, 4, 5 and 6 the current flows twice as much as through edges 7, 8, 9, 10, 11 and 12. It is clear that the resistance of the circuit will change the most if the conductor carrying the most current is removed from the circuit. From the wire cube you need to remove one of the edges 1, 2, 3, 4, 5 or 6.

Make sure of this by calculating the resistance of the corresponding circuits.

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Some tips. The resistance of the cube between vertices A and B is easy to calculate. To do this, you just need to note that, due to symmetry, vertices 1, 2 and 3 have the same potential. Therefore, the resistance of the circuit will not change if they are connected together. Of course, vertices 4, 5 and 6 also have the same potential and can also be connected. After this, you will receive a chain (Fig. 9), the resistance of which is very easy to calculate. If we take the resistance of one edge equal to 1 ohm, then the resistance of the chain will be equal to 5/6 ohms. It is not difficult to find the resistance of the circuit when one of the edges 1, 2, 3, 4, 5 or 6 is removed. Here you can also find equivalent points, connecting which you get a circuit of series and parallel connections. But before finding the resistance of the circuit after removing one of the ribs 7, 8, 9, 10, 11 or 12, read the article by A. R. Zilberman “Transformation of Electric Circuits” in «Kvant» No. 3 for 1971. You will need the method explained in this article.

F60

An image of a candle is visible in a flat mirror. What will happen to it if a plane-parallel glass plate is placed between the mirror and the candle?

Having drawn the course of several rays, it is not difficult to verify that after placing a plane-parallel glass plate between the candle and the mirror, the image of the candle will approach the mirror (Fig. 10). Blue lines show the path of the ray without a plate, red lines - with a plate.

F61

A cube sliding without friction on a smooth horizontal floor hits a vertical glass with one of its side faces. The coefficient of friction between the cube and the glass is k. At what angle to the glass will the cube bounce if before colliding with it it was moving in a direction making an angle α with the glass?

When the cube is removed from the glass, two forces act: the force N of the normal reaction of the stack and the friction force Ftr (Fig. 11). Let us decompose the speed of the cube before and after colliding with the stack into components parallel to these forces. The velocities v1 = v sinα and u1 are perpendicular to the glass, and the velocities v2 = v cosα and u2 are parallel to it. In the direction perpendicular to the glass, between the cube and the glass, as usual, an absolutely elastic impact occurs, as a result of which the component of the speed of the cube, perpendicular to the glass, changes to the opposite: u1 = −v1 = −v sinα. It means:

Fig. 10.
Fig. 11.

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It means that the component of the cube’s momentum perpendicular to the wall changes to $2mv\sin\alpha$ ($|u_1-v_1| = 2v\sin\alpha, m - \text{masca кубика})$.

If the time of keeping the cube with the wall is denoted by $\tau$, and the average value of the reaction force of the wall is equal to $N_{cp}$, then, according to Newton’s second law, the change in the component of the cube’s momentum perpendicular to the wall is equal to the force impulse $N_{cp}$, that is

$$2mv\sin\alpha = N_{cp}\tau.$$

If the friction force did not act on the cube, then the component of the cube’s momentum parallel to the wall would not change and the cube would bounce off the wall at the same angle $\alpha$ at which it moved towards it before the impact. However, due to the action of the friction force in our case, the component $v_2$ of the speed of the cube also changes. Let us first assume that the average friction force $F_{\tau p,cp} = N_{cp}k$ acting on the cube is such that during the interaction of the cube with the wall, the component of the cube’s velocity parallel to the wall does not have time to decrease to zero. In this case, according to Newton’s second law,

$$mu_2 - mv_2 = -F_{\tau p,cp}\tau,$$ or $mu_2 = mv\cos\alpha - N_{cp}k\tau$.

But $N_{cp}\tau = 2mv\sin\alpha$, therefore

$$mu_2 = mv\cos\alpha - 2kmv\sin\alpha,$$

or

$$u_2 = v(\cos\alpha - 2k\sin\alpha).$$

Now let’s find the angle at which the cube will bounce:

$$tg\beta = \frac{|u_1|}{u_2} = \frac{v\sin\alpha}{v(\cos\alpha - 2k\sin\alpha)} = \frac{tg\alpha}{1 - 2ktg\alpha}.$$

This answer was received by V. Goncharenko from Maikol, Krasnoarsk region, M. Gilburg from Borislav, Lviv region, N. Smirnov from Gorky, V. Zakass from Usolye-Sibirsky, Irkutsk region, I. Gritsik from David-Gorodok, Brest region, A. Senyavin from Moscow, A. Anishchenko from Sarapula, UASSR, and Kudek from Rzhev.

However, this is not a complete solution to the problem. This was noticed by readers A. Vlasov from Valdai, B. Shparberg from Mogilev on the Dnieper, A. Zhukov from Kirovsk, Donetsk region and I. Berman from Chernivtsi.

Obviously, our solution is valid if $u_2 > 0$, that is, $2ktg\alpha < 1$.

At $tg\alpha \rightarrow \frac{1}{2k}$ $tg\beta \rightarrow \infty$ and $\beta \rightarrow \frac{\pi}{2}$. What does this mean?

If $tg\alpha = \frac{1}{2k}$, then $u_2 = v(\cos\alpha - 2k\sin\alpha) = v\left(\cos\alpha - \frac{1}{tg\alpha}\sin\alpha\right) = 0$.

During the interaction of the cube with the wall, the component of the cube’s velocity parallel to the wall becomes zero. But how will a collision with a wall occur if $tg\alpha > \frac{1}{2k}$?

In this case, the cube will slide relative to the wall not during the entire impact, but until the component of its velocity parallel to the wall becomes zero. Together with it, the friction force acting on the cube will also go to zero, and the cube will bounce off at an angle $90^\circ$ to the wall.

So,

$$\beta = \begin{cases} \arctg\left(\frac{tg\alpha}{1 - 2ktg\alpha}\right) & \text{при } tg\alpha < \frac{1}{2k}, \\ \frac{\pi}{2} & \text{при } tg\alpha \geq \frac{1}{2k}. \end{cases}$$

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A thin, non-stretchable rope consists of two parts. The mass per unit length of one of the parts is $\rho_1$, and the other is $\rho_2$. The rope encloses a very light hoop of radius $R$, the mass of which is negligible compared to the mass of the rope enclosing it, the ends are attached to the floor (Fig. 12). A piece of rope with mass per unit length $\rho_1$ rolls with a speed $v_1$. At what speed will the hoop roll along the second section of the rope?

As the hoop moves from one section of the rope to another, the potential energy of the rope changes. Indeed, when the hoop moves along a section of rope with linear density $\rho_1$, the potential energy of the rope bulb is equal to $m_1 g R = \rho_1 2 \pi R g R = 2 \pi \rho_1 R^2 g$ (the center of mass of the bulb is at its center). When the hoop moves along a section of the rope with linear density $\rho_2$, it is equal to $m_2 g R = 2 \pi \rho_2 R^2 g$. The correspondence changes and the kinetic energy of the rope flask. Since there are no losses, we can use the law of conservation of energy.

However, for this we need to know the kinetic energy of a flask of mass $m$ moving without slipping along the plane with a speed $v$. You can use the result obtained in the article by A.K. Kikon, Rotational motion of bodies》（Kvant No. 1, 1971):

$$K = \frac{mv^2}{2} + \frac{I \omega^2}{2} = \frac{mv^2}{2} + \frac{mR^2 \omega^2}{2} = \frac{mv^2}{2} + \frac{mv^2}{2} = mv^2.$$

But we will get this formula using a simple trick.

A flask moving without piercing at any moment of time rotates around the instantaneous center of rotation - the point of contact with the plane (Fig. 13). Angular speed of rotation of the bulb $\omega = \frac{v}{R}$. This means that the speed of the point $A$ bulb is perpendicular to the segment $OA$ and is equal to $OA \cdot \omega$. From $\triangle OAD$ $OA = 2R \cos \alpha$. Therefore $v_A = 2R \cos \alpha = 2v \cos \alpha$. Similarly, the speed of point $B$, diametrically opposite to point $A$, is $v_B = 2v \cos (90^\circ - \alpha) = 2v \sin \alpha$. Kinetic energy of small areas with masses $\Delta m$ near points $A$ and $B$

$$K_{A+B} = \frac{\Delta mv_A^2}{2} + \frac{\Delta mv_B^2}{2} = \frac{1}{2} \Delta m (4v^2 \cos^2 \alpha + 4v^2 \sin^2 \alpha) = 2\Delta mv^2.$$

$K_{A+B}$ is independent of $\alpha$. This means that any other pair of diametrically opposite points on the bulb has exactly the same energy. Therefore, the kinetic energy of the entire flask is $mv^2$. Let us now write the law of conservation of energy:

$$m_1 v_1^2 + m_1 g R = m_2 v_2^2 + m_2 g R;$$

from here

$$v_2 = \sqrt{\frac{m_1}{m_2} v_1^2 + g R \left(\frac{m_1}{m_2} - 1\right)} = \sqrt{\frac{\rho_1}{\rho_2} v_1^2 + g R \left(\frac{\rho_1}{\rho_2} - 1\right)}.$$