Kvant 1971, No. 8 · Pages

Popular science physics and mathematics
quantum
8
1971
magazine
Academy
Sciences of the USSR
And
Academy of Pedagogical
Sciences of the USSR

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In the room:

Perfect numbers 1 I. Ya. Depman

Some cosmic aspects of radioactivity 9 E. Rutherford

From memories of Professor Rutherford 17 P. L. Kapitsa

Mathematical circle Inversion and Apollonius’ problem 23 A. P. Savin

Laboratory «Kvant» Four experiments with vision 29 J. Greee

Problem «Kvant» Problems 33

Solutions to problems M45, M50—M60; F66—F71 35 N. B. Vasiliev, I. Sh. Slobodetsky

This terrible cosmic cold 51 A. L. Stasenko

Chess and mathematics notes 55 E. Ya. Gik

Reviews. Bibliography Is our eye right? 60 V. A. Leshkovtsev

Collector’s Corner Stamps dedicated to the creator of nuclear physics 62 A. V. Altykis

Answers, directions, solutions 63

«Kvant» for younger schoolchildren A few more problems by Vasya Smekalkin 3rd cover page
Game with dominoes 4th cover page

V. M. Rosenutler

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Pythagoras (VI century BC). Portrait from a fresco by Raphael. The personality of Pythagoras already became popular in ancient times. He founded a mathematical and philosophical school, and many remarkable scientists of Greece called themselves followers of Pythagoras.

Many learned societies were supposed to have members, often simply due to prejudice, the reasons for which were forgotten long ago.

Ancient mathematicians were surprised by a special property of these two numbers: each of them is equal to the sum of all its own divisors:

$$6 = 1 + 2 + 3,$$
$$28 = 1 + 2 + 4 + 7 + 14.$$

Before Euclid, only these two perfect numbers were known, and no one knew whether there were other perfect numbers or how many such numbers there could be. The great founder of geometry deals a lot with the study of the properties of numbers; Of course, he could not help but be interested in perfect numbers. Euclid proved that any number that can be represented as the product of the factors $2^{p-1}$ and $2^p-1$, where $2^p-1$ is a prime number, is a perfect number. If we substitute $p=2$ into Euclid’s formula $2^{p-1}(2^p-1)$, we get $2^{2-1}(2^2-1)=6$ - the first perfect number, and if

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Pierre Fermat (1601 - 1665). A French mathematician, a lawyer by profession, he studied mathematics “in his spare time.” Along with Descartes, he is the founder of analytical geometry. One of the creators of number theory; I worked a lot on prime numbers. Among his most interesting results is Fermat’s little theorem.

first prove that the numbers $2^p$ - 1 with the values ​​$p$ indicated by them are real primes.

It is not at all difficult to calculate any of them, but to find out whether all these numbers are prime or not was far beyond the limits of human beings.

It remains unknown whether Mersenne was right or wrong.

No need, it was discovered that the cheerful Intaine Cataldi, a former professor of mathematics in Florence and Bologna, who first gave a method for extracting square roots, also studied Japanese perfect numbers to save his soul. His notes indicated the values ​​of the sixth and sevento perfect numbers, which he found almost a hundred years before Mersenne:
8 589 869 056 (sixth number),
137,438,691,328 (seventh number).

It turned out that both of these numbers coincide with those pointed out by Mersenne: $2^{16} \cdot (2^{17} - 1)$ and $2^{18} \cdot (2^{19} - 1)$.

It remained unproven whether these numbers were really perfect; This requires that the factors $2^{17} - 1$ and $2^{19} - 1$ be prime.

St. Petersburg academician, founder of modern mathematics, continuous calculator, friend of Lomonosov, the great Leonard Euler managed to find a new theorem about tanistic and mysterious perfect numbers. He proved that all even perfect numbers have the form indicated by Euclid. But what form odd perfect numbers should have and whether they can exist at all remains unknown to this day.

Euler proved that the first three numbers mentioned by Mersenne: $2^{17} - 1$, $2^{19} - 1$ and $2^{31} - 1$ are indeed prime.

The sixth and seventh perfect numbers discovered by Cataldi and later by Mersenne turned out to be correct. And the mysterious secret of how they managed to find them remained forever in history. Until now, only one “explanation” for this riddle has been proposed - it was still perfect: they were helped by the natural providence, it suggested to its chosen one the correct values ​​of two perfect numbers.

Thus, the eighth perfect number, which corresponds to $p = 31$ in Euclid’s formula, is
2 305 843 008 139 952 128.

Again, for a century, this number remained the best known of the perfect numbers, but during this time the mathematician managed to find a new method by which one can determine whether the number $2^{p} - 1$, where $p$ is a prime number, is prime or not, without performing direct calculations. It turned out that not all of Mersenne’s predictions were correct. He correctly predicted the value $p = 127$, but numbers with values ​​$p = 67$ and $p = 257$, Mersenne asked, were not perfect.

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numbers. She checked 42 numbers in two hours, the smallest of which had more than 80 digits! All these numbers turned out to be composite. The machine discovered a new perfect number by the evening of January 30:

$$2^{520} \cdot (2^{521} - 1) \quad (p = 521);$$

The thirteenth perfect number turned out to be a composite number of 314 digits.

The machine found the fourteenth perfect number on the same day by midnight. After searching and testing thirteen more Euclidean numbers, she found the prime number $2^{607} - 1$, which in the decimal system has only one hundred and eighty-three digits, and the corresponding perfect number

$$2^{606} \cdot (2^{607} - 1) \quad (p = 607).$$

The fourteenth perfect number has 366 significant digits.

The machine found the fifteenth perfect number only in June 1952. She was very busy and could devote only her free time from more important matters to the problem of perfect numbers.

Continuing with the Japanese new prime numbers, she proved the primeness of $2^{1279} - 1$ and found a perfect number of seven hundred and seventy digits:

$$2^{1278} \cdot (2^{1279} - 1) \quad (p = 1279).$$

The sixteenth and seventeenth perfect numbers were discovered in October 1952. By this time the machine had found two more Euclidean primes: $2^{2203} - 1$ and $2^{2281} - 1$ and calculated two corresponding perfect numbers:

$$2^{2202} \cdot (2^{2203} - 1) \quad (p = 2203),$$

consisting of only one thousand three hundred twenty-seven digits, and

$$2^{2280} \cdot (2^{2281} - 1), \quad (p = 2281),$$

which has 1373 digits.

The eighteenth perfect number was discovered in September 1957 by the Swedish mathematician G. Riesel. Using an electronic calculating machine, he established the prime number in five and a half hours

$$2^{3217} - 1$$ and got the eighteenth perfect number:

$$2^{3216} \cdot (2^{3217} - 1) \quad (p = 3217).$$

It has about 2000 digits.

The Ponski of subsequent perfect numbers required more and more calculations. But computer technology was constantly improving, and in 1962 two new perfect numbers were found, and in 1965 three more. This number corresponds in Euclid’s formula to the values ​​$p$, equal to the corresponding 4253, 4423, 9689, 9941 and 11213. The perfect number $2^{11212} \times (2^{11213} - 1)$ has 3376 digits. Of course, only thanks to such an assistant as a computer, man was able to establish that this huge number is perfect.

That’s all that the Lodi have learned about perfect numbers for almost two millennia.

The history of the Japanese perfect numbers clearly shows how greatly the machine of human possibility increases. However, according to Edmund Landdu, one of the leading experts in the field of number theory:

“…Two problems remain unresolved to this day:

— Is there an infinite number of four perfect numbers? - Don’t know.

— Is there an infinite number of four perfect numbers? “I don’t even know if there is one such number.”

There is hardly any need to add anything to this.

Exercises

1. Prove that a number of the form $2^{k-1}(2^k - 1)$, where $2^k - 1$ is a prime number, is perfect.
2. Let us denote by $\sigma(n)$, where $n$ is a natural number, the sum of all divisors of the number $n$. Prove that if the numbers $n_1$ and $n_2$ are mutually prime, then $\sigma(n_1 \cdot n_2) = \sigma(n_1) \cdot \sigma(n_2)$.
3. Let $n$ be an even perfect number. Then $\sigma(n) = 2n$. Represent $n$ as $2^{k-1}b$, where $k \geq 2$ and $b$ are odd, and prove that $b = (2^k - 1)c$. Prove further that $c = 1$ and $2^k - 1$ are prime numbers.

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Some cosmic aspects of radioactivity

E. Rutherford

August 30, 1971 marks the 100th anniversary of the birth of Ernest Rutherford. It is difficult to name a person who has done more to advance our knowledge of the structure of the atom than Ernest Rutherford, the great experimenter of the 20th century.

While studying the radioactive decay of uranium, Rutherford discovered α- and β-rays and established that the first of them are a stream of fast helium nuclei, and the second are a stream of extremely fast electrons. He also showed that radioactive decay leads to the transformation of chemical elements. Through a series of brilliant experiments, Rutherford built a planetary model of the atom, being the first to realize that at the center of the atom there is a heavy nucleus around which electrons move. He created the basis of modern ideas about the structure of matter. In 1919, Rutherford carried out the first artificial nuclear reaction, during which one natural gas, nitrogen, was converted into another natural gas, oxygen.

The rapid development of nuclear physics, the founder of which was Ernest Rutherford, revolutionized natural science and led to the fact that the role of science in the life of society increased immeasurably.

We are printing today an abridged translation of a paper given by Ernest Rutherford at the Royal Astronomical Society of Canada on April 3, 1907. The paper was published in the journal of that society for May-June 1907.

The report is printed in abridged form. Translation from English and preparation of the publication was carried out by Yu. M. Tsipenok.

The topic of my speech in a completely natural way is divided into two parts: one is a consideration of the properties of radioactive substances themselves, the other is the prevalence of radioactive substances in the surface of the Earth and in our atmosphere.

A distinctive feature of radioactive substances in general is their ability to spontaneously and continuously emit radiation of a special type, releasing bodies, and in some cases also emitting light rays. In the case of radioactive substances such as uranium or thorium, these properties manifest themselves, if not indefinitely, then at least over a period of time varying over millions of years. In the case of radium, the duration of activity is shorter, but also changes over thousands of years. This learning is spontaneous and completely resistant to chemical or physical influence.

I’ll use a radioactive element as an example for the sake of it, but keep in mind that most other active elements have similar radioactive properties. Radium emits three types of rays called α-, β-, and γ-rays. α-rays have a low ability to pass through matter. A thin layer of writing paper is enough to completely

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absorb α-radiation. These rays turned out to consist of heavy atoms of matter carrying a positive electric charge; they emit from active substances at a speed of about 30,000 km per second. In all likelihood, the α-particle is a charged atom of helium *).

β-rays are more penetrating than α-rays and carry a negative electrical charge. They eject from radioactive matter at a speed significantly greater than the speed of emission of alpha particles. Some beta particles are emitted at speeds close to the speed of light. Their rest mass is only one thousandth of the mass of an α particle, and in fact they are identical to the electrons produced during a cathode discharge in a vacuum tube.

γ-rays have extraordinary penetrating ability, easily passing through 5-10 cm of iron. It has now been established quite reliably that γ-rays are waves similar in character to the well-extracted X-rays, but they have greater penetrating ability.

What all these types of radiation have in common is the ability to act on a photographic plate and cause phosphorescence **) in certain types of matter. But from a measurement point of view, their most important property is the ability to discharge onto electrified bodies. This property undoubtedly leads to the most revealing method for studying radioactive substances, and therefore we will consider it in more detail. Let’s take an ordinary good insulated electric scooper and charge it so that its golden leaves spread widely. It is well known that under normal conditions and good insulation the leaves converge extremely slowly, and after a few minutes the leaves will seem almost motionless. Now let’s place a radioactive substance on the electrode of the electric juice. The sheets will immediately begin to quickly converge, which is due to the loss of charge on the electrified system. Regardless of whether the electroscope has a positive or negative charge, the loss of charge occurs at the same rate. The mechanism that causes the discharge of the electric juice has been studied in great detail, and it is now known that this effect is due to the fact that the radioactive radiation surrounding the electric juice has an impact, that is, a certain amount of carriers of negative and positive charges - ions - are formed in it. The ions are mixed in an electric field. If, for example, the electroscope is charged positively, then the negative ones will move towards the charged system. The effect of electric juice discharge is due to the attraction of a large number of negatively charged ions to a positively charged conductor. When the radioactive substance is removed, the rapid merging of the sheets stops immediately.

This ability of radiation to influence them, for example, light. An example would be the original light dial (Note ped.)

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until a state of stable equilibrium occurs. In this state, the formation of new quantities of emanation is compensated by the decay of the emanation due to its further radioactive transformation. Then boil the solution over it, mix it with the emanation, introduce it into the electroscope and mix the speed of movement of the gold leaves.

If the rate of discharge has dropped to half its original value after four days, this is definite proof that radium emanation is present in the electroscope.

In this way, you can easily change the amount of radium in the solution in the range of up to $10^{-11}$ e.

The emanation of radium is unstable and decays into another substance, which behaves like a solid and has perfect yin radioactive properties.

The inner surface of a vessel containing radium emanation is covered with an invisible coating of radioactive substance. If the emanation is quickly removed with a stream of air, this so-called “active sediment” settles inside. The activity of this sediment is not constant and quickly subsides over time.

After a few hours, activity decreases in the geometric program, halving in about 28 minutes. In the case of active sediment, the activity of napolin decreases in 11 hours; in the case of active - in 34 minutes. The formation of this characteristic active precipitate from each emanation provides another opportunity to distinguish the presence of thorium, radium or actium individually.

An interesting property of active sediment, which, as we will see, plays a significant role in the analysis of the radioactive state of the atmosphere, is that in an electroscope the sediment is contained on a negatively charged conductor.

The active sediment particles somehow acquire a positive charge, tend to the negative electrode and adhere to it.

Radioactive state of the atmosphere

I have briefly discussed some of the most important properties of radioactive bodies, which have proven very useful in solving the problem of the abundance of radioactive substances in the earth and atmosphere. The pioneers in this area were Elster and Geitel, teachers of the gymnasium in Wolfenbitt, Germany.

Geitel in 1900 noticed that the opening of air has the property of an arbitrary weak discharge of the electroscope. He showed that this phenomenon is caused by the formation of negative and positive ions in the air. In search of a possible cause for this phenomenon, Elster and Geitel proved that it could be associated with the presence of radioactive substances in the atmosphere. Then they tried to carry out an extremely daring experiment. I already point out that the emanation from radioactive bodies converts active residue, which can concentrate on negatively charged wiring. If there is emanation in the atmosphere, then the active deposit must collect on a negatively charged conductor placed in the atmosphere.

Elster and Geitel strung an insulated wire 20 or 30 m long (Fig. 1) outside the laboratory window. Using an electrical machine, the wire was charged to a negative potential of several thousand volts. After a few hours, the wire was quickly cleaned, welded around a frame attached to the electroscope, as shown in figure2, and the discharge rate of the electroscope was changed. The golden leaf quickly fell away. The magnitude of this effect is independent of wire materials and has been convincingly shown to be completely

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caused by the presence of a radioactive substance on the surface of the wire.

We must now consider the question of how to determine what radioactive substance was concentrated on the wires. The presence of part of solid bodies such as radium, uranium or thorium in the atmosphere is impossible. However, emanations from radium, thorium and active gases are gaseous, and if these substances are present in the earth, then it is quite reasonable to expect that some of the emanation released by them may enter the atmosphere. The question of what kind of emanation this is can be answered in full definition. If there is only an emanation of radium in the air, then the active residue collected on the wires should decay at a rate characteristic of radium, that is, a few hours later the activity decreases exponentially, halving within 28 minutes. It turned out that active sediment from the air will disintegrate almost exactly at this rate. Therefore, we can conclude that there is an emanation of radium in the atmosphere.

Preliminary experiments have shown that the radioactivity of the atmosphere is due to the presence of radium emanations. Bumstead in New York and Blanc in Italy, however, have convincingly shown that in these areas there is some amount of thorium emanation in the air. This can be verified by testing

in the best possible way. The wire is installed outdoors, charged negatively and left for several days. After removing the wires, part of the activity, due to the active deposit of radium, quickly decreases. After 5 or 6 hours there is still activity, but the decline is much slower, decreasing by half every 11 hours. This decay rate is characteristic of the active sediment of thorium and indicates that there is also an emanation of thorium in the atmosphere. Blank has recently shown that in some parts of Italy 70 per cent of the activity must be attributed to thorium, but in most other places it appears that the effect is due to radium.

From the above, we can conclude that the atmosphere above the surface of the Earth contains large quantities of emanations of radium and thorium, as well as active substances that arise during their transformations right there in the atmosphere. The magnitude of their arbitrary onization is small, but can be precisely changed by a device specially designed for this purpose by Ebert. Using a fan driven by a motor, a constant flow of air is passed between two concentric cylinders. The inner insulated cylinder is charged and connected to the electroscope (Fig. 3). When pro-

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Fig. 3.

When infused air passes between the cylinders, ions settle on the inner cylinder and reduce its charge. The discharge rate is a measure of the number of ions per cubic centimeter of outside air. One cubic centimeter of open air usually contains about 1500 eta; the magnitude is not constant. Compared to the number of neonized molecules, the number of atmosphereons present is extremely small. Each cubic centimeter of air contains about $4 \times 10^{19}$ molecules, so on average only an infinitesimal fraction of the air molecules are neonized at any one time. Although it is very minor and requires a sensitive electroscope to measure it, it still plays a very important role in the electrical state of the atmosphere.

Prevalence of radioactive substance in the earth’s crust

Elster and Geitel long ago noticed that caves and cellars in which the same influence remained for a long time. This was replaced by greatly reducing the charge of a negatively charged wire placed in these enclosed areas. It was assumed that if this occurs as a result of radium emanation from the soil, then the amount of emanation should increase in an enclosed space. This led to research into the group’s activities. In all of the problems, radioactivity was detected, and in many, the presence of radium emanation.

There seems to be no doubt that soil is a source of emanation in the atmosphere. This emanation is released from the soil through the process of diffusion or from water sources, and is then carried by the wind. Due to the high rate of transformation of thorium and active emanations compared to radium emanations, one should expect the predominance of radium emanations in the atmosphere. Therefore, it is unexpected that large amounts of thorium emanation were discovered in a number of areas. The emanation of radium must continue through the soil, disintegrating over time, and it is to be expected that most of the emanation present in the air comes from a thin layer of earth, not more than a few decimeters thick.

Inner body of the Earth

So, it is natural to assume that there is a very large amount of radium and other radioactive substances distributed over the surface of the Earth. This radioactive materia must constantly give tello to the Earth, and there seems to be greater inter-consistency with the influence of this body effect and its influence on the problem of transmission of the inner tello of the Earth.

All radioactive substances continuously release tello. In one hour, radium releases an amount of tell sufficient to melt a quantity of ice that exceeds the weight of the radium itself. Over the course of a year, a kilogram of radium releases as much tell as the amount produced by scraping 100 kcal of good coal. This release of tell by radium and all other radioactive substances is in fact a secondary property - a consequence of the bombardment of an ac-

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OF THE SUBSTANCE BY α-PARTICLES EMISSED FROM ITS OWN MASS.

β- and γ-RAYS CREATE ONLY A SMALL PART OF THE TOTAL TELL EFFECT. THE THERMAL EFFECT OF Radium Vseda SHOULD BE PROPORTIONAL TO THE AMOUNT OF Radium IN THE EARTH’S CRUST AND SHOULD TAKE PLACE INDEPENDENT OF HOW RADIUM IS DISTRIBUTED THROUGHOUT THE EARTH’S CRUST. Consequently, there must be a CONSTANT INCLUDE OF TELLIUM INTO THE EARTH, DUE TO THE PRESENCE OF RADIAUM AND OTHER RADIOACTIVE SUBSTANCES IN IT. But the THERMAL EFFECT RESULTING FROM A GIVEN AMOUNT OF Radium cannot EXIST INFINITELY. Radium IS TRANSFORMED INTO OTHER SUBSTANCES, AND IT SEEMED THAT AFTER 20,000 YEARS ONLY A SMALL PART OF IT SHOULD REMAIN. HOWEVER THIS IS NOT SO. IT IS NOW DEFINITELY ESTABLISHED THAT RADIUM IS FORMED FROM URANIUM AND THAT IN OLD MINERALS THE AMOUNT OF RADium IS ALWAYS PROPORTIONAL TO THE AMOUNT OF URANIUM.

THE TRANSFORMATION OF URANIUM IS CONSIDERABLY SLOWER THAN THE TRANSFORMATION OF RADium, AND IT WILL LIKELY TAKE A PERIOD OF AT LEAST A THOUSAND MILLION YEARS WHILE THE URANIUM IS DISTRIBUTED. Consequently, DUE TO THE FORMATION OF RADIA DURING THE DISTRIBUTION OF RANIUM, THE AMOUNT OF RADIA IN THE EARTH WILL BE MAINTAINED AT A CONSTANT LEVEL FOR A PERIOD OF HUNDREDS OF MILLIONS OF YEARS.

Now WILL BRIEFLY DISTRIBUTE THE THEORY OF THE ORIGIN OF THE INNER BODY, THE AGE OF THE EARTH. IT IS KNOWN THAT THE TEMPERATURE OF THE EARTH GRADUALLY INCREASES AS IT ROTATES DEPTH - ON AN AVERAGE BY 1 ° C FOR EVERY THIRTY METERS IN THE DEPTH. ACCORDING TO DATA ABOUT VOLCANIC ACTIVITY, IT IS KNOWN THAT THE EARTH’S INSIDE HAS A HIGH TEMPERATURE.

WHEN SOLVING THE PROBLEM ABOUT THE AGE OF THE EARTH, KELVIN PREPOL stated that the EARTH ORIGINALLY HAD THE TEMPERATURE OF ROCK MELTING AND WAS GRADUALLY COOLED, RADIATING SOLID FROM THE SURFACE INTO SPACE. SOME QUANTITIES REQUIRED FOR DISTRIBUTION THROUGHOUT

THIS BODY WERE INSUFFICIENTLY KNOWN, BUT KELVIN CONCLUDED THAT, UNDOUBTEDLY, NO MORE THAN 100 MILLION YEARS HAVE PASSED SINCE THE EARTH’S SURFACE BECAME TEMPERATURALLY SUITABLE FOR PLANTS AND ANIMALS.

THERE IS NO DOUBT AS TO THE GENERAL CORRECTNESS OF THE CONCLUSIONS BASED ON THE ASSUMPTIONS MADE BY KELVIN. LET’S HOWEVER LET’S CONSIDER THESE PRE-HAULBS. IT IS ASSUMED THAT THE EARTH IS AN ISOLATED COOLING IN THE SPACE OF THE BODY, HAVING NO SOURCES OF THE BODY FROM THE MOMENT OF COOLING. HOWEVER, THE DETECTION OF SUBSTANCES SIMILAR TO RADIUM IN THE EARTH, WHICH EMMIT TELLO AT HIGH SPEED, IMMEDIATELY LEADS TO THE THOUGHT THAT THE EARTH IS NOT JUST A COOLING TELLO, AN ATTACK IN WHICH TELLO IS CONSTANTLY BORN.

IN 1902 I MADE SOME CALCULATIONS TO DETERMINE WHAT AMOUNT OF RADIAUM (OR OTHER RADIOACTIVE SUBSTANCE), HOMOGENEOUSLY DISTRIBUTED IN THE EARTH, WOULD BE NECESSARY TO CHANGE THE TEMPERATURE OF THE LAYERS OF THE EARTH TO REACH THE SURFACE WITH THE DEPTH REMAINED CONSTANT. IN SUCH A CASE, THE BODY COMING FROM RADIUM MUST ACCURATELY COMPENSATE THE CONSUMPTION OF THE BODY RADIATING INTO SPACE. THE CALCULATIONS ARE SIMPLE AND THEY SHOW THAT A UNIFORM DISTRIBUTION OF ONLY 1.7·10^{-13} e per 1 cm^{3} WOULD BE SUFFICIENT.

R. J. Strutt HAS RECENTLY MADE A SERIES OF SYSTEMATIC OBSERVATIONS ON THE LIMIT OF Radium CONTENT IN TYPICAL CRUSTAL ROCKS AND OBTAINED EXTREMELY INTERESTING RESULTS. Strutt TOOK A NUMBER OF TYPICAL ROCKS, BOTH VOLCANIC AND SEDIMENTARY, FROM DIFFERENT LOCATIONS, AND DETERMINED THE RADIA CONTENT IN THESE ROCKS. IT TURNS OUT THAT THE CONTENT OF RADIA IN THE EARTH’S CRUST IS NOT ONLY SUFFICIENT FOR SUPPORTING THE EARTH’S BODY INSIDE, BUT IT IS EVEN MORE THAN REQUIRED IN THE CASE IF WE ASSUME THAT IT IS EVEN DISTRIBUTED ALL OVER

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Fig. 4.

the mass of the Earth. The latter assumption should lead to the conclusion that the Earth is constantly warming up, not cooling, a conclusion that can hardly be taken with confidence. In order to get around this difficulty, Sturt suggested that the radioactive substance is not distributed evenly inside the Earth, but only in its thin surface layer. If we assume that radium is evenly distributed in this layer, then it can be easily shown that the thickness of this layer is about 70 km.

Assuming that radium is contained in a surface layer 70 km deep, it is easy to calculate the temperature of the Earth. The calculation results are shown in figure 4.

At a depth of 70 km, the temperature should increase to approximately 1500 °C. The temperature inside the Earth mass below 70 km should be approximately constant and equal to 1500 ° C.

I hope you can now understand how the study of radioactivity has profoundly changed our view of the Earth’s interior. The new conclusions are by no means final, but what has been done casts doubt on the validity of the old theories about the origin of the Earth and the changes in its internal body.

TASKS
ON PAPER
IN THE CAGE

1. How many squares?

The knots of a piece of checkered paper form a cross, as shown in figure.

How many squares can you draw with vertices at the selected nodes?

(For details, see page 50.)

2. How many rectangles?

At figure you see a square piece of checkered paper. Let’s count how many squares with vertices can be formed

at the nodes of this sheet and with sides parallel to its sides.

Obviously, the number of possible “unit” squares is $n^2$. Squares of size $2 \times 2$ will be found $(n-1)^2$. Indeed, the upper left corner of such a square can be in any

(Continued on page 54.)

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known even to his students. Sometimes, only from individual hints that emerged in a conversation with him, one could discern that he was trying something, but it didn’t work out. He did not like to talk about his work projects and was more willing to talk only about what had already been done and dado results…

…Rutherford was an experimenter and in this respect he resembles Faraday. He used little formulas and had little recourse to mathematics. Sometimes, trying to derive a formula in his reports, he got confused and then simply wrote the result, noting:

• If everything is deduced correctly, then it will work out.

But he owned the experiment exclusively. We can say that he “saw” the phenomenon he was working on, even if it occurred in an immeasurably small nucleus.atoma…

…The peculiar nature of Rutherford’s thinking could easily be seen in conversations with him on scientific topics. He loved it when they told him about experiments, but in order for him to serve with interest - and from his expressive face it was immediately clear whether he was drying with interest or searching - it was necessary to talk only about basic facts and ideas, without going into technical details, which did not interest Rutherford. When I had to bring him drawings of a high-power pulse generator for producing strong magnetic fields for approval, he would put the sink down, not paying attention to the fact that it lay upside down in front of him, and said: “This drawing does not interest me, you just indicate the principles on which this machine works.” He grasped the main idea of ​​the experiment very quickly, right away. This gave birth to me, especially in the first years of my stay at Cambridge, when, due to my lack of knowledge of the English language, I still spoke so poorly that I could not clearly talk about my ideas and experiences, and despite this, Rutherford quickly grasped the idea and always gave a very interesting assessment.

Rutherford willingly talked about his experiments and loved to show off his installations and experiments. He liked to accompany the story with drawings; For this purpose, he always had several small pencil weights in his pocket. He held the pencil in a special way, it always seemed to me - a very awkward way, somehow with the ends of three fingers. He always drew with a trembling hand; the drawing was simple, consisting of a small number of strokes made with great pressure. Quite often the point of the pencil would break, then another weight would be taken out of the pocket…

…I really loved Rutherford’s lectures; I listened to the physics course that he taught to students as a Cavendish professor. I learned little from this course that was new for myself, since by that time I already knew physics quite well, but Rutherford’s approach to physics taught me a lot. Rutherford read with great enthusiasm, he hardly used mathematics, he described phenomena communally with diagrams and conducted lectures with clear but spare gestures, from which it was clear how concretely and figuratively he thought. But what was interesting to me about his lectures was that he changed the topic beforehand. According to the plan, he was supposed to read about one thing, but then, by analogy, his thought switched to another language, community related to some new experiment made in the field of radioactivity, and he began to talk with enthusiasm about what was currently occupying him. At the same time, his assistant had the worst situation: Rutherford had to offer him to do a demonstration that was not part of the personal lecture plan…

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Rutherford visiting Capizza. First on the right is E. Rutherford, second is F. Aston, third is P. L. Capizza.

…Rutherford’s most remarkable quality as a teacher was his ability to direct work, support the scientist’s endeavors and correctly evaluate the results obtained.

The greatest thing he valued in his students was independence of thinking, initiative, and individuality. It must be said that Rutherford used every opportunity to bring out his individuality in a person.

I remember, early in my work at Cambridge, I once said to Rutherford: “We have X working on a hopeless idea and wasting time, equipment and process.” “I know,” Rutherford replied, “that he is working on a hopeless problem, but this is his own problem, and if the work does not work out for him, then it will teach him how to mouse on his own and will lead to another problem, which will already have an experimental solution.” That’s how it turned out later.

He was ready to sacrifice a lot just to cultivate independence and originality of thinking in a person, and if they manifested themselves, he surrounded him with care and especially encouraged his work…

…Rutherford believed that a novice scientist should not be given technically difficult work. For a novice worker, even if he is talented, he needs success, otherwise the necessary disappointment in his abilities may occur. If a student has success, then he must be fairly assessed and celebrated.

Once, in one of his open conversations, Rutherford told me that the most important thing for a teacher is to learn not to depend on the success of his students, and this becomes difficult over the years! This deep truth

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made more of an impression on me. The main characteristic of a teacher should be generosity. Undoubtedly, Rutherford knew how to be generous, this, apparently, is the main secret of the fact that so many great scientists came out of his laboratory, it was always free and good to work in his laboratory, there was a good business atmosphere…

…Rutherford showed exceptional concern for his students. His view of the students was, schematically, like this - he said:

“If a young scientist works for me and after two years of work comes to me and asks what I should do next, I advise him to quit working in the field of science, or if a person after two years of work does not know what to do next, he cannot become a scientist.”

Essentially, he never posed the question so sharply and, under one proposal or another, he always found the opportunity to provide his unsuccessful students with a place either somewhere in an industrial laboratory, or a teaching position in some school or university…

…From this image of Rutherford that I have drawn for you, you will immediately see a man with a large, strong temperament. This temperament was expressed in everything, especially in cases. Rutherford could not stand dishonesty in work and mutual relations between people. If any student showed even the slightest dishonesty in anything - whether it was that he misrepresented the results of his work, or that he did not mention the source of his ideas and tried to present the work he produced as his own, when in fact he got the idea for it elsewhere - such a person lost all interest for Rutherford …

Rutherford’s workplace in the Cavendish laboratory.

…He treats people extremely caringly, especially his students. Arriving to work in his laboratory, I was immediately struck by this care. Rutherford does not allow you to work longer than six hours in the laboratory, and on weekends he does not allow you at all. I protested, but he said:

— It’s completely enough to work until 6 o’clock, the rest of the time we need to think. Bad people are those who work too much and think too little…

…I later saw that the time spent on separate tasks was completely repaid by the acquired energy…

…Once Rutherford called me into his office, and I found him reading a letter and roaring with his open and infectious laughter. It turned out that the letter was from students of some Ukrainian high school. They informed him that they had organized a physics circle and were going to continue his fundamental work on the study of the atom nucleus, asking him to become an honorary member and send reprints of his scientific works. When describing Rutherford’s achievements and discoveries made in the field of nuclear physics, they used physiological terminology instead of physical one. Thus, the structure of atoma in the description of the students acquired the properties of a living organism, which made Rutherford laugh. I explained to Rutherford how this distortion could have occurred. Apparently, the schoolchildren themselves translated the letter and used a dictionary, and in Russian, unlike English, the word “kernel” has two meanings. Rutherford said that he assumed so, and responded to the guys with a letter in which he thanks them for the high part of their election and sends reprints of his works…

…The death of Professor Rutherford is a very heavy blow for scientists all over the world. In it, science lost the greatest pioneer of physical research since Faraday. Throughout his life, fruitful in scientific discoveries, he worked on the most fundamental problems of modern atom theory.

He can be considered not only as the creator of a new chapter in science, but also as the creator of a whole new science - nuclear physics…》.

Lord Rutherford’s coat of arms. The cross on the shield is formed by exponential radioactive decay curves. Above is the New Zealand kiwi bird (Ruserford was born in New Zealand).

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V ALL-UNION PHYSICS OLYMPIAD

On April 15-20, the 5th All-Union Physical Olympiad for schoolchildren was held in Novosibirsk. 550 schoolchildren took part in it - winners of Resursikan and regional Olympiads and winners of the 1970 Weight-Union Olympiad.

On April 16, the first round of the Olympiad took place - the theoretical one. The participant was offered tasks: 5 tasks each for students in grades 9 and 10, and 4 tasks for students in grades 8. 5 hours were allotted to solve the problems.

On April 17, the jury that checked the works announced the results of the 1st round. Those participants who completed the theoretical round more successfully than others were admitted to the second round of the Olympiad - the experimental round, which was held on April 18.

On the evening of April 18, an interview took place; Students who were the best in the final tasks of the 1st and 2nd round of the Olympiad took part in it.

On April 19, schoolchildren visited Akademgorodok. They visited the laboratories of the Institute of Nuclear Physics, the Institute of Hydrodynamics, the Institute of Semiconductor Physics, and the Novosibirsk State University, where they met with scientists from the Siberian Branch of the USSR Academy of Sciences.

On April 19, the jury summed up the final results, and on April 20, at the closing ceremony of the Olympiad, the names of the winners were announced.

The following participants of the Olympiad were awarded first degree diplomas: in 8 grades - Igor Korenanov (school No. 23 in Dnepropetrovsk), Aleksadr Zuev (school No. 1 in Perevalsk, Voroshilovgrad region); in 9th grade - Alexander Sakharov (school No. 47 in Kaliningrad). Serey Provotorov (school No. 45 in Leningrad), Boris Kuzmin (school No. 18 in Moscow), Lev Vaidman (school No. 45 in Leningrad); in 10 grades - Andrey Pervukhin (school No. 51 in Sverdlovsk), Aleksandr Snegirev (Osipovskaya secondary school in Chaussky district, Mogilev region), Vladimir Golovizin (school No. 16 in Kirov).

The following participants of the Olympiad received second-degree diplomas: in 8 classes - Alexander Levin (school No. 34 in Sverdlovsk), Aleksadr Teokharov (school No. 11 in Almalyk, Tashkent region), Vadim Mirny (school No. 2 in Moscow), Alexander Turchin (school No. 145 in Kyiv); in 9 classes - Serey Lyagushiin (school No. 23 in Dnepropetrovsk), Serey Katsur (school No. 34 in Chisinau), Igor Lobov (school No. 170 in Krasnodar), Georgy Polyakov (school No. 3 in Chelyabinsk), Mikhail Dezhurko (school No. 3 in Pinsk, Brest region); in 10 grades - Vladimir Nazaykinsky (school No. 2 in Moscow), Serey Parkovsky (school No. 145 in Kyiv), Boris Yakovson (school No. 116 in Odessa), Alexander Pukhovsky (school No. 21 in Arkhangelsk), Igor Artyukhov (school No. 18 in Moscow), Yuri Boedanov (school No. 40, Simferopol), Valery Fradkov (school No. 27, Kharkov).

Third degree diplomas were awarded to ten 8th grade participants, fourteen 9th grade participants and 12 10th grade participants.

74 Olympiad participants received certificates of encouragement.

The organizing committee of the Olympiad, the jury, employees of the Siberian Branch of the USSR Academy of Sciences, the Novosibirsk Regional Department of Public Education put a lot of effort into ensuring that the Olympiad was held clearly, in an organized manner, and brought benefits to the participants.

23

MATHEMATICAL CIRCLE

INVERSION
and the task of Apollonius

A. P. Savin

HOW TO CATCH A LION IN THE DESERT?

Observer in empty presence in a round cage. Lion is in a cage. An inversion is performed. Then the lion will be in the cage, and the presence will be outside it.

From graduate folklore
Moscow State University of the 20s

Geometry translated from Greek means land surveying. It owes its birth to ancient surveyors, architects, and builders. They do not have the sophisticated instruments that their descendants are armed with; they made do with the simplest means: a compass, a ruler and a measuring tape. Naturally, the ancient mathematicians used precisely these means for geometric constructions, rejecting only the measuring tape as an imprecise tool.

Sometimes you are amazed at how many different construction problems they could solve using only these simple tools.

It seemed that there were no construction problems that could not be solved with the help of a compass and a ruler.

But such problems were found. Many mathematicians puzzled over the problems of “trisection of an angle” and “doubling of a cube”. Only at the end of the last century was their unsolvability proven *). However, even now there are still people trying to trisection an angle using only a compass and a ruler.

Of the problems that can be solved with the help of a compass and a ruler, one of the most interesting is the problem of the outstanding ancient Greek geologist

*) Problem on trisection of an angle: “Indicate a method for dividing any angle into three equal parts”; cube doubling problem: “Find an edge of a cube whose volume is twice the volume of a given cube.” The proof of the unsolvability of these problems can be found in the book “Encyclopedia of Elementary Mathematics”, vol. IV, “Geometry”, M., Fizmatiz, 1963, pp. 220-223.

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meter of Apollonius from Perga *), which is formed as follows:

It is known that the solution to this problem was contained in the work of Apollonius “On Touches,” but the work itself was utter. Subsequently, Apollonius’ problem gave rise to numerous mathematical studies; it was also addressed by outstanding mathematicians such as L. Euler and I. G. Lambert.

Currently, there are many different solutions to this problem, one of them will be given here. However, the purpose of this article is not only to tell you about the solution to Apollonius’ problem, but also to equip you with the “inversion method” used in this solution.

Inversion

About the limit. Let a circle of radius r with a center at a point O be given. The inversion with respect to this circle is called the following transformation of the plane: each point M of a plane different from the point O is associated with a point M’ lying on the ray OM at a distance $\frac{r^2}{OM}$ from the point O. The point O is called the center of inversion, and the number $r^2$ is the degree or coefficient of inversion.

*) Apollonius (c. 260-170 BC) was born in the city of Perga, one of the Greek colonies in Asia Minor, worked in Alexandria for a long time, and then returned to Perga. His writings on the concepts of values ​​formed the basis for the study of algebraic curves (curves defined by equations), and the method he used led Ekarta and Fermat to the creation of analytical geometry.

Inversion is also called transformation by inverse radii, as well as symmetry around the circle. It was introduced in 1830 by the German mathematician L. Magnus, but is found in the works of Apollonius himself, for example in the essay “On Plane Geometric Places.” Since its rebirth, inversion has proven to be a powerful tool in mathematical research.

Let’s consider a number of properties of inversion.

1. Points on the circle we have chosen transform into themselves during inversion; points lying inside the circle go to external points (except for point O), and external points - to internal ones.

2. If, during inversion, the figure Ф goes into the figure Ф’, then the figure Ф’ goes into the figure Ф.

3. During inversion, points lying on a line passing through the center of inversion go to points P, and points P to point O. Then any two straight lines would intersect at point P. Probably, now it will be easier for you to understand the following property of inversion.

4. A straight line that does not pass through the center of inversion becomes a circle passing through the center of inversion.

Proof. Let us drop the peredicle from point O onto the straight line (Fig. 1). Let us find for point P—the base of the peredicle—the point P’ at which it passes during inversion. Let now M be an arbitrary point of a given line and M’ be the point at which it passes during inversion. Then from the correspondence

$$OP \cdot OP' = r^2, \quad OM \cdot OM' = r^2$$

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Solution to Apollonius’ problem

So, the path on the plane is given by three circles… Stop! Take a look at figure 3. How many more ways can three circles be arranged, even if we cut them to either touch or have no common points at all? What if we consider the possibility of intersections? Should we really consider each case separately? Of course not! Let us first consider at once all cases in which at least two of the given circles touch (it makes no difference whether it is external or internal). Let us assume that it is interesting for an auxiliary circle of arbitrary radius with a center at the point of contact of these circles. Then, by property 7, the tangent circle will turn into parallel lines, the third circle into a circle or straight line. The required circle should be obtained by inverse of the circle tangent to the specified parallel lines and the circle (or straight line). And the problem “to construct a circle tangent to two given parallel lines and a given circle (or straight line)” can be solved by each of you within a few minutes. The construction, of course, must be done using a compass and a ruler. Now let us note that with the help of only a compass and a ruler we can easily construct the circle or straight line into which any given circle or straight line goes when inverted with respect to each line of the circle. To do this, you need to select some three points on it, construct the points at which they pass (this is what you are dealing with), and then draw a circle (or straight line) through the resulting three points.

From all that has been said, the following construction proceeds:

a) construct parallel lines and a circle (or straight line) into which the given circles go when inverted with respect to the auxiliary circle with the center at the point of contact;

b) construct circles (there may be several of them) tangent to the obtained lines and the circle (or straight line);

b) on the three initial given circles, mark those points that go into the found points of contact;

d) for each triple of points, restore the original circles.

Fig. 3. Three circles on a plane can be arranged in many ways.

Fig. 5. Inverter Posel.

A device that performs inversion_transformation. There is more than one such device. They are called inverters. figure 5 shows the Posel inverter, consisting of six hinged rods, $OA=OB$, $AM=AM'=BM=BM'$. This inverter works as follows: point $O$ is fixed; if point $M$ describes a certain line, then point $M'$ describes the line obtained by inversion

versions from the original one with the center of inversion at point $O$ and the inversion coefficient equal to $AO^2 - AM^2$. Following such an inverter on your own is not at all a difficult task, even for schoolchildren who do not attend the “skillful hands” club. If we add to this inverter a rod $O'M$ such that $O'M=OO'$ and a point $O'$ are fixed, then when rotating the rod $O'M$ the point $M'$ will describe a straight line. Such inverters are used in technology to convert circular motion into rectilinear motion.

Even simpler is the Garth inverter (Fig. 6). It consists of only four hinged rods. The figure that these rods form is called an antiparallelogram. A characteristic property of such a quadrilateral is the condition: $AB=CD$ and $BC=AD$. If some point $O$ of the segment $AB$ is fixed on the plane, mark points $M$ and $M'$ on the rods $AD$ and $BC$, lying on a straight line passing through the point $O$ and parallel $AC$, then when point $M$ moves along some figure, point $M'$ will move along the figure obtained from the previous one using inversion with the center at point $O$ and a coefficient equal to

$$\frac{AO \cdot BO}{AB^2} (AD^2 - DC^2).$$

Now try to solve several problems.

1. Three circles are drawn through two points, each of which passes through the points. What will they turn into during inversion with the center at one of these points?
2. What will three parallel straight lines become during inversion?
3. Construct a circle passing through two given points and tangent to a given line.
4. Prove that the Posel and Garth inverters actually perform the indicated inversion.

29

FOUR EXPERIENCES WITH VISION

J. Greg

In the 10th and 11th issues «Kvant» over the past year, we have already talked about some experiments with black-and-white and color vision (see articles by G.I. Kosourov “Don’t believe your eyes” and “Orange sky”). Now we will tell you about several more worthy experiments with vision. With them, given at the beginning, the features of our vision are used - the ability to see in the dark and retain the memory of the images seen for some time. The last experiment demonstrates another nice visual illusion. All experiments are taken from the book by James Gregg, which we will talk about on page 60 in the same issue of our magazine. The publication was prepared by V. A. Leshkovtsev.

About test 1. Adaptation to darkness

On a bright sunny day, when you enter an unlit room from the street, it seems to you that it is deep night in the house. But soon you are already perfectly oriented, despite the relatively low light: your eyes have adapted - adaptation to the dark has occurred. As a result of adaptation, the color of the eye to light increases a million times. Under ideal conditions, an eye that is fully adapted to darkness can notice the light from a single candle 20 km away.

Observe the adaptation process very simply: in the evening, go from a well-lit room to a dark one. Notice how long it takes you to read the numbers on a watch with a luminous face or to read a newspaper headline if there is little light coming from a window or a slightly open door.

Very weak light is better seen if you look at a little bit of it. So, a dim star in the sky will seem brighter if you look a little away from it. Check it out when you get a chance.

Let us present a simple experiment using the adaptation of the eye to darkness.

Take a piece of black cardboard measuring 20x25 (cm). On both sides of the center at a distance of 5 cm, glue a white circle and square to the cardboard. Hold the cardboard like a book while reading (Fig. 1).

Have your partner suddenly turn off the lights. Your eyes are gradually

Fig. 1.

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will adapt to the dark and you will see white spots (on invisible cardboard). Calculate how much time will pass from the moment the lights are turned off until the moment you can distinguish between a white circle and a square. Take the cardboard in your hands (before the lights go out), but turn it away from you and rotate it further so that you don’t know where the white circle and square are located; when the light is turned off, turn the cardboard towards you to the side where the figures should be. Please note that the circle is visible better if you look at the square rather than at it. Can you oblige, why?

If you have carefully read the beginning of the description of this experiment, then you can correctly interpret its results.

O torture 2. Followers

The number of successors you can observe is endless. The term “after-time” means that the image remains after the discharger that caused it ceases to act. A sequence is called positive when it contains the same distribution of bright and dark areas as the discharger itself, and imprinted when the distribution of bright and dark areas is the opposite.

Nanomore interesting are the induced afterproducts. The induction effect is caused by something additional to the main discharger, such as the background against which the view is viewed successively; this background can be either a pattern, a gray field, or a colored field. First, it always shows the main discharger - figure, color. Then watch looking at some background. It is easiest to detect the induced sequence on a neutral gray background. figure2 shows figures with which you can get a good follow-up. Such figures should be glued onto gray cardboard. Leave only one figure in your field of vision and look at it for 30-60 seconds. Then look at the gray background. There should not be too much light in the room (dark lighting). You may see a bird flying on the ceiling, on the wall, on your palm, or your eye projects simultaneously onto any surface that you are looking at.

Cut out any images from colored paper. Experiment with shapes of different colors. Look at such a figure, move your gaze to the gray background - and you will see an additional color to the original one. Thus, you will become acquainted with one of the basic principles of forming induced sequences.

Take a background of the same color as the first discharger. Look at the bright red square for a minute, and then move your gaze to a piece of red paper. What does one after the other look like?

Leave the same main discharger, but take the background green-blue, that is, the color complementary to red. How does the current sequence differ from the previous one and from the one that turned out on a gray background? Try different colors and different combinations for the main discharger and for the background.

A pattern consisting of concentric circles (Fig. 3) is a very strong discharger, easily causing a sequence. When looking at it, effective movements and rolling appear. Such effectiveness is observed in his consistent manner. Follow this figure with a diameter of 30-50 cm. Install in a hundred

31

Fig. 3.

there is a large gray screen for deconstructing the afterimage. If your figure is colored, the effect will be especially impressive.

Experience 3. Inertia of vision

The horse running on the movie screen, of course, is in no way dead. She is not dead on the television screen either. And in general, the “moving” objects in the frame are motionless - the frames themselves move, but at such a speed that the human eye does not replace their change. You can see movement on the screen because human vision has a certain inertia. In other words, any visual sensation persists for some time after the image that caused it disappears. Therefore, if the images change often enough, the visible image of the previous image smoothly passes into the subsequent image, and under certain conditions they can be seen simultaneously.

figure 4 shows a tattoo trope—a device that, by rotating it, can demonstrate the inertia of vision. If you rotate the tattoo trope quickly enough, figure on one side will align with figure on the opposite side. On one side of the cardboard square, glue figure birds, and on the other side - figure cells (Fig. 5). Bottom needle one end into the edge of the cardboard -

32

Place the end of the ruler in the same position where you placed the sheet of paper. Moving the ruler slightly, find such a distance from the projector that any part of the image is clearly focused on the white end of the ruler. Now get started: quickly move the ruler up and down, up and down.

Observe that those standing behind the projector see the full image of the color figure in the air at the place where you move the ruler. This happens because individual pieces of the image, directly visible only at those moments when they are projected onto the ruler, are retained by the eye for some time, enough for you to have time to move the ruler to show the next piece of the frame.

In the same way, a television picture is obtained. Only there is not a ruler, but an electron beam showing horizontal lines of the image one after another. And the whole picture is obtained thanks to the inertia of vision.

About torture 4. Upside-down cube

Using 12 wires or round sticks, make a cube frame as shown in figure 6. The effect will be better if all the material is shown

Fig. 6.

in matte black color. First make two quadrangular sides of the frame, and then fasten them with four sticks. Attach a longer stick - a holder - to one of the corners. Everything can be done with glue and then reinforced with black thread.

Hold the cube in front of you so that its far corner (at figure 6 is a solid arrow) is visible approximately in the middle of one of the sides of the cube (the line of sight at figure 6 is shown by a dotted arrow). Close your other eye.

Look, without looking away, at the far corner of the cube. Suddenly you will see how the cube “turns inside out” so that the far corner appears to be facing forward, the closest one. When the stick is rotated slowly, the cube “flips” in the direction opposite to the direction of rotation. Move the stick back and forth, and you will see how strange the upside-down cube behaves.

Get lost first to learn to see everything properly. It is better to conduct the experiment against the backdrop of a peephole wall. If this doesn’t work, try changing the background to rotate the cube, tilt it, or slightly deform it.

With a little fiddling, you will observe the effect of a complete absence of ivo. Remember that it only works when you look with one eye.

This effect is called the von Horibostl cube irony. When both eyes work together, they see depth so well that they cannot be fooled by this cube, but one eye will definitely make a mistake in determining which part of the cube is closer. This iriznyaya is one of the most interesting.

Some people have the effect of receiving it when observing it with their eyes. The main thing is to fix your gaze with both eyes on one (far) corner of the cube.

33

TASKS

Solutions to problems from the “Kvant Problem Book” can be sent no later than one and a half spaces after the corresponding issue of the journal goes out of print. The solution to each problem must be written on a separate sheet(s); at the end of each decision you must indicate your last name, first name and patronymic; six-digit postal code, home address; the class and school you attend.

M96. It is known about five positive numbers that if from the sum of any three of them we subtract the sum of the remaining two, then the difference will be positive. Prove that the product of all ten such differences does not exceed the square of the product of the given heel of numbers.
S. T. Berkolaiko

M97. In the trapezoid $ABCD$ with bases $AB=a$ and $CD=b$ there is a segment $A_1B_1$ connecting the midpoints of the diagonals. In the resulting trapezoid $A_1B_1CD$, the segment $A_2B_2$ is again drawn, connecting the midpoints of the diagonals, and so on (Fig. 1). Is it possible for a number to appear twice in the sequence of segment lengths $AB, A_1B_1, A_2B_2, \dots$? Will this sequence be monotonic (increasing or decreasing)? Is she aiming for some kind of limit?
A. L. Rosenmal

M98. Prove that in the table

1
1 1 1
1 2 3 2 1
1 3 6 7 6 3 1

where each number is equal to the sum of the three above it, each line (starting from the third) has an even number. Does each line (except the first two) contain a number divisible by three?

M99. In a triangle, $ABC$ side $AC$ is the largest. Prove that for any point $M$ of the plane $AM+CM$ is at least $BM$. In what cases is equality possible?

N. B. Vasiliev

Fig. 1.

Fig. 2.

M100. Prove that the sum of 45 numbers $tg 1^\circ + tg 5^\circ + tg 9^\circ + \dots + tg 173^\circ + tg 177^\circ$ is 45.
V. P. Beshkarev

F108. Ten ants decided to remove a straw from the table that was lying on it. How should they build if the force with which each of the ants can pull away the straw is slightly less than $1/10$ the frictional force acting on the straw when it is shuffled around the table? An ant cannot lift a straw either.
G. L. Komkin

F109. In experiment, it was found that the temperature of 142 °C ice water in a light state, suspended in the middle of the room, rose by 4 °C in half an hour. When there is ice in the vessel, it took 10 hours to pull it out. Based on this experiment, what is the specific heat of fusion of ice? The specific heat of water is 1 cal/e.

F101. If the voltmeter is connected in parallel to the upper resistance $r_1$ (Fig. 2), then it will show 6 $\theta$, if in parallel to the lower resistance $r_2$, then 4 $\theta$, and if it is connected to points $A$ and $B$, then it will show 12 $\theta$. What is the actual voltage drop across the resistances?

F111. There is a homogeneous cord with an explosive substance. The speed of propagation of the explosion reaction along the cord is $v$, the speed of propagation of the blast wave through the air is $c$. Find the shape of the line along which you need to place the cord so that the waves from all points of the cord arrive at a given point simultaneously.

Is it possible to do the same with an explosive surface and get a descending spherical wave with a high energy density?

F112. There is a stack of $k$ plane-parallel plates whose refractive indices are $n_1, n_2, \dots, n_k$. The thickness of each plate is $d$. How much will a ray incident on a plate with a refractive index of $n_1$ at an angle of $\alpha$ to the surface of the plate be displaced after passing through a stack of plates?

35

SOLUTIONS

In this issue we publish the solution to problems M50–M60 and the end of the solution problem M45.

M50

The vertices of a regular $n$-gon are shown in several colors (each with one color) so that points of the same color serve as vertices of a regular polygon. Prove that among these polygons find equal ones.

Suppose that the vertices of some regular $n$-gon are removed by coloring so that all identically shown vertices form different regular polygons: $m_1$-gon, $m_2$-gon, $m_3$-gon, …, where $m_1 < m_2 < m_3 < ...$. The smallest of these numbers $m_1$ will play a special role in our reasoning, and we will simply denote it by $m$.

Let us draw vectors from the center of the $O$ $n$-gon to all its vertices; Let’s denote them in order: $a_1, a_2, a_3, ..., a_n$ (Fig. 1a). Then

$$\vec{a}_1 \vec{a}_2 = \vec{a}_2 \vec{a}_3 = \ldots = \vec{a}_{n-1} \vec{a}_n = \vec{a}_n \vec{a}_1 = \frac{2\pi}{n}.$$

Here and below, by $\vec{ab}$ we denote the magnitude of the (smallest non-negative) angle at which the vector $a$ must be rotated counterclockwise so that it coincides with the vector $b$; always $0 \leqslant \vec{ab} < 2\pi$.

We will say that vectors $c_1, c_2, c_3, ..., c_q$ on the plane drawn from the point $O$ form a regular system if they have equal lengths and form equal angles with each other:

$$\vec{c}_1 \vec{c}_2 = \vec{c}_2 \vec{c}_3 = \ldots = \vec{c}_{q-1} \vec{c}_q = \vec{c}_q \vec{c}_1 = \alpha > 0.$$

We will need the following almost very significant fact (its proof is given at the end of the solution).

Lemma. The sum of the vectors forming a regular system is zero.

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Fig. 1. Imagine that the colored dots on figure $a$ are beads strung on a wire, arranged in a spiral. We look at this spiral from above, and therefore the two red beads $A'$ and $A''$ project for us into one point $A$. If we now fasten one of the beads - $A'$, and $A''$ against the wire for another 2 turns, then we will arrive at figure $b$, at which the angles between adjacent vectors have increased 3 times.

Before using the lemma, let’s do the following transformation with our vectors. We will assume that the vector $a_1$ goes to one of the vertices of the $m$-gon. Let’s simply denote it by $a$ and count all angles from it (in the positive direction, that is, counterclockwise). Let us denote by $b_i$ the vector that is obtained from $a$ by rotation by an angle $m \rightarrow aa_i$ (for each $i=1,2,\dots,n$). When processing $a_i \rightarrow b_i$, which we just constructed*) (at figure 1, $\delta$ such a mapping is depicted for $m=3$), the angles between the vectors increase by $m$ times; more precisely: if $m \rightarrow a_i a_j < 2\pi$, then $\rightarrow b_i b_j = m \rightarrow a_i a_j$. Therefore, this mapping again transforms the regular system of vectors with angle $\alpha < \frac{2\pi}{m}$ into a regular system of vectors. In particular, the sum of all vectors $b_1, b_2, \dots, b_n$ is equal to zero.

Let’s calculate this sum in another way: we first find the sum of vectors corresponding to the vertices of each color, and then add all these sums. Let’s first consider all the vectors $a_i$, look at the vertices of the $m_h$-gon, where $m_h > m$. This is a regular system of vectors with angle $\frac{2\pi}{m_h} < \frac{2\pi}{m}$. After mapping, it will again form a correct system, and the sum of the resulting vectors will be equal to zero.

Now consider the vectors, look at the vertices of the $m$-gon. The angle between adjacent vectors is equal to $\frac{2\pi}{m}$ in this case, and after mapping all these $m$ vectors coincide with the $a$ vector, so their sum will be $m$ (at figure 1 this is the case with red and black vectors).

So, we have a contradiction: on the one hand, the sum of all $b_i$ is equal to zero, on the other hand, it is equal to $m$. Therefore our assumption is incorrect.

*) For those who are familiar with complexes of numbers, note that if the point $O$ represents zero on the complex plane, and the end of the vector $a$ represents one, then this mapping is a raising to the power $m: z \rightarrow z^m$.

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Proof of the lemma. A physicist would say that this statement is obvious from the symmetry messages. One way to turn these messages into rigorous evidence is as follows. Let’s assume that $c_1 + c_2 + \dots + c_n = s$. Let’s rotate all vectors $c_1, c_2, \dots, c_n$ around point $O$ by angle $\alpha$. Then, of course, their sum will also rotate through the angle $\alpha$. But according to the condition, the entire system of $n$ vectors after rotation by an angle $\alpha$ will coincide with itself ($c_1$ will fall into place $c_2, c_2$ - into place $c_3, \dots, c_n$ - into place $c_1$). Therefore, the vector $s$ after rotation through the angle $\alpha$ should not change. This can only be the case if $s = 0$.

It can be proven that a regular system of $q$ vectors in the general case consists of $q = dr$ vectors drawn from the point $O$ to the vertices of a regular $r$-gon, so that at each vertex there are $d$ equal vectors (if we include here “doublegons”—pairs of opposite vectors, corresponding to $m = 2$); Moreover, $\alpha$ can be equal to any of the numbers $\frac{2\pi l}{r}$, where $l$ and $r$ are coprime, $0 < l < r$. We left this remark at the end, since formulaically in the proof we dispensed with a detailed explanation of how the correct system of vectors $b_i$ is structured. But, of course, understanding the proof “in essence”, it is useful to find out this.

The idea of ​​the solution to the problem presented here was proposed by A. Livshits (Leningrad). Having carefully understood this solution, we can prove that any partition of the vertices of a regular $n$-gon into several sets corresponding to (not necessarily different) regular polygons can be obtained in this way: first, partition the $n$-gon into several $m$-gons (for this it is necessary that $n$ is divisible by $m$), then partition one of the resulting regular polygons again into several regular polygons (see Fig. 1) and so on several times. For readers who want to understand this in more detail, we also advise you to consider the connection between our “one-dimensional” problem and the “two-dimensional” problem M3, the formulation of which was discussed in «Kvant» No. 7 for 1970 (pp. 54-55) and, first of all, to establish the equivalence of our problem M50 to the following: prove that the set of all integers can be divided into a finite number of arithmetic products (infinite in both directions), the differences of which are distinguished in pairs.

M51

Prove that if the product of three positive numbers is equal to one, and the sum of these numbers is strictly greater than the sum of their reciprocals, exactly one of these numbers is greater than one.

Let $a, b, c$ be the given numbers. Then $abc = 1$ and $(a - 1)(b - 1)(c - 1) = abc - ab - bc - ac + a + b + c - 1 = a + b + c - \frac{1}{a} - \frac{1}{b} - \frac{1}{c} > 0$, therefore, of the three numbers $(a - 1), (b - 1), (c - 1)$, two are negative and one is positive (all three cannot be positive, since $abc = 1$).

M52

The five segments are such that any three can be used to form a triangle. Prove that at least one of these triangles is acute.

A triangle can be formed from the segments $x \leq y \leq z$ if $z < x + y$, and it will be acute if $z^2 < x^2 + y^2$. Suppose that there are five segments $a_1 \leq a_2 \leq a_3 \leq a_4 \leq a_5$ such that any three form a non-acute triangle. Then

$$a_5^2 \geq a_4^2 + a_3^2, \quad a_4^2 \geq a_3^2 + a_2^2, \quad a_3^2 \geq a_2^2 + a_1^2,$$

and therefore

$$a_5^2 \geq 2a_3^2 + a_2^2 \geq 3a_2^2 + 2a_1^2 \geq 2a_1a_2 + a_1^2 = (a_2 + a_1)^2,$$

that is, $a_5 \geq a_2 + a_1$ and, therefore, it is impossible to form a triangle from the segments $a_1, a_2, a_5$.

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M53

In triangle ABC, through the midpoint M of side BC and the center O of the inscribed circle in this triangle, draw a straight line MO, which intersects the height AH at point E. Prove that the segment AE is equal to the radius of the inscribed circle.

Let us denote the lengths of the sides of the triangle opposite the vertices A, B, C by a, b, c, respectively. We can assume that b>c (b≠c, since otherwise the lines MO and AH would not intersect, but coincide). Let us drop the perpendicular OP=r to BC (Fig. 2). Then

$$MC = \frac{a}{2}, \quad PC = \frac{a+b-c}{2}, \quad HC = \frac{a^2+b^2-c^2}{2a}.$$

$$\frac{EH}{OP} = \frac{HM}{PM} = \frac{HC-MC}{PC-MC} = \frac{2HC-a}{b-c} = \frac{a^2+b^2-c^2-a^2}{a(b-c)} = \frac{b+c}{a},$$

$$\frac{AE}{r} = \frac{AH}{r} - \frac{EH}{r} = \frac{a+b+c}{a} - \frac{b+c}{a} = 1.$$

Another, geometric solution to this problem is based on the following lemma: if we extend the radius PO to the diameter PQ of the inscribed circle and draw the straight line AQ to the intersection at point R on the side BC, then the segments BP and CR will be equal. We leave it to the reader to prove and use this lemma.

M54

Two equal rectangles are located so that their contours intersect at eight points. Prove that the area of ​​the common part of the rectangles is greater than half the area of ​​each of them.

Let the lengths of the sides of the rectangles a and b. Note that on the sides of each of the rectangles there are exactly two points of intersection with two adjacent sides of the other. (It is easy to prove that if there are 8 points of intersection in total, then each side must contain at least two points and that the intersection of the sides of one rectangle with two parallel sides of another is impossible). Let A and C be the points at which the sides of different rectangles, equal to a, intersect; B and D are the points where sides equal to b intersect. Then, obviously, the segment AC serves as the bicsettrix of the angle between sides of length a passing through point A (to prove it, it is enough to drop perpendiculars from point C to these sides and consider the pair of equal triangles formed in this case). Similarly, BD is the bicsetris of the angle between sides of length b passing through point B. Therefore, AC⊥BD, and therefore the area

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and check that for each individual $l$ the sums of terms of the form $X^{k-l}y^l$ on the right and left sides of the equality are equal to $^*$) (we do not write the coefficient $C_k^l$):

| | Numbers with an odd sum of digits | Numbers with an even sum of digits |
| :— | :— | :— |
| Sum of terms of the form $X^k$ | $5 \cdot 10^{n-1} \Sigma A^k + 5 \cdot 10^{n-1} \Sigma B^k$ | $5 \cdot 10^{n-1} \Sigma A^k + 5 \cdot 10^{n-1} \Sigma B^k$ |
| Sum of terms of the form $X^{k-l}y^l$ $1 \leq l \leq k-1$ | $\Sigma A^{k-l}b^l + \Sigma B^{k-l}a^l = \Sigma b^l \Sigma A^{k-l} + \Sigma a^l \Sigma B^{k-l} = s_l (\Sigma A^{k-l} + \Sigma B^{k-l})$ | $\Sigma A^{k-l}a^l + \Sigma B^{k-l}b^l = \Sigma a^l \Sigma A^{k-l} + \Sigma b^l \Sigma B^{k-l} = s_l (\Sigma A^{k-l} + \Sigma B^{k-l})$ |
| Sum of terms of the form $y^k$ | $5 \Sigma b^k + 5 \Sigma a^k$ | $5 \Sigma a^k + 5 \Sigma b^k$ |

Note that our initial calculation for $n=2$ can be written using similar notation as follows:
$$\Sigma (A+b) + \Sigma (b+a) = 5\Sigma A + 5\Sigma a + 5\Sigma b + 5\Sigma a, \quad \Sigma (A+a) + \Sigma (B+b) = 5\Sigma A + 5\Sigma a + 5\Sigma B + 5\Sigma b$$
at $k=1$ only the first and last terms corresponding to $l=0$ and $l=k$ will remain).

It is easy to see that the statement of the problem is valid not only in decimal, but also in any other number system with the base $d$, where $d$ is an even number (think where in our solution the parity of the base $d=10$ is used). If we take $d=2$, we get the following curious series of equalities:
$$\begin{array}{l} 1+2=3 \\ 1+2+4+7=3+5+6 \\ 1^2+2^2+4^2+7^2=3^2+5^2+6^2 \\ 1+2+4+7+8+11+13+14=3+5+6+9+10+12+15 \\ 1^2+2^2+4^2+7^2+8^2+11^2+13^2+14^2=3^2+5^2+6^2+9^2+10^2+12^2+15^2 \\ 1^3+2^3+4^3+7^3+8^3+11^3+13^3+14^3=3^3+5^3+6^3+9^3+10^3+12^3+15^3 \end{array}$$

M56

Four ones and five zeros are written in random order on the circle. Then a zero is written between two identical numbers, a one is written between different numbers, and the original numbers are erased. Prove that no matter how many times we repeat the process, we will never get a set of $N$ zeros. Indeed, suppose that at some $t$th step we first received a set of only zeros. Then at the $(t-1)$th step all $N$ digits were the same, and not all were equal to 0, so they were all equal to 1, and at the $(t-2)$th step every two adjacent digits should be different, but since $N$ is odd, then such an arrangement of 0 and 1 on the circle does not exist.

At the same time, we answered (so far only for odd $N$) the question that remained unanswered in the solution M19 («Kvant» No. 11, 1970, pp. 37-38); in this problem, zeros are called “resting cells”, ones are called “excited cells”, and the correct

*) We do not need to know the formulas for calculating $C_k^l$ (the so-called “binomial coefficients”). The only important thing is that these are numbers that do not depend on the variables $X$ and $y$.

**) Those who solved this problem before may have noticed that we swapped the words “ones” and “zeros” in the wording. This, of course, does not change the existence of problem M19 and M56, which will be discussed below.

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the transitions are exactly the same. In these terms, our result is formulated as follows: if $N$ is odd, and at the initial moment not all cells are excited (and not all are at rest), then the excitation will never die out.

What will happen with other $N$? To find out this, we use the following lemma (it can be proved by induction on $k$; it is valid for both the problem on a line and for a circle); if at $t=0$ in place $2^{k-1}-M$ there are identical numbers on either side of a given number $a$, then after $t=2^k$ steps there will be a zero in place $a$, and if e are different, then one. It immediately follows that if $N=2^m$, where $m$ is a natural number, then from any set in $N$ steps a set of only zeros will be obtained.

Let now $N=2^m L$, where $L$ is odd. Let’s follow the changes that occur every $T=2^m$ steps. For each $L$ digits spaced $2^m$ apart (“at the vertices of a regular $L$-gon”), these changes will be exactly the same as if only these $L$ digits were placed around the circle, the same process was repeated with them, and we followed each step. From here it is not difficult to obtain the answer in the most general case: the “all zeros” state is achieved if and only if the initial sequence of 0 and 1 was periodic with period $2^m$ (see Fig. 4).

M57

a) Find a number $k$ that is divisible by 2 and 9 and has a total of 14 divisors (including 1 and $k$).

b) Prove that if you replace 14 with 15, then the problem will have several solutions, but if you replace 14 with 17, there will be no solutions at all.

Let $M=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_r^{\alpha_r}$ be the decomposition of the number $M$ into prime factors; for the definition we will assume that $\alpha_1 \geqslant \alpha_2 \geqslant \dots \alpha_r \geqslant 1$. The divisors of the number $M$ have the form $p_1^{\beta_1}p_2^{\beta_2}\dots p_r^{\beta_r}$, where each $\beta_i$, independently of the others, can belong to the values 0, 1, …, $\alpha_i$, and therefore the total number of divisors $M$ (including 1 and $M$) is equal to $(\alpha_1+1)(\alpha_2+1)\dots(\alpha_r+1)$ (see the solution to Problem 1 in the article “Combinatorics” in «Kvant» No. 1, 1971, p. 19).

a) 14 can only be factored one way: $14=7 \cdot 2$, so if $M$ has 14 factors, then $r=2$, $\alpha_1=6$, $\alpha_2=1$. We also know that 3 is included in the expansion $M$ with an exponent greater than 1, and 2 - with an exponent greater than 0. It follows that $M=3^6 \cdot 2=1458$.

b) For 15 divisors, the possible answers are $M=3^2 \cdot 2^4$ and $M=3^4 \cdot 2^2$, and 17 divisors can only have numbers of the form $M=p^{16}$, where $p$ is a prime.

M58

There are three bottoms on the plane, intersecting at one point. A dot is marked on one of them. It is known that the primaries are bissertrices of a certain triangle, and the marked point is one of its vertices. Construct this triangle.

First decision. Suppose that $O$ is the intersection point of bissertrices $AK$, $BL$ and $CM$ of triangle $ABC$ (Fig. 5); Then

$$\angle MOB = \angle OCB + \angle OBC = \frac{1}{2} (\angle ACB + \angle ABC) = \frac{\pi}{2} - \angle CAK,$$

that is, $\angle CAK = \frac{\pi}{2} - \alpha$. Therefore, if the lines $AK$, $BL$, $CM$ and the point $A$ are given, then by constructing angles equal to $\varphi = \frac{\pi}{2} - \alpha$ on one and the other side of the ray $AO$. We will find the required vertices $B$ and $C$ (provided that $\varphi > 0$.

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Fig. 5. Fig. 6.

φ<β and φ<γ). It is also necessary to prove that the constructed triangle ABC has lines BL and CM along the bisectors. Angles ABL and ACM are easy to calculate—they are equal to the correspondence of $\frac{\pi}{2} - \beta$ and $\frac{\pi}{2} - \gamma$. The difficulty lies only in proving that $\Rightarrow LBC = \frac{\pi}{2} - \beta$ and $\Rightarrow MCA = \frac{\pi}{2} - \gamma$ (although it is clear that their sum is $\alpha$); it can be preflighted, for example, like this: if $\Rightarrow LBC < \Rightarrow ABL$, then the bisector of the angle ABC intersects the segment AO, therefore the bisect of the angle ACB also intersects it, and therefore $\Rightarrow MCB < \Rightarrow ACM$, so the sum $\Rightarrow LBC + \Rightarrow MBC$ is less than $(\frac{\pi}{2} - \beta) + (\frac{\pi}{2} - \gamma) = \alpha$. In the same way it can be shown that the case $\Rightarrow LBC > \Rightarrow ABL$ is impossible.

Second solution. Let us construct the points $A'$ and $A''$, symmetrical to the given point $A$ with respect to bisectors not passing through $A$. It is clear that both points $A'$ and $A''$ must lie on the line BC - on the side of the desired triangle ABC (or on its extension). By drawing a straight line through $A'$ and $A''$, we will thereby find the required points B and C (Fig. 6).

Note that although the second solution is more effective, with this approach it is more difficult to write down the conditions under which the problem has a solution *). These conditions are: $\alpha < \frac{\pi}{2}$, $\beta < \frac{\pi}{2}$, $\gamma < \frac{\pi}{2}$ (since $\alpha + \beta + \gamma = \pi$, they can also be written like this: $\alpha + \beta > \frac{\pi}{2}$, $\beta + \gamma > \frac{\pi}{2}$, $\gamma + \alpha > \frac{\pi}{2}$). If they are fulfilled, then the solution is unique. Think about how you can obtain these conditions for each of the above solution methods.

M59

There are several weights with weights 1 e, 2 e, 3 e,…, n e. They need to be divided into three piles of equal weight. At what $n$ was this possible?

Let us show that this can be done if and only if $n > 3$ and one of the numbers $n$ or $n+1$ is divisible by 3 (that is, with $n$ equal to 5, 6, 8, 9, 11, 12, 14, 15, …).

*) It may turn out that two of these lines are bisectors of external, and not internal, angles of the constructed triangle; this does not eliminate us.

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The necessity of these conditions, since the total weight of the mushroom $1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$ must be divided by 3.

For the proof, it is enough first that partitioning is possible with $n$ equal to 5, 6, 8 and 9 (this is shown in Fig. 7). All other values ​​of $n$ that interest us are obtained from these four by adding a certain number of sixes, and any group of six consecutive integers $a+1, \dots, a+6$ is easily divided into three pairs of numbers giving equal sums: $a+1$ and $a+6$, $a+2$ and $a+5$, $a+3$ and $a+4$.

Fig. 7.

M60

Let’s consider all natural numbers whose decimal notation includes only the digits 1 and 0. Divide these numbers into two groups so that the sum of any two different numbers from the same group contains at least two units in its decimal notation.

Let us assign to one group all the numbers whose notation has an odd number of ones, and to the other group all the numbers whose notation has an even number of ones. Then it is clear that two numbers cannot fall into one group; the cession is excellent only in one digit, therefore such a division satisfies the requirements of the problem (units appear in those digits of the sum where one number has a 1 and the other does not).

Many readers have provided reasoning to prove that the described partition:

first group: 1, 10, 100, 111, 1000, 1011, 1101, 1110, …
second group: 11, 101, 110, 1001, 1010, 1100, 1111, 10001, …

the only one that satisfies the conditions of the problem.

More complete solutions to the problems discussed above were sent to us by V. Didenko from cela Vesely Kut, Odessa region, E. Turkevich from Chernivtsi, A. Rubin from cela Ukladovka, Vinnitsa region, A. Slinkin from Moscow, Y. Tomsinsky from Leningrad, M. Perelmuter from Kyiv, O. Khudaverdyan from Yerevan.

M45

(Continued. For the beginning of the solution, see «Kvant» No. 7, p. 30.)
We just have to prove this fact.

Theorem. From 2$p$-1 any integers we can choose $p$, the sum of which is divisible by $p$ (an arbitrary prime number).

We present here the proof suggested by S. Voronin (Moscow).

It is clear that in the proof we can consider all numbers “modulo $p$”, that is, we are only interested in which of the remainders 0, 1, 2, …, $p-1$ gives this or that number when divided by $p$. It is convenient to imagine (especially when talking about “addition modulo $p$”) that these $p$ remainders are considered in a circle (Fig. 8).

Lemma. Let data $r$ integers $b_1, b_2, \dots, b_r$; $0 < b_i < p$ for all $i=1, 2, \dots, r$ and $0 < r < p$ (prime). Then from these numbers it is possible to compose at least $r+1$ sums that give different remainders when divided by $p$ (in this case, it is allowed to be the sum of the “empty set of terms”, which is considered equal to zero, the sum of one term, of two,…, of all $r$ terms).

Proof. For $r=1$ this is obvious: the sums give remainders 0 and $b_1$. Suppose this is true for $r=k<p-1$ and false for $r=k+1$, and we arrive at a contradiction. Let the sums of $k$ terms $b_1, b_2, \dots, b_k$ give $k+1$ different remainders 0, $s_1, \dots, s_k$. Then,

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Fig. 8. To find the sum of $a+b$ modulo $p$, you need to count $b$ units from $a$ counterclockwise (this figure, corresponding to $p=11$, shows the sum 8+8, which modulo 11 is equal to 5).

Since after adding $b=b_{k+1}$ the number of different sums should not increase, all sums

$$0+b, s_1+b, \ldots, s_k+b$$

(modulo $p$) are contained in the set

$$\{0, s_1, \ldots, s_k\}.$$

In other words, if you add $b$ to any element of this set, you will again get an element of the same set (try to imagine such a set on figure 8). Thus, this set depends on and contains the elements 0, $b$, $2b$, $3b$, $\ldots$, $(p-1)b$. But it is clear that all these elements are different (in absolute value): the difference

$$ib-jb=(i-j)b,$$

where $0<i-j<p$ and $0<b<p$ cannot be divided by $p$ since $p$ is prime. Thus, we have proven that the set $\{0,s_1, \ldots, s_k\}$ contains all $p$ distinct elements, although we assumed that

$$k+1<p.$$

The lemma is proven.

Proof of the theorem. Let

$$a_1 \leqslant a_2 \leqslant \ldots \leqslant a_p \leqslant a_{p+1} \leqslant \ldots \leqslant a_{2p-1}$$

— remainders from dividing the given $2p-1$ numbers by $p$. Let’s consider $p$ more such numbers:

$$a_{p+1}-a_2, a_{p+2}-a_3, \ldots, a_{2p-1}-a_p.$$

(*)

If any one of them is zero, for example

$$a_{p+1}-a_{l+1}=0,$$ is then $a_{l+1}=a_{l+2}=a_{l+3}=\ldots=a_{l+p}$, and the sum of the corresponding $p$ numbers is divisible by $p$. It remains to consider the case when all numbers (*) are different from zero.

Let’s find the remainder $x$ when the sum $a_1+a_2+\ldots+a_p$ is divided by $p$. If $x=0$, then everything is clear. If $x\neq 0$, then, using the lemma, we can compose from the differences (*) sums that give the remainder $p-x$ when divided by $p$. By adding the appropriate differences to $a_1+a_2+\ldots+a_p$ and making obvious cancellations, we get the sum of $p$ terms dividing by $p$.

Using the result problem M45, it is not difficult to establish that the statement “From any $a$ integers we can choose $b$ numbers whose sum is divisible by $c$” (where $a, b, c$ are specific natural numbers) is true if and only if $b$ is divisible by $c$ and $a \geq b+c-1$.

N. B. Vasiliev

In this issue we publish solutions to problems F66—F71.

F66

To the battery with e. d.s. 9 V with an unknown internal resistance, an amperor and a voltmeter (Fig. 9) are connected in series. The resistances of the devices are unknown. If a resistance is connected in parallel with the voltmeter (its value is also unknown), then the ampere reading is doubled, and the voltmeter reading is doubled. What was the voltmeter reading after connecting the resistance?

The voltmeter shows the presence of voltage on itself, and the amperor shows the flow through it. According to Ohm’s law for a complete circuit e. d.s. battery is equal to the sum of the voltage in the sections of the circuit: $U$ - on the voltmeter and $U_1$ - on the internal resistance of the source and the ampere:

$$E=U+U_1.$$

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Fig. 9. Fig. 10.

After connecting the resistance, the voltage drop across the voltmeter and the resistance connection in parallel with it became equal to $U' = \frac{U}{2}$ (voltmeter readings). At the same time, the voltage drop across the source resistance and ammeter doubled and became equal to $U'_1 = 2U_1$. Indeed, the resistance of this section has not changed, but the current flowing through it has doubled. Pozotomu

$$E = \frac{U}{2} + 2U_1.$$

(2)

Solving equations (1) and (2) together, we find that $U = 6\ \theta$. Therefore, after connecting the resistance, the voltmeter showed voltage

$$U' = \frac{U}{2} = 3\theta.$$

The correct solution was sent by N. Fedin (Omsk), S. Ivanov (Moscow), E. Gubarev (Moscow), V. Sytsev (Alga Aktobe region), N. Smirnov (Gorky), A. Anishchenko (Sarapul, Udmurt Autonomous Soviet Socialist Republic), A. Braverman (Perm), A. Senyavin (Moscow), V. Pomatsev (Gorky), V. Kalinichev (Moscow), M. Praeger (Tomsk).

F67

The buffer device (Fig. 10) consists of a rod $A$, a spring $B$ placed on the rod, and a guide bushing $C$. The bushing can move inside a channel made in a solid wall $D$. When the bushing moves between its outer surface and the wall, a constant friction force $F_{Тр}$ acts. The bushing rod and the spring move along the rod without friction.

A ball of mass $m$ is placed on the end surface of the rod $A$, having a speed $v_0$ before the collision. At what speed will the ball fly away?

The masses of the rod, spring and bushing are negligible compared to the mass of the ball. Spring stiffness coefficient $k$.

When the ball is removed, the rod begins to move, releasing the spring. In this case, a force $F = kx$ acts on the bushing from the side of the deformed spring, where $x$ is the deformation of the spring. If the initial speed of the ball is high enough that the deformation of the spring reaches a maximum value $x_{\max}$, at which the spring acts on the bushing with a force equal to the friction force of the bushing against the wall. After this, the bushing begins to move, and during the movement of the bushing the spring will not deform: the force pushing it off does not change and is equal to $F_{Тр}$. Since when the bushing moves, a friction force acts on it, the kinetic energy of the system turns into heat, and after some time the ball and rod are left. Then the spring begins to straighten. This, of course, will not cause the bushing to move - the force with which the action.

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a spring blows on it, less than $F_{\tau p} = kx_{\max}$, since $x < x_{\max}$. So there will be no further energy losses in the system. When the spring straightens, its potential deformation energy $W_{\pi} = \frac{kx_{\max}^2}{2}$ will turn into the kinetic energy of the ball.

Reception, the potential energy of the ball is zero.
Let us now write down the law of conservation of energy

$$\frac{mv^2}{2} = \frac{kx_{\max}^2}{2},$$

where $v$ is the speed of the ball after it bounces off the wall.

Considering that $x_{\max} = \frac{F_{\tau p}}{k}$, we get $v = \sqrt{\frac{F_{\tau p}^2}{km}}$.

At a sufficiently high speed $v_0$, the speed of the ball that bounces off the rod does not depend on its initial speed. The system “eats” the entire energy reserve above the value $W = \frac{kx_{\max}^2}{2}$.

Let us now find out what “sufficiently high speed $v_0$” means. This is the speed at which the force acting from the spring on the bushing reaches a value of $F_{\tau p}$, and, therefore, the deformation of the spring reaches a value of $x_{\max} = \frac{F_{\tau p}}{k}$. It is clear that for this the initial kinetic energy of the ball must be greater than the value $\frac{kx_{\max}^2}{2} = \frac{F_{\tau p}^2}{2k}$:

$$\frac{mv^2}{2} > \frac{F_{\tau p}^2}{2k}.$$

This condition gives us the limit of “sufficiently high speed”:

$$v_0 > \sqrt{\frac{F_{\tau p}^2}{km}}.$$

How will a collision occur if the initial speed of the ball is less than this value or equal to it?

In this case, the spring deformation is less than $x_{\max}$ and the spring will not move the bushings. This means that there will be no loss of energy and the ball will bounce back, having the same kinetic energy, and therefore the same speed, as before the collision. So,

$$v = \begin{cases} v_0 \text{ при } v_0 \leqslant \sqrt{\frac{F_{\tau p}^2}{km}}, \\ \sqrt{\frac{F_{\tau p}^2}{km}} \text{ при } v_0 > \sqrt{\frac{F_{\tau p}^2}{km}}. \end{cases}$$

The correct solution was sent by P. Mkrtchyan (Yerevan), Kutsek (Rzhev), N. Fedin (Omsk), I. Grishik (David-Gorodok, Brest region, BSSR), A. Vyrodov (Gagra), P. Zaitsev (Gagra), M. Premer (Tomsk).

F68

The stacks of the vessel in which the gas is located have a temperature of $T$. Gas temperature $T_1$. In which case is the gas pressure on the sides of the vessel greater: when the sides of the vessel are colder than the gas ($T < T_1$), or when the temperature ($T > T_1$)?

The temperature of a gas is determined by the average kinetic energy of its molecules: $\frac{mv^2}{2} = \frac{3}{2} kT$ ($k$ is Boltzmann’s constant). This means that than

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The higher the temperature of the gas, the greater the average speed of movement of its molecules, that is, the greater the average momentum of the molecule.

If the temperature of the wall coincides with the temperature of the gas, then the molecule, hitting the wall, changes its momentum $p_0$ to $-p_0$. This means that the change in momentum is $2p_0$.

When $T > T_1$, the gas heats up. This means that gas molecules bounce off the wall with a higher speed than they fly into, and therefore with a greater impulse (Fig. 11, a). The change in momentum in this case will be greater than $2p_0$.

If $T_1 < T$, then the gas cools, that is, the gas molecules bounce off the wall with less momentum than they hit it (Fig. 11, b). It is clear that in this case the change in the momentum of the molecule is smaller than in the case of $T_1 > T$. Since, in accordance with Newton’s II law, the change in momentum is proportional to the average force acting on the molecule from the side of the wall, and in accordance with Newton’s III law, the average force acting on the molecule is numerically equal to the average force acting on the wall, then at $T_1 < T$ the gas pressure on the wall is greater than at $T_1 > T$.

The correct solution was sent by V. Chupin (Torzhok), N. Fedin (Omsk), V. Kalintsev (Moscow), S. Ivanov (Moscow).

The shape of the liquid surface must be such that the reaction force N of the surrounding liquid acting on an element of the liquid volume with mass $\Delta m$ at the surface is directed perpendicular to the surface. Otherwise, the force N will have a component directed along the surface of the liquid, and the liquid will not be able to be in equilibrium.

Let us now look at how the force N is directed. It must balance two forces: the force $F_1$ of the Earth’s attraction, directed towards the center of the Earth, and the additional force $F_2$ of the ball’s attraction, directed towards the center of the ball (at figure 12 the force $F_2$ is made disproportionately large compared to the force $F_1$). The force N must be numerically equal and opposite in direction to the forces $F_1$ and $F_2$ equal to Q. This means that it is deviated to the right from the vertical and the surface of the water above the ball is convex.

Let us estimate the magnitude of the “swelling” of the water surface. Consider a point B lying at a small distance x from the vertical AO passing through the center of the ball ($x \ll r$, where r is the radius of the ball), O is the center of curvature and BO = R is the radius of curvature of the water surface.

$Q \approx F_1$ (since $F_2 \ll F_1$), and $Q \sin \alpha = F_2 \sin \beta$. Since the angles α and β are small ($x \ll r$, $x \ll R$), then

$$\sin \alpha \approx \mathrm{tg} \alpha \approx \frac{x}{R}$$

Pozny can be considered that

$$Q(x/R) = F_2(x/r)$$

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or
$$R = r \frac{Q}{F_2} = r \frac{F_1}{F_2}.$$

But
$$F_1 = \gamma \frac{\left( \frac{4}{3} \pi R_3^3 \rho \right) \Delta m}{R_3^2} = \frac{4}{3} \pi \gamma r R_3 \rho \Delta m$$
($R_3$ is the radius of the earth, $\rho$ is its density), and
$$F_2 = \gamma \frac{\frac{4}{3} \pi r^3 (\rho_1 - \rho_0) \Delta m}{r^2} = \frac{4}{3} \pi \gamma r (\rho_1 - \rho_0) \Delta m$$
($\rho_1$ is the density of the ball, $\rho_0$ is the density of water). That’s why
$$R = r \frac{\frac{4}{3} \pi \gamma R_3 \rho \Delta m}{\frac{4}{3} \pi \gamma r (\rho_1 - \rho_0) \Delta m} = R_3 \frac{\rho}{\rho_1} \approx R_3 \frac{\rho}{\rho_1}$$
(since $\rho_1 \gg \rho_0$). Substituting the values ​​$R_3 = 6,4 \cdot 10^6 m$, $\rho = 5,5 e/cm^3$ and $\rho_1 = 11,3 e/cm^3$ here, we find $R \approx 3,2 \cdot 10^6 m$.

Assuming that the swelling extends to a distance $x$ from the point $A$ (this estimate is highly dependent!), we find its height
$$h = x \lg \alpha \approx \frac{r^2}{R}.$$

If the radius of the lead ball is 1 m, then $h \approx 3 \cdot 10^{-7} m$.

This means that the swelling effect cannot be observed experimentally even with the help of integration of methods. After all, the length of light is several thousand angstroms ($1 \text{Å} = 10^{-10} m$), that is, greater than the magnitude of the rise.

The correct solution was sent by $H$. Fedin (Omsk).

F70

A cyclist easily develops a traction force of 10 kg. The force does not exceed 5 kg. It would seem that in a few hours a bicycle can reach escape velocity. However, this has not yet been possible. Why?

The force with which a bicycle can press on the pedals is not constant and depends on the speed of the bicycle (Fig. 13). It is maximum at $v = 0$ and becomes equal to zero at a speed at which the pedal rotation period becomes equal to the cycling reaction speed. After this, the cycling speed can no longer increase.

Many readers argue this way: the power that a bicycle develops cannot exceed a certain maximum value. But

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$N = F \cdot v$, so the speed value cannot become greater than

$$v = \frac{N_{\text{max}}}{F - F_{\text{trp}}}.$$

This, of course, is true and is due precisely to the fact that the dependence of the force with which a person presses on the pedals on the speed of movement is as shown in figure 13.

F71

There is a uniformly charged segment $AB$. What is the direction of the electric field created by this segment at point $C$: along the median of triangle $ABC$? according to his bissertika? in height? on none of these lines?

The field voltage is a vector, and in our problem there are only four selected directions: along the median, bissertic or altitude of the triangle $ABC$ and parallel to the segment $AB$. The field direction vector must be directed in one of these directions. It is clear that it cannot be parallel to the segment $AB$: the direction of the field of the charged segment is equal to the sum of the directions of the fields of the parts of the segment, and they are all directed towards the segment or away from it (depending on the sign of the charge of the segment).

The height of the triangle $ABC$ is also not suitable: for example, if the point $C$ is located away from the segment, at a distance much greater than the length of the segment, then it is clear that the direction of the field at the point should be directed towards the segment $AB$, and not perpendicular to it.

So, we are left with the median and bissertis. the directionality $E$ of the field of a segment is equal to the sum of the directions of the fields of its parts. In order for the vector $E$ to be directed along the median, it is necessary that segments of equal length $\Delta x$, located at equal distances from the middle of the segment, create the same field at point $C$.

Only in this case the sum of vectors $\Delta E$ will be directed along the median of the triangle $ABC$. However, this is impossible: segments of equal length have the same charge $\Delta Q = \sigma \Delta x$ (σ is the linear charge density, that is, the charge per unit length of the segment), and the distances to the point $C$ are different.

This means that the vector $E$ is directed along the bisector of the triangle $ABC$.
Let’s make sure of this. Let us find the directionality of the field of the segment $\Delta x$ such that the straight line drawn from it to the point $C$ forms an angle $\alpha$ with the altitude of the triangle and $\Delta x \ll r$ (Fig. 14). $\Delta E = \frac{\Delta x \sigma}{r^2}$. If the segment $\Delta x$ is visible from the point $C$ at an angle $\Delta \alpha$, then $\Delta x_1 = r \Delta \alpha$ (the angle $\alpha$ is small, so $tg \frac{\Delta \alpha}{2} = \frac{\Delta \alpha}{2}$, and $\frac{\Delta x_1}{2} = r tg \frac{\Delta \alpha}{2}$) and $\Delta x = \frac{\Delta x_1}{\sin(90^\circ - \alpha)} = \frac{r \Delta \alpha}{\cos \alpha}$. Therefore $\Delta E = \frac{r \Delta \alpha \sigma}{r^2 \cos \alpha} = \frac{\Delta \alpha \sigma}{r \cos \alpha}$. But $r \cos \alpha = h$.
That’s why
$$\Delta E = \frac{\Delta \alpha \sigma}{h}.$$

The field voltage of the segment, which is visible from the point $C$ at an angle $\Delta \alpha$, does not depend on the angle $\alpha$. All segments that are visible from point $C$ at the same angle create a field of the same size in it. But for any segment $\Delta x$ located to the left of the bisector $CD$ of triangle $ABC$, one can find a segment to the right of it such that it is visible from the point $C$ at the same angle $\Delta \alpha$ as the segment $\Delta x$, and the angle $\beta$ is formed by the primo connecting it with point $C$, the same as the segment $\Delta x$. Since these segments create the same field at point $C$, the vector $\Delta E$ of the sum of vectors $\Delta E_1$ and $\Delta E_2$ of the directional fields of the segments is directed along the diagonal of the rhombus built on the vectors $\Delta E_1$ and $\Delta E_2$, that is, along the bisector $CD$. This is true for any angle $\beta$, that is, for any pair of segments. This means that the directivity vector of the field of the segment $AB$ is directed along the bisector of the triangle $ABC$.

The correct solution was sent by N. Fedin (Omsk) and I. Grichik (David-Gorodok, Brest region).

I. Sh. Slobodetsky.

To “Problems on squared paper”

1. Add two dots to the cross from each of the four ends. The number of possible squares with vertices at the highlighted nodes will increase by 10 (four small red squares (see Fig.]), four squares whose pattern is colored black; two squares, one of which is highlighted blue).

The cross in the problem can be considered as an additional row of nodes to the central square, allowing only one square. It allows the construction of $10 \cdot 8 + 1 = 81$ different squares.