Kvant 1972, No. 3 · Pages

KEANT
3
MARCH
1972
POPULAR SCIENTIFIC PHYSICS AND MATHEMATICS JOURNAL
USSR ACADEMY OF SCIENCES and USSR ACADEMY OF PEDAGOGICAL SCIENCES

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

Editorial collection:

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

On the first page of the cover there is an illustration of one of the proofs of the Pythagorean theorem. The proof is given in the article by V. N. Berezin “Pythagorean Theorem” (see page 18).

In figure, pictured on the left, all three triangles surrounded in yellow are equal. The height in one of them is highlighted in blue. Due to the fact that the selected angles are equal, they are vertical, and the red segment lies on the same straight line as the blue one. (This is the key to another way to prove the Pythagorean theorem, located on the fourth page of the cover).

Head of the editorial office L. V. Chernova. Chief artist A. I. Klimanov
Art editor O. N. Yakovleva. Proofreader V. P. Sorokina
Publishing house “Nauka” Main editorial office of physical and mathematical literature
117071, Moscow, V-71, Leninsky Prospekt, 15. Tel. 234-08-11, 234-07-93

Submitted for printing on 11/X 1971. Signed for printing on 28/I 1972. Paper 70×100¹/16
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MANUSCRIPTS ARE NOT RETURNED

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FOUNDED
IN
1970

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

PUBLISHING HOUSE SCIENCE
MAIN EDITORIAL OFFICE OF PHYSICAL AND MATHEMATICAL LITERATURE

IN THE ROOM:

3 To the 200th anniversary of the death of Isaac Newton A. Einistein
6 Dynamic programming M. I. Reitman
14 Inert mass Ya. A. Smorodinsky
18 Pythagorean theorem V. N. Berezin
22 Why wires are pulled by L. G. Aslamazov
24 Surrounding the Ten A.P. Savin
28 Z planietry - theorem, in I. M. Yael
stereometry - an unsolved problem

LABORATORY “QUANTA”
34 Artificial solar shading R. Wood

PROBLEM “QUANTUM”
36 Competition Winners «Kvant»
38 Problem M131—M135; F133—F137
40 Solutions to problems M89—M95; F100—F105

APPLICANT PRACTICUM
50 Learn to work with logarithms A. Ya. Margupis, B. A. Radunsky
Options for entrance exams in mathematics 1971
56 Impulse. Law of conservation of momentum I. A. Zaitsev
Options for entrance exams in physics in 1971 at the Moscow Institute of Physics and Technology V. E. Belonuchkin

REVIEWS, BIBLIOGRAPHY
65 A book about the most massive natural phenomena I. E. Evgeniev

INFORMATION
66 State Prizes of 1971 V. A. Leshkovtse
69 ANSWERS, GUIDELINES, SOLUTIONS

COLLECTOR’S CORNER
72 Stamps dedicated to Isaac Newton A. V. Altykis

“QUANT” FOR JUNIOR SCHOOLCHILDREN
(3 pages region)
MIXTURE (pages 27, 33, 55)

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To the 200th anniversary of his death
Isaac Newton
A. Eystein

March 1972 marks the 245th anniversary of the death of the great English physicist and mathematician Isaac Newton. In this regard, we post a short article by another great physicist, Albert Einstein, published in 1927. The article is given with slight abbreviations. The full text of the article was published in Russian in the fourth volume of the collected scientific works of Einstein (Nauka publishing house, 1967).

It is not difficult to characterize Newton’s scientific activities, which forever ensured his special place in the history of human spiritual development. Newton was the first to try to formulate elementary laws that determine the time course of a wide class of processes in nature with a high degree of completeness and accuracy. His law of motion, together with the law of gravity, governs the movement of all celestial bodies, occurring under the influence of forces of mutual acceptance. Thus, Newton fulfilled the dreams of the ancient material philosophers - Democritus and Epicurus, who believed that there should be a fundamental interconnection of all physical phenomena without exception. After these successes, was there any doubt left that the development of all material phenomena in general produces the necessary regularity, which could be compared with the winding of a clock. In addition, it became obvious that the processes of thinking must be inextricably linked with the material processes occurring in the brain, and therefore the idea became inevitable that the thinking and desires of humans and animals should be based on the same strictly causal laws. Thus, Newton expressed his worldview as a whole.

Now we know that gravity is not the only force operating in nature. Gravity cannot explain the cohesion forces in bodies, electrical forces, or light. However, Newton’s theory of motion, apparently, can serve as a completely sufficient foundation for understanding any physical process, if we assume that between particles of matter, in addition to gravitational forces, forces of a completely different kind also act. This expansion of the theory of motion was begun by Newton himself, who applied it, for example, to the theory of light.

Thus, Newton laid the foundations of that set of laws of nature that allows the laws of all phenomena. Newton believed that this could be achieved by reducing any processes to the movements of particles interacting with each other. This program lasted until the second half of the 18th century and proved its fruitfulness in the field of physics.

Let’s try to penetrate a little into the laboratory of Newtonian thought.

The field of activity for the great systematizer that Newton was was prepared by Galileo and Kepler. Galileo discovered that “unregenerated motion”

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The movement of the body is straight and uniform. In this case, by “unperturbed motion” of a body we mean the motion of a body that is not affected by other bodies. In this state the law of inertia. It can be formed as follows: the direction of movement and speed of the body remain constant, and soon there are no external influences on the body, called forces. Galileo also discovered that on the surface of the Earth the speed of a freely falling body increases by equal amounts at equal intervals of time.

Newton posed a general question: how does the speed of a free body change under the action of an arbitrary given force? This is a much more general problem compared to the one considered by Galileo, or the current village in its size and direction can arbitrarily change over time. The answer to this is related to the consideration of voluntary movement; it must contain the general law of motion. This problem can be solved using Galil’s solution to the problem of the division of a free body under the influence of gravity, but it requires a new mathematical apparatus, specially created for this purpose by Newton, namely differential and integral calculus. Newton can be compared to therefore, whose poems are so subtle that they can only be written in a new language, which must be created by the poet himself.

What, strictly speaking, is the speed of movement of a body representing a material point? Let us imagine the arbitrary movement of such a body.

Let’s consider two moments of time, separated mathematically accordingly.

Fig. 1.
Acceleration is determined by speed in the same way as speed is determined by a given movement. At each moment of time, the speed is set using a vector. Let’s imagine that the speed of a body at some point in time is given by a vector L having a certain length and direction. After a small arbitrary amount of time t, the speed will change, i.e., the new speed vector, M, will have some other length and a different direction. Let’s imagine that the origins of the vectors L and M are intersected at one point (Fig. 2). Then the ends of the vectors L and M will not coincide. For example, the segment S - T connecting the ends of the vectors will represent the change

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DYNAMIC
PROGRAMMING

Pharaoh’s staircase

The pharaoh’s palace was famous for its Russian style: pearl curtains, walls decorated with amber, gold dishes - there’s just so much to list. But what struck most of all those who were allowed into the throne room of the palace was the golden staircase that led to the throne. What this staircase looked like in cross-section, if you convert the ancient Egyptian measures into actual ones, you can see on figure 1.
Archaeologists did not have to rack their brains for long to understand that the steps of the staircase were so low, everything was explained simply: the presented pharaoh did not want his subjects to see how hard it was for him to climb to the throne, they could lose respect, and who knows how this could end?
It was then that he ordered the leading jewelers to make a staircase with such low steps, no higher than 1.5 inches.
But time passed, and the old pharaoh died. His young son took over the kingdom.
The young pharaoh had heard before how the courtiers took turns at his father’s naive cunning. And although he himself soared to the throne, jumping three steps, the rumors continued. And the young pharaoh decided to put an end to him by building up the ladder. To do this, he called the court jewelers and the treasurer and told them:

• Listen to me! I command you to build the ladder so that it has at most four steps. Arrange them wherever you want, but make sure there are no more than four of them!
  “But sir,” the treasurer timidly stepped forward, “where did you get so much gold?” Here I pinned it on a piece of paprius (and the treasurer showed what is shown on the figure staircase with a dotted line). The staircase is 1 m wide, or 10 dm. This means that to increase it you will need [3×1.5+1.2×(5+1)+1×1+1×3+0.8(3+4)+1.2×11]×10 = 345 dm³ of gold! Alas, sir, so much will not be found in our treasury, devastated by the war with Efyupne.
• Despicable! You cast a shadow on my power! If in 7 days the ladder is not ready, I myself will drown you in the Holy Nile, and the jewelers will be sold at work. And don’t think about meddling with your stupid drawings again!
  The dejected treasurer left the pharaoh and fell back to his friend. There was no more skillful person when it came to planning a paradise or calculating land allotments. Although the friend was the priest of the capital’s temple, he was heard to be a prudent and practical man. Having learned about the trouble, he asked:
  — How much gold is left in the treasury?
• 200 dm³.
• Not too much! Or maybe you can get by with less if you plan the steps differently?

“I tried,” the treasurer sighed, “but there are so many varnat, and time is running out.”
“Okay, friend,” said the priest. - Let me think and come back tomorrow.
When the grief-stricken treasurer trudged up to the priest the next day, he greeted him with a satisfied smile.

• Tell me, friend, can I count on the remaining gold if I get by with less?
• Oh gods! - exclaimed the treasurer, not believing his luck. “You’ll get even more work and the same number of measures of grain!”
• So look! - and the priest handed him a sheet of panirus, where there was another project for building a staircase.
  — This project only requires 171 $\partial m^3$ gold. And the remaining 29 $\partial m^3$ are mine!
  “But tell me, O greatest of computing, how did you find such a varnat?” I tried this and that, but it didn’t work for me!
  “Listen,” said the priest. — First, I analyzed the case when stairs with a different number of steps are replaced by just one step (Fig. 2*)). To make the numbers simpler, I will only calculate the area $f$ of the figure between the solid and dotted lines, and add a constant factor, the width of the stairs, later. Non-refundable? So, if the original staircase had consisted of only one step,

*) The priest looked at the stairs from the first few steps of the pharaoh’s ladder.

then there would be nothing to increase:
$f_1(1)=0$.
Formula
$f_1(2)=1,5\times 3=4,5 \partial m^2$
shows that the ladder is replaced by one step, and at first there were two (the first two steps of the pharaoh’s ladder). We can similarly find
$f_1(3)=1,5\times 3+0,7\times (3+4)=9,4$.
Well, then
$f_1(4)=9,4+1\times (3+4+1)=17,4$
$f_1(5)=17,4+1,2\times (3+4+1+5)=33$.
“All this is understandable,” the treasurer returned, “but why do I need one step?” After all, Pharaoh asked for four! And the staircase now has not five of them, but much more!

• Take your time! — the priest smiled. - Let there be more new steps now - two. Let’s now calculate the minimum areas for two steps. $f_2(1)$ does not make sense; we cannot replace one step with two. To replace two steps with two new ones, you don’t have to build them up:
  $f_2(2)=0$.
  Here $f_2(3)$ is a little more difficult to count (Fig. 3):
  $f_2(3)=\min \{(f_1(1)+0,7\times 4), (f_1(2)+0)\}$.
  The values ​​in curly brackets correspond to two varnates: it is necessary

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increase either the second or the first step, and the min sign indicates that you need to take the smallest of them. Let us substitute the values ​​$f_1(1)$ and $f_1(2)$ into brackets, found to be equal:
$$f_2(3) = \min\{(0+0,7\times4), (4,5+0)\} = 2,8.$$
Now it is already clear that if there are only three steps in the leaf, and their number needs to be reduced to two, then the second step needs to be increased.
— What if there were 4 steps?

• Then
  $$f_2(4) = \min\{(f_1(1) + 0,7\times4 + 1\times(4+1)), (f_1(2) + 1\times1), (f_1(3) + 0)\} = \min\{(0+7,8), (4,5+1), (9,4+0)\} = 5,5$$
  that is, if there were 4 steps in the leaf, and we replace them with two, then the first one will best finish on the second old one (Fig. 4).
  “Wait, I want to make sure I understand,” the first is his treasurer. - So, to find $f_2(5)$, you need to act like this:
  $$f_2(5) = \min\{(0+7,8+1,2\times(4+1+5)), (4,5+1\times1+1,2\times(1+5)), (9,4+1,2\times5), (17,4)\} = \min\{(19,8), (12,7), (15,4), (17,4)\} = 12,7.$$
  “The question is, the wise ones look after the state treasury,” they used.

Fig. 2.
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priest. - And then
$$f_2(6) = \min\{(0+19,8 + 1\times(4+1+5)), (4,5+8,2+1\times(1+5+3)), (9,4+1,2\times5+8), (17,4+1\times3), (33+0)\} = \min\{(0+32,8), (4,5+17,2), (9,4+14,0), (17,4+3), (33+0)\} = \min\{(32,8), (21,7), (23,4), (20,4), 33\} = 20,4.$$
We could go on and find $f_2(7)$, but this is not necessary, as you will soon see. Let us now move on to the case when the ladder is replaced by three steps.
Let’s start by finding $f_3(4)$, that is, the most reasonable option for increasing the first four steps. After all, $f_3(1)$ and $f_3(2)$ are meaningless (you shouldn’t replace one or two steps with a tray), and $f_3(3)=0$ - increasing three steps to three does not require any consumption of gold. To find $f_3(4)$, let’s look at where the second rung might end. Of course, either on the 2nd or 3rd old one. Means,
$$f_3(4) = \min\{f_2(2)+1\times1), (f_2(3)+0)\} = \min\{0+1), (2,8+0)\} = 1.$$
Therefore, building up 4 steps so that there are 3 of them, you need to complete a new second step.

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actions - addition and multiplication (except for a certain fee to the priest, where subtraction was required), and they were well known in Ancient Egypt. Well, a little more common sense is used. Meanwhile, a mathematical apparatus is used here that became widespread only about twenty years ago; it’s called dynamic programming.

Let’s think about why we were able to find the optimal solution without considering all the possible options for building a staircase (and there are a lot of them, and the priest really wouldn’t have enough time to study them completely). The fact is that at each step we break the general problem into a number of simpler ones.

Let it be necessary to build up a staircase that has $N$ steps, so that it has $n$, where $N \gg n$, that is, $N$ is much larger than $n$. Let us assume that we already know the optimal distribution of steps when there are $j$ in the new staircase, where $j \leq n$. Let $S_j$ be the number of the old steps on which the $j$th new one ends. Obviously, this $j$th step begins on the $(S_{j-1}+1)$th old one. The additional area of ​​the staircase, shown in figure 6 by hatching, will depend on $S_{j-1}$ and $S_j$, that is, on where the $j$th step begins and ends; let’s denote it by $g(S_{j-1}, S_j)$. Let $f_j(S_j)$ be the minimum area for a staircase of $S_j$ steps replaced by $j$ steps, and $f_{j-1}(S_{j-1})$ be the minimum area of ​​a staircase of $S_{j-1}$ steps replaced by $j-1$ steps. Let us establish the relationship between these quantities, that is, we will find such values ​​of the variables $S_j$ that give the smallest value of the additional area:

$$g(S_1, S_2) + g(S_2, S_3) + \dots + g(S_{n-1}, S_n), \quad S_n = N.$$

Let’s introduce the function

$$f_j(S_j) = \min \left[ g(S_1, S_2) + \dots + g(S_{j-1}, S_j) \right],$$

where $j=1, 2, \dots, n$. Note that $g(S_{j-1}, S_j)$ is independent of $S_1, \dots, S_{j-2}$. Therefore, $g(S_{j-1}, S_j)$ can be taken out as a sign of minimization along $S_1, S_2, \dots, S_{j-2}$ (similar to how paraboly can be taken along the $OY$ axis - the lowest point of paraboly will not change from this):

$$f_j(S_j) = \min \left\{ g(S_{j-1}, S_j) + \min \left[ g(S_1, S_2), \dots, g(S_{j-2}, S_{j-1}) \right] \right\}.$$

But we denoted the second term in the formula in brackets by $f_{j-1}(S_{j-1})$. From here we get

$$f_f(S_f) = \min \left\{ g(S_{j-1}, S_f) + f_{j-1}(S_{j-1}) \right\}.$$

This is R. Bellan’s equation, which underlies dynamic programming. Capacity in carrying out numerical calculations, and you will see that we get them

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At every step. It should be understood as follows: to find the minimum, one should assign $S_j$ all possible values, starting from 1, and for each, find and fill in the value $S_{j-1}$, which delivers the smallest value $f_j(S_j)$. And then, working until the last value $j=n$, you need to “back away” and go back and find all the optimal values ​​$S_j$.

Introduction to the method and a little history

Let us pay attention to the fact that in the described method no conditions are imposed on the form of the function $g(S_{j-1}, S_j)$. In this way, dynamic programming differs favorably from linear programming *), in which it is required that all the dependencies under consideration be linear and the variables change continuously. Therefore, R. Bellan’s equation and the above method were now used very widely. But, of course, not to satisfy the whims of the pharaohs. Without it, try to find the optimal match for the weights of the stages of a space rocket, determine the best principle for unifying parts, and build a schedule that would have the least amount of downtime! You will find yourself in a desperate position as a treasurer, overwhelmed by the abundance of possible options. Of course, using dynamic programming, you can almost

*) See, for example, the article: A. V. Katok, Economics and linear inequalities, «Kvant» No. 3, 4, 1971.

they don’t count. This method is well suited for calculation on electronic computers; the method is easy to program and does not contain any “pitfalls” that, unfortunately, are found in linear programming.

The idea of ​​the dynamic programming method has been around for a long time. To some extent, it was also used by C. Maclaurin (1698-1746). However, it finally took shape in the works of the American mathematician R. Bellan and is associated with his name.

Here we have shown the application of dynamic programming to a simple scale problem. The thought immediately arises: how applicable is this method to more complex problems? It must be answered with some caution. The fact is that this creates a specification difficulty, which was called the “curse of dimensionality.” You have to try so many different possibilities and remember so many results that the method loses its positive features.

Another application

The sounds of life either grew threateningly or moved away, but it was clear that everything was final. In his absence, the pharaoh sat in front of his shagar, surrounded by dejected courtiers. A bloody messenger ran up staggering:

• Sovereign! Nomad reinforcements are approaching! We are dying!

• We’ll have to retreat. We need to see how to cross it,” the pharaoh squeezed out with a groan, his eyes on the map. - But which way should we go? Tell me! “You are skilled in calculations and have already helped me once,” he turned to the treasurer.

• Oh, Osiris! - exclaimed the treasurer. - But that was a completely different task! Now ask your commanders!

• I don’t believe them. These fools predicted victory for me.

“Now I’m lost,” thought the treasurer, peering at the map (Fig. 7).

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Fig. 7. The names of the points are replaced by numbers in circles, and the numbers between them show the length of the transitions in days.

Suddenly it dawned on him: “But the priest’s thinking can be used here too!”

Let’s first consider the path from battle site 1 to the nearest points 2, 3 and 4, the numbers of which we will write in each circle on the left (Fig. 8), and on the right we will show the length of the path (in red). For example, it takes 8 days to reach point 4.

Now consider the entire path from points 2, 3, 4, without paying attention to point 1 (Fig. 9). Some paths have already led to the goal. So, in paragraph 5 two ways were given. Let’s leave the best of them (it took 18 days), and exclude the worst by marking it with a minus sign. At the same time, we eliminate the path to point 7 (it took 21 days), to point 6 (it took 24 days) and to point 2 (at the first step we had the best path in 2 days). There are only 3 points left to check: 3, 4, 5.

Analysis of Figure 10 shows that the best path takes 16 days and leads through points 2, 3, 5.

Note that at the last stage the treasurer was no longer interested in how the army got to points 3, 4 (compare with the ladder problem!). That is, the best further path does not depend on how we ended up at any point. This general rule is called the “optimality reason” and is the basis of Dianic programming.

Exercises

1. For the school workshop, we first decided to buy 9 different lathes with a cost of 10, 20, 40, 60, 75, 100, 130, 160,

Fig. 10.

190 rub. Each of the subsequent machines in this row can replace any previous one, but not the other way around. For example, a machine for 100 rubles. can form the same parts that form a machine for 40 rubles. However, the school director objected:

— Too many types of machines! It will be difficult for us to service them - after all, each requires its own spare parts. Let’s buy 9 machines, but only four different types, and so that the purchased machines have no less capabilities than the original 9 machines.

— What types of machines will we take?
— We’ll choose them so that we can pay for all 9 smaller machines. Evgeniy Ivanovich, can you help us choose the type of machines? - he turned to the mathematics teacher.

What types of machines did the teacher suggest you choose and how long did they last?

2. What the director proposed to do in the previous task is called unified. In practice, the unified approach usually provides economic benefits. But how can this benefit be calculated?

Let there again be 9 machines on the side 16, 18, 19, 22, 23, 24, 24, 28, 33 rubles. It is required to compose unified series from them, as was done earlier, considering that combining machines in series due to

Fig. 11. Travel costs: minibus - 10 kopecks, metro - 5 kopecks, bus - 5 kopecks, trail bus - 4 kopecks, tram - 3 kopecks.

simplification of maintenance results in a reduction in costs per series according to the following table:

Table

Number of machines in a series
1 2 3 4 5 6 7 8 9

Cost reduction
0 12 16 18 20 20 21 21 21

3. “Well,” Nina said to her older brother Victor. — I told you that you don’t need to buy ice cream! How do we get to grandma now? We only have 30 kopecks left.

“Wait,” Victor objected. - Firstly, it was you who offered to buy ice cream. Secondly, diagnostic programming will help us now. We took it in school math class.

And he quickly began to draw a route diagram on paper (Fig. 11). What path should Victor and Nina (ten years old) take to get to their grandmother?

4. Evening was approaching, and Petya still did not know what was assigned in geometry. You can find out this, for example, from Lena (her house is 12 minutes away), but Petya doesn’t know the apartment number. Vasya knows him, but it’s a 10-minute walk to him and another 6 from Vasya to Lena. You can go to a pay phone (it’s nearby, but there’s always a 3-minute queue) and call Yura. But Yura is from a different class, and he can only give Lena or Zinna’s address (she lives 6 minutes walk from Petya, and only she knows Lena’s phone number and can call her). Although no, it could also be Sasha, who lives 4 minutes from Yura, and Yura knows where. There is also a neighbor at the desk, Vitka, he never knows what is asked, but you can call him and ask him to run to Lena (from him it’s an 8-minute walk to Lena), and then call him again. Yes, and you can also ask Zinna to be escorted to Lyusa (this will take 5 minutes), she also knows the task. Finally, you can look for Lena without knowing the apartment (it will take 4 minutes to find one), or ask Vitka to call Zhenya - let him go to Lena (4 minutes to her) or to Lyusa (6 minutes to her).

How can Petya know the task quickly if we assume that time is spent only waiting and walking, and all conversations happen instantly?

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Starting to study physics, we immediately encounter an important, fundamental concept: the mass of a body or the mass of the materials of a point.

What is mass?

Answering questions related to basic concepts in physics is always difficult. Newton (1642-1727), in his classic work “Mathematical Principles of Natural Physification,” began his presentation of mechanics with the definition of mass. He, however, wrote “quantity of materials” together with the word “mass” and defined it as follows: the quantity of materials is a measure of such, established depending on its density and volume.

This definition, unfortunately, says little. In fact, in order to define mass in this way, we must first define density, and we cannot do this unless we introduce new concepts.

Mass physicalizes in different ways, and we do not always realize that mass, strictly speaking, characterizes different properties of the body. There are at least three different cases in which we can think of mass, and in each case we have a different concept in mind.

I. When buying something in a retail store, we talk, for example, about a kilogram of sugar, knowing that the mass of different portions of sugar adds up (as they say, the mass is additive). Here we study the quantity of substances by mass.

II. When meeting at a place with an expensive sign that says “10 tons,” the driver knows that it is forbidden for a car with a total weight of more than 10 tons to pass through this place. In this case, the mass changes the force of attraction to the earth.

III. When calculating the trajectory of the proto-accelerator, we write down an equation of motion that relates acceleration to mass. In this case, the mass acts as a measure of the inertia of the protor—as an inert mass.

The fact that in all three cases we can use the same value is an amazing property of nature, which we want to talk about here.

Under general conditions, all three cases are indeed described by the same mass. This statement is only true as long as we consider physical bodies consisting of particles that move at speeds much less than the speed of light.

If the speed of particles is high, such as for protons and neutrons that make up the atomic nucleus, then the situation changes. The mass of the system will no longer be equal to the sum of the masses of its parts (as they say, the mass will cross to be an additive quantity). Acceleration is not proportional to force and the very concept of mass becomes complex.

We will consider methods for determining mass not in order of their importance, but in a way that is convenient for consistent presentation.

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Quantity of substances

Let’s start with a simple question. What does it mean when we buy a kilogram of sugar or take 0.5 grams of aspirin? We are, of course, not interested in the root that can be given to this amount of sugar or aspirin. We are not interested in the strength of their attraction to the Earth.

We are interested in a certain amount, the number of servings of sugar or aspirin, and not at all in mass. We simply use the fact that a kilogram can be determined by weighing. In this case, the edentulousness of the mass and its proportionality to the force of earthly tension is used. In this case, this mass can be, for example, a piece of sugar, so you can simply count the mass of any amount of sugar. The only condition is that all the pieces of sugar are the same.

How do you know if two lumps of sugar are the same? We can count how many sugar molecules are in each piece (this is difficult to do, but let’s say we can do it). If both pieces contain an equal number of molecules, then their mass is the same. But at the same time, we assume that all sugar molecules are the same, that is, we will use a new physical property - the identity of the molecules. By postulating this property and the edentulous nature of the mass, we are able to compare the masses of different portions of the same substance.

This way you can set a mass scale for sugar, for oil, for copper, for iron. However, we still need to have a way to compare the masses of samples made from different substances. How, for example, can we compare the masses of copper and iron balls?

This can be done by studying the stability of elastic balls.

This is exactly what Christan Huygens (1629-1695) did even before Newton discovered his laws *).

*) These messages appeared in print only after the death of Huygens in 1703 in the memoir “On the Movement of Bodies and Centennials.” But the results were obtained in 1652.

Huygens considered the central collision of two elastic balls of equal mass.

Let us assume that two elastic balls of equal masses fly towards each other with the same absolute speed (Fig. 1). After the collision, the balls will scatter in different directions. Each next, that the speeds with which the balls fly apart are also the same in absolute value. Why?

The speed of the flying masses of balls and the speed of their merger. Consequently, none of the balls has priority over the other, and their movement must remain symmetrical. This means that they will fly apart at the same speed. And vice versa, from the fact that the balls fly away at the same speeds, it follows that their masses are equal.

In fact, the fact that the velocities after the collision are determined entirely by the initial speed and masses of the balls, and does not depend, for example, on the size of the balls, temperature or some other quantity, does not follow from anywhere. Only because experience has confirmed such a precedence can we use the outlined method of comparing the masses of bodies made of different materials.

Huygens’s method makes it possible to compare what can be called mass scales for different substances. Huygens himself did not develop this idea. He reasons simply: “…I will consider bodies made of one substance, or we will accept that the size of bodies is determined by their weight…”. By “magnitude” Huygens meant “mass”.

Fig. 1.

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Law of conservation of momentum

Let us write the law of conservation of momentum in the case of a collision between two bodies (momentum is equal to the product of mass and speed $p = mv$). Let two bodies before the collision have impulses $p_1$ and $p_2$. Their impulses after the collision are denoted by $p'_1$ and $p'_2$.

Then the law of conservation of momentum $p_1 + p_2 = p'_1 + p'_2$.

Noting the masses and velocities of bodies with indices 1 and 2, we can write

$$m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2.$$

or

$$m_1(v_1 - v'_1) = -m_2(v_2 - v'_2).$$

Denoting the absolute values ​​of the vector change in speed through $|\Delta v_1|$ and $|\Delta v_2|$, we obtain

$$\frac{m_1}{m_2} = \frac{|\Delta v_2|}{|\Delta v_1|}.$$

Thus, by changing the velocities of moving bodies, their masses can be compared. This is how the French mechanic Barré de Cen-Benan (1797-1886) defined the concept of mass.

Mass can also be determined from the acceleration ratio:

$$\frac{|a_1|}{|a_2|} = \frac{m_2}{m_1}$$

The connection between the mass ratio and the acceleration ratio can be taken as a way to compare masses, as is done *).

This definition of mass becomes more incorrect as soon as velocities approach the speed of light $c$. The matter can be improved if in the definition of momentum we replace the speed $v$ with the “relativistic” speed $u$ according to the formula

$$u = \frac{v}{\sqrt{1 - \frac{v^2}{c^2}}}$$

and write together the differences $v_1 - v'_1$ and $v_2 - v'_2$, respectively $u_1 - u'_1$ and $u_2 - u'_2$. The mass $m$ in this case is called the rest mass.

**) This definition can be obtained from the previous one, if there is a difference in speed at two very close points in time.

However, in this form, the formula is correct only for bodies that move at low (compared to $c$) speeds.

barks in the textbook. Just remember that this method is not the only one.

Centripetal acceleration

We have already said that Huygens, before Newton, understood some laws of motion (by studying the laws of motion of elastic balls). How well Huygens understood mechanics is evident from the fact that he was the first to correctly formulate the laws of circular motion and introduce the concept of centripetal acceleration. After all, if we know how to measure force (for example, using Hooke’s law), then we can change the mass without the help of gravitational forces, changing the centripetal acceleration of the rotating body.

Huygens himself considered motion in a coordinate system associated with a rotating body. In this system, it is necessary to take into account the centrifugal force of inertia, so Huygen’s formulation was as follows: “When two equal bodies move in equal circles with unequal speed, but both are uniform, then the centrifugal force of the faster one is related to the centrifugal force of the slower ones, as the squares of their speeds.”

Standard mass

The standards of length and time are now accepted for a wave of a certain line in the spectrum of krypton and the oscillation frequency of a hydrogen maser. Such standards are better than artificial ones, since they can be reproduced in any good laboratory, and they cannot be lost or deteriorated. Why can’t we use, for example, the mass of a proton as an eton of mass? We know that this is almost exactly what is done in chemistry and nuclear physics. The atomic mass unit equal to $1/12$ mass of the carbon isotope $C^{12}$ is accepted as mass units. By measuring the proton mass in these units, very precise values ​​were obtained.

$$M_p = 1,00727661 \text{ (8) ат. ед.}$$

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The number in brackets means that the possible inaccuracy in this value is estimated at 8 units in the last, eighth digit.

Thus, the proton mass is known with very high accuracy. Unfortunately, this greater precision is only achievable in atomic units. The proton mass in common units (kg) is much less known:

$$M_p = 1,672614 \times 10^{-27} \text{ кг.}$$

Here a possible error affects already the sixth character! Although this is great accuracy, it is still insufficient to approve the proton as a mass standard.

Why does this happen? The difficulty is that physicists are not able to calculate with sufficient accuracy the number of protons in a given amount of hydrogen (for example, 1 kilomole). This is due to the fact that the number of molecules in 1 kilomole is relatively poorly changed:

$$A = 6,02217 \times 10^{26} \text{ молекул/кмоль.}$$

The possible error here is almost 7 millionths of the measured value. This is too bad for a standard! Only with greater measurement accuracy can the proton take its place as a natural standard of mass.

Mass defect

Let’s talk a little more about the aditivity of mass.

If all bodies consisted of identical parts, some kind of corpuscles (or monads, as the ancients called them), differing from each other only in density - the number of corpuscles per unit volume, then the question of the mass of bodies would be solved simply. The masses of all bodies would be determined by the number of such corpuscles, and this would be the end of all difficulties. Could this be exactly what Newton was thinking about when he believed that the density of a body is a simpler concept than its mass?

But the world is more complicated. Atoms are built from different particle elements: electrons, protons, neutrons. Therefore, from molecules to their constituent parts, we will only replace the question of comparing the masses of different atoms with the question of comparing the masses of different elements of particles.

In addition, as is known from the theory of relativity, the masses of protons and neutrons cannot simply be added to each other when these particles are part of the nucleus. When a nucleus is processed, energy $E^*$ is released and the mass of the nucleus decreases in comparison with the sum of the masses of all protons and neutrons by amounts

$$\Delta m = \frac{E}{c^2}.$$

It is impossible, strictly speaking, even in principle to determine the mass of atoms from the number of particles. The mass in this case is not additive, that is, the mass of the nucleus is not equal to the sum of the masses of its constituent particles. The aditivity of mass is valid as long as we divide the body into parts that almost do not interact with each other, for example, when we saw an iron bar.

Let us estimate, for example, the error that we will make if we assume that the mass of hydrogen atoms is equal to the sum of the masses of the proton and the mass of the electron. In order to remove an electron—to ionize an atom—you need to expend 13.5 electron volts of energy. (In other words, it is necessary to apply a potential difference of 13.5 points to the proton-electron system.) 1 electron-volt is equal to 1.6·10^{-19} joules. From here we get

$$\Delta m = 2,4 \cdot 10^{-35} \text{ кг.}$$

Let’s compare this number with the proton mass of 1.7·10^{-27} kg. We see that the error (if we consider the mass of hydrogen atoms as the sum of the masses of a proton and an electron) is very small and will affect only the eighth digit. This amendment can be neglected. For nuclear reactions, this effect makes a sufficiently large contribution and cannot be neglected.

*) $E$ there is, of course, also energy that must be imparted to the nucleus so that it “breaks apart” into protons and neutrons.

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V. N. Berezin

The fate of other theorems and problems is peculiar…How to declare, for example, the exclusive attention of mathematicians and mathematics lovers to the Pythagorean theorem? Why were many of them not content with already known evidence, but found their own, bringing the five relatively obvious quantities to several hundred?

When it comes to the Pythagorean theorem, the non-educational begins already with its name. It is believed that it was not Pythagoras who first formulated it. It is also believed that he gave proof of it. If Pythagoras is a real person (some even doubt this!), then he most likely lived in the 6th-5th centuries. BC e. He himself did not write anything, called himself a philosopher, which means, in his understanding, “striving for wisdom,” and founded the Pythograean Union, whose members studied music, hypnastics, mathematics, physics and astronomy. Apparently, he was also a magnificent organ, as evidenced by the following legend relating to his stay in the city of Croton: “Pythagoras’s first appearance before the people in Croton began with a speech to the young men, in which he so strictly, but at the same time so captivatingly outlined the designations of the young men that the elders in the city asked not to leave them without honor. In this second speech he pointed to legality and purity of rights as the foundations of the family; in the next two he addressed children and women. The consequence of the last days of speech, in which he especially condemned Russia, was that thousands of precious dresses were delivered to the temple of Hera, or not a single bridegroom dared to show them on the street anymore…” However, even in the second century of our era, that is, 700 years later, real people lived and worked in the field, outstanding scientists who were clearly under the influence of the Pythogrean Union and treat with great respect to the fact that, according to legend, Pythagoras is created.

There is also no doubt that interest in the theorem is caused both by the fact that it occupies one of the central places in mathematics, and by the satisfaction of the authors of the proof, who overcame difficulties, about which the Roman poet Quite Horace Flask, who lived before our era, well said: “It is difficult to express well-known facts well.”

Initially, the theorem established a correspondence between the areas of squares built on the hypotenuse and legs of a right triangle: a square built on the hypotenuse is equal to the sum of squares built on the comets. This theorem is proved in this formulation using the drawings shown on the first page of the cover. Here, in the upper left figure, a right triangle is highlighted with dashed lines, on the legs and hypotenuse of which squares are constructed, on the hypotenuse

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tenuze - outward, on the legs - internal of the triangle. The sides of these squares are continued wherever one of the squares overlaps the other. In this case, several triangles, trapezoids and one head square were formed. Equal figures are painted the same color. The fact that the triangle formed from the red trapezoid and the yellow triangle is equal in size (moreover, symmetrical) to the triangle formed from the purple triangle and the green trapezoid is highlighted by the fragment in the upper right corner.

In the lower part figure on the legs of the right triangle (white) the same squares are constructed externally. Along the way, in one of them the violet-green triangle is replaced by an equal-sized red-yellow one. It is not at all difficult for Telery to show that a figure composed of two squares built on the sides of a right triangle is equal in size to a square built on the hypotenuse of this triangle. To do this, let us once again replace the green trapezoid together with the purple triangle with a red trapezoid with a flat triangle and replace that the resulting figure turns out to be equilateral with the square built on the hypotenuse of this right triangle. This also proved the Pythagorean theorem.

Here’s another proof using equivalence. At figure 1, the segment colored green is equal to one of the quadets of the right triangle located at the bottom of the drawing, and the red triangle is equilateral and right. Having proven that the angle between the two cuts - on the left and below - is straight, we see how from the parts of this figure, which is the union of a square and an equilateral right triangle, we can add either two squares or one. Moreover, in the second case, the side of the square is equal to the hypotenuse of this very private triangle. The hinge mount at figure 2 shows that the right triangle representing half of the red has been cut off and rotated 135° relative to the “hinge” point. On figure 3 uses figure 1. Again near the hinge points, the separated triangles are rotated 135° each.

How to carry out the proof of the theorem in the case when the smaller of the quadets is more than half of the larger one, you can easily determine for yourself.

Modern geometry prefers the arithmetic formulation of the Pythagorean theorem, namely: if the sides of a right triangle are changed by the same

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Fig. 4.

scale, then the square of the number expressing the hypotenуzy is equal to the sum of the squares of the numbers expressing the legs. Briefly: the square of the hypotenуzy is equal to the sum of the squares of the catats. We will now carry out two proofs that also use the formulation.

figure 4 shows a square with four equal right triangles highlighted on it. It was from this figure that the Indian mathematician Bhaska-Acharna based his proof in the 12th century.

Let the side of the large square (also the hypotenуza of the right triangle, colored here in yellow) be equal to $c$. Let also two of its katat be equal to $a$ and $b$, respectively. Then, in accordance with the drawing, $(a-b)^2 + \frac{4ab}{2} = c^2$, that is, $c^2 = a^2 + b^2$. Consequently, if the triangle is right-angled, then the sum of the squares of its real catatas is equal to the square of the hypotenуzy.

At figure 5 one of the three given right-angled triangles is enclosing. All three triangles are pairwise similar. This is the key to the proof, either the areas of similar figures, built respectively on the catatas and hypotenуze of a given right triangle, are in the same ratio as the areas of the squares built on these catatas and hypotenуze. In other words, with the help of figure we obtain the equality $ka^2 + kb^2 = kc^2$, where $a$ and $b$ are the catata of the enclosing triangle, $c$ is its hypotenуza, $k$ is a number equal to the ratio of the area of ​​the enclosing triangle to the area of ​​the square built on its hypotenуze. Reducing both sides of the equality by $k$, we obtain, as a consequence, the Pythagorean theorem.

Fig. 6.

Fig. 7.

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Even the ancient Greeks noticed that a string stretched in the wind sometimes begins to sound young - to sing. Perhaps the Elov harp was already known then, called so by the name of the wind god Eolla. The Aeolian harp consists of a frame on which several strings are stretched; it is placed in a place where the strings are set in motion by the wind. Even if you limit yourself to one string, you can get a whole range of different tones. Something similar, but with less variety of tones, will pass through when the wind moves telegraph wires.

For quite a long time, this phenomenon and many others associated with the flow of air and water around bodies were not explained. Only Newton, the founder of modern mechanics, provided the first scientific approach to solving such problems.

According to the law of resistance to the motion of bodies in a liquid or gas, discovered by Newoton, the resistance force is proportional to the square of the speed: $F = k\rho v^2S$.

Here $v$ is the speed of the body, $S$ is the area of ​​its cross-section perpendicular to the direction of the speed, $\rho$ is the density of the liquid.

Later we found out that Newton’s formula is not always correct. In the case when the speed of motion of a body is small compared to the speed of the tellan motion of molecules, Newton’s law of resistance is no longer valid. When moving slowly, the resistance force is proportional to the speed of the body, and not its square, as with fast movement. Such a situation will arise, for example, when small drops of rain move in an area or when sediment settles in a glass. However, in modern technology with its rapid speeds, Newton’s law of resistance is usually valid.

It would seem that since the laws of resistance are known, the humming of wires or the singing of an el harp can be explained. But that’s not true. After all, if the strength of resistance was constant, then the wind would simply stretch the string, and not excite its sound.

What’s the matter? To explain the sound of a string, those simple ideas about the force of resistance that we have just discussed will not be enough. Let’s discuss in more detail some types of fluid flow around a stationary body (this is more convenient than considering the movement of a body in a stationary fluid, and the answer, of course, will be the same).

Look at figure 1. This is a case of low fluid velocity. The streamlines follow the cylinder (a section is shown at figure) and smoothly continue behind it. This resistance flow is laminar. The resistance force in this case owes its transmission to internal friction in the fluid - gravity and is proportional to $V$. The speed of the fluid at any place, as well as the resistance, does not depend on time (stationary flow). This case is of no interest to us.

No look at figure 2. The flow speed increased, and liquid whirlpools - vortices - appeared in the area behind the cylinder. The developments in this case will no longer completely determine the nature of the process. Changes in momentum, occurring not on a microsoptic scale, but on a scale comparable to the size of the body, begin to play an increasingly important role. The resistance force is proportional to $V^2$.

And finally, at figure 3 the flow speed increased slightly, and the vortices lined up in regular chains. Here it is, the key to explaining the riddle! These chains of vortices, periodically falling off the surface of the string, excite its sound.

The phenomenon of the correct arrangement of vortices allowing a streamlined body was first studied experimentally by the German physicist Benard at the beginning of our century. But it was only thanks to the work of Karman that soon followed that such a movement, which at first seemed so figurative, received an explanation.

Fig. 4.

After the name of this scientist, the system of periodic vortices is now called the Karman path.

However, as the speed increases, the vortices have less and less time to melt into a larger area of ​​liquid. The vortex zone becomes narrow, the vortices mix, and the flow becomes chaotic and irregular (turbulent). True, at very high speeds, recent experiments have revealed the appearance of some new periodicity, but its details are still not clear.

It may seem that the Karman vortex street is just a beautiful natural phenomenon with no practical significance. But that’s not true. Power line wires also sway when exposed to constant wind force due to vortex shearing. In places where wires are attached to frames there are significant services that can lead to destruction. Tall chimneys sway in the wind.

However, the most widely known is definitely the Tacoma Coastal Sway in America. This downtime lasted only a few months and collapsed in the fall of 1940. figure 4 shows the view of the place during the oscillations. The vortices were torn off from the supporting structure of the roadway of the place. After extensive research, the places appeared again, only the wind-blown surfaces had a different shape. Thus, the reason causing the vibration of the place was eliminated.

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Geskanma environment

A. P. Savin

Playing tic-tac-toe on checkered paper has long become one of the favorite pastimes of schoolchildren and students, since they always have a pen or pencil and a piece of checkered paper at hand.

In problem M91 a new version of this game was proposed.

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

Explore similar games in which the second player is allowed to place not three, but only two or only one polyk in one turn. What will be the result here if the partners play correctly: will the players be able to “lock” the crosses (and what is the cleanest move that the crosses can “hold out”) or can the game continue indefinitely?

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

The actions of the crosses are very similar to the actions of the ten trying to avoid being surrounded, which is why we called this game “surrounding the ten.”

The simplest version of this game is this: the beginner puts a cross, the second one puts three tiles on any three free fields, then the first one puts

Fig. 1. Fig. 2. a) Fig. 6)

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one more cross on the free field adjacent to the one where the cross already stands, and the second - three more regiments on the free fields. And so on, each time the crosses are placed on the free fields adjacent to the one where the crosses already stand, and the shelves are placed on any free fields. The beginner is asked to put as many crosses as possible, and the second one is asked to do this.

In this version, our readers solved a lot of the problem. They showed that a beginner can put no more than 7 crosses. figure 1 shows the regimental strategy on the first move that achieves this result. Indeed, after this the beginner can put a cross on one of the five fields. It is not difficult to see that only by placing it on the south-eastern field, the first can hope to put seven crosses in a fair game, because with the response of the regiments shown in figure 2, a, he can only hope to put crosses on the fields marked with dots, and of the two fields marked with red dots, he will be able to put a cross on no more than one. How the second one should place the shelves is clear from figure 2, 6.

If the beginner puts a second cross on another field, then after the second move, one of the situations shown in figure 3 occurs, where there are only 4 fields (they are also marked with dots) on which the beginner can put a cross if the second one plays correctly. It consists in cutting off all the advancement fields from the cross point placed together, similar to how it was done in the first case (Fig. 2, 6).

Let us now try the conditions of the regiment problem. Let’s allow the second one to place not 3 tiles each, but only 2. It is interesting that if from the very beginning he tries to place shelves on the cells adjacent to the crosses, then the second one will not be able to delay the movement of the crosses. Look at figure 4. Here the shelves cannot place crosses moving up (red signs - first move, black - second, green - third).

This is clear, since with each move the regiments cover only two of the three fields leading up.

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It would seem that here the polyknei will no longer be able to surround the crosses, and especially if the second one puts only one polykne on each cross. The last statement is contained in the vast majority of letters.

What if the second one tries to organize a defense far from the first cross? Will he be able to at least put the crosses up? It turns out yes! And even if the second one answers each cross with only one answer. This strategy of the second (we will also need it for the final solution of the problem) is called “reflecting a love attack.”

So, we are trying to keep the crosses moving upward. Look at figure 5. It would seem impossible not to miss the crosses above the horizontal line, because after six moves the crosses can end up on any of the 13 squares marked with dots, and the second one can only place 6 crosses during this time. Now let’s look at figure 6. The game has begun, and with each move the crosses threaten fewer and fewer fields behind the horizontal line.

After the second move there will be one square, after the third - nine, after four - seven. But the second one manages to place the polikki in even moves so that on the threatened fields they stand across the field. (Try to prove on your own that with any game of the first, the second can do this.)

Now, after the fifth move of the crosses, one of the two situations shown in figure 7 will result. In case a) the second one can place polykki in the two remaining unfilled cells in any order, and in case b) he first puts the polyk in the middle cell, and then, depending on the move of the first, to the left or to the right, and the path is blocked for the crosses.

So, the “love attack” has been repulsed! the second one may not miss the crosses

above the sixth horizontal line (from the one where the first cross is).

It would seem that the problem has already been solved. After all, in the same way we can not miss crosses in any of the directions, that is, limit, for example, all the crosses inside a certain square. However, this conclusion

premature, we did not teach the possibility of “attacking the corner.”
If the cross is on the diagonal of the defended square, then it threatens almost twice as many fields outside the square as in a “frontal attack” on the side of this square, for example, on figure 8 it threatens, as before (Fig. 5), thirteen fields on each side, and a total of 25 fields.
Are the zeroes really not able to withstand this attack? The ability to take the sides of a large square as a defense plazar comes to the aid of the zeros. Let’s take a larger square (much larger than figure 9) and draw a square in it, spaced six squares from the side of the first one. Let’s take the center of the square in the cell where the first cross was placed. As long as the crosses are inside the smaller square, the side of the square is not threatened by any attack, and while the crosses are moving towards the side (or corner) of the inner square, the crosses will be able to strengthen their positions in the corners of the large square, for example, placing 11 zeros in each corner (Fig. 9), thereby the “attack on the corner” will be meaningless, and we have learned to repel an attack on the side if it starts from a distance of six cells from the side. To put 11 zeros in each corner (in fact, you can get by with fewer), you will need an inner square with a side of 87 cells, and therefore an outer square will have a side of 99 cells.
We leave it to the readers to precisely formulate the “winning” strategy of the zeros and explore other variants of this game (for example, to find out at what $p$ and $q$ zeros they cannot “lock” the crosses if $p$ crosses and $q$ zeros are placed in one move).

Exercise

1. On sides $AB, BC, CA$ of triangle $ABC$ points $C_1, A_1, B_1$ are selected respectively. Prove that the circles circumscribed around the triangles $ABC_1, BCA_1, CAB_1$ intersect at one point. Generalize this statement to the case of six points chosen on the sides of the teradir.

2. Given the pyatulnik $ABCDE$. A circle is drawn through the ends of each side and the intersection point of two adjacent sides. Prove that these circles intersect at a fifth point (different from the vertices of the fifth point) lying on the same circle.

3. Through the point $P$ lying inside $BAC$, two equal circles are drawn, one of which touches the line $AB$, and the other touches the line $AC$.

4. Two circles touch (on the same side) line $l$ at point $C$, and line $m$ is parallel to $l$ and intersects both circles at points $S, T, X, Y$ (sequentially). Prove that the point $C$ lies on the bissertrix of one of the two angles formed by pairs of tangents to the circle at points $S, T$ or $X, Y$.

5. A triangle is inscribed in a circle, and the square lengths of its sides are proportional to the lengths of perpendiculars drawn from opposite vertices to a fixed tangent to the circle. Prove that one of the sides lies on the diameter of the circle.

6. Two circles of radii $R$ and $r$ are arranged so that the square of the distance between the centers of the circles is $R^2 + 14Rr + r^2$. Prove that there exists an infinite multitorus of circles tangent in pairs to each other and to both given circles.

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In tabletry it is a meorem, in stereometry it is an unsolved problem.

I. M. Yaglom

Issue 6 of the magazine «Kvant» for 1971 contained the following problem (problem M89):

Prove that in any convex polygon, except a parallelogram, you can also choose three sides, which, when extended, form a triangle that encloses the given polygon. (For example, at figure 1, where the polygon is outlined with black lines, the three red lines satisfy the required condition, but the three blue lines do not.)

This problem has many different solutions. However, before we tell her decision, let’s talk about why she is interesting.

On figure 2 you see triangle ABC and three smaller triangles $Ab_1c_1$, $Ba_1c_2$ and $Ca_2b_2$, similar to ABC and resulting from ABC by slopes to points A, B and C - the vertices of the original triangle. Triangles $Ab_1c_1$, $Ba_1c_2$ and $Ca_2b_2$ are not only similar to ABC - they are ABC-ready, that is, similar to ABC and located parallel to it. (I later learned that the segments corresponding to each other, say, in similar triangles ABC and $Ab_1c_1$ are figuratively parallel to each other.) It is clear that if the triangles $Ab_1c_1$, $Ba_1c_2$ and $Ca_2b_2$ are slightly smaller than the original triangle ABC, then they completely cover triangle ABC.

figure 3 shows parallelogram ABCD and four smaller parallelograms $Ab_1c_1d_1$, $Ba_1d_2c_2$, $Cb_2a_2d_3$ and $Da_3b_3c_3$, resulting from

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Fig. 2.

Parallelograms corresponding to a given parallelogram ABCD that can be covered by a parallelogram ABCD. In fact, if the side AB of the parallelogram ABCD is equal to c, then any segment belonging to ABCD parallel to AB will not exceed c; Therefore, in preparing an ABCD and less than ABCD parallelogram, every parallel segment AB will be less than c. It follows from this that both vertices A and B of the parallelogram ABCD cannot be covered by (any!) parallelogram prepared by ABCD or smaller than it. Thus, it is established that we will have to cover each vertex of the parallelogram ABCD with another parallelogram, less than ABCD and ready ABCD; therefore, the total number of covering ABCD less than ABCD less will be at least 4 (according to the number of vertices ABCD). In the same way, it is established that the total number of triangles covering triangle ABC, preparatory to ABC and smaller than it, cannot be less than 3 (since each vertex of ABC will have to be covered with its triangle).

The problem of how many polygons ready for a given (convex) polygon M and fewer are sufficient to cover M was posed in 1960 by the Kshshnev mathematicians I. Ts. Gokhberg and A. S. Markus (somewhat more recently, in the mid-50s, a related problem was

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ki books; here we will show how the proof of this theorem can be derived from the result of Problem M89.
In fact, let ABC be a triangle appearing in the conditions of this problem, containing inside itself (different from a parallelogram!) a convex polygon M ≡ A₁A₂A₃…Aₙ, and side AB of the triangle contains the segment A₁A₂, side BC - segment AₖAₖ+1 and side CA - segment A₁A₁+1 (Fig. 5).
Let us now select some point O inside M and connect it with some points P, Q and R of the segments A₁A₂, AₖAₖ+1 and A₁A₁+1. When segmented, OP, OQ and OR will cut M into three parts m₁, m₂ and m₃. Now let’s reduce the polygon M to point A. If the polygon M₁ obtained from M in this way (see Fig. 5) differs little from M, then it will definitely completely cover m₁ - to prove this, it is enough to compare the boundaries of the polygons M₁ and m₁ (carry out this reasoning carefully yourself!).
Similarly, it is shown that with prepared M and smaller M polygons M₂ and M₃, obtained from M by slopes to points B and C, it is possible to completely cover parts m₂ and m₃ of the polygon M. So, M can be covered with trigons (polygons M₁, M₂ and M₃), economically M and smaller M.
We have proven that for each parallelogram of a convex polygon M, the number of ready-made M and smaller M polygons that can be completely covered by M is no more than 3. We can also prove that for any such polygon (not degenerating into a line segment) this number is exactly equal to 3 (see Problem 1 on page 32).

Let us now return to problem M89. Here is one of its simplest solutions. Note that if a convex polygon M is not a triangle or a parallelogram, then it has two non-parallel sides that do not have a common vertex *). Continuing them to the point of intersection, we obtain a convex polygon M₁ with a side number smaller than M, containing M (Fig. 6). If this polygon M₁ is again not a triangle or a parallelogram, then we can treat it as we did with M - and so on until we get a triangle or a parallelogram containing M, the sides of which, very indo, are obtained as a continuation of the side M.
If the last polygon we arrive at in the process of our transformations of the original polygon M is a triangle, then problem M89 is thereby solved (Fig. 7). If in the end we arrive at a parallelogram.

*) If a polygon M has five or more sides, then this follows from the fact that no convex polygon has three or more parallel sides; for a quadrilateral other than a parallelogram, this follows from the definition of a parallelogram.

Fig. 4.

31

leloerm ABCD, but the original polygon M itself is not a parallelogram, then some of the vertices A, B, C and D of the parallelogram will not be a vertex of M. Let, for example, point A not be a vertex of M (Fig. 8). Then the distance between the vertex K close to A, which belongs to the segment AB, and the side KL starting from this vertex K, which does not belong to the straight line AB. Since the polygon M is convex and one of its sides belongs to the segment AD, the side KL can only go as shown in figure 8 (since the entire polygon M must lie on one side of KL). But in this case, when the lines BC, CD and KL, coinciding with the crown sides of the polygon M, define a triangle that satisfies the condition problem M89.

So, the only polygon for which there is no required triangle is a parallelogram.

So, we see that the Gochberg-Marcus theorem is not particularly difficult - the most difficult point in its proof can be considered the solution problem M89. But from I. Ts. Gokhberg and A. S. Markus (as well as from F. Levy) comes a stepeometric version of the same problem:

For each convex polygon M, indicate the smallest number of elements of M (that is, similar to M and located parallel to M*) polygons that can completely cover the polygon M - and this problem turns out to be imperfectly more difficult than the corresponding planning problem.

It is easy to see that for the tetrahedron T (triangular paramid) on

*) Two (convex) polygons M and μ are called economically if they are similar (that is, they are essentially polygons of the same type and all the dimensions of one of them are proportional to the corresponding sizes of the second), and each segment AB of the polygon M corresponds in the polygon μ to a segment ab parallel to AB.

32

the number of preparatory and smaller terrazds that can be covered by T is equal to four; for cube K the corresponding number of smaller cubes is equal to eight (prove it!). It is easy to see that the number we are interested in in the posed stereometric problem can take any value between 4 and 8 (see problem 2 at the end of the article). There is a proposal that we cannot accept any other values ​​for this number (and it is equal to eight only in the case when the polyhedron in question is paralleled!). However, proving this statement is still only possible in one direction (see problem 3 at the end of the article). The proof (or refutation) that the number in question cannot in any way be greater than eight, while a problem is being compiled, which no one has been able to solve - and this despite the fact that the polyhedron M, the polyhedron M and less than M are trying to solve this problem.

The branch of geometry, to which the problems considered in this note relate, arose relatively recently - just some 15-20 years ago. This section is called combinatorial geometry; it is characterized by the simplicity of the conditions of large_instances of the problem under consideration (often understandable to any schoolchild!), but, unfortunately, the solutions to these problems are, as a rule, quite complex. Today, dozens of books and hundreds, if not thousands, of scientific articles are devoted to combinatorial geometry. Some books available to schoolchildren are listed at the end of this note.

All the problems posed below relate to combinatorial geometry; they are relevant to the topic of this note. Asterisks are removed from problems whose solution is not known to the author of the note; Here I would especially like to draw attention to problem 6, which seems to me very interesting (but perhaps also very difficult!).

Exercises

1. Prove that no convex polygon (cannot be expressed as a line segment) can be covered by two polygons equal to the given one and smaller than it.

2. Let us denote the smallest number of a polygon M and the smallest number of its polygons that can be completely covered by M by s(M); then s(T)=4, where T is a tetrahedron, and s(K)=8, where K is a cube.

Also indicate three polygons A, B and Γ, such that

s(A)=5, s(B)=6, s(Γ)=7.

3. Prove that no (not expressed in a flat polygon) convex polygon M can be covered by a third of a polygon M, ready for a given M and less than M.

4. Prove that every convex polygon can be covered by the following convex polygon M:

a) a triangle with sides a, b and c (the required number, of course, may depend on the values ​​of a, b and c);

b) rectangular with sides m and n;

b*) parallelogram with sides m and n and angle α≤90°?

6*). From the result of Problem 4 it follows that the minimum number less than a convex polygon M and similar polygons that can cover it is always equal to 2 or 3. Characterize the class of those polygons M for which this number is equal to 2.

Literature on combinatorial geometry

1. G. Hadwiger, G. Debrunner, Combinatorial geometry of the plane, M., “Science”, 1965.

2. V. G. Boltyansky, I. Ts. Gokhberg, Dividing figures into smaller parts, M., “Nauka”, 1971.

3. V. G. Boltyansky, I. Ts. Gokhberg, Theorems and problems of combinatorial geometry, M., “Nauka”, 1965.

4. I. M. Yaglom, On combinatorial geometry, M., “Knowledge”, 1971.

33

After considering approximate numerical solutions of the equation of calculated powers using the methods of trials, chords, tangents and integration, in the final elective lesson on the topic: “Approximate solutions of the equation,” students tried to explore the following method for numerically solving the equation.

Let the equation $f(x) = 0$ have a single root $x_0$ on the segment $a \leq x \leq b$, and let the function $f(x)$ be continuous on this segment and take values ​​of different signs at its ends. Let us draw chords $AC$ and $BC$ for the graph of the given function on the segment $[a, b]$.

Let us write down the equation of the line passing through the points $A[b; f(b)]$ and $C(x_0; 0)$:

$$\frac{f(b) - y}{f(b) - 0} = \frac{b - x}{b - x_0},$$

from which it is located

$$y = f(b) - \frac{b - x}{b - x_0} f(b).$$

Similarly, we write the equation of the line passing through the points $B[a; f(a)]$ and $C(x_0; 0)$:

$$\frac{f(a) - y}{f(a) - 0} = \frac{a - x}{a - x_0},$$

Graph of the function $y = f(b) - \frac{b - x}{b - x_0} f(b)$

where

$$y = f(a) - \frac{a - x}{a - x_0} f(a).$$

To find the coordinates of the intersection point of these lines, we solve the system

$$\begin{cases} y = f(b) - \frac{b - x}{b - x_0} f(b), \\ y = f(a) - \frac{a - x}{a - x_0} f(a). \end{cases}$$

Writing out the whole thing 收支

$$f(b) - \frac{b - x}{b - x_0} f(b) = f(a) - \frac{a - x}{a - x_0} f(a),$$

$$f(b) \left[ 1 - \frac{b - x}{b - x_0} \right] = f(a) \left[ 1 - \frac{a - x}{a - x_0} \right],$$

$$f(b) \cdot \frac{x - x_0}{b - x_0} = f(a) \cdot \frac{x - x_0}{a - x_0},$$

$$\frac{f(b)}{f(a)} = \frac{b - x_0}{a - x_0},$$

we find

$$x_0 = \frac{bf(a) - af(b)}{f(a) - f(b)}.$$

A formula has been obtained for the root of this equation $f(x) = 0$! Learning from the unexpected result with great excitement and enthusiasm, they even exclaimed: “Eureka!”

But after a few minutes, an alarming silence was restored in the audience. Everyone was stunned and everyone asked: “How can this be?” Prominent scientists - mathematicians Abel and Evariste Galya for a comprehensive proof that for all equations above the fourth degree it is impossible to create formulas for exact solutions.

Where was the mistake made?

Y. M. Kleiman

34

When studying the solar corona, it is desirable to have an artificial corona with properties as close as possible to the properties of a real ECLIPSE. The device described below is designed specifically for this purpose. The artificial crown of a real sill on the real one that it gives rise to anyone who has witnessed a total solar eclipse. The resemblance of a real sill is that it is worth adding a few details in order to create a purely natural solar eclipse that is difficult to distinguish from the genuine one. The difference is that the glow columns are straight and not crooked, as in the photographs. The greenish-blue color of the sky and the characteristic bile white of the crown are well reproduced. An artificial eclipse in lecture demonstrations gives the audience a vivid impression of the beauty of the phenomenon. Figure and the photographs do not give such an impression. Anyone who views the installation described below will be rewarded with a red image of a solar eclipse.

All that is required is a six-candle incandescent lamp and a glass container. The size of the tank does not matter much; for example, a glass aquarium is good for this purpose. The reservoir will be filled with clean water, to which a spoon or two of alcohol-based mastic solution is added. The mastic immediately precipitates, and the water becomes like milk.

The wires leading to the lamp are passed through a short glass tube. A lamp (Fig. 1) is attached to the end of the tube using sealing wax. Care should be taken to ensure that the connection is tight, otherwise water will enter the tube. Five or six strips of foil are placed on the lamp at intervals of 0.5-1 mm. The thickness of the strips is approximately the same. Put it down

lean on opposite sides of the lamp. The rays passing between them form columns of glow. The number of stripes, their width and location, which give the desired effect, can be easily determined experimentally.

A metal disk is attached to the bulb of the lamp using soup or any glue that does not dissolve in water. The diameter of the disk is slightly larger than the diameter of the lamp; it blocks direct light from the lamp and corresponds to the dark disk of the Moon. The entire system is lowered into the aquarium, with the lamp in a horizontal position and the metal disk placed near the face stack (Fig. 2). It’s a good idea to include a rheostat in the lamp, then you can adjust the light intensity. When the current is turned on, you can see a beautiful corona caused by the scattering of the lamp’s light on small private mastics suspended in water. The asymmetrical lamp filament produces non-uniform illumination, and this enhances the effect. If the glow columns are too sharply defined or too wide, this can be easily corrected by replacing the foil strips.

However, the picture of the background is not yet perfect; the color of the “sky” is too light and comparatively bright. If you add some bluish-green dye to the mixture, the “sky” takes on a tanny color and the crown appears more clearly. By adding paint, you can ensure that the “sky” is heavily colored, without changing the color of the crown in the slightest. This circumstance

Fig. 3.

very surprising, since both the color of the “sky” and the color of the crown are enhanced by the same means.

So, an excellent reproduction of the circumsolar atmosphere has been obtained, a crown of pale gold color with a pearly sheen, with a distinguishable structure. Protruding hazy columns of glow are visible, fading against the bluish-green background of the “sky”.
Thanks to non-uniform lighting, not only the columns are visible, but also the dark areas. The effect is enhanced if the eyes are half-closed.

Figure 3 is taken from a photograph of one of these artificial shades. Unfortunately, much of the fine detail captured on the negative was lost during printing.

Published article in the journal Nature for 1901.
The publication was authorized by G. A. Sorokin.

36

We have already reported (see «Kvant» No. 1 for 1970 and 1971) that schoolchildren, who are given especially original and complete solutions to the problems “Task «Kvant», will receive the right to participate in the field of three Vesegoznoy Olympids along with the winners of regional and city Olympids. Over the course of 1971, the editors received more than one and a half thousand letters with solutions and selected the authors of the correct and most interesting solutions.

Below we publish a list of schoolchildren in grades 7–10 who received the right to participate in the Moscow and Leningrad city Olympids, as well as in the region, regional and resource (in the Autonomous Soviet Socialist Republic and co-resource without regional division) Olympids.

The following are allowed to participate in the Mathematical Olympiad:

A. Blyukhin - Kiselevsk, Kemerovo region, northern sh. 27;
S. Buubatryn - Mold. USSR, Glodyany settlement, northern sh. 2;
L. Bratsisky - Frunze, s.sh. 61;
A. Brevdo - Minsk, s.sh. 10;
A. Butsay - Izyaslav, s.sh. 5;
A. Viner - Gorodkovka village, Vinnytsia region, northern sh. 1;
A. Volber - Leningrad, s.sh. 239;
G. Vysotskaya - Krasnoyarsk, s.sh. 81;
A. Golber - Moscow, s.sh. 2;
A. Gordenko - village of Poltavenskoye, Kushevsky district, Krasnoarsk Territory, northern sh. 37;
A. Grigoryan - Baku, s.sh. 211;
V. Gritsevich - Chortkiv, Ternopil region, northern sh. 1;
E. Gundar - Voroshilovgrad, s.sh. 17;
L. Dachevi - Leningrad, s.sh. 38;
A. Zhumadildaev - Alma-Ata, school - int. 56
V. Zaporozhskaya - Sositsa, Cherngov region;
A. Zaslavsky - Kalinin, s.sh. 20;
I. Ienatyev - village Sarmanovo TASSR;

M. Illartsionov - Voronezh, s.sh. 58;
I. Kashapov - Moscow, Faculty of Mechanics and Mathematics, Moscow State University;
Yu. Kisin - Staraya Russa, s.sh. 1;
T. Kislitsina - p. Sharanga, Gorky region;
L. Knizherman - Moscow, s.sh. 259;
S. Konyazin - Sapatov, s.sh. 19;
V. Kopylov - Ufa, s.sh. 7;
V. Korenyako - Voronezh, s.sh. 58;
H. Kornienko - Minsk, s.sh. 50;
R. Kosoy - Odessa, s.sh. 116;
A. Karimov - Moscow, Faculty of Medical Sciences at Moscow State University
A. Kumukov - Baku, s.sh. 151;
L. Lazareva - Leningrad, s.sh. 298;
V. Lodinov - Moscow, s.sh. 91;
V. Luzin - Moscow, Faculty of Mechanics and Mathematics, Moscow State University;
S. Lyagushin - Dnepropetrovsk, s.sh. 23;
A. Makariev - Lvov, s.sh. 14;
G. Makarov - Zlagoust, s.sh. 25;
A. Merkuryev - Leningrad, fms;
M. Perelmuter - Kyiv, s.sh. 145;
M. Praeger - Tomsk, s.sh. 51;
A. Pukhalsky - Moscow, s.sh. 2;
A. Redchenko - p. Novopetrovka, Belopolsky district, Sumy region, Kuyanovskaya northern highway;
M. Rozov - Minsk, s.sh. 50;
A. Sboev - village Medvedok, Nolinsky district, Kirov region;

V. Servakh - Frunze, p. w. 61;
Sh. Slenoy - Chernivtsi, s. w. 5;
B. Slenenko - Chelyabinsk, p. w. 106;
A. Slinkin - Moscow, p. w. 2;
V. Terentyev - Pavlovo, Gorky region, p. w. 9;
Y. Tomsinsky - Leningrad, p. w. 211;
A. Udaltsov - Kaliningrad, Moscow region, p. w. 8;
V. Faymelevy - Armavir, p. w. 1;
N. Famkulin - Kazan, p. w. 131;
G. Filkovsky - Baku, p. w. 134;
O. Khudaverdyan - Yerevan, fms;
E. Chasovnikov - p. Parygino, Zyryanovsky district, V-Kazakhstan region;
A. Chernyak - Minsk, p. w. 50;
A. Shamaev - Moscow; With. w. 179;
A. Sherstyuk - Nikolaev, p. w. 2;
R. Shtsadov - Lyubertsy, Moscow region, p. w. 6;
F. Schmidel - Moscow, p. w. 444;
I. Shparlinsky - Moscow, p. w. 2;
V. Shuvaev - Elektrostal, Moscow region, p. w. 14.

Allowed to participate in the physical competition
V. Avseykov - Sevastopol, p. w. 43;
A. Alexandrov - Glazov, p. w. 14;
P. Amosov - Yaroslavl, p. w. 12;
S. Araslanova - Nizhne-Ivkinskaya northern highway. Kumensky district, Kirov region;
V. Belov - Vologda, p. w. 8;
A. Blyukhin - Kiselevsk, Kemerovo region, p. w. 27;
L. Bratsinsky - Frunze, p. w. 61;
A. Bratsikovsky - Moscow, p. w. 62;
I. Bratovskaya - Usolye-Sibirskoye, village. w. 2;
A. Budnikov - Slavyansk, Donetsk region, p. w. 11;
A. Buzuluchkov - Novosibirsk, p. w. 130;
S. Galakhvaridze - Tbilisi, fms;
S. Gorbulin - Krasnovardeyskoye, Stavropol Territory, village. w. 1;
E. Gromak - Novokuznetsk, p. w. 102;
M. Dzik - p. Kratovo, Aromashevsky district, Tomen region;
L. Dachevi - Leningrad, p. w. 38;
E. Dolgov - Moscow, p. w. 681;
G. Doldze - Tbilisi, fms;
V. Dolmatov - Tashkent, p. w. 147;
M. Zabekinskaya - Leningrad, p. w. 211;
S. Zapesochny - Uzhgorod, village. w. 3;
O. Zaumyslova - Moscow, p. w. 842;
N. Zykov - Saratov, p. w. 13;
A. Istomin - Podolsk, p. w. 13;
Y. Itmin - Rechnitsa, Gomel region, village. w. 5;

M. Kachelson - Magnitogorsk, p. w. 53;
Yu. Kisin - Staraya Russa, p. w. 1;
T. Kislitsna - p. Sharanga, Gorky region;
L. Knizherman - Moscow, p. w. 259;
L. Kozan - Chernivtsi, s. w. 35;
M. Kolodochki - Moscow, p. w. 612;
P. Konev - Plesk, Ivanovo region, p. w. 3;
V. Korenyako - Voronezh, p. w. 58;
S. Korpilov - Grozny, p. w. 2;
V. Korotkikh - Novokuznetsk, p. w. 11;
V. Komenov - Moscow, p. w. 215;
V. Krylov - Ivanovo, p. w. 6;
S. Kuzmin - Zhitomir, p. w. 8;
V. Kulii - Ivanovo, Brest region, village. w. 2;
V. Kuusk - Rzhev, p. w. 4;
G. Levin - Kuibyshev, p. w. 135;
Yu. Lurie - Grozny, p. w. 1;
A. Mamula - p. Dybintsy, Boguslavsky district, Kyiv region;
Y. Marzanov - h. Rodnikovsky, Krasnoarsk Territory, village. w. 9;
M. Maron - Kuibyshev, p. w. 11;
I. Mats - Makeeva, p. w. 22;
L. Menkova - Alexandrov, Vladimirov region, village. w. 4;
V. Nadezhko - Leningrad, p. w. 239;
V. Nedelsky - Leningrad, p. w. 30;
A. Panfilov - Sverdlovsk, p. w. 43;
Yu. Polonsky - Zelenodolsk TASSR, p. w. 3;
S. Pomashev - Ulyanovsk, p. w. 48;
M. Preeder - Tomsk, p. w. 51;
L. Rudtser - Kharkov, p. w. 118;
L. Safullin - p. B. Saby TASSR;
A. Semenov - Frunze, p. w. 9;
I. Sidorov - Moscow, p. w. 738;
G. Simonenkov - Kaunas, p. w. 10;
A. Slepyshev - Krivoy Rog, p. w. 95;
M. Solomonov - Kshinev, p. w. 34;
E. Sychulov - Krasny Uralets village, Kurgan region;
A. Udaltsov - Kaliningrad, Moscow region, p. w. 8;
V. Faymelevy - Armavir, p. w. 1;
G. Fast - Karaganda, p. w. 21;
G. Furman - Chernivtsi Ukrainian SSR, p. w. 35;
M. Khomenko - Vitebsk, p. w. 29;
O. Khudaverdyan - Yerevan, fms;
E. Chasovnikov - p. Parygino, Zyryanovsky district, V-Kazakhstan region;
S. Chekmarev - Moscow, p. w. 160;
G. Shepenko - D.-Gorodok, Brest region, village. w. 2;
A. Sherstyuk - Nikolaev, p. w. 2;
R. Shtsadov - Lyubertsy, Moscow region, p. w. 6;
I. Yurchenko - Kyiv, p. w. 145.

38

QUANTA PROBLEM
Tasks

Please send solutions to problems from the “Problem Book «Kvant»” no later than one and a half months after the corresponding issue goes out of print. On the envelope after the address of the journal, write the solutions to which problems you are sending (for example, 117071, Moscow, V-71, Leninsky Prospekt, 15, “Science”, editorial office of the journal «Kvant», “Problem book «Kvant» M131, F134). At the beginning of the letter, please include your last name, first name, middle name, six-digit zip code, and address (as well as your school and grade level).

Star excellent more difficult tasks.

M131. Prove that the four points at which the bissertis of the angles between the extensions of opposite sides of a cyclic quadrilateral intersect its sides are the vertices of the rhombus (Fig. 1).

Muram Urtembaev, 10th grade student, school No. 56, Almaty

M132. Let $n$ numbers $x_1, x_2, \ldots, x_n$ be written around a circle, each of which is equal to $(+1)$ or $(-1)$, and the sum of $n$ pairwise reactions is 0 (as in the given)

M93, p. 42) and in general for each $k=1, 2, \ldots, n-1$ the sum of $n$ pairwise opposites of numbers separated by $k$ places is equal to 0 (that is, $x_1x_3+x_2x_4+\ldots=0, x_1x_4+x_2x_5+\ldots=0$, etc.); example for $n=4$ dan in Fig. 2).

a) Prove that $n$ is the square of an integer.

6)* Is there such a set of $n$ numbers for $n=16$?

(We do not know the complete solution to the question: for what $n$ does this set of numbers exist.)

M133. One of the simplest multicellular organisms, the Volvox algae, is a spherical shell composed mainly of heptagonal, hexagonal and pentagonal cells (that is, cells with seven, six or five connected; each

39

Three cells converge at the “vertex” (Fig. 3). There are specimens that have both quadrangular and octagonal cells, but biologists have noticed that if there are no such “non-standard” cells (with less than five and more than seven sides), then the pentagonal cells are always exactly 12 larger than the heptagonal ones (there can be several hundred or even thousands of cells in total). Can you explain this fact?

V. Maresin

M134. What set of points fill the centers of gravity of triangles whose three vertices lie respectively on the three sides AB, BC and AC of a given triangle ABC?

L. G. Makarov

M135*. Prove that for every natural number $n>1$ the identity
$$\sin x \sin \left( x + \frac{\pi}{n} \right) \sin \left( x + \frac{2\pi}{n} \right) \times \cdots \times \sin \left( x + \frac{(n-1)\pi}{n} \right) = c_n \sin nx,$$ where $c_n$ is some number (depending on $n$), and find $c_n$.

V. Maresin

F143*. The plane lands on the aircraft carrier’s gunfire at a speed of 100 km/h. Having caught the braking rope, the plane runs 50 m until it comes to a complete stop. Determine the overloads during landing if the elasticity coefficient of the rope does not change as it stretches. Pilot weight 70 kg.

F144. What must be the ratio of the triangle to the floor in order for it to stand as shown in figure 4? The length of the thread holding the rod is equal to the length of the rod.

I. A. Zaitsev

F145. An ideal gas first transitions from state 1 ($P_1, V_1, T_1$) to state 2 ($P_2, V_1, T_2$). Then from state 2 the gas slowly and adnabatically (without body supply) transitions to state 3 ($P_3, V_3, T_3$). It is known that during the transition 2→3 the gas performs work equal to the amount of body imparted to the gas during the transition 1→2.

Fig. 3.

Show that $T_3 = T_1$. Draw processes 1→2 and 2→3 on the VT plane.

F146. There is a battery with e. d.s. 100 V and internal resistance 2 ohm. At the load you need to get a voltage of 20 V, and when the load resistance changes from 50 to 100 ohm, the voltage on it should change by no more than 2%. Come up with a simple circuit to power the load and calculate the parameters of this circuit.

A. R. Zilberman

F147 *. In the Rutherford and Bohr model of the atom, electrons spin around the nucleus in specific circular orbits. When moving from one orbit to another, closer to the nucleus, the atom emits a photon. What is the energy and frequency emitted by a hydrogen atom during the transition of an electron from an orbit of radius $r_1=2,1\cdot 10^{-8}$ cm to an orbit of radius $r_2=5,3\cdot 10^{-9}$ cm?

S. M. Kozel

40

Solutions

In this issue we publish solutions to problems
M89—M95; F100—F107

M89
The solution to this problem is contained in the note by I. M. Yaglom on page 28.

M90
Prove that if $x_1 < x_2 < x_3 \dots$ are natural numbers, then
$$\frac{\sqrt{x_2-x_1}}{x_2} + \frac{\sqrt{x_3-x_2}}{x_3} + \dots + \frac{\sqrt{x_n-x_{n-1}}}{x_n} < 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n^2} + \frac{1}{n^3} + \frac{1}{n^4}.$$

Let us first assume that $x_n \leq n^2$. Then
$$\frac{\sqrt{x_i-x_{i-1}}}{x_i} \leq \frac{x_i-x_{i-1}}{x_i} = \frac{1}{x_i} + \frac{1}{x_i} + \dots + \frac{1}{x_i}.$$
$$x_i - x_{i-1} \text{ слагаемых}$$

The last sum obviously does not exceed the sum containing the same number of terms
$$\frac{1}{x_{i-1}+1} + \frac{1}{x_{i-1}+2} + \dots + \frac{1}{x_i} + \frac{1}{x_i-1} + \frac{1}{x_i}.$$

No then
$$\sum_{i=1}^{n} \frac{\sqrt{x_i-x_{i-1}}}{x_i} < \frac{1}{x_0+1} + \dots + \frac{1}{x_1} + \frac{1}{x_1+1} + \dots + \frac{1}{x_i} + \dots + \frac{1}{x_{n-1}+1} + \dots + \frac{1}{x_n} < \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n^2} < 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n^2}.$$
Now suppose that among the numbers $x_0, x_1, x_2, \dots, x_n$ there are greater than $n^2$. If $x_i > n^2$, then
$$\frac{\sqrt{x_i-x_{i-1}}}{x_i} < \frac{\sqrt{x_i}}{x_i} = \frac{1}{\sqrt{x_i}} < \frac{1}{n}.$$
Thus, which term with a value $x_i$ greater than $n^2$ is less than $\frac{1}{n}$.
Consequently, the sum of all such terms (there are no more than $n$) is less than 1. But the sum of the remaining terms, as shown above, is less than $\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n^2}$, which completes the proof of inequality(*)

Yu. I. Ionia

41

M91
The solution to problem M91 is linked to a separate note “Environment of ten” on page 24.

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

This problem has attracted the attention of many of our readers. Here is her detailed solution.
Let’s number all the days: 1, 2, 3, …, 90.
Let’s find out how many times Petya went swimming. He did this on the first day, and then every other day. This means that he went swimming on days with odd numbers (Fig. 1). There are 90:2=45 such days. Consequently, he went swimming 45 times.

In the same way, you can find out how many days there were when he went to the store, and how many when he solved problems. He went to the store on those days whose numbers, when divided by 3, give a remainder of 1. (There are 90:3 = 30 such days (Fig. 2)). And I solved problems on those days whose numbers, when divided by 5, give a remainder of 1. (There are 90:5=18 such days (Fig. 3)).

However, in order to find out how many days there were when Petya (a) swam and went to the store, (b) swam and did math, (c) went to the store and did math, and (d) swam, went to the store and did math. In order not to solve which of these problems separately, we need such a general lemma.

Lemma. Among the natural numbers less than N, the remainder q when divided by n gives $$\frac{N-1-q}{n} + 1$$ numbers (q≠0). (Recall that [x] denotes an integer

Fig. 1.

Fig. 2.

Fig. 3.

Fig. 4.

Fig. 5.

part of a number x: the most whole number not exceeding x.)

Proof of the lemma. The remainder q when divided by n is given by the numbers

q, q+1·n, q+2·n, …, q+kn, …

Since we are only interested in numbers less than N, we need to find an integer k such that q+kn<N≤q+(k+1)n. Let us rewrite this inequality as follows:

$$kn \leq N-1-q < (k+1)n,$$

and finally like this: $k \leq \frac{N-1-q}{n} < k+1$. Therefore, $k = \frac{N-1-q}{n}$, and the total number of numbers less than N and giving a remainder q when divided by n is $$\frac{N-1-q}{n} + 1$$ (the one appears due to the number $q=q+0\cdot k$).

Now it’s easy to get answers to questions a, b, c and d. Indeed, each of these questions can be reformulated as follows: How many natural numbers less than 91 leave remainder 1 when divided by n_i, where n_i depends on the question: $n_a=6$, $n_b=10$, $n_c=15$ and $n_d=30$. Substituting these numbers into our formula, we get that it was:

42

15 days when Peta swam and went to the store;
9 days when Peta swam and solved problems;
6 days when Peta went to the store and solved problems;
3 days when Peta swam, went to the store and solved problems.
Now we can already calculate how many “boring” and how many “pleasant” days there were. In total, Peta spent 90 days in the village. For 3 days he was doing three things, 15+9+

Fig. 6.

Fig. 7.

Fig. 8.

42

+6=30 days - at least two things and 45+30+18=93 days - at least one thing. Therefore (see Fig. 4, 5) 90-93+30-3=24 days he did nothing. So, there were 24 “boring” days. The number of “pleasant” days is calculated in the same way: in total Peta swam for 45 days, he swam and did something else for 15 + 9 = 24 days, and finally, he did all three things for 3 days. Finally, there were 45—24+3=24 “pleasant” days.

M93
Each of the numbers $x_1, x_2, \dots, x_n$ is equal to the plane or minus one. It is known that
$$x_1x_2+x_2x_3+\dots+x_{n-1}x_n+x_nx_1=0.$$
Prove that $n$ is divisible by even.

Since the sum $x_1x_2+x_2x_3+\dots+x_nx_1$ is equal to $n$ terms, then $n$ is even—the number of minus ones in this sum is equal to the number of the plane. It remains to show that the number is minus one. To do this, let’s multiply all the terms
$$x_1x_2 \cdot x_2x_3 \cdot x_3x_4 \cdot \dots \cdot x_nx_1=x_1^2x_2^2 \cdot \dots x_n^2=1.$$
To Pozny, the number of minus ones is indeed even and $n$ is divisible by 4.

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

Let us assume that such a polyhedron exists. Let $\alpha_1, \alpha_2, \dots, \alpha_k$ be the plane angles of its faces. Let’s calculate in two different ways what their arithmetic mean is equal to. Let’s group the angles by faces. Then, since each face has at least four angles, the arithmetic mean of its angles is at least 90°. From here it immediately follows that the arithmetic mean of all angles is not less than 90°. Let’s group them now by vertices. Since each vertex is adjacent to at least four angles, and the sum of all angles at a vertex is less than 360°, then the arithmetic mean of the angles at a vertex is less than 90°*). But then the arithmetic mean of all angles is less than 90°. And, as we proved above, it is no less than 90°. The resulting contradiction proves that such convex polyhedra do not exist.

Another solution to the problem can be obtained based on Euler’s formula**). Let

*) In this place we use the convexity of the polyhedron. As some of our readers have noticed, it is possible to construct a non-convex polyhedron with a “hole”, in which each face is a quadrilateral and four edges meet at each vertex (see Fig. 6).

**) You can read about Euler’s formula in the book: R. Courant and G. Robbins, What is mathematics?, Enlightenment, 1967.

43

In a ring of vertices, $P$ are edges, and $\Gamma$ are faces of the polyhedron. Then for a convex polyhedron (and in general for a polyhedron without “through holes”) the following equality is true:

$$B - P + \Gamma = 2.$$

In our polyhedron, each vertex has at least four edges. Therefore $P \geq \frac{4B}{2}$. (We must divide in half because each edge is adjacent to two vertices.) Similarly, since each vertex is adjacent to at least four edges, $P \geq \frac{4F}{2}$.

Therefore, $2P \geq 2(B + \Gamma)$, that is, $P \geq B + \Gamma$, and according to Euler’s formula, $B + \Gamma = P + 2$. Thus, from the existence of a polyhedron with the indicated properties, it would follow that $P \geq P + 2$, which is contradictory.

M95

A trapezoid was marked on the board, the middle line $EF$ was drawn in it and the perpendicular $OK$ was dropped from the point $O$ of intersection of the diagonals to a large base. Then the trapezoid was erased. How to restore the drawing using the preserved segments $EF$ and $OK$ (Fig. 7)?

Construction of a trapezoid. Let us draw a straight line $l$ through the point $K$, parallel to the segment $EF$. Let’s find the middle $M$ of the segment $EF$. Let’s find the point $N$ of intersection of line $l$ and line $OM$. Let us construct a parallelgram with vertices $E$, $N$ and $F$, for which the segment $EF$ is a diagonal. Let’s denote its fourth vertex by $P$. Let us draw a line $l'$ parallel to $l$ through it. Let us draw straight lines through the points $O$, parallel to the sides of the parallelgram $ENFP$. The points at which these lines intersect lines $l$ and $l'$ are the vertices of the trapezoid.

Our construction is based on the fact that the straight line $OM$ divides the bases of the trapezoid in half. We leave the proof of this simple fact and the proof of the correctness of the construction to the reader. (In addition, we took advantage of the fact that the middle line of the triangle is parallel to its side.)

Another solution to the problem can be obtained based on the fact that the point $Q$ of intersection of the extended lateral sides on the line $OM$ and the ratio of the distances to the lines $l$ and $l'$ for the point $O$ Q are the same. Using this, after the lines $l$ and $l'$ have been constructed, you can construct the point $Q$ and, connecting it with the points $E$ and $F$, find the vertices of the trapezoid.

Study. Let us examine various cases of the location of the segments $EF$ and $OK$.

1. Point $O$ lies on the segment $EF$. Then there are no solutions (if the point $O$ is the midpoint of the segment $EF$, then the line $OM$ can be drawn almost arbitrarily - any

Fig. 11 a.

A trapezoid is obtained only if the segment OM is shorter than the segment MN (Fig. 9, a, b). If the segments OM and MN are equal, then you get a triangle (Fig. 10). If OM is greater than MN, then a “self-intersecting trapezoid” is obtained (Fig. 11, a, b).

The correct solutions to the problems were sent by:
N. Bashkina (Kurgan-Tube) M92; S. Buu-bamrin (Glodyany MSSR) M92; A. Alyaev (Pachelmy, Penza region) M92, M95; A. Bueay (Izyaslav) M93, M95; V. Gamolin (Odessa) M93—M95; V. Goncharenko (Maykop) M95; V. Afanasyev (Kolomna) M95; O. Volkov (Sverdlovsk) M95; O. Viner (village of Gorodovka, Vinnytsia region) M92, M95; A. Dovzhkov (Leningrad) M95; V. Zubarev (Arkhangelsk) M95; I. Bramovskaya (Usolye-Sibirskoye) M92; A. Zaslavsky (Kalinin) M95; G. Zlatkus (Sovetsk, Kaliningrad region) M92; V. Zaporozhskaya (village of Sosnitsy, Chernigov region) M92, M95; A. Golbere (Moscow) M92—M95; V. Korenyako (Voronezh) M92, M94, M95; A. Nikolaev (Moscow) M92—M94; V. Gritsetsich (Chortkiv, Ternovol region) M93, M94; T. Kislitsa (village Sharanga, Gorky region) M95; S. Belchko (Dokuaevsk, Donetsk region) M95; Yu. Kisin (Staraya Russa) M95; R. Kosoy (Odessa) M92, M95; L. Knizhnerman (Moscow) M92; L. Lazareva (Leningrad) M92; V. Lozinov (Moscow) M93; T. Casanov (Tbilisi)

Fig. 11 b.

M95; A. Gritsoryan (Baku) M93, M94; M. Illarinov (Voronezh) M92, M93, M95; A. Sboev (pss. Medvedok, Kirov region) M92, M93, M95; A. Kolchtsov (Gorky) M92, M95; A. Kutsukov (Baku) M92, M93, M95; A. Slinin (Lazovaya, Kharkov region) M94; Yu. Polonsky (Zelenodolsk) M95; V. Nadezhko (Leningrad) M95; L. Rudtser (Kharkov) M95; V. Terentyev (Pavlov, Gorky region) M92; G. Makarov (Zlatoust, Chelyabinsk region) M93, M95; V. Slepchenko (Chelyabinsk) M92, M93, M95; S. Muklytsin (Novosibirsk) M92; I. Porumchikova (Sokolniki, Tula region) M92; M. Rozov (Minsk) M92—M95; M. Preeer (Tomsk) M94—M95; E. Turkevich (Chernivtsi) M92—M95; P. Sereev (Grozny) M95; A. Slikin (Moscow) M92; A. Shmaev (Moscow) M92, M93, M95; A. Udaltsov (Kaliningrad, Moscow region) M92, M93, M95; O. Khudaverdyan (Yerevan) M93, M95; R. Shtsanov (Lyubertsy, Moscow region) M92, M93; G. Fi lkovsky (Baku) M92, M95; Yu. Yakushiin (Lomonosov) M92; E. Chasovnikov (village of Parygino, Zyryanovsky district) M92; I. Yurgel (Glubokoe BSSR) M92; S. Chekmarev (Moscow) M93; N. Famkulin (Kazan) M93, M95; A. Chernyak (Minsk) M92, M93, M95; V. Malchenko (Slavyansk-on-Kuban) M92; R. Soltsas (Vilnius) M95; P. Sukhov (Saratov) M95; A. Sherstyuk (Nikolaev) M93, M95; L. Reznik (Leningrad) M92, M95.

L. Limanov

46

F101

A ball is suspended from a hanging very thin spring with stiffness $k$. At first the spring is not stretched. What is the maximum speed the ball reaches while moving? The mass of the ball is $m$.

Let’s write down the equation of motion of the ball. Two forces act on it: the gravitational force $mg$ and the spring tension force $F = kx$, where $x$ is the deformation of the spring. According to Newton’s second law

$$mg - kx = ma.$$

From this equation it follows that the acceleration of the ball $a$ is initially directed downwards and decreases, then becomes equal to zero at $x = \frac{mg}{k}$, and then returns again, but is directed upwards.

The speed of the ball returns while its acceleration is directed downward, and decreases when it is directed upward. Therefore, the speed of the ball is maximum at $x = \frac{mg}{k}$.

In order to find it, we write down the law of conservation of energy. If the ball dropped to a distance $x$, then its potential energy decreased by $mgx$, and its kinetic energy increased by $\frac{mv^2}{2}$ (previously it was equal to zero). A fraction of the potential energy of the ball, equal to $\frac{kx^2}{2}$, went to increase the potential energy of deformation of the spring. That’s why

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

Hence, for $x = \frac{mg}{k}$ we get

$$v_{\text{max}} = g \sqrt{\frac{m}{k}}.$$

F102

Oxygen and helium were mixed equally in a hermetically sealed vessel. Then a hole was made in the wall of the vessel. What is the composition of the molecular beam emerging from it?

The energy of gas molecules depends only on its temperature. Since the gases in the vessels are in equilibrium and their temperatures are the same, then

$$\frac{m_{\text{O}_2}v_{\text{O}_2}^2}{2} = \frac{m_{\text{He}}v_{\text{He}}^2}{2},$$

where $v_{\text{O}_2}$ and $v_{\text{He}}$ are the average speeds of movement of gas molecules, $m_{\text{O}_2}$ and $m_{\text{He}}$ are their masses. From

46

Fig. 12.

this formula follows that

$$\frac{v_{\text{O}_2}}{v_{\text{He}}} = \sqrt{\frac{m_{\text{He}}}{m_{\text{O}_2}}}.$$

For simplicity, we will assume that gas molecules can move only in three mutually perpendicular directions and the speeds of all molecules are the same and equal to the average speed. Then one sixth of the molecules of each gas moves towards the hole. During time $\Delta t$, those gas molecules that were initially at a distance no greater than $v\Delta t$ from the hole fly out of the vessel. That is, these are oxygen molecules located in volume $v_{\text{O}_2}\Delta t \cdot s$ and hydrogen molecules located in volume $v_{\text{He}}\Delta t \cdot s$ (s is the area of ​​the hole). The total number of such oxygen molecules is $N_{\text{O}_2} = v_{\text{O}_2}sn_{\text{O}_2}\Delta t$ and gel molecules $N_{\text{He}} = v_{\text{He}}sn_{\text{He}}\Delta t$, where $n_{\text{O}_2}$ and $n_{\text{He}}$ are the number of oxygen and gel molecules per unit volume. If there are equal numbers of molecules of both gases in the vessel, then $n_{\text{O}_2} = n_{\text{He}}$ and

$$\frac{N_{\text{O}_2}}{N_{\text{He}}} = \frac{v_{\text{O}_2}}{v_{\text{He}}} = \sqrt{\frac{m_{\text{He}}}{m_{\text{O}_2}}} = \frac{1}{2\sqrt{2}}.$$

If the masses of gases in the vessel are equal, then $m_{\text{O}_2}n_{\text{O}_2} = m_{\text{He}}n_{\text{He}}$ and

$$\frac{N_{\text{O}_2}}{N_{\text{He}}} = \frac{v_{\text{O}_2}}{v_{\text{He}}} \cdot \frac{n_{\text{O}_2}}{n_{\text{He}}} = \sqrt{\frac{m_{\text{He}}}{m_{\text{O}_2}}} \cdot \frac{m_{\text{He}}}{m_{\text{O}_2}} = \frac{1}{16\sqrt{2}}.$$

F103

Two series of shots are fired from the cannon, tilting the barrel at angles $30^\circ$ and $40^\circ$ to the horizon. In what case will shell hits be more concentrated if the spread is caused by inaccurate aiming, and not by the spread of initial projectile velocities? Air resistance should be considered negligible.

Let us first find the dependence of the further flight of the projectile on the angle of its departure. Let’s write it down

47

for this, the kinematic equations of projectile motion. If the speed of the projectile is equal to $v_0$, and the angle at which the projectile flies out of the ordium is equal to $\alpha$, then in the vertical direction along the axis $y$ the projectile moves with acceleration $a=-g$ and initial speed $v_{0y}=v_0 \sin \alpha$. Therefore, the change in coordinate $y$ of the projectile is determined by the equation

$$y = v_0 t \sin \alpha - \frac{g t^2}{2}.$$

Horizontally along the $x$ axis, the projectile moves uniformly with a speed $v_x=v_0 \cos \alpha$. That’s why

$$x = v_0 t \cos \alpha.$$

At the moment of falling to the ground $y=0$ and $v_0 t \sin \alpha - \frac{g t^2}{2} = 0.$

From here you can find the time of movement of the projectile. It is equal to $t = \frac{2v_0 \sin \alpha}{g}$. Now substituting this expression into the equation of horizontal motion, we find that the projectile flight range is equal to

$$x_{\max} = \frac{2v_0^2 \sin \alpha \cos \alpha}{g} = \frac{v_0^2 \sin 2\alpha}{g}.$$

This is the expression for the maximum at sin $2\alpha=1$, that is, at $\alpha=45^\circ$.

Let’s draw a graph of the dependence of the projectile’s flight range on $\alpha$ (Fig. 12). For small deviations $\Delta \alpha$ of the projectile departure angle, the dependence graph $x_{\max}(\alpha)$ can be replaced by a straight line - a tangent to the actual graph $x_{\max}(\alpha)$. The greater the slope of this tangent, the greater the error in the range of the projectile for the same error $\Delta \alpha$ in the angle of departure of the projectile.

From figure 12 it is clear that at small projectile departure angles the angle of inclination of the tangent to the horizon (to the $0\alpha$ axis) is greater than at large ones. This means that at small projectile departure angles, the change in range when changing the departure angle is greater than at large ones, and with an arrow with a barrel angle of $40^\circ$, the accuracy of projectiles will be higher than at an angle $30^\circ$.

The same result can be obtained using formulas. Let the error in the projectile departure angle be small and amount to $\Delta \alpha$. Then the error $\Delta x$ in the projectile range is

$$\Delta x = \frac{v_0^2}{g} \sin 2\alpha - \frac{v_0^2}{g} \sin [2(\alpha + \Delta \alpha)] = 2 \frac{v_0^2}{g} \cos(2\alpha + \Delta \alpha) \sin \Delta \alpha \approx 2 \frac{v_0^2}{g} \cos 2\alpha \cdot \Delta \alpha.$$

Thus, $\Delta x$ is proportional to $\Delta \alpha$, but the coefficient of the proportional depends on $\alpha$. The smaller the angle $\alpha$, the larger it is.

F104

In a closed container there are several drops of liquid of different sizes. What will happen to him after a long time?

The saturated vapor pressure at the surface of the drop depends on its radius. Let’s show it.

Let us imagine a closed vessel in which there is a capillary tube, a non-wettable liquid. The pressure at point $B$ is greater than the pressure at point $A$ (Fig. 13) by amounts $\rho_{\pi}gh$:

$$P_B = P_A + \rho_{\pi}gh,$$

where $h$ is the difference in liquid levels in the capillary and in the vessel, $\rho_{\pi}$ is the vapor density.

Since the liquid film in the capillary is in equilibrium under the action of two forces: the tension surface force equal to $2\pi r\sigma$ (σ is the tension surface coefficient, r is the capillary radius) and the fluid pressure force at a depth $h$ equal to $\rho_{\pi}gh \cdot \pi r^2$ ($\rho_{\pi}gh$ is the liquid density), then $2\pi r\delta = \pi \rho_{\pi}ghr^2$.

From here

$$h = \frac{2\sigma}{\rho_{\pi}gr}$$

And

$$P_B = P_A + \frac{2\sigma}{r} \frac{\rho_{\pi}}{r_{\pi}}.$$

Thus, the smaller the radius of the drop, the greater the pressure above the surface of the liquid. This means that if steam is saturated above the surface of a small drop, then above the surface of a larger drop it will not be saturated. This will cause steam to condense on the large drop, thereby affecting the steam pressure at the surface of the small drop. This in turn causes the small droplet to evaporate. As a result, after some time only one large drop remains in the vessel.

The same result can be obtained from the most general reasoning. Let us assume that there is liquid in the vessel (the surface of the liquid is flat) and there is a drop.

49

From the second equation it follows that this triangle is right-angled and the angle between the electric velocity vectors is 90°

F107

In the circuit shown in figure 16, the voltmeter changes the voltage drop across the resistance $R=300$ com. What should the voltmeter’s resistance be so that its readings differ by no more than 2% from the actual value $u_R$? Internal resistance of the source of claims.

In the absence of a voltmeter, resistances $R$ and $R_1$ are switched on in series. Therefore, the voltage drop across resistance $R$ is equal to $u_R = \frac{R}{R+R_1} E$. When a voltmeter is connected, it will change the voltage drop in the section consisting of parallel-connected resistances $R$ and voltmeter resistance $r$. Voltmeter shows values

$$u' = \frac{Rr}{R+r} E$$

According to the condition, the voltmeter reading should differ by no more than 2% from the value $u_R$. That’s why

$$\frac{u_R - u'}{u_R} \leq 0,02.$$

Substituting the expressions for $u_R$ and $u'$ into this equation, we find that the resistance of the voltmeter should not exceed 3675 kΩ.

Correct solutions were sent by: V. Avseikov (Sevastopol) $\Phi 103$, $\Phi 105$—$\Phi 107$; A. Alexandrov (Glazov UASSR) $\Phi 100$; P. Amosov (Yaroslavl) $\Phi 100$; S. Araslanov (Nizhne-Ivkino, Kirov region) $\Phi 103$, $\Phi 107$; V. Belov (Vologda) $\Phi 100$—$\Phi 105$, $\Phi 107$; A. Blokhin (Kiselevsk, Kemerovo region) $\Phi 106$, $\Phi 107$; L. Brachinsky (Frunze) $\Phi 101$; A. Bratkovsky (Moscow) $\Phi 103$, $\Phi 104$, $\Phi 106$; A. Budnikov (Slavyansk, Donetsk region) $\Phi 101$; A. Buzulukov (Novosibirsk) $\Phi 107$; Y. Givi Dolidze (Tbilisi) $\Phi 101$; V. Glem-

$$R_1 = 100\text{ ком}$$

Botsky (Yaroslavl) $\Phi 101$, $\Phi 102$; E. Gromak (Novokuznetsk) $\Phi 103$, $\Phi 106$; M. Dzik (village of Kratovo, Armavir district, Tyumen region) $\Phi 101$, $\Phi 103$; A. Dovspevich (Moscow) $\Phi 101$; E. Dolov (Moscow) $\Phi 100$, $\Phi 101$; E. Domov (Moscow) $\Phi 03$, $\Phi 106$; G. Zaitsev (Gagra) $\Phi 101$, $\Phi 103$, $\Phi 106$, $\Phi 107$; S. Zanetsochny (Uzhgorod, Transcarpathian region) $\Phi 101$; N. Zykov (Saratov) $\Phi 101$; O. Zamova (Moscow) $\Phi 100$; A. Ivanov (Lgov, Kursk region) $\Phi 102$; Y. Imin (Cutter of Gomel region) $\Phi 101$; A. Istomin (Podolsk, Moscow region) $\Phi 103$, $\Phi 105$; T. Casanov (Tbilisi) $\Phi 103$, $\Phi 106$; A. Karimov (Ufa) $\Phi 101$; T. Kislicina (village of Sharanga, Gorky region) $\Phi 105$, $\Phi 107$; Yu. Kisin (Staraya Russa) $\Phi 103$; L. Knizherman (Moscow) $\Phi 100$—$\Phi 103$; L. Kodan (Chernivtsi) $\Phi 101$; M. Kolodochkin (Moscow) $\Phi 101$; S. Korillov (Grozny) $\Phi 101$; V. Kolomitsev (Rostov-on-Don) $\Phi 101$; P. Konev (Plees, Ivanovo region) $\Phi 101$; V. Kamanov (village of Tsalka, GSSR) $\Phi 101$; V. Komenov (Moscow) $\Phi 101$; V. Krylov $\Phi 100$; S. Kuzmich (Zhitomir) $\Phi 103$, $\Phi 105$, $\Phi 107$; V. Kulii (Mogilno village, Brest region) $\Phi 101$; V. Kuusk (Rzhev) $\Phi 100$, $\Phi 101$, $\Phi 104$, $\Phi 106$; G. Levi (Kuibytsev) $\Phi 101$; V. Lukashin (Moscow) $\Phi 101$; Yu. Lurie (Grozny $\Phi 101$; A. Mamula (village of Dobintsy, Boguslavsky district, Kyiv region) $\Phi 101$, $\Phi 102$, $\Phi 107$; Yu. Murzanov (village Rodinkovsky, Krasnodar Territory) $\Phi 101$; M. Maron (Kuibytsev) $\Phi 100$, $\Phi 101$; I. Mats (Makeevka) $\Phi 100$; L. Menkova (Alexandrov, Vladimir region) $\Phi 101$; V. Nadezhko (Leningrad) $\Phi 101$, $\Phi 107$; A. Nikolaev (Sverdlovsk) $\Phi 107$; S. Pomashey (Ulyanovsk) $\Phi 100$, $\Phi 101$; Yu. Polonsky (Zelenodolsk TASSR) $\Phi 101$; M. Preer (Tomsk) $\Phi 100$, $\Phi 101$, $\Phi 105$, $\Phi 106$; L. Rudtser (Kharkov) $\Phi 101$, $\Phi 105$; L. Safulin (village B. Gardens of the TASSR) $\Phi 107$; P. Sergeev (Grozny) $\Phi 100$, $\Phi 101$, $\Phi 103$, $\Phi 106$, $\Phi 107$; I. Sidorov (Moscow) $\Phi 100$, $\Phi 101$; G. Simonenkov (Couns) $\Phi 106$; M. Solomonovich (Chisinau) $\Phi 101$; E. Sychudov (village of Krasnodar region) $\Phi 101$; M. Taits (Moscow) $\Phi 101$; O. Tatarinskaya (Almaty) $\Phi 101$; I. Turishche (Kropotkin, Krasnodar Territory) $\Phi 100$, $\Phi 102$, $\Phi 103$, $\Phi 105$, $\Phi 107$; A. Udaltsov (Kaliningrad, Moscow region) $\Phi 105$; V. Faimelevich (Ramavir) $\Phi 100$—$\Phi 102$; G. Fast (Karagana) $\Phi 104$; I. Fedin (Omsk) $\Phi 100$—$\Phi 103$, $\Phi 105$, $\Phi 107$; G. Furmanov (Chernivtsi, Ukrainian SSR) $\Phi 101$; M. Khomenenko (Vitebsk) $\Phi 102$; O. Khudaverdyan (Yerevan) $\Phi 103$, $\Phi 107$; S. Chekmarev (Moscow) $\Phi 100$—$\Phi 103$, $\Phi 105$, $\Phi 107$; E. Chasovnikov (parychino village, Zyknovsky district) $\Phi 101$; S. Chernikov (Sempalatinsk) $\Phi 101$; Zh. Shenemko (D.-Gorodok, Brest region) $\Phi 103$, $\Phi 104$, $\Phi 107$; A. Sherstyuk (Nikolaev Ukrainian SSR) $\Phi 107$; R. Shzaanov (Lyubertsy, Moscow region) $\Phi 101$; V. Shuvalov (Electrostal) $\Phi 101$; I. Yurenko (Kyiv) $\Phi 101$.