Kvant 1970, No. 3 · Pages

Popular science physics and mathematics
quantum
3
1970
magazine
Academy
Sciences of the USSR
And
Academy of Pedagogical
Sciences of the USSR

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The abyss has opened and is full of stars;
The stars have no number, the abyss has no day.
M. V. Lomonosov

N. Y. VILENKIN

SECRETS
INFINITY
(ZENON, DEMOCRITUS, ARCHIMEDES)

The finite person does not dare to dissolve the finite as something given and accessible to his attractive intuition.

Carl Friedrich Gauss

Infinity Nothing has moved the human mind so wisely.

David Gilbert

A MOMENT OF ETERNITY

Interest in very large numbers dates back to ancient times. The Egyptians, in order to say about something that it was very large, resorted to figurative comparisons. The text of one tomb of the priest of the god Ai (XIV century BC) says:

“May he (God) reward you with anniversaries like the number of the sand of the seashore, measured by iron, and like the measure of the sea, determined by jayet, or the weight of a mountain weighed on scales, or the feathers of birds, or the leaves of trees.”

It was difficult for the Egyptians to express their thoughts differently, since they did not sufficiently develop a system of numerical notations. But about 5000 years ago, the sexagesimal number system * appeared in Ancient Babylon, and Babylonian mathematicians were able to easily handle large numbers. One of the ancient Babylonian tables lists, for example, all the divisors of the number $60^8+10\cdot 60^7=195\ 955\ 200\ 000\ 000$.

Very large numbers are also spoken of in Indian legends about Buddha. According to one of them, he was tested in numbers even in childhood, and, moving from one size to another, he must reach numbers that are expressed by the number of grains of sand of ten lakh rivers (lakha = 100,000) such as the Ganges. And then came even larger numbers and, finally, the number “with the help of which the rich calculate their product and

*) The reader interested in the ancient Babylonian number system can familiarize themselves with it in the book by M. Ya. Vyudsky “Arithmetic and Algebra in the Ancient World.”

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a l and higher surfaces (having a minimum thickness). For example, a cone, according to Democritus, consisted of circles with minimal thickness, and a pyramid consisted of the same polygons. He imagined the ball to consist of pyramidal needles, the tops of which are located in the center of the ball. Aristotle wrote:

“According to Democritus, the ball also cuts, since it is also angular.”

This approach allows Democritus to solve many problems that geometers faced. So, he was the first to calculate the volume of a cone and a pyramid, he was the first to calculate the volume of a ball and to calculate the area of ​​a sphere. This is apparently how he dealt with the pyramid.

First he proved the theorem that two pyramids with equal bases and equal heights have the same volume. Here there were no additions for him - pyramids with the same height consisted, according to Democritus, of an equal number of corresponding triangles, and their volumes were the sum of the “volumes” of the triangles (after all, according to Democritus, triangles were material and had a minimum thickness). Since the areas of the bases of the pyramids were the same, the “volumes” of the corresponding triangles in both pyramids coincided, and therefore the volumes of the pyramids coincided. And then he divided the triangular prism into three pyramids, each having the same height and equal bases (see figure). In this way, Democritus “proves” that the volume of a pyramid is equal to one third of the volume of a prism having the same height and the same base. And the equality of the volumes of a pyramid and a cone, having the same height and the same base area, was proven quite simply - they consisted of the same number of figures of the same “volume”.

It is also clear how Democritus approached the problem of the volume of a sphere. Since the ball consists of needle-pyramids, then the volume of the ball is equal to the sum of the volumes of these pyramids. And he already knew how to find the volume of each pyramid. Summarizing these volumes, he comes to the answer: “The volume of a sphere is equal to one-third of the product of the radius of the sphere and the surface area of ​​the sphere.” He was unable to calculate this area.

ROUGH PYRAMID

Democritus added remarkable results in determining the volumes of geometric bodies. But most of his contemporaries responded with his methods. The ideas of Democritus were too different from the results of direct experience. After all, even dividing a segment in half was a very difficult gap in Democritus’s geometrics. A segment consisting of an odd number of weekly debts could not be divided in half: the average weekly debt went to one of the halves, and this half became larger than the other. It was impossible to separate

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Aristotle’s Vatel, medieval scholastic Thomas Bradvarin, “all scientific systems are true only insofar as they are not based on the assumption that a continuous state is made of weekly ones.” And back in the 17th century, the Parlament of Paris threatened with the death penalty all those who opposed the opinions of Aristotle. And in part - to all supporters of atomism.

EXHAUSTION METHOD

So, mathematicians had to respond to both Democritus’ method of weeks and infinite processes - Zeno’s criticism undermined any confidence in the results obtained by using the concept of infinity. As a result, a difficult situation arose - the methods of Democritus, although through loose calculations, provided the correct formulas for calculating areas and volumes; now it was necessary to develop a new procedure for calculating geometric quantities, in which not a word would be said about infinity, or infinitesimals, or weeks. This procedure was created in the 4th century. BC e. Euclid’s predecessor was the mathematician Eudoxus, who developed the “method of exhaustion” (or, otherwise, “exhaustion”), in which many see the ancestor of the modern method of limits. This method was widely used by Euclid in his Elements.

For example, here is how Euclid proves that the area of ​​two circles is the same as the square of their diameters:

$$\frac{S_1}{S_2} = \frac{d_1^2}{d_2^2}.$$

This may not be true. Then $$\frac{d_1^2}{d_2^2} = \frac{S_1}{S},$$ where $S$ is either less than or greater than $S_2$.

Let us first assume that $S < S_2$. Let us inscribe a square in a circle with diameter $d_2$ and construct regular inscribed polygons, successively doubling the number of sides. It is easy to prove that after a few steps the difference between the area of ​​the circle and the area $P_2$ of the inscribed polygon

became than $S_2 - S$, that is, $S_2 - P_2 < S_2 - S$, and therefore $P_2 > S$. Now let’s inscribe in a circle with diameter $d_1$ a polygon similar to that inscribed in a circle with diameter $d_2$; Let’s denote the area of ​​this polygon by $P_2$. Since the areas of similar regular polygons are related as the squares of the diameters of circumscribed circles, then

$$\frac{P_1}{P_2} = \frac{d_1^2}{d_2^2}.$$

But we admitted that $$\frac{d_1^2}{d_2^2} = \frac{S_1}{S},$$ and therefore

$$\frac{P_1}{P_2} = \frac{S_1}{S}$$

or, due to the properties of proportions,

$$\frac{P_1}{S_1} = \frac{P_2}{S}.$$

But $P_1 < S_1$ (the area of ​​the inscribed polygon is less than the area of ​​the circle), and $P_2$, as just shown, is greater than $S$. So, the left side of the equality (*) is less than one, and the right side is greater than one, which is absurd. This means that the inequality $S < S_2$ is impossible.

In the same way, it is proved that the inequality $S > S_2$ is impossible (do the proof yourself!).

So, the assumptions $S < S_2$, $S > S_2$ are determined and remain the only possible one: $S = S_2$, that is, $$\frac{d_1^2}{d_2^2} = \frac{S_1}{S_2}.$$

But if for a circle the result could be easily guessed, and the calculations were not too complicated, then calculating the volume of a cone or the parameters of the source was a complex and confusing matter. However, it was still possible to cope with the cone and the parameter - after all, the answer was known at the time of Democritus, and Eudoxus only had to clarify Democritus’s messages, replacing the division of parameters into weekly given triangles with differences according to the source method.

But when it was necessary to look for the volumes of figures from which Democritus did not deal

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had, for example, a vertical ellipsoid or a vertical parabolide, the situation was completely bad - it was unknown for what value to prove that the volume could not be greater or less than this value. In other words, the mode calculus method serves only to prove already found results, but not to discover new ones.

AND LAZINESS CAN BE USEFUL

However, mathematicians who lived after Eudoxus were not too eager to discover new formulas. The center of scientific thought at this time moved from Greece to Egypt, where, under the tutelage of the Ptolemies in the Alexandrian Museum, scientists studied the sky, systematized previously accumulated mathematical knowledge, wrote books in honor of the high patrons of science, and copied ancient manuscripts. It must be said that the Alexandrian scientists did not so little, only their work was more of an epigone character. They developed what was previously available rather than selling new paths.

One of the few exceptions in the quiet family of Alexandrian scientists was a relative of the Syracusan king, Archimedes, who came to study from distant Syracuse. Unlike most scientists of that time, he was interested not only in high theory, but also in the practical applications of science, invented a screw pump, which was then used for many centuries, made mechanical toys, and later, when his hometown was besieged by the Romans, built military machines. It is clear that Archimedes became interested in mechanics, the balance of levers, the floating of bodies, etc. And mechanics has been separated from pure science since the time of Plato and was considered something akin to craft. Therefore, mechanics did not carefully monitor the rigor of reasoning, and mechanics widely used atomism in their research, splitting the wedges into atoms-wedges having a common vertex with the whole wedge, water outlets - into a number of “atomic” circles of small thickness, etc. *).

And in many of Archimedes’ works he encountered a curious combination of the method of exudation with reasoning drawn from mechanics. He learned to apply the concepts of the center of gravity and leverage to the solution of geometric problems and in this way obtained new results that were inaccessible to the “pure mathematics” of that time**). And since the results of mechanics were ultimately based on atomistic reasoning, then in fact Archimedes in these works relied on atomism. But obviously he didn’t write anything about it. The good mathematical tone of the time seemed to require that scientists present their results without referring to the anathematized and den of Democritus. Or maybe Archimedes had not yet read Democritus - this is indicated by some details of his works.

From the great works of Archimedes, which reached the 16th-17th centuries, mathematicians could not find out how he agreed, what results he needed to prove. They only had to follow the bends of Archimedes’ thoughts, only at the very end becoming convinced of his correctness. But in some passages quoted by Heron and other scholars, there was mention of some message from Archimedes to Eratovthenes. In this message, Archimedes wrote “about reasoning based on the method, but which is not yet proof for the text of Archimedes’ works. But to be sure of this, it was necessary to have

*) It is curious that even now engineers and physicists prefer to talk about elementary planes, infinitely thin layers, etc., without resorting to the precise, but somewhat complicated language that mathematicians used.
**) See about this in V. F. Kagan’s book “Archimedes”.

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in your hands the full text of the sequel. As often happens, a happy accident helped.

In 1906, a private assistant professor at St. Petersburg University, Popadopulogo-Keramevs, found some Bogoslav treatise in the library of one of the non-Rusalem monasteries. Since parchment was very expensive in the Middle Ages, they usually took ancient books, washed or erased pagan texts from them, and wrote the life of some holy martyr. The hand that interested Popadopul was also of the same origin. The assistant professor was very weak in mathematics and was not too interested in the remnants of the erased text (fortunately, the monk who wrote the treatise was lazy and only erased the text, and did not erase it). He gave only a small excerpt from the ancient manuscript. But for the famous Danish source of mathematics, Heiberg, this excerpt was enough to establish that the monk washed away the text of Archimedes. Among the many works read by Heiberg was the last one by Eratovfenu.

In this message, Archimedes pays tribute to the merits of Democritus, saying that he was the first to establish the theorems on the volume of a cone and a pramid, without, however, giving them proper proof. And the method with which Archimedes conducted a preliminary study of the problem turned out to be close to the methods of Democritus. He decomposes cylinders, cones and balls into extremely thin circles, proves the position he needs for one of them, notes that the conclusion must be true for the rest, and finally says: “Since the whole body is made up of such circles and is completely filled with them, the statement is true for the whole body.” But this coincides almost word for word with the reasoning of the atomists!

Archimedes well understood the value of his method and hoped that with the help of this method “one of the current or future researchers… will find other theorems that have not yet occurred to me.”

Unfortunately, the posture of Archimedes turned out to be unknown to precisely those mathematicians who needed it most - the creators of the calculus of infinite malch, Kepler, Galil, Cavalieri. But they, using individual hints, reconstructed Archimedes’ train of thought.

The story of how atomistic ideas were revived and how modern mathematical analysis was created will appear on pages «Kvant».

LITERATURE

1. Vander Waerden, Awakening Science. Fizmatgiz, 1959.
2. M. Ya. Vygodsky, Arithmetic and algebra in the ancient world. Science, 1967.
3. I. A. Geiberg, Natural science and mathematics in classical antiquity.
4. V.F. Kagan, Arrifmet. Gostekhidat, 1951.
5. S. Ya. Lurie. The theory of infinite malch among the ancient atomists. Publishing house of the USSR Academy of Sciences, 1935.
6. S. Ya. Lurie, Arrifmet. Publishing house of the USSR Academy of Sciences, 1935.

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LAW OF INERTIA,
HELIOCENTRIC
SYSTEM
AND DEVELOPMENT OF SCIENCE
M. Y. AZBEL
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The nature of the development of science is dramatic. But over a great historical distance, even large masks on the canvas of science merge, contradictions become invisible, and the development of science seems consistent, strictly logical and always directed in one direction. At a short distance, science is truly understandable only to professionals, but today’s decrees seem unconditionally true. Therefore, in order to see the nature of the development of science, the flourishing and this also happened - the extinction of its individual fields *), to see the drama of it, its characters, its convictions, one must sweat into that science that has already become history and is understandable to the end - it has turned from the arena of beats into a school truism. Then, perhaps, you will be able to imagine the present day of science in a new and more correct way.

It is very convenient to trace in this regard the path that led to the discovery of the law of inertia. In a school presentation, this law looks very simple and primitive, and it is not clear why the genius of Galileo, Descartes and Newton was needed to discover it.

Meanwhile, the law of inertia is one of the greatest and, in accordance with this, one of the deepest laws of physics.

One of the greatest, or it is universal, being valid for both electrons and galaxies. He was not shaken by any of the natural revolutions of the 20th century: neither the theory of relativity, nor quantum mechanics.

The reason for such stability of the law of inertia is its depth. It is therefore valid for all objects without exception, because it is associated not with a specific property of objects, but with a fundamental property of space and time: with

*) For example, science about the concepts of cutting or about the exact (in radix) solution of algebraic equations.

homogeneity and fabrication of space (that is, with the equivalence of any places and directed in empty space) and homogeneity of time (that is, the equivalence of any moments in time).

In fact, if a material point moving by inertia, that is, not interacting with the rest of the world, changed, for example, the direction of its movement, only one thing could be known: the non-equivalence of the direction in space - the direction in which the material point turned would be somehow different from all the others.

A question for supporters of the evaluative nature of the law of inertia: how will a material body move without interacting with anything? Well, let’s say, how would the movement of our Earth occur if the attraction of the Sun suddenly disappeared? Would its rotation around its axis be maintained or not? *)

In answering this question, it is worth bearing in mind that a number of answers to the more complex question of the free movement of a wolf have been given as a result of the work of such great mathematicians as Lagrange and Euler.

Isn’t it true that even after one question asked, the law of inertia begins to seem not only naively obvious? To put this into print, it is worth asking us when and under what conditions (a physicist would ask: in what reporting systems) the law of inertia operates. Well, for example, imagine a billiard ball lying on the frictionless floor of a tram. No forces act on the ball, and yet the observed driver will notice that from time to time (when?) the ball begins, subject to the law of inertia, to move unevenly across the floor. What’s the matter here?

Please note that such constructions are by no means exercises in scholasticism. An example of this is the concepts

*) Answers to all data questions posed in the article are at the end of the magazine.

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absolute and relative peace. Everyone knows that there is no absolute peace, but the relative peace of the interconnected gas surprisingly conditions the task of flying at near-light speeds due to the collision of happy gases.

But let’s return to the law of inertia. It’s not just not obvious. He is the first crazy law in the history of science. The afterlifer has never seen uniform motion on Earth without the action of an external force. Absolutely all experiments, without a single exception, convince us: without forces there is no movement. It is no coincidence that before Galileo it was believed that speed is proportional to the acting force. And this is the correct law for the usual speeds of that time. But it’s just very unrestrained! What does the coefficient of proportionality in it depend on? Generations of scientists can work (and are working!) to study this coefficient, especially since with increasing speed the law itself is conditioned, the very dependence between the speed of movement and the applied force. And it is not surprising: after all, the force lies, strictly speaking, outside the boundaries of mechanics (when friction occurs, the body is released).

But how can one manage to remove this always approaching force? How to abandon absolute value and the natural path of ever new, ever more subtle and complex experiments?

The answer of history is unexpected and received. The law of inertia, like all the basic laws of mechanics - Newton’s laws, was born not on Earth, but in heaven. And who knows when it would be open if our sky, like the sky of Venus, was always covered with clouds? Moreover, the law of inertia was discovered as a result of an error! Galileo, who discovered the law, believed that by inertia a body can only move in a circle!

To understand how this happened, we will have to follow the history, which lasted 25 centuries, of the formation of the heliocentric system, which was opposed by such geneticists as Aristotle and Archimedes. Let’s try to present in modern language - and therefore, inevitably changing somewhat - the arguments of the disputants.

The Earth was first deprived of immobility back in the fifth century BC. This was done by Philolaus, one of Pythagoras’ students. He believed that the Earth, together with the Sun and planets, revolves around a certain central fire.

A century later, Philolaus had a brilliant opponent - Aristotle *). He responded to Philolaus’s hypothesis as contradicting the existence of fixed stars. After all, if the Earth is moving, all the stars should appear to be moving to an observer on Earth! It took 23 centuries to refute this objection - 3 of them after the death of Copernicus!

In the 3rd century BC. Aristarchus of Samosko proposed a system identical to the future Copernican system: the Earth and all the planets rotate in circles around the motionless Sun. Archimedes and Apollonius of Perga became opponents of Aristarchus, or Aristarchus’ system contradicts astronomical observations - in the 2nd century BC. this was strictly proven by Hipparchus. Both scientists, Aristarchus and Hipparchus, were both right and wrong: the Earth and the planets revolve around the Sun, but not in circles, but in ellipses! It would seem to be true

*) The ancient Greeks also tried to answer the question of why celestial bodies violate the general rule and do not fall to Earth. The Greeks attached them to crystal spheres because a sphere could not fall onto its own center without breaking! Seam of crystal hemispheres of the Milky Way. The idea seems naive, but it is worth remembering that in the twentieth century, Nobel laureate F. Dyson proposed the idea of ​​high civilizations creating an opaque sphere around the central light to fully utilize its energy.

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Ptolemy’s world system.

one step left. But not only Copernicus, but also Galileo failed to take this step. Circles as planetary orbits seemed absolutely indisputable: they were the only perfect curves. André Bonard, in Greek Civilization, published in 1959, writes: “It is a pity that the scientists who objected to Aristarchus did not come to Kepler’s discovery. But the prejudice about the superiority of circular motion is firmly rooted.”

The words of the ancients about the obligatory nature of orbits seem ridiculous. Let us translate, however, these words into perfect language. One of the leading ideas of modern physics, as already mentioned, is the idea of ​​homogeneity and production of space. The basic laws of physics follow from this idea. But if the planet’s orbit is not circular, then the seen direction of the greatest elongation of the orbit appears in space, which turns out to be something different from all other directions! The fact that the movement of planets is determined by their prehistory was still incomprehensible.

Therefore, Ptolemy (2nd century AD) proceeds from the geocentric system and, introducing a complex picture of the circle

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Copernican world system.

gov orbits, explains all observational facts. Moreover, he gets the opportunity to imagine the movement of the planets and even eclipses of the Sun and Moon. This struck the imagination of the ancients so much that his work received the name “Almagest” - the greatest. Ptolemy became the first astronomer in history who managed to reconcile astronomical observations with his theory. Great achievement! But, perhaps, it was precisely this that overshadowed the elucidation of the true motion of the planets for many centuries: after all, everything is so good, the theory coincides so perfectly with experiment!

17 centuries pass, and history repeats itself! The canon of the Frausenburg Cathedral, Nicolaus Copernicus, seeks only to bring the Ptolemaic system in accordance with the observations of his time and replaces: the Ptolemaic system can be somewhat simplified if we assume that the Earth and the planets revolve around the Sun, but still in a complex system of circles. And again, like Aristarchus, Copernicus has an opponent - one of the greatest astronomer-observers of all times, Tycho Brahe. He responds to the Copernican system for approximately the same reason as Hipparchus

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Kepler’s model of the Universe, which he originally proposed.

systems. After many years of flawless attempts to find the harmony of the world in the system of regular polyhedra that enclose the orbits of the planets, Kepler came to the great law: the planets move around the Sun, but they move in ellipses!

This law reflected not only Kepler’s readiness to express an insane idea that rejected the views of twenty centuries. Kepler was also extraordinarily lucky. Due to the enormous distance of the planets (especially heavy ones) from each other, their orbits are close to an ellipse - one of the three curves well known at that time (ellipse, hyperbola, parabola).

Kepler was the first to establish the laws of planetary motion. How he arrived at his second law: in equal periods of time, the planets cover equal areas - remains almost incomprehensible. After all, interral calculus does not yet exist, and the calculation of areas was not only a science, but an art: each curve required its own approach. Where did the idea of ​​creating unusual areas even come from? Probably, Kepler’s early passion will also have an impact - the specific methods for calculating areas that he invented in his work “On the Stereometry of Wine Barrels” (a genius from an applied problem is created by science!), and an almost mystical belief in the existence of unified laws of nature *), and, believe it, just a love of calculations for the sake of calculations, and what is called genius **).

Kepler’s irreconcilable scientific opponent was his contemporary Galileo. He was the first to raise the question of what moves the planets, causing their orbits to bend. Galileo considered the influence of the Sun at such a distance mystical, as well as linking the tides with the Moon. How do the planets know that in the next hour they must cover the same area as in the previous hour?

*) Kepler was a deeply religious man and left wealth for astronomy, reassuring himself that “God can be glorified not only through wealth.” But God removed from the astronomy textbook written by the pious Kepler, and the book was immediately included in the list of prohibited books by the cataloguer. This is the frequent result of an honest descent into truth.

**) It is impossible not to name at least some of Kepler’s works: the third fundamental law of planetary motion, following Newoton, as the basis for deriving the law of universal gravitation (!); explanation of tides; the idea of ​​light pressure (!), creating comet tails; prediction of the satellites of Mars and astroid; establishing one of the most important provisions of differential calculus - the conditions for the maximum of a function.

Kepler’s second law: $S_1 = S_2$.

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what? *) A fundamentally new formulation of the question was required, because the grafted question itself was incorrect. And Kepler’s answer: “The planets are moved by the world soul,” of course, could not satisfy Galileo. But since there is no force acting on the planets, Galileo reasoned, then such movement is a natural property of the planets. Galileo called it inertia and very convincingly proved the erroneous statement: the reason for the rotation of the planets is inertia, motion by inertia can only be motion in a circle, all points of which are equal (!).

Galileo considered motion in a straight line to be excluded: the transition over time from one point to another was familiar because of their non-equivalence; in addition, in this case the body would never reach the final goal (in modern language - an equilibrium stable state), and nature never sets itself inaccessible goals (that is, an equilibrium state occurs sooner or later).

This reasoning led Galileo to a contradiction not only with Kepler, but also with himself: after all, he had previously proven that motion in a circle is associated with a force acting towards the center! The contradiction seemed insoluble, and Descartes made the situation even more dramatic by showing that motion by inertia is motion in a straight line.

What moves the planets?

*) It is worthwhile to compare this question with Rutherford Bohr’s question: how does an electron know what part of the light it should emit when jumping to a new orbit?

There seemed to be no way out. But in the year of Galileo’s death, Newoton was born. He returned to Kepler, to his laws, and on their basis created an exact science of the motion of bodies.

Until now, it has only been about the evolution of ideas, about disputes between the knights of thought: Philolaus and Aristotle, Aristarchus of Samos and Hipparchus, Copernicus and Tycho Brahe, Kepler and Galileo, disputes in which it became clear that the truth had not yet been born. But scientists are living people, with their passions, pride, pride and ability to make mistakes even in small things. Galileo was convinced of the impossibility of leaving the Earth and did not notice that his formulas gave the first cosmic velocity *). He explained the tides by the movement of the Earth around the Sun (I so wanted to be the first to point out the phenomenon that proved this movement!), but the period of the tides was twice as different from his calculation and coincided with Kepler’s results. Galileo declared Kepler’s coincidence an accident **), and he himself naively and helplessly appealed to the employees that somewhere in the Sargasso Sea the tide period was “correct”. But at the same time he is trying to connect the irregular tide period in the Mediterranean Sea with its depth - and creates a new field of science: the study of surface waves and waves in closed basins! Ga-

*) By the way, how can I use the formula $S = \frac{gt^2}{2}$ to obtain an expression for the first escape velocity!

**) Let us remember Bohr: “The coincidence of a stupid theory with experiment does not prove anything; there are countless such theories.”

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Lily discovered the independence of the period of oscillation of a pendulum from amplitude - and immediately, carried away, mistakenly extended this law to any amplitude; discovered the constancy of the earth’s acceleration - and immediately concluded that it is independent of the distance to the Earth! This is how the founder of almost all areas of mechanics was mistaken *).

Thus, in a thick mixture of pronunciation and error, without rights and wrongs, science developed. But at the same time, everyone we were talking about were geniuses: their works have been in use for several centuries; they have advanced humanity many decades into the future (and what, if not the number of man-years saved, determines the degree of talent?); it took more time to understand and master their ideas - a great truth is rarely born simple and understandable **).

The presented history of the development of science, of course, is extremely simplified and written in crude masks, sometimes with a popular condensation of facts. But it is clear that development does not proceed along the straight line of a telephone pole, but rather along the complex curve of the tree of science, the branches of which are inclined at different angles to the truth and are gradually straightened to rise to a new language of knowledge.

*) About the existence of geniuses. Newton, out of scientific conscientiousness, did not publish his law of universal gravitation for 20 years, trying to derive Kepler’s laws from it, and then spent many years arguing about nature with Hooke; did not publish his discovery of differential and integral calculus, trying to find a general formula for the integral, and then proved his nature in a dispute with Leibniz. For Newton lived only in science, throughout his life he was lonely and could not sacrifice his only joy in life - weak, although weak his discoveries would have been enough for many.

**) Here, for example, is the multiplication rule from a 15th century manuscript: 7×8=? 7+8-10=5, then 5x10=50, now 10-7=3, 10-8=2 are defined and these results are variable: 3x2=6. Finally, 50+6=56, so 7x8=56.

To whom
What
It’s clear

The outstanding Russian geometer and teacher A.K. Vlasov, having presented the problem of the intersection of two lines defined by his equation at a regular lecture on analytical geometry, added:

— Two students, having become familiar with this issue for the first time, were talking among themselves.

One said: “Now I understand why a system of two first-degree equations with two unknowns generally has one solution. This is because the two lines intersect at one point.”

Another replied: “That’s when I finally understood why two lines intersect at the same point.”

Another replied: “That’s when I finally understood why two straight lines intersect at one point.”

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Parquets made of regular polygons

A. N. KOLMOGOROV

WHAT IS PARQUET

The simplest, but also the most boring parquet is obtained if the plane is divided into equal squares as shown in figure 1, a. Here two squares either have a common side, a common vertex, or no common points at all.

We will call parquet any covering of a plane with regular polygons, in which two polygons have either a common side, a common vertex, or no common points at all.

You should probably see a parquet made of regular octagons and squares (Fig. 2, a). A beautiful parquet can be made from regular hexagons, squares and equilateral triangles (Fig. 2, b).

Parquet makes a pleasant impression if it is sufficiently symmetrical. A figure is called symmetrical if it can be superimposed on itself in a “non-attributive” way (i.e., without a maxim, when all the points remain in their place).

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For example, at Fig. 2, 6, rotating the entire grid of vertices and sides that makes up the parquet of hexagons, squares and triangles 60° around the center of one of the hexagons, we get the same grid of vertices and sides. The center of each hexagon of this parquet is the “center of the sixth order symmetry”*).

Problem 1. Find all symmetry centers of the 4th, 3rd and 2nd order of the parquet shown at Fig. 2, a.

WHAT IS CORRECT PARQUET
From a symmetry point of view, our definition of parquet is not too far off. It allows parquets that do not have any symmetry. Taking a common parquet of hexagons (Fig. 1, c), you can “spoil” it by subdividing some of the hexagons into six triangles. It is easy to understand that the result will be “parquet” again in the sense of our definition. But it can be proven (try it!) that by subdividing, for example, three hexagons, as shown in figure 3, and leaving all the rest undivided, we get a parquet floor that is completely devoid of symmetry. To install ugly, insufficiently symmetrical parquet floors, we use the following definition:

A parquet is called regular if it can be superimposed on itself in such a way that any given vertex of it can be superimposed on any other given vertex of it.

Task 2. Prove that the parquets presented in Figures 1 and 2 are correct and build as many correct parquets as possible on your own.

*) A point O is called the nth-order symmetry center of a certain figure if, when this figure is rotated around O by $\frac{360^\circ}{n}$, it overlaps itself.

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MAIN OBJECTIVE

It turns out that all the variety of correct parquet floors can be described. If the length $h$ of the side of the parquet polygons is given, then there is a finite number of different (non-overlapping) regular parquets. I don’t want to tell you exactly how much.

Listing them all and thereby answering the question about their number is the main task that you have to solve.

SOME GUIDELINES

The solution to the natural problem begins with the use of parquet tops. A regular $n$-gon has $n$ exterior angles (Fig. 4), the sum of which is equal to four right angles (see this for yourself). Therefore, each angle of a regular $n$-gon is equal to

$$\alpha_n = 2d - \frac{4d}{n} = 2\left(1 - \frac{2}{n}\right)d.$$

Polygons with a sum of angles equal to 4d must converge at the vertex of the parquet. So,

$$\alpha_3 = \frac{2}{3}d, \alpha_4 = d, \alpha_6 = \frac{4}{3}d, \alpha_8 = \frac{3}{2}d$$

and for the parquets shown in Fig. 1 and 2, we have:

$$4\alpha_4 = 4d,$$

$$6\alpha_3 = 4d,$$

$$3\alpha_6 = 4d,$$

$$\alpha_4 + 2\alpha_8 = 4d,$$

$$\alpha_3 + 2\alpha_4 + \alpha_6 = 4d.$$

In the general case, denoting by $m_n$ the number of applications to the vertex of $n$-gons, we should obtain

$$\sum m_i\alpha_i = 4d,$$

(1)

where in the sum we include terms with those numbers $i$ for which $m_i > 0$ and $\alpha_i = 2\left(1 - \frac{2}{i}\right)d$.

Our first task is to find all solutions to equation (1) with integers $m_i > 0$. Equation (1), reducing by 2d, can be conveniently written in the form

$$\sum m_i\left(1 - \frac{2}{i}\right) = 2.$$

For each solution to equation (2), one must use the corresponding

31

stretch completely differently. At the center, the tension of the ball is obviously zero, near the surface it is maximum, and between the center and the surface it changes, as mathematical calculations show, according to the law of direct proportionality: the farther from the center, the greater the tension. (The graph of changes in the tension of a homogeneous ball depending on the distance to the center is shown at figure 2.)

Even more curious is the case of a ball having a sospherical cutout in the middle. The famous science fiction novel by V. A. Obruchev “Plutonia” tells about travels into the bowels of the Earth, and our playa

neta turns out to be hollow. A red-hot body was discovered in the center of the travel cavity. Unlike the author of the novel, let us assume that there is no body at the center of the Earth and that the Earth is a homogeneous ball with an empty spherical core. It can be proven that, whatever the radius of the “cut”, inside the cavity at any point (and not just in the center!) tension is completely absent. This, in particular, suggests that weightlessness should dominate in it, which, however, is explained by a different reason than weightlessness during space flights.

figure 3 shows a graph of changes in the gravity of a hypothetical hollow Earth depending on the distance to the center. Inside the massive spheroidal layer, the tension returns as it moves away from the center and reaches a maximum on the surface. Off-surface field

gravity is the same as that of a material point located in the center of this hollow ball.

The shape of the earth is quite close to a sphere. But our planet is not homogeneous: the density of its constituent rocks at different points is different. The Earth’s core has a much higher density. It is plausible to assume that the density of earth’s rocks at equal distances from the center of the Earth is the same. In other words, the Earth can be, to a first approximation, imagined as a state of homogeneous spherical layers laid out into each other. Such a body is said to have spheroidal symmetry in its density distribution. It can be proven

It is clear that such a ball, like a homogeneous one, hides any particle outside its surface or on it as if all its mass were concentrated in the center.

But the real Earth is not a ball, but a so-called spheran. Our planet can be considered as a sphere attached to the stripes with a radius equal to the equatorial radius of the Earth (Fig. 4,a). The magnitude of the applied is 21 km. On the other hand, the same spherad can be imagined as a ball with a radius equal to the polar radius of the Earth (6357 km), to which mass is added on the sides - the so-called “equatorial bulge” (Fig. 4, b). The thickness of the bulge at the equator is 21 km. The attraction of any material particle by spheran can be considered as a state of two parts: from the attraction of the central ball of radius 6357 km, depicted in Fig. 4,b,

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ONCE UPON (the matter arose at the beginning of our century), at a meeting of the academic council of the university, the head of the theology department spoke, who stated that physics students were very unserious about his subject, inattentive in lectures, and an outrageous example was given.

“I picked up one at the lecture, who himself did not push and did not give to others, and said to him: “Young man, I was just explaining what divine power is. Repeat! And he says: “Divine power is the product of divine mass and divine acceleration, the product will include the divine squared, therefore, to obtain divine power, only one of the collaborators must be divine.

This historical anecdote well illustrates the well-known rule: if, during calculations and transformations with naming, the dimensions of the left and right sides of the equation (or inequality) are different, then there is an error in the solution.

This rule often helps to remember the exact spelling of half-written formulas or physical laws. So, knowing that the formula for the period of oscillation of a mathematical matrix includes length (in cm) and acceleration of gravity (in cm/sec²), we can agree that to obtain the period in seconds, we need to divide the length by the acceleration and take the square root:

$$T = k \sqrt{\frac{l}{g}}.$$

We had to put $k$ because we don’t remember the formula exactly, and the dimensionless coefficient cannot affect its correctness. (Of course, dimension theory won’t help you remember that $k = 2\pi$.)

If you want to get acquainted with the “secrets of dimension” in more detail, we recommend that you read the small and accessible brochure by B. Yu. Kogan “Dimension of a physical quantity” (Nayka publishing house, 1968)

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ness, we will use both si-
nonima: both energy and mass.

• What about philosophers?
  — Philosophers are different from philosophers.
  Why look for problems where there are none, why monitor the use of words, where the true meaning is not in words, but in exact relationships…
  People with a practical mindset are concerned about something else: is it true that the energy released in the decay of elementary particles is much greater than nuclear energy?
  Yes. For example, one of the cycles of nuclear reactions that give energy to stars coincides with the transformation of four protons into a gel nucleus. Their mass is 0.94·4=3.76 GeV, and the mass of the gel is 3.73 GeV, therefore, 0.03 GeV is released - less than 1% of the total energy. And in the decay of the π° meson into photons, the entire mass of the meson (100%) is converted into energy.

• So, we have before us a source of energy more powerful than a thermonuclear reaction?

• Not at all. The rarity and instability of such mechanisms is a hindrance. Next to the accelerator there are at least 10¹⁹ times fewer of them than protons. In addition, they die instantly; they cannot be accumulated. And most importantly, they must be created, spending energy on this exactly equal to that released during their decay; protons are always ready - these are hydrogen nuclei. In a thermonuclear reaction, we transfat the energy reserves accumulated by nature; decays of π° mesons, at best, only return to us the energy spent on their creation.

“Then what’s the use of them?”
— Tolk from mezons, hyperons, etc. in another: they provide clues to the structure of the world…

TASKS

1. Calculate the momentum of the proton in MeV/c, where c is the speed of light, if it is known that the kinetic energy of the proton is 500 MeV.
2. Calculate the acceleration of an electron moving into an accelerator along the lines of direction of a uniform electric field with E = 25 kV/cm, at the moment when its kinetic energy is T = 0.5 m₀c², where m₀ is the rest mass of the electron.
3. What mass of hydrogen is converted into energy when a 20-megaton hydrogen bomb explodes? The equivalent of trinitrotoluene is 10³ cal/2.
4. How much mass does a 100-watt light bulb emit (in the form of tell and light) over the course of a year?
5. How to determine the rest mass of a system of moving particles? Is it equal to the sum of the masses of the particles at rest?
6. How long does a person have to ride on the magnitude to convert it into energy? When a person cranks the teacher as hard as he can, he produces 0.5 net horsepower (373 bhp). The efficiency of the human body is about 25%, that is, 75% of food is burned, turning into heat, and only 25% of food goes into useful work.
7. How much mass does the Earth receive from the Sun in the form of light during the year if 1.4 km of light energy arrives per square meter of surface perpendicular to the direction of the sun’s rays?
8. What is the phonon impulse? Show, based on the laws of conservation of energy and momentum, that a free electron cannot absorb a phonane.

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WHY DO WE SAY THIS?

sinus
chorda

1. WHERE DID THE WORD COME FROM?
   “SINUS”

The word “sine” is of Latin origin. If we look in the Latin-Russian dictionary, we see the following meanings of this word: 1) curvature, curvature, bending, convexity; 2) loose, pocket, fold of that (ancient Roman clothing) on ​​the chest; 3) dress, clothing; 4) breasts, hugs; 5) tender love, care;

6. middle, center; 7) refuge, shelter; 8) bay, bay; 9) vladina, deepening, failure. The word “sinus” is well known in the meaning of “sinus”, “cavity”. However, none of these many meanings have anything to do with sine in trigonometry. Where did this term come from?

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44

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2. WHERE CAME FROM
   MATHEMATICAL TERM
   “ROOT”

The word “root” is a Russian word, but the mathematical meaning of this word has a very long and interesting history.

Ancient Greek scientists - Pythogoreans (5th century BC) associated geometric concepts with numbers and depicted the product of two and three factors in the form of rectangles and rectangular parallelepipeds, the sides and edges of which are equal to som factors. Therefore, they call the product of two and three som factors of the corresponding “flat” and “solid” numbers, and the product of two and three equal som factors - “square” and “cubic” numbers, respectively. It is this terminology that is used in the varithmetic books of Euclid’s Elements (3rd century BC), a remnant of this terminology is our terms “square” and “cube” for numbers of the form $n^2$ and $n^3$.

According to Pythogorean terminology, the root of a “square” number is called “side” (in Greek “pleeura”), which also means “shell”, from which “pleeuritis” is an inflammation of the membrane of the lung, and “base” (in Greek “basis”), from which our word “base”.

When in the 5th century A.D. e. The Alexandrian scientific school perished, and the Alexandrian astronomer Paulos fled to India, the term “base” was translated into Sanskrit by the word “pada”, meaning base, as well as the root of a plant.

In the 8th century, when translating the “siddhantas” of Indian scholars into Arabic, the translator understood the words “pada” as “root” and translated it into the Arabic word “jizr”, denoting the root. In parallel, in Arabic mathematical literature there was another term for the root of a number - “dill” (“side” or “edge”), a translation of the Greek word “pleevra”: if the word “jizr” was used for square roots and roots of quadratic equations, then the word “dill” was used for roots of calculus powers: mathematicians writing in Arabic called the cube root “edge of the cube”, the root of the 4th degree - “edge” square-square”, the root of the 5th degree is “edge of a square-cube”, the root of the 6th degree is “edge of a cube-cube”, etc.

In the 12th century, when translating Arabic terms into Latin, the word “jizr” was translated by the word “radix”, also denoting a root (from this word, which also denoted a “root vegetable”, our word “radish” comes from), and the word “dill” was translated by the word “latus”, also denoting a side and a rib. In Magnitsky’s “Arithmetic” the word “radix” was left without translation - “radix”, and the word “latus” was translated by the word “side”, but later both of these terms were coined by the Russian translation of the word “radix” - the word “root”. However, sometimes the term “radical” is also used with this word, which also comes from the Latin term “radix”.

B. A. ROSENFELD

49

However, it was carried out not on all problems, but on four that were best solved.

The task was chosen in such a way that a geometry lover who did not want to bother with any formal calculations could get an A, but in order to prove his “right” to neglect technical transformations - this is also one of the components of mathematical art - he had to overcome special geometric difficulties. In the same way, an “algebraist” or “analyst” who dislikes geometry could spare himself the trouble of understanding a relatively complex geometric configuration, but “in return” was obliged to solve a non-standard problem of a formal nature.

To get a grade of “3” it was enough to solve two problems. Despite this not very strong criterion, the number two turned out to be internal - 224*). Some unsuccessful works were written, unfortunately, by medal winners and graduates of mathematical schools, including Moscow ones.

But what is especially unpleasant is that most bad works contain errors that cannot be called anything other than ugly. The imagination of the authors of such works is too great to be able to list all these “pearls.” However, we will still mention one that surprised even seasoned examiners: in one of the works on the drawing for the problem where I decided on the common chord of two circles, these circles (different!) also had a common arc, and outside this arc they could diverge in different directions.

All this means that not all applicants to colleges know mathematics even at an elementary level; many of them have significant problems in mathematical processing. And it’s not so much about the breadth, but about the depth of knowledge, in the mind to note that both Muscovite and nonresident applicants did about the same about the mathematics exam (written and oral).

Option

1. Solve the equation
   $$\sqrt{1 - \log_2 x^2} \cdot \log_2 x = -1.$$

2. Solve inequality
   $$2 \cos x (\cos x - \sqrt{8} \mathrm{tg} x) < 5.$$

3. The rectangle $ABCD$ from the sides $a$ and $2a$ lies at the base of the rectangle, all of whose lateral edges are equal to $\sqrt{3}$. Find the cross-sectional area of ​​the straight line passing through the diagonal of the base $BD$ parallel to the edge $AS$.

4. A straight line is drawn through the end $A$ of the common chord $AB$ of two circles, intersecting the first circle at point $C$, and the second circle at point $D$. Prove that the intersection point of the tangents drawn through points $C$ to the first circle and through points $D$ to the second circle lies on the same circle with points $B$, $C$ and $D$.

5. Prove that if
   $$3x^2 - 31x + 80 < 0,$$
   That
   $$\cos \frac{3}{6-x} < 0.$$

Solutions

Task 1. Since $\log_2 x^2 = 2 + \log_2 x$ and $\log_2 x^4 = \frac{1}{\log_2 x^2} = -\frac{1}{4 + \log_2 x}$ are given, the equation is reduced to the form
$$\sqrt{3 - \log_2 x} = -4 - \log_2 x.$$
Denoting $3 - \log_2 x$ by $y$, we obtain the equation
$$\sqrt{y} = y - 1.$$

Surprisingly, the solution to even such a simple equation was not

50

many applicants can do it. Everyone agrees to square both sides and obtain a quadratic equation with roots $y_{1,2} = \frac{3 \pm \sqrt{5}}{2}$, but then different “points of view” are revealed; Some people don’t even think about the fact that with such a solution, roots may appear; others know that this can happen, but do not know why, and, as a rule, are confident that those roots that are not included in the permissible value range will be constructed; Still others try to check the roots by directly substituting them into the equation - this method is correct in principle, but is convenient only for checking “good” roots, and $y_1$ and $y_2$ are not very “good”.

However, it is absolutely inconceivable that since the left side of equation () is non-negative, then of the numbers $y_1$ and $y_2$ the root will be only that for which the right side is also non-negative. Therefore, the only root of equation () is $y = \frac{3 + \sqrt{5}}{2}$, and to find $x$ we get the equation $\log_x 2 = \frac{-(9 + \sqrt{5})}{2}$, which implies $x = 2^{-\frac{2}{9 + \sqrt{5}}}$.

We emphasize: when solving, we completely dispensed with finding the ODZ of the original equation. Meanwhile, recently there has been a widespread prejudice among applicants that the solution to any equation or equality must be indicated starting with the calculation of the ODZ. This opinion, however, has no theoretical basis, and in practice it does a disservice: those who followed this path had to solve not only this equation, but also the logarithmic equation $-1 - \log_x 2x^2 \geq 0$, which he himself invented - a problem no less difficult. In general, there is no strictly defined rule whether it is necessary or not to find the ODZ at the beginning of the decision, and already here, when choosing a path, the applicant can show how much he can think concretely and independently.

Problem 2. We will not solve this inequality in detail. Opening the brackets, we easily arrive at the quadratic inequality for sin $x$:

$$2\sin^2x + 4\sqrt{2}\sin x + 3 > 0,$$

solving which we obtain the simplest trigonometric inequality $\sin x > -\frac{\sqrt{2}}{2}$. But… this is precisely what most applicants failed to cope with. We will not analyze the data and answers, which are often completely meaningless, and simply draw the attention of future applicants to such tasks. The solution method is sufficiently described in the textbook and in many manuals, and you just need to be aware of it. Let us also note that even those who correctly solved the simplest inequality did not always remember that from the resulting solutions it was necessary to discard the values ​​$x$ for which tg $x$ does not make sense.

Problem 3. It turned out to be too “treacherous” for those applicants who are accustomed to solving problems only on regular geometric bodies. They were unable to reject patterns and erroneously imagined a given geometric configuration. Thus, many people believed “out of habit” that the height of $\triangle BKD$ is the segment $KO$ (Fig. 1).

But everyone knows that geometric equation is only an auxiliary means of solving the problem, and the facts “seen” in the drawing still need to be strictly confirmed.

51

show. But, unfortunately, many people know this purely abstractly, but in practice they neglect the proofs, sometimes limiting themselves to phrases like “this is true by the theorem of three perpendiculars” even in the case when it is completely unclear which perpendiculars we are talking about.

Meanwhile, a small attempt to prove the perpendicularity of $KO$ and $BD$ immediately leads to opposition. Indeed, if this were so, then the diagonal $BD$ would be perpendicular to $SO$ and $KO$, that is, to the plane $ASC$, and therefore to the diagonal $AC$. But the diagonals of the base, as is absolutely clear from the condition, are not perpendicular.

And to correctly construct the height at $\triangle BKD$, it is necessary from point $K$ to carry out a perpendicular to the plane of the base of the paramid, prove that its base $O_1$ lies on $AC$ (having previously proven that the height of the paramid passes through the point $O$ of intersection of the diagonals), and from point $O_1$ to carry out a perpendicular to $BD$. The segment $KL$, according to the theorem of three perpendiculars, will have a height of $\triangle BKD$. To complete the task, one more statement will come up - the parallelism of $OK$ and $AS$, and this also needs to be proven. After this, the computational part of the problem proceeds without difficulty.

Problem 4. Among the proposed problems in planology, this turned out to be one of the most difficult for applicants: out of 93 who solved the corresponding option, only 3 people completely solved it. Meanwhile, it has a very simple and, most importantly, perfect natural solution in two replacement books by D. Polya “How to solve a problem.”

Indeed, what does it mean that the point $K$ (Fig. 2) lies on the indicated circle? This means that a circle can be circumscribed around the quadrilateral $BCKD$. When can a circle be drawn around a quadrilateral? When the sum of its perpendicular angles is equal to $180^\circ$. Therefore, from the conditions of the problem it is necessary to extract some statement about the equality of angles.

What theorems about the equality of angles can come to mind in connection with this problem? Only theorems on angles between tangents and chords and inscribed angles. Based on these theorems, the angles marked identically on figure 2 are equal. And now one cannot help but report that the sum of the opposite angles of the quadrilateral $BCKD$ at the vertices $B$ and $K$ is equal to the sum of the angles $\Delta CKD$, that is, $180^\circ$. Of course, it already follows from this that the sum of the other two opposite angles is also equal to $180^\circ$, and the required statement is proven.

The failure of the majority of those who solved this problem is apparently simply explained by the fact that schoolchildren in the 9th and 10th grades largely forget the simplest planimetric theorems and do not bother to fill this gap in the process of direct preparation for exams. It is quite likely that 8th grade graduates would cope with this task with greater success than applicants who are wise in general knowledge.

Problem 5. Tasks of this type turned out to be of medium difficulty, and most of those who took on their solution completed it more or less correctly. It must be said, however, that many people did not take on this task, although it is simpler than others. Here, apparently, the “psychological unpreparedness” still manifests itself.

52

ity" to the unusual form of the problem, although recently quite a lot of such problems have been considered in the literature for schoolchildren.

Here is the solution to this problem: if $x$ satisfies the first inequality, then $5 < x < \frac{16}{3}$, hence $\frac{2}{3} < 6 - x < 1$ and, therefore, $3 < \frac{3}{6 - x} < \frac{9}{2}$. Since $\frac{\pi}{2} < 3 < \frac{9}{2} < \frac{3\pi}{2}$, the angle $\frac{3}{6 - x}$ lies in either the second or third quadrant, and therefore its cosine is negative.

So, we have presented a solution to our problems, some difficult, some easy. You can evaluate their difficulty, and in turn, your strength, by solving problems from the two options below.

Options

1. Solve the equation
   $$\log_{x+3}(3 - \sqrt{1 - 2x + x^2}) = \frac{1}{2}.$$

2. Solve inequality
   $$6\sin^2\frac{x}{2}\mathrm{ctg}^2x \leqslant 2\cos^2x + 3\mathrm{ctg}^2x.$$

3. At the base of the base $SABC$ lies $\Delta ABC$ with an angle $C=90^\circ$ and an acute angle $30^\circ$. The lateral edges of the prim are inclined to the plane of the base at an angle of $45^\circ$, the height of the prim is $H$. Find the cross-sectional area of ​​the prim by a plane passing through the point $C$ parallel to $AB$ and dividing the faces $ASB$ into parts of equal area.

4. Prove that if a line connecting opposite vertices of an inscribed quadrilateral passes through the point of intersection of tangents drawn at two other vertices, then the product of opposite sides is equal.

5. Prove that if $\frac{x^2 - 5x + 6}{x^2 - 11x + 30} < 0$, then sin $x$ and cos $x$ have different symbols.
   (MGPI named after V.N. Lenin)

6. In a regular triangular triangle, take the side of the base equal to $a$, the angles between the edges at its vertex equal to $\alpha$ ($\alpha \leqslant 90^\circ$). Determine the area of ​​the section drawn through the sides of the base perpendicular to the opposite side edge.

7. Solve the equation
   $$\lg^2 x + \lg^2 (x + 10) = 2\lg 11.$$

8. Solve the equation
   $$\sin 3x \cdot \sin x + 1 = 0.$$

(Ivanovo Pedagogical)

UNITY IS STRENGTH

“While algebra and geometry each developed in its own way, its progress was slow and its applications limited.

But when these sciences united, they energetically supported each other and quickly marched towards perfection.”

Lagrange
NOT ENEMIES, BUT FRIENDS

“It is a mistake to think that rigor in proof is the enemy of simplicity. For example, you can apply the fact that the strict method is at the same time simpler, easier and more accessible. Every effort in the rigor of rigor is directed towards finding the simplest methods of proof.”

Gilbert
4, 5 and 6

4 identical balls can be placed so that each one touches the other three.

5 identical balls can be placed so that each one touches the other four.

Is it possible to arrange 6 identical unsharpened pencils so that each one touches the other five?

53

SOMETHING ABOUT RADICALAX

A. G. MORDKOVICH

“And now,” said the Examiner, “I want to offer you one more task, a very simple one.” Here it is:

Find the sides of a right triangle if it is known that the projection of one kamet on the sinomendus is 6 cm, and the projection of another kamet is 1.5 cm less than the length of this kamet.

… A few minutes later the Applicant goes to the Examiner together.

• Well, how did you solve the problem? - asked the Examiner.

• Yes! - answered the Entrant.

• And what happened?
  — The problem has no solution.

• And you can prove it?

• Certainly! Consider triangle ABC (see figure); C is the top of a right angle, CH is the height. Let AC = x, then by condition AH = = x—1.5, and BH = 6.

54

We use the known relation
$$CH^2 = BH \cdot AH$$ or $$CH^2 = 6(x - 1,5).$$

Applying the Pythagorean theorem to the right triangle $ACH$, we obtain

$$\sqrt{AC^2 - CH^2} = AH,$$

that is

$$\sqrt{x^2 - 6(x - 1,5)} = x - 1,5.$$

“Well, so far everything is correct,” the Examiner noted, “although we must say that the solution path chosen was not the most rational.”

“And then,” continued the Entrant, “there are simple calculations left:

$$\sqrt{x^2 - 6x + 9} = x - 1,5; \quad \sqrt{(x - 3)^2} = x - 1,5;$$

$$x - 3 = x - 1,5; \quad -3 = -1,5.$$

Since the last equality is false, the problem has no solution.

“Unfortunately, I must disappoint you,” said the Examiner, “the problem has a solution.” It can be solved, for example, like this: from the known relation $AC^2 = AB \cdot AH$ we have $x^2 = (x + 4,5) (x - 1,5), 3x = 4,5 \cdot 1,5$ and $x = 2,25$, that is, $AC = 2,25$ cm. Then $AB = 6,75$ cm, and $BC = 4\sqrt{2}$ cm.

And, catching the Applicant’s puzzled look, the Examiner added:

• You are apparently interested in where the error is in your reasoning. Well, let’s figure it out.

WHERE IS THE ERROR?

“You remember,” the Examiner began, “that I, although with a reservation, approved your geometric reasoning.” So the error could only have occurred in what you called “simple calculations.” More precisely, an error was made when extracting the square root. You did not teach that, according to the meaning of the problem, the quantity $\sqrt{(x - 3)^2}$ must be positive. And in general, in the field of real numbers, the sign $\sqrt{}$ implies the arithmetic, non-negative value of the root. So, $\sqrt{16} = 4$, and you cannot write that $\sqrt{16} = -4$ or $\sqrt{16} = \pm 4$. Therefore, it cannot be said that $\sqrt{a^2} = a$ is always: for $a \geq 0$ this equality is true, but the number $a$ can also be negative! Simplify, for example, $\sqrt{(-3)^2}$:

The applicant wrote:

$$\sqrt{(-3)^2} = \sqrt{9} = 3.$$

• Right. In other words,

$$\sqrt{(-3)^2} = -(-3).$$

And in general, if $a < 0$, then $\sqrt{a^2} = -a$.

Thus,

$$\sqrt{a^2} = \begin{cases} a, если a \geq 0, \\ -a, если a < 0. \end{cases}$$

“I understand,” exclaimed the Applicant, “this can be briefly written as follows: $\sqrt{a^2} = |a|$, because the model $a$ is defined in exactly the same way:

$$|a| = \begin{cases} a, если a \geq 0, \\ -a, если a < 0. \end{cases}$$

• Absolutely right. It’s exactly the same in your problem: you can’t write

55

$$\sqrt{(x-3)^2} = (x-3), \text{ потому что } (x-3) \text{ может быть и отрицательным числом.} \text{Если воспользоваться тем, что } \sqrt{(x-3)^2} = |x-3|, \text{ то приdem к управлению } |x-3| = x-1, 5.$$

Next we should consider two cases: 1) $x-3 \geq 0$; 2) $x-3 < 0$. You only considered the first case, so the conclusion that the problem has no solution was made prematurely. In the second case $|x-3| = -(x-3)$. Then from the equation $-(x-3) = x-1, 5$ we obtain $x = 2, 25$, that is, $AC = 2, 25$ cm.

“Everything is clear,” said the Entrant.
In this case, we will consider our conversation finished. To beg your pardon, let me give you the following advice:

Don’t forget that $\sqrt{a^2} = |a|!$

We will discuss this with the Applicant and the Examiner, but we will take the Examiner’s advice into account. In particular, we will use it when solving the following example.

Example 1: Simplify an expression

$$\sqrt{\frac{a+x^2}{x} - 2\sqrt{a}} + \sqrt{\frac{a+x^2}{x} + 2\sqrt{a}}, \text{ если } x > \sqrt{a}.$$

Solution. We have

$$\sqrt{\frac{a+x^2-2x\sqrt{a}}{x}} + \sqrt{\frac{a+x^2+2x\sqrt{a}}{x}} =$$

$$\frac{\sqrt{(V\bar{a}-x)^2}}{\sqrt{x}} + \frac{\sqrt{(V\bar{a}+x)^2}}{\sqrt{x}} = \frac{|V\bar{a}-x|}{V\bar{x}} + \frac{|V\bar{a}+x|}{V\bar{x}}.$$

Since by condition $x > \sqrt{a} \geq 0$, then $|V\bar{a}-x| = -V\bar{a} + x$, $|V\bar{a}+\bar{x}| = \sqrt{a} + x$, and this expression is transformed as follows:

$$\frac{-V\bar{a} + x + \sqrt{a} + x}{\sqrt{x}} = 2\sqrt{x}.$$

A FEW WORDS ABOUT THE PROPERTIES OF RADICALS

When solving examples of actions with radicals, one has to use various properties of radicals. So, in example 1 we used, and quite carefreely, the well-known property:

$$\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{b}.$$

However, it should be taken into account that this equality occurs only in the case when $a \geq 0$ and $b > 0$. If $a$ and $b$ are negative numbers, then the notation $\frac{\sqrt{a}}{b}$ does not make sense in the field of real numbers. In example 1, on our part, no error occurred, since $(V\bar{a}-x)^2 > 0$, $(V\bar{a}+x)^2 > 0$ and, in addition, by condition $x > 0$. But, in any case, from the above we can draw the following conclusion: you must be very careful when revealing the root (of the degree) of a product or a fraction. In the general case, when the sign of $a$ and $b$ is unknown, it is advisable to carry out the following reasoning:

$$\sqrt{\frac{a}{b}} = \sqrt{\frac{|a|}{|b|}} = \sqrt{\frac{|a|}{|b|}}.$$

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So, if the numbers $a$ and $b$ have the same symbols, then

$$\sqrt{\frac{a}{b}} = \frac{\sqrt{|a|}}{\sqrt{|b|}}$$

and similarly

$$\sqrt{ab} = \sqrt{|a|} \cdot \sqrt{|b|}.$$

Example 2: Simplify

$$\sqrt{\frac{a-\sqrt{x}}{a+\sqrt{x}}} + \sqrt{\frac{a+\sqrt{x}}{a-\sqrt{x}}} - \sqrt{\frac{16}{a^2-x}},$$ if $x = 4(a-1)$.

Solution. Let us express, on the contrary, $a$ through $x$: $a = \frac{x+4}{4}$, and substitute it into this expression. It is converted like this:

$$\sqrt{\frac{x-4\sqrt{x}+4}{x+4\sqrt{x}+4}} + \sqrt{\frac{x+4\sqrt{x}+4}{x-4\sqrt{x}+4}} - \sqrt{\frac{16^2}{(x+4)^2-16x}}$$

$$= \frac{|\sqrt{x}-2|}{|\sqrt{x}+2|} + \frac{|\sqrt{x}+2|}{|\sqrt{x}-2|} - \frac{16}{\sqrt{x^2}-8x+16}$$

$$= \frac{(x-4\sqrt{x}+4) + (x+4\sqrt{x}+4)}{|(\sqrt{x})^2-4|} - \frac{16}{|x-4|} = \frac{2x+8}{|x-4|} - \frac{16}{|x-4|} = 2\frac{x-4}{|x-4|}.$$

Note also that $x > 4$ if $a > 2$, and $x < 4$ if $1 \leqslant a < 2$.

Final answer:

2 if $a > 2$,

• 2 if $1 \leqslant a < 2$.

We used - and this is often useful - the following property of the module: for all $a$ and $b$

$$|a| \cdot |b| = |ab|, \quad |a|^2 = a^2.$$

And now we will dwell on one more important property of the root.

Example 3: Simplify

$$\sqrt[10]{\frac{1}{2}(19+6\sqrt{10})} \sqrt[5]{3\sqrt{2}-2\sqrt{5}}.$$

Solution. People often think like this. Using the well-known property of the root, you can show the root of the second factor by multiplying by 2 and at the same time squaring the subordinate expression; we get

$$\sqrt[10]{\frac{1}{2}(19+6\sqrt{10})} \sqrt[5]{3\sqrt{2}-2\sqrt{5}}$$

$$= \sqrt[10]{\frac{1}{2}(19+6\sqrt{10})} \sqrt[10]{(3\sqrt{2}-2\sqrt{5})^2}$$

$$= \sqrt[10]{\frac{1}{2}(19+6\sqrt{10})} \sqrt[10]{19^2-(6\sqrt{10})^2} = \sqrt[10]{1} = 1.$$

But there is a mistake here! The correct answer is minus one. The point is that $3\sqrt{2}-2\sqrt{5} < 0$ because $(3\sqrt{2})^2 = 18 < (2\sqrt{5})^2 = 20$.

Therefore, it is not true that $\sqrt[5]{3\sqrt{2}-2\sqrt{5}} = \sqrt[10]{(3\sqrt{2}-2\sqrt{5})^2}$, but it is true that $\sqrt[5]{3\sqrt{2}-2\sqrt{5}} = -\sqrt[5]{2\sqrt{5}-3\sqrt{2}} = -\sqrt[10]{(2\sqrt{5}-3\sqrt{2})^2}$.

Example 4: Simplify $\sqrt[8]{(\pi^2-10)^4}$.

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Here, too, it would be incorrect to simply divide the indicator of the root and radical expression by 4:

$$\sqrt[8]{(\pi^2 - 10)^4} = \sqrt{\pi^2 - 10}$$

(the result does not make sense because $\pi^2 < 10$). The following equality is true:

$$\sqrt[8]{(\pi^2 - 10)^4} = \sqrt{10 - \pi^2}.$$

As a result, we come to the following conclusion:

Using the basic properties of radicals in cases where you are not sure that the root contains a positive number, you need to ensure that, firstly, the result obtained makes sense (if necessary, the modulus sign will help us out) and, secondly, that the result obtained has the same sign as the first analogous expression.

TASKS

1. List the conditions under which the following equalities are true ($k$ and $n$ are natural numbers):

a) $$\sqrt[n]{-a} = -\sqrt[n]{a};$$
e) $$a \cdot \sqrt[n]{b} = \sqrt[n]{a^n b};$$
6) $$\sqrt[n]{a^n} = a;$$
g) $$\sqrt[n]{a} = \sqrt[n]{a^k};$$
b) $$(\sqrt[n]{a})^n = a;$$
h) $$\sqrt[k]{\sqrt[n]{a}} = \sqrt[k]{a};$$
d) $$\sqrt[n]{a \cdot b} = \sqrt[n]{a} \cdot \sqrt[n]{b};$$
i) $$(\sqrt[n]{a})^k = \sqrt[n]{a^k};$$
e) $$\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{n \sqrt[n]{b}};$$
j) if $a < b$, then $$\sqrt[n]{a} < \sqrt[n]{b}$$.

2. Simplify the following expressions:

a) $$(\sqrt{\frac{a+1}{a-1}} + \sqrt{\frac{a-1}{a+1}}) \cdot (\sqrt{\frac{a+1}{a-1}} - \sqrt{\frac{a-1}{a+1}});$$
6) $$\sqrt{\frac{3y+x^3}{2x}} + \sqrt{3xy} - \sqrt{\frac{3y+x^3}{2x}} - \sqrt{3xy};$$
b) $$\frac{(4a^2m^{-2} + a^{-2}m^2 - 4)^{-1/2}}{2ma (m^2 - a^2)^{-1/2}} \cdot \left(2 \sqrt{m^4 - \frac{a^2}{m^{-2}}} - \frac{2a^2}{\sqrt{1 - a^2}m^{-2}}\right);$$
d) $$\frac{a + \sqrt{2 + \sqrt{5}} \cdot \sqrt[12]{(9 - 4\sqrt{5})^3}}{\sqrt[3]{2 - \sqrt{5}} \cdot \sqrt[6]{9 + 4\sqrt{5}} - \sqrt[3]{a^2} + \sqrt[3]{a}}.$$

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WRITING

EXAM
in physics is conducted in many universities across the country,
but only in three of them,
in addition to oral
There is also a written exam.
It’s done
at the Moscow Physico-Technical Institute,
as well as in physics departments
Moscow and Novosibirsk universities.
A written assignment usually contains 4-5 tasks.
It will respond to its execution
at the Physico-Technical Institute
5 o’clock
at Moscow State University and Novosibirsk State University - 4 hours.
Below are the options,
offered to applicants last year
to the Physico-Technical Institute and NSU.
Options
Faculty of Physics, Moscow State University
will be published
in the next issue of the magazine.

Option 1 (MIPT)

1. Two particles - one of mass $m$, flying with a speed of $v$, the other of mass $2m$, flying with a speed of $2v$, directed perpendicular to $v$ (Fig. 1) - are subject to forces of equal magnitude and direction for some time. By the time the forces cease, the right side turns and begins to move in the opposite direction at a speed of $2v$, as shown in figure. At what speed did the second part begin to move?

2. A vessel with a volume of 1.1 l contains 100 $e$ adsorbent and hydrogen at a temperature $t = -193^\circ$ C and a pressure $P = 0,2$ am. The adsorbent at this temperature absorbed hydrogen in an amount equal to $\frac{1}{50}$ the weight of the adsorbent itself. Determine the pressure in the vessel if it is heated to $t_1 = 31^\circ$ C, when all the hydrogen molecules leave the adsorbent. Adsorbent density $1 \frac{e}{cm^3}$.

3. Determine the resistivity of the metal if it is known that in each cubic

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4. An observer examines a distant object using an astronomical telescope (Kepler telescope). Lenses with focal lengths $F_1 = +30\text{ cm}$ and $F_2 = +5\text{ cm}$ are used as the object and eyepiece of the telescope. An observer sees a clear image of an object if the distance between the object and the eyepiece of the scope is between $L_1 = 33\text{ cm}$ and $L_2 = 34,5\text{ cm}$. At what distances are absences observed ranging from $L_1 = 33\text{ cm}$ to $L_2 = 34,5\text{ cm}$?

Option 2 (MIPT)

1. The flywheel of radius $R = 20\text{ cm}$ is mounted on a fixed axis of radius $r = 2\text{ cm}$. The friction force between the max wheel and the axle is constant and equal to 100 kg. To make it easier to remove the max wheel from the axis, a force $F = 8\text{ kГ}$ is applied to its rim, creating a torque relative to the axis (Fig. 2). With what minimum force $N$ do you need to pull the max wheel along the axis in order to remove it?

2. The pressure gauge consists of three glass tubes (Fig. 3). There is a pipe at the bottom of the pressure gauge, and oil in the left elbow above the pipe. How much does the pressure $P$, monitored by the pressure gauge, change if the oil level in the thin open tube rises by $\Delta h = 10\text{ mm}$? The ratio of the density of the pipe and oil is $\frac{\rho_1}{\rho_2} = 14$, and the ratio of the cross-sections of the tubes is $\frac{S_3}{S_2} = 2$, $\frac{S_2}{S_1} = 50$.

3. A flat capacitor with horizontal plates is connected to a battery with an emf. With. $U$ and placed in a vessel, which is gradually filled with kerosene ($\varepsilon = 2$). Write it down as a formula and graph the dependence of the field strength in the center of the capacitor on the thickness of the kerosene layer inside it. The distance between the plates of the capacitor is $d$.

4. The distance between an object and its direct and twice magnified image obtained using a thin lens is $L$. Find the optical power of the lens and find out the nature of the image (real or minimal).

Option 3 (NSU)

1. A square grid of uniform wiring consists of four identical square houses (Fig. 4). The resistance of one side of the cell is $r$. The current enters one corner of the grid and exits the opposite corner. Find the resistance of the entire grid.

2. Between the piston and the house of a cylindrical vessel filled with air, there is a partition with a hole closed with a stopper (Fig. 5). The pressure to the right of the septum is the same. When the piston moved, the plug flew out due to a pressure difference

Fig. 5.

nia $\Delta P$. At this moment the piston was stopped. Find the steady state pressure. The temperature did not change during the entire process.

3. Masses $m_1$ and $m_2$ move along a smooth table, fastened by an insoluble thread of length $l$. At some moment the speed of mass $m_1$ is equal to zero, and the speed of mass $m_2$ is equal to $v$ and is perpendicular to nothing. Find the tension of the niggas.

4. Inside a charged sphere whose radius is $b$ and charge $q$, there is a grounded conducting sphere of radius $a < b$ (Fig. 6). The centers of the sphere coincide. Find the direction of the electric field outside the large sphere at a distance $r$ from the center.

Option 4 (NSU)

1. Two identical balls are connected by a fixed rod of length $l_0$. The system is located on a horizontal plane and rotated. The initial speed of each of the balls is $v_0$, the coefficient of friction on the plane is $k$, the acceleration due to gravity is $g$. How many revolutions will the system make before stopping?

2. How much work must be done to induce a hydrogen atom, that is, to drive an electron ($e=1,6 \cdot 10^{-19}$ cool $=4,8 \cdot 10^{-10}$ CG SE) away from a proton to a very large distance? The diameter of a hydrogen atom is $d \approx 10^{-8}$ cm. Express the result in electron volts ($1 \text{ Эθ} = 1,6 \cdot 10^{-19} \text{ Эж}$, dielectric constant of vacuum in the SI system $\varepsilon_0 = 8,8 \cdot 10^{-12} \text{ ф/m})$.

3. $P$ grams of water are poured into the glass, and a heated glass with thin walls is placed on top, turned upside down. To what minimum temperature $T_1^\circ \text{K}$ must the glass be heated so that after obstructing it to the ambient air temperature $T_0^\circ \text{K}$ all the water will be drawn into it (Fig. 7)? Atmospheric pressure $p_0$, cross-sectional area of ​​the glass $S$, height $l$, water density $ρ$. The volume of water is less than a glass. Neglect evaporation, surface tension and expansion of the glass itself.

4. A horizontal rod of weight $P$ and length $l$ slides without friction along two vertical rods connected downward by a capacitor embone $C$ (Fig. 8). A uniform magnetic field $H$ is perpendicular to the plane of descent of the rod. Find the acceleration of the rod, neglecting the electrical resistance of the formed circuit.

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The evening mathematics school at the Faculty of Mechanics and Mathematics of Moscow State University (MSH) has been operating for the seventh year. Its main goal is to promote the mathematical development of schoolchildren and expand their mathematical horizons. VMS continues the traditions of mathematical circles that have been working at Moscow University for more than thirty years, and, in essence, is itself a free mathematical circle, covering more than 800 schoolchildren. Admission to all VMS classes is free, and you can join the work of the school throughout the year.

Once a week, 6-9th grade students gather in 26 VMS groups. Classes are conducted in groups by graduate students and students of the Faculty of Mechanics and Mathematics of the university, students of the Moscow Pedagogical Institute. V.I. Lenin, Institute of Steel and other universities. In addition to solving interesting problems, the classes cover various mathematics issues that go beyond the scope of the high school curriculum. Classes are often conducted in the form of a game. Bot example. The manager says: “A cowboy walked into the bar and asked for water. Instead of answering, the owner grabbed his tights and shot at the ceiling. The cowboy thanked him and left. What’s the matter? Are you missing data? Well, I’m ready to truthfully answer some of your questions that apply to this case. Just keep in mind that questions should begin with the words: “Is it true that…” And I will only answer “yes” or “no.” After twenty minutes of questioning and probing, it dawns on someone: “The cowboy got a bone stuck in her throat and jumped out of fright!!!”

Every week, the leading Moskovs give lectures on problems of modern mathematics to students in grades 8-9.

One of the forms of work of the VMS is the “home Olympics”. Every week the children will be given the conditions of 4-5 problems, which will be given two weeks to solve. Written decisions will be shared with the group leadership for review. At the end of the cycle, consisting of 20 tasks, the jury, selected from the group leaders, sums up the results of the competition. At the competitions (with a mark!), all children must demonstrate an understanding of methods for solving competitive problems.

Classes at the Higher School of Music have a good effect on students’ studies. Thus, both Moscow, which hosted the section in the 1969 International Mathematical Olympiad as part of the Soviet team, were students of the Higher School of Music. Many current students of mechanical engineering and physics at Moscow State University and the Institute of Physics and Technology once studied at the Higher School of Medicine. Among the former students of the VMS there are students - chemists, biology, economists, psychologists…

In conclusion, we present some of the competitive problems of the Evening Mathematical School for grades 8-9. Most of them are also available to students in grades 6-7.

Tasks

1. There are 7 identical coins, but among them 5 coins are real and 2 are fake. Real coins weigh 10 e, and counterfeit coins weigh 9.8 e. What is the smallest number of weighings on a cup scale without mushrooms that must be done in order to be sure to identify counterfeit coins?

2. Sasha drew a parallelogram ABCD, marked point M - the middle of side BC, point N - the middle of side CD and went for a walk. The harmful Lila approached the drawing and erased everything except points A, M, N. Help Sasha restore the drawing, that is, find points B, C and D.

3. There are 10 people, dogs and flies in the room. Each person has 2 legs, each dog has 4 and each flour has 6 legs. Everyone in the room has 46 legs. How can this be (Find all possibilities.).

4. Also find two natural numbers whose sum of squares is 16,000. (Indicate all solutions.).

5. The country of Resolventa has a network of cities. Between any two of them there is a direct railway connection, so that from any city to any other you can get along a straight line without passing through other cities. Draw a diagram of this country’s railway communication so that there are as few road crossings as possible. (No more than two roads can intersect in one place.)

6. Given points A and B on the plane. Find the locus of points C on the plane such that triangle ABC is acute.

7. Prove that no power of 2 has four identical digits. Find the power of the number 2 that ends the triangle with the same digits.

8. Serpent Gornych has 2000 heads. The fabulous rich man can knock off 33, 21, 17 or 1 head with one mesh blow, but at the same time the Snake will lose the corresponding 48, 0, 14, 349 heads in return. If all the heads are cut off, no new heads will grow back. Will the rich man be able to satisfy the Serpent?

A. A. Orlov, A. L. Rosenal

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---

running_head:
folio:

To the article
“The law of inertia,
gel-centric system
and development of science”

1. If the acceptance of the Sun disappeared,
   the Earth’s center of gravity would begin to move tangentially to its orbit. The rotation of the Earth around its axis would be preserved.

2. The ball begins to roll on the floor of the tram when acceleration appears.

3. A body thrown horizontally with a speed $v$ from a height $h$ above the Earth, in a short time $t$ will travel a distance $x=v \cdot t$ in the direction of the speed and in a perpendicular direction $\Delta h=\frac{gt^2}{2}$, ending up at a certain height $h_1$ above the Earth. If $h=h_1$, then the body rotates around the Earth in a circular orbit. By calculating $h_1$ and putting $h=h_1$, we find that $v$ must be equal to $\sqrt{gR}$.

To the article
“Energy and momentum of fast particles”

1. $p=\frac{1}{c}\sqrt{T(T+2m_0c^2)}=$
   $=1090\ M_3\mathrm{/c}$.

2. $a=\frac{eE}{m_0}\frac{1}{\left(1+\frac{T}{m_0c^2}\right)}=1,3\cdot 10^7\ m/\mathrm{cek}^2$.

3. $0,93\ \mathrm{k2}$.

4. $3,6\cdot 10^{-5}\ e$.

5. The rest mass of the system of a part is defined as the sum of their energies (and not the rest masses) in a coordinate system in which the momentum of the system, that is, the sum of the momentum of the part, is equal to zero.

6. $4\cdot 10^6$ years.

7. The Sun is at a distance of $1,4\cdot 10^{11}\ m$ from the Earth. This means that 1.4 kW of energy is distributed in a spherical angle of $\frac{1\ m^2}{2\cdot 10^{22}\ m^2}=5\cdot 10^{-23}$ sterdian. In total, the Sun emits energy $1,4\ \mathrm{kw}\frac{4\pi}{5\cdot 10^{-23}}=3,5\cdot 10^{23}\ \mathrm{kw}$, or mass $4\cdot 10^{11}\ e$, in 1 second.

8. $p=\frac{E}{c}=\frac{hv}{c}\left(h-\text{постоянная Планка}, v-\text{частота фотона}\right)$.

To the article
“Entrance exams in mathematics”
Option I

1. $x_1=\frac{9-\sqrt{29}}{2},\ x_2=\frac{-3+\sqrt{5}}{2}$;

2. $2k\pi<x\leq 2k\pi+\frac{2\pi}{3}$;
   $-\frac{2\pi}{3}+2l\pi\leq x<2l\pi$;

3. $H^2\left(1-\frac{\sqrt{2}}{4}\right)$.

Option II

1. $\frac{a^2}{4}\sqrt{4\cos^2\frac{\alpha}{2}-1}$;

2. $x_1=-11,\ x_2=1,$
   $x_3=-5+\sqrt{14},\ x_4=-5-\sqrt{14}$;

3. $x=\frac{2k+1}{2}\pi$.

To the article
“Something About Radials”

1. a) – b) If $n$ is odd, then for any $a$;
   if $n$ is even, then a) for $a=0$; b) at $a\geq 0$.
   d), e) If $n$ is odd, then for any $a$ and $b$; if $n$ is even, then for $a\geq 0$, $b\geq 0$.
   d) If $n$ is odd, then for any $a$ and $b\neq 0$;
   if $n$ is even, then for $a\geq 0$, $b>0$.
   g), h) If $kn$ is odd, then for any $a$;
   if $kn$ is even, then for $a\geq 0$.
   i) If $n$ is odd, then for any $a$; if $n$ is even, then for $a\geq 0$.
   j) $n$-any, $a\geq 0$.
2. a) $a$ (with $a<-1$ or $a>1$);

6. $\frac{\sqrt{6xy}}{x}$ if $x\sqrt{x}\geq \sqrt{3y}$;
   and $x\sqrt{2}$ if $x\sqrt{x}<\sqrt{3y}$;

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b) $m$, if $\frac{a}{m^2 - 2a^2} > 0$,
and $(-m)$ if $\frac{a}{m^2 - 2a^2} < 0$;
d) $(-3\sqrt{3}\bar{a} - 1)$.

By the way
“Written Physics Exam”
Option 1

1. 2.5 $v$. Use Newton’s second law in the form $\Delta(mv) = F\Delta t$.

2. $P_1 = \frac{1}{50} m_a \delta \frac{RT_1}{\mu(V - V_{a\Delta})} + P \frac{T_1}{T} \approx 25,35 \text{ am}$.
   $T$ and $T_1$ are absolute gas temperatures. It is convenient to solve the problem using the Mendeleev-Clapeyron equation: $PV = \frac{m}{\mu} RT$.

3. $\rho = \frac{2m}{Ne^2\tau}$. The current through the sample is $i = NeSv$,
   where $S$ is the cross section of the sample, and $v$ is the average electron velocity. During the time from 0 to $\tau$ in the field $E$ the electron speed changes from 0 to value $\frac{eE}{m}\tau$. The average speed is $\frac{1}{2}\frac{eE}{m}\tau$.

4. $x_{\text{max}} = \frac{F_2(L_1 - F_1)}{F_1 + F_2 - L_1} = 45 \text{ cm}$,
   $x_{\text{min}} = \frac{F_2(L_2 - F_1)}{F_1 + F_2 - L_2} = 7,5 \text{ cm}$.

The image of the object, obtained in the form of the first lens, is viewed through the second lens, and the minimum image given by the second lens should appear in the field of sharp visibility with the naked eye.

Option 2
$$N = \sqrt{\frac{F^2 \tau P - F^2 \frac{R^2}{r^2}}{2}} = 60 \kappa \Gamma.$$ The friction force must be equal to the equal force $N$ and the force $F_1 = F \frac{R}{r}$, directed tangentially to the axis.

2. $\Delta P = \Delta h \left[ \left(1 - \frac{S_2}{S_1}\right) \frac{\rho_2}{\rho_1} + \frac{S_1}{S_2} + \frac{S_1}{S_3} \right] = 1 \text{ mm Hg}$.

The change in height of the oil column is $\Delta x_1 = \Delta h - \Delta h \frac{S_1}{S_2}$, and that of mercury is $\Delta x_2 = \Delta h \frac{S_1}{S_2} + \Delta h \frac{S_1}{S_3}$.

3. $E = \frac{\varepsilon U}{\varepsilon(d - x) + x}$ at $x \leqslant \frac{d}{2}$
   and $E = \frac{U}{\varepsilon(d - x) + x}$ at $x \geqslant \frac{d}{2}$
   ($x$ is the thickness of the kerosene layer). The graph of $E = E(x)$ is shown in figure. The system can be considered, for example, as two capacitors connected in series.

4. Miniature image, $D = \frac{1}{2L}$.

Option 3

1. $\frac{3}{2} r$. The corners and center of the grid are equipotential. Therefore, they can be connected by a conductor without resistance, without changing the distribution of current in the circuit and, therefore, without changing the resistance of the circuit.

2. $P = \frac{2P_0(P_0 \pm \Delta P)}{2P_0 \pm \Delta P}$. You can use Dalton’s law (the pressure in a vessel is equal to the sum of the partial pressures of gases filling parts of the vessel) and the Boyle-Marnott law. The ± sign corresponds to the moving in or out of the piston.

3. $\frac{m_1 m_2 v^2}{(m_1 + m_2) l}$. The system rotates in a circular center of mass, which is located at a distance $x = l \frac{m_2}{m_1 + m_2}$ from the ball with mass $m_1$. Therefore, it is most convenient to solve the problem in a coordinate system associated with the center of mass. Centripetal acceleration is imparted to the balls by the tension force of the thread.

4. $E = \frac{q}{r^2} \left(1 - \frac{a}{b}\right)$. The charge of the internal sphere can be found from the condition that the potential of this sphere is equal to zero (that is, the work spent on transferring a unit positive charge from infinity to the sphere): $\frac{q_1}{a} + \frac{q}{b} = 0$.

Option 4

1. $\frac{v_0^2}{2\pi l_0 gk}$. The work done by the friction force $2F_{\text{Tp}} \cdot 2\pi R \cdot n$ is equal to the initial kinetic energy of the system $2\frac{mv_0^2}{2}$.

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2. $A = \frac{e^2}{d} \approx 14$.

The total energy of an electron in an atom is

$$W = E_{\text{киин}} + E_{\text{пот}} = \frac{mv^2}{2} - \frac{2e^2}{d} = \frac{e^2}{d} - \frac{2e^2}{d} = -\frac{e^2}{d}.$$

3. $T_1 = \frac{T_0}{\left(1 - \frac{P}{p_0 S}\right)\left(1 - \frac{P}{\rho g l S}\right)}$.

4. $a = \frac{mg}{m + H^2 l^2 C}$.

E.m.f. induction occurring in the rod is $E = Hlv$. The charge of the capacitor is $Q = EC = CHlv$, therefore the current in the circuit is $I = \frac{\Delta Q}{\Delta t} = HlCa$ ($a$ is the acceleration of the rod). The magnetic field acts on this current with a force $F = IHl = H^2 l^2 aC$. Now we need to write down the equation of motion of the rod $mg - F = ma$ and solve it for $a$.

To trem in dews in physics

1. The easiest way is to use a calormeter.
2. In the coordinate system associated with the dock, at the moment the chalk hits the dock, it has a speed of $v_1 = u + (-v)$. Since the friction force acting on the chalk and on the dock is directed along the vector $v_1$, it cannot change the direction of the speed of the chalk relative to the dock. When moving, only the relative speed changes. This means that the track on the dock is a straight line making an angle $\alpha$ with the direction of movement of the dock such that

$$tg \alpha = \frac{u}{v}.$$

3. The resistance of the light bulb from a flashlight is small - a few. The resistance of the entire garland is several hundred. Finger resistance is several thousand. When connected in series, the direction differences in the sections of the circuit are proportional to the resistances of the sections; therefore, on your finger, if you put it into the cartridge, you will have almost the entire direction of the network.

To the question “4, 5 and 6”
Can. See figure.

1. Three weigh-ins. It is necessary not only to show in what order the coins should be weighed in order to avoid friction by weighing, but also to prove that two weighings are not enough.

2. You can take advantage of the fact that the middle of the segment $MN$ lies on the diagonal $AC$ and divides it in a ratio of 3:1.

3. Three solutions (not counting one “degenerate” solution, when the number of people is 0).

4. Four solutions. Find out what remainders the number $x^2 + y^2$ gives when divided by 4 depending on the four $x$ and $y$.

5. Nine intersection points.

6. See figure:

7. For example, $2^{39}$ ends with three eights.

8. No. Each time the number of goals changes to a multiple of 3.

30

ENCRYPTION

Decipher the given text, where each stolnik of three characters corresponds to a specific letter of the Russian alphabet. Note that the arrangement of signs in the table is related to the number of the letter in the alphabet.

ANSWER TO THE CROSSWORD,
PUBLISHED IN NO. 2

Horizontal:
4. Sekushaya; 9. Order; 10. Similarity; 11. Focus; 14. Sinus; 15. Number;
16. Cone; 19. One hundred; 20. Side; 21. X; 24. Neper; 25. Line; 28. Formula; 29. Theorem; 30. Segment.

Vertical:

1. Method; 2. Beam; 3. Law; 5. Monom; 6. Lyapunov; 7. Cosine; 8. Minus;
2. period; 13. Sector; 17. Segment; 18. Axiom; 22. Lemma; 23. Number;
3. Tangent; 27. Arrow.

Amendments:
to the crossword published in issue 1 of the magazine
Printed Must be
Horizontal:
i j
Vertical:
n. Prime number n. Square
to an article like the second issue figure on page 30 is turned over.

PRICE 30 kopecks.
INDEX 70465

kvasht 3