Kvant 1970, No. 4 · Pages

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human has discovered a lot
strange in nature and will open
even more, increasing the
his power over her…"
V. I. Lenin

1870-1970

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In the room:

Philosophical ideas of V. I. Lenin and the development of modern physics
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Matchings and transport networks
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Problem book “Quenta”
25

Hinton’s Flaky World
28

Written exam in physics at MSU Physics
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Humanities students take mathematics
35

Geometric meaning of some inequalities with two variables
41

14 days in Bucharest
49

12 days in Czechoslovakia
52

Seriously + popular
55

Checkers by E. Lasker
59

Answers, directions, solutions
61

Number pyramid. Another encryption

I. K. Kikopin
M. I. Bashmakov
V. P. Lishchevsky
K. N. Baransky,
A. V. Ustinova
A. S. Kuzchtsev,
V. A. Uspensky
A. Ya. Margulis,
B. A. Radunsky
I. S. Petrakov
A. Klimov,
N. Kondratiev,
A. Chernochan
V. N. Berezin,
M. L. Smolyansky

3rd cover page

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The nature of our actions provides proof of the correspondence (Ubereinstimmung) of our perceptions with the objective (gegenständlich) nature of the perceived things*).

This brings up the next question. The history of science knows many examples indicating that the ideas of scientists that are clearly incorrect from a modern point of view nevertheless led to “the ruin of our actions.”

As an example, take the old theory of magnetism, in which a magnetized piece of steel was viewed as a magnetic dipole, consisting of two magnetic poles or “magnetic charges” (similar to an electric dipole). Using this idea, physicists create a magnetostat, on which all practice and technology of using magnets is based. This practice quickly confirms the theory of “magnetic charges”. Meanwhile, it is now well known that no magnetic charges actually exist. And when they now talk about magnetic poles, they always stipulate that this is a fictitious concept.

It seems that the criterion of practice cannot serve as a reliable basis for ascertaining the truth of our ideas about a particular subject? This, of course, is not a separate question.

One of the greatest modern physicists, Richard Feynman, trying to figure out what the philosophical interpretation of a physical law is, illustrates the significance of this issue with the following example.

“Let those who insist that the only thing that matters is the agreement between theory and experiment, imagine a conversation between a Mayan astronomer and his student. Maya was able to imagine with amazing accuracy, for example, the time of an eclipse, the position of the Moon, Venus and other planets in the sky. All this was done with the help of arithmetic… They did not have the slightest idea about the rotation of celestial bodies.

Imagine that a young man comes to our astronomer and says: “This is what came to my mind. Maybe all this is believable, maybe these are balls of stone… and their movement can be calculated completely differently.” Further, having learned that the young man has not yet reached the point of calculating the movements of the planets, the Mayan astronomer will answer him that we can already accurately calculate eclipses, so there is no need to bother with your ideas.

As we see,” Feynman concludes, “it is not an easy task to decide whether or not it is worth thinking about what lies behind our theories.”

This is indeed no easy task. After all, if we accept unconditionally the statement that the criterion of truth is practice, then such a statement, as can be seen from the examples given, can lead to stagnation in science.

This difficult task was brilliantly solved by V.I. Lenin.

Arguing that “the point of view of life and practice must be the first and fundamental point of view of the theory of knowledge,” Lenin adds the following significant paragraph.

“…of course, we must not forget that the criterion of practice can never, in its very way, expose or refute the position of the Maya and his student. Maya was able to imagine with amazing accuracy, for example, the time of an eclipse, the position of the Moon, Venus and other planets in the sky. All this was done with the help of arithmetic… They did not have the slightest idea about the rotation of celestial bodies.

*) Agnosticism is a philosophical movement that affirms the irreducible limitations of human knowledge.

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Fig. 1, a
Fig. 1, 6

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identical light sources turn out to be dark (Fig. 2), while under the action of each, the paradoxical fact is that both the wave and quantum theories of light are correct, and the formula $E=hv$ establishes a connection between these contradictory theories.

This challenge to “common sense” reached its climax in 1924, when the dual concept of “wave” and “particle” was extended by theoretical reasoning to electrons and atoms. In other words, the atom and the electron, which since their discovery have all the properties of particles, must also behave like waves. And very soon experiments confirmed these conclusions.

Physicists had to understand the situation more deeply, which led to the creation of modern quantum mechanics (or, as it was previously called, wave mechanics).

The foundations of quantum mechanics were laid in 1925-1927. A significant contribution to its development was made by the German physicist Werner Heisenberg with his formulation of the so-called uncertainty correspondence, in which

circle of which a wide-ranging philosophical discussion flared up. This correspondence led many famous physicists to the camp of philosophical idealists.

At first glance, the correspondence of the following is completely harmless:

$$\Delta x \Delta p \geq \frac{h}{2\pi}.$$

Here $\Delta x$ is the uncertainty (inaccuracy) of the particle coordinate, $\Delta p$ is the uncertainty of the momentum (or speed) of the particle, and $h$ is Planck’s constant. Translated into ordinary language, this means that it is impossible to accurately know the location and speed of a particle at the same time. In other words, if we try to fix a particle in a certain place, then our attempt will lead to the fact that we will not be able to determine where and at what speed it will then fly. On the contrary, if we make a particle move very slowly at a certain speed, then we will not be able to indicate where it is, that is, the particle will appear diffuse. From here it is easy to make the “simplest” conclusion of a clearly idealistic nature that human knowledge is fundamentally limited, since we are not given the opportunity to answer such a simple question. Moreover, from here we can draw a conclusion, idealistic in its meaning, that events in the world are unpredictable, that is, principles are violated.

Indeed, we are accustomed to the fact that classical mechanics allows us to foresee the future movement of a body if its initial position and speed, as well as the forces acting on it, are known. All mechanics are based on this. For example, the successes of astronautics are based on the fact that, knowing the launch site of the rocket (initial coordinates) and setting the extracted initial speed, we can, according to the laws of classical mechanics, foresee in advance where it will be.

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there will be a package at any given time.

The situation is different with a particle that obeys quantum mechanics. Since we, in accordance with the uncertainty relation, cannot accurately indicate its coordinates and speed, then, obviously, we cannot predict what they will become in the future. It is not difficult to agree that all this is a consequence of the fact that particles (electrons, atoms, neutrons, etc.) have wave properties.

In fact, let us consider the following roughly schematized experiment, which, nevertheless, is very close to actually carried out experiments. Let’s imagine (Fig. 3) that electrons emitted by some source, for example, incandescent wiring, fly through two branches in the screen. From each answer, electrons can fly in all directions. Behind the screen we can move the electron counter parallel to the answer screen. The electron counter allows recording of every electron that enters it. This means that the number of electrons can be directly counted.

It is natural to assume that each electron entering the counter has gone through one of two reactions. Therefore, if we count the number of electrons that entered the counter through the counter with the second closed, then do the same with the electrons that entered the counter through the counter with the first closed, then we can rightly estimate that the number of electrons that entered the counter through both branches will be equal to the sum shown by the counter. This requires common sense. We are sure that this would be the case if we fired a machine gun through an armored shield with two arms, behind which in some place there was a box of sand where the leaking bullets would get stuck. There is no doubt that the number of bullets that fall into a box of sand with open holes in the shield is equal to the sum of the number of bullets that fall into the same box through each of the holes separately (of course, over the same period of time, say, an hour).

But when we “shoot” electrons, this doesn’t work! Moreover, it may turn out that a counter installed in the proper place, which registers the same number of electrons as they pass through each answer separately (when one of the answers is closed), will not register a single electron when both answers are open (Fig. 3).

The question naturally arises: how can this be?

It is not difficult to see here the same phenomenon that is observed when a screen is illuminated by two identical light sources, that is, the phenomenon of integration.

Let us now try to answer the question why quantum mechanics is incomprehensible. This will help us find the roots of the idealistic conclusions that some philosophers and physicists have come to when analyzing the current situation in physics.

The main difficulty in understanding the physical phenomena discussed above lies in the fact that when analyzing them we try to use the same concepts to which we are accustomed in higher life. Centuries of human experience have led

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to the fact that a person considers for himself understandable what he can imagine in the form of a geometric or mechanical image. This experience, the practice of studying the world around us, led to the creation of a number of understandable ones, with the help of which the real world is reflected in the human brain. But until the 20th century, mankind was concerned only with macroscopic bodies moving at relatively low speeds. Macrosoplasts are bodies that can be seen, determined their shape and size (create a geometric image) and study their movement (create a mechanical image). Corresponding concepts were also adapted to this experience.

Newtonian mechanics, developed in relation to the motion of macroscopic bodies, established that the mechanical state of a body is uniquely determined by its coordinates and impulse (or speed, if the mass of the body is considered constant).

But now we have moved to the world of objects subject to the movement of macroscopic bodies, and it has been established that the mechanical state of a body is uniquely determined by its coordinates and impulse (or speed, if the mass of the body is considered unchanged).

Not only have we moved to the world of objects subject to the movement of macroscopic bodies, but we have established that the mechanical state of a body is uniquely determined by its coordinates and impulse (or speed, if the mass of the body is considered constant).

If experience shows that these objects have both the properties of particles and the properties of periodic waves, then in reality they are not waves or particles, but must be something else, “a unit of opposites.”
And indeed, if we carefully consider the experimental evidence that the electron is a common mechanical part, we will see that this is very indirect evidence. Most likely, only the deep conviction of physicists in the atomic structure of matter led them to the conclusion that experience with electrons (and atoms too) indicates that they are particles in the general mechanical sense.

Further, Einstein could not renounce the “mechanical” point of view on the movement of the electron. He often shook his head and said: “But God can’t guess “heads or tails” to decide where the electron should move.”

In our opinion, the certainty of the real existence of atoms, electrons, and protons can compete with the certainty of the Copernican system in astronomy. But this does not mean that we can imagine these objects as a reduced model of astronomical bodies. We still cannot imagine them; they are like nothing else. And nature is not to blame for this!

It would, of course, be an expression of the highest degree of philosophical idealism to believe that nature must adapt itself to the human mind so that it can figuratively imagine all its objects. It is precisely because we cannot imagine quantum objects in the form of geometric and mechanical images that we consider them incomprehensible.

But if this is so, then what grounds do we have to assume that the state of motion of quantum objects should be determined by the same quantities, that is, coordinates and momentum, which determine the state of physical bodies? In a word, the question “what are the coordinates and implus at a given moment in time?” in relation to quantum objects - an illegal question. After all, not every question is legitimate. For example, it is impossible to answer the question “what color is the Pulkovo meridan?” But it does not follow from this that human knowledge may be limited!

To the same extent, apparently, the question “what are the coordinates and impulse of the electron?” is illegitimate.

Lenin’s words, which he said on a similar occasion, can best be applied to the current situation in quantum mechanics: “The movement of bodies is transformed in nature into the movement of that which is not a body with constant mass, into the movement of that which is an unknown charge of an unknown electricity in an unknown ether - this dialectic of material transformations,

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2. First
   proof.
   7 Tasks.
   MATCHING
   AND
   TRANSPORT
   NETWORKS
   m.i. BASHMAKOV
3. Theorem
   Gzyla o
   saturation.
4. Algorithm
   Ford -
   Folkerson.
5. Networks
   and streams.
6. Theorem
   about matchmaking.
7. Weddings
   and graphs.

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There can only be two cases:
Case 1. For some $k$ ($1 \leq k < n$) there is a group of $k$ boys who are friends (all together) with exactly $k$ girls.

Let us denote the set of these young men by $IO_k$, and the set of their girlfriends by $Д_k$. For the group of boys $IO_k$ the induction sentence was completed. At the same time, their brides exhaust the entire set of $Д_k$ (unevenly $k$ girls). Let’s throw these boys and girls out of consideration and prove by contradiction that the condition of the theorem is satisfied for the remaining $n - k$ boys.

Let us select from these $n - k$ young men a company containing $r$ young men. Let’s assume that the number of their friends $s$ turned out to be less than $r$. Let’s get back on track and unite these $r$ young men with the $IO_k$ group. There are $r + k$ young men in the new group. Since the set $D_k$ contains all the girlfriends of the boys from $IO_k$ (and, possibly, some other girlfriends of the $r$ boys from the displayed company), then these $k + r$ boys had $k + s$ girlfriends at the very beginning. According to the initial condition $k + r \leq k + s$. Therefore, $r \leq s$. The result is a contradiction with proposal $s < r$. This means that in this case the theorem is true.

Case 2. Any group of $k$ boys ($1 \leq k < n$) is friends with at least $k + 1$ girls.

In this case everything is simple. We marry any young man to any of his girlfriends. Let’s exclude this pair from consideration. Let us prove that the conditions of the theorem are satisfied for the remaining $n - 1$ boys. Let’s take any group of $k$ young men ($k \leq n - 1$). At the very beginning they had at least $k + 1$ girlfriends. Then the bride of the chosen young man gets here, then there are still $k$ girls left. This completely proves the theorem.

Belief, you had to spend some conditions to understand this proof. But, in addition to the fact that the proof is not simple, it suffers from an important drawback - it does not show a simple way to find out whether the condition of the matchmaking theorem is satisfied for each specific case, and if it is, then how to choose a bride, you need to go through all the non-fair subsets (and there are $2^6 - 1 = 63$) sets of young men and for each of these subsets of young men check that they have enough girlfriends. But even if we do this and make sure that, in principle, all weddings can be arranged, we will not receive any instructions from here about who to marry whom. True, in this example $a$ you undoubtedly easily managed to select the required six pairs (there are even several possible options), but when the sets are very large, then random selection only takes a lot of time. It is especially important to have a clear and economical selection rule in order to entrust the solution of problems of this type to a computer.

We will build some algorithm *), which will not only re-prove the matchmaking theorem, but also provide a quick way to arrange the necessary matchmaking, and if this is impossible, then indicate the option for the maximum number of weddings.

3. WEDDINGS AND COURTS

A system of points that are somehow connected by arrows is called a directed graph*). In this case, the points are called the vertices of the graph, and the segments are called its lines or edges. Examples of directed graphs are shown in the discussion already figure 1, as well as in Figures 3 and 5. In the case of Figure 1, the set of all vertices of the graph was divided into two sets - girls and boys - so that the edges go from the points of one set to the points of the other. In such a situation

*) Sequence of actions to solve a certain type of problem. Muscovites were promised an algorithm, Leningraders - an algorithm. Currently the second page of pages.

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entrance
exit
A
entrance
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exit
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in the direction opposite to the direction of the arrows).

This path at figure 5.6 is highlighted with a red line.

Now we will do the following: in the selected red path we will make the free arrows saturated and vice versa (Fig. 5,8), i.e. we will change the flow on the red line so that it takes on each of its edges the value 1 together with 0 and 0 together with 1*). It is easy to see that the new function is again a flow, i.e. at intermediate vertices the number of incoming and outgoing saturated arrows is again the same - the balance is not violated. At the same time, the face of the flow increased by one unit.

So in the first case we could increase the flow. If this operation can be done again, we will repeat it as long as possible.

2. THE OUTPUT IS NOT MARKED (Fig. 5,2). Our method cannot increase the flow. But it does not yet follow from this that it is maximum. It is possible that our method of increasing is bad, and we can increase the flow in another way. Let us prove that this is not the case, i.e. we will prove that the existing flow is maximum. Let’s consider the set of all unplaced vertices. By condition, this set includes the output. Let’s count the number of arrows leading to this set, i.e., those whose left end is placed, but the right end is not. It is clear that the magnitude of any flow is not greater than this number. At the same time, for our flow all these arrows are saturated, since otherwise the right end could be placed flat. In addition, all arrows coming from the set of unplaced vertices (that is, those whose left end is not placed and the right end is placed) are free, otherwise the left end could be placed in a minute. This means that the size of the original flow is equal to the number of arrows leaving in

6. GAYLE’S THEOREM
   ABOUT SATURATION

Using the Ford-Falkerson algorithm, we will now formulate and prove the central theorem of the theory of transport networks. The formulation will use some natural economic terms, but you will feel how close it is to the matchmaking theorem.

Let us be given a transport network. Consider an arbitrary set of arbitrary vertices $A$ (i.e., containing neither input nor output). Let us define the capacity of this set $A$ as the number of arrows going to $A$. Let’s denote this number

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via $c(A)$. Let us define by the total requirement of the set $A$ the number of edges leading from $A$ directly to the exit. Let’s denote this number by $d(A)$.

Let us study the following question: under what conditions does there exist a flow that saturates all output edges, i.e., one whose values ​​on all edges leading to the output are equal to one. It is clear that such a flow should be maximum.

Gale’s theorem. For any number, there is a flow that saturates all output arcs; it is necessary and sufficient that for any set $A$ its total demand does not exceed the capacity:

$$d(A) \leq c(A).$$

Proof. As with the matchmaking theorem, the necessity is almost a turn. Indeed, if we can construct a flow passing along all edges leading to the exit, then for any set $A$ it also passes along the edges leading to the exit only from $A$. At the same time, the number of flows leading to the sets $A$ cannot be greater than the number of edges leading to $A$, i.e. the numbers $c(A)$. But this amount of flow must completely leave $A$ along all the edges leading primo to the exit, and maybe along some others. From this it is clear that $d(A) \leq c(A)$.

Adequacy. Let there be maximum flow through our network. Let us prove that it passes along all output arcs. Let’s label the vertices using the Ford-Folkerson method and consider the set of unlabeled vertices. Let’s remove the output from this set. We denote the resulting set by $A^*$. The ribs leading to the exit can be of two types.

Firstly, these are also edges whose end lies in $A$, i.e., is a labeled vertex. They will be saturated, since according to the method of constructing the flow, all edges whose left end is placed, but the right end is not, are saturated.

Secondly, these are the edges leading to the exit from the set. Since $A$ together with the exit forms the set of all unlabeled vertices, then all $c(A)$ edges leading to $A$ will be saturated, and all edges leading from $A$ not directly to the exit will be free.

*) The set $A$ may turn out to be empty - our reasoning will still be correct. But in this case the validity of Gale’s theorem is immediately visible. Why?

This means that the entire flow of value $c(A)$ that entered $A$ must exit $A$ directly to the exit. But from $A$ only $d(A)$ each lead to the exit, and we know that $d(A) \leq c(A)$. Therefore, all these edges are saturated and $d(A) = c(A)$.

You probably noticed that the conditions of Gale’s theorem correspond to the conditions of the wedding theorem. Try to deduce this theorem yourself as a consequence of Gale’s theorem. To do this, you need to make a transortunal network from the bipartite graph of acquaintances between boys and girls, connecting the input with all the girls, and the output with all the boys (it is then convenient to assign a weight $N$ to each arrow connecting the girl with the boy, where $N$ is greater than the number of boys).

What is much more important for us now is not that we have obtained yet another proof of the matchmaking theorem, but that we have a simple algorithm for constructing maximum flows.

For the matchmaking problem, the algorithm naturally formulates the flow as follows: for each girl who is not a bride, all her friends are placed flat, then in a minute - the brides of these friends, then the rest of the friends of these brides are placed on the flat, etc.; if an unoccupied young man turns out to be placed, then the number of weddings can be increased by making replacements along the chain drawn (like the red line at Fig. 5.6) along the marks from the unoccupied girl to the unoccupied young man.

The arrangement of weddings for all young men is nothing more than the construction of a flow that saturates all the exit arcs; if this is not the case, then the Ford-Folkerson algorithm provides the maximum possible new interesting algorithm for solving a number of difficult problems in finite mathematics.

written off in the novel. You can easily imagine this two-dimensional world by placing a few coins on the table. Let one of them, say, a metal ruble, represent the Sun, while the other, smaller coins, represent the planets rotating around it. The force of gravity in this two-dimensional world behaves differently than in our space. The only difference is that it changes in inverse proportion to the distance between the bodies, and not the square of the distance, as in the world we are accustomed to.

Astria rotates around an axis, like any planet. For example, in which it rotates is called east, the opposite is called west. Hinton seemed to place his flat actriyans (inhabitants of Astria) on the “rim” of the planet. They can mix and remain in the plane of the Astria disk. Trees grow and houses stand in the same plane. Therefore, when encountering a tree, the actor must either climb over it or surround it. In order for one to pass the other, two inhabitants of the planet must move either one under the other, or jump over each other, as two acrobats would do on a tight rope.

To see something behind him, the actor must bend down and stand on his head or use a mirror. Since the second method is more

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4. Two flat mirrors $3_1$ and $3_2$ (Fig. 2) are installed close to each other at an angle $a=15^\circ$ to the vertical. A thin converging lens is placed at a distance of $L=8,3$ cm from the mirror connection line so that its main optical axis is perpedicular to the mirror connection line. Between the mirrors and the lens on the main optical axis of the lens at a distance of $R=2$ cm from the mirror connection line there is a point light source $H$. Determine the distance between the images of the light source that are obtained using the lens. The focal length of the lens is $f=8$ cm. The partition $K$ protects the lens from rays and light from the light source.

5. A ball $A$ of mass $m$, fixed on a long weightless spoke, forms a pendulum, which can oscillate, curing around the axis $O$ (Fig. 3). The ball $A$ is deflected to a height $h$ and released without initial velocity. At the bottom point of its trajectory, the ball $A$ is contained with two balls $B$ moving towards it, the mass of which is $M$. The ball $B$, like the ball $A$, is fixed on a long weightless knitting needle, moving around the axis $O_1$. Immediately before the collision, the velocities of the balls are directed along the lines connecting their centers. What speed $v$ must the ball $^1$ $B$ have before the collision so that the ball $A$ rises to the same height $h$ after the collision? The impact is absolutely elastic. Friction in wasps and ovozdux should not be taken into account.

Option 2

1. A small lead ball of volume $V=0,02$ $cm^3$ falls uniformly in water. How much heat will be released when the ball moves down 6 m? The density of lead is $\rho=11,3$ $e/cm^3$. Accept $g=10$ $m/cek^2$.

2. A thin converging lens with a focal length $f=10$ $cm$ is cut in half along the diameter and its halves $J-L$ (Fig. 4) are mixed relative to each other along the main optical axis at a distance $l=10$ $cm$. At a distance $a=30$ $cm$ from the upper half of the lens, a mirror is installed on the main optical axis, shown in figure 5. With what acceleration will the cart begin to move along the floor if a force $F=8$ $cm$, directed vertically upward, is applied to the free end of the rope. Friction coefficient between load and trolley $k=0,6$. Accept $g=10$ $m/cek^2$.

3. A low trolley weighing $P=12,5$ $k\Gamma$ can move on a horizontal floor without friction. There is a load on the cart weighing $P_1=10$ $k\Gamma$. A rope is attached to the load, throwing a weightless block mounted on the cart, as shown in figure 5. With what acceleration will the cart begin to move along the floor if a force $F=8$ $cm$, directed vertically upward, is applied to the free end of the rope. Friction coefficient between load and trolley $k=0,6$. Accept $g=10$ $m/cek^2$.

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TASK OPTIONS FOR THE WRITTEN EXAM*)

Faculty of Psychology

1. The internal point $A$ of a ball of radius $r$ is connected to the surface of the trey ball by straight segments of length $l$ and drawn at an angle $a$ to each other. Find the distance of point $A$ from the center of the ball.

2. Find all positive numbers $a$ for which all distinct non-negative values ​​$x$ satisfying the equation
   $$\cos[(5a - 9)x] = \cos[(9a + 17)x]$$
   and arranged in ascending order form an arithmetic progression.

3. Prove, without using decimals, that
   $$\log_2 5 > \log_5 32.$$

4. The distance between $A$ and $B$ is 7 km. Two pedestrians simultaneously embroidered towards each other and met equals an hour later. If the first one walked twice as fast as he actually walked, and the speed of the second one was 2 km/h more than his actual speed, then by the time they met, the second one would have covered most of the distance. Which pedestrian has the fastest speed?

5. Prove that
   $$8x^2 + y^2 + 11z^2 + 4xy - 12xz - 5yz > 0,$$
   if $x^2 + y^2 + z^2 > 0$.

3 tasks
from other options

a) A diameter $AB$ is drawn into a ball of radius $r$ and three equal chords $AC, AD$ and $AF$ are drawn at an angle $a$ to each other. Find the volume of the body bounded by the planes of triangles $ACD, ADF, ACF, BCD, BDF$ and $BCF$.

6. From $A$ to $B$ a raft floats along the river. At the same time that the raft began its journey from $A$ to $B$, a boat sailed towards it from $B$ to $A$, which met the raft no earlier than 2 hours later, and then arrived at $A$, spending less than three hours and twenty minutes on the entire journey. Will the raft manage to travel from $A$ to $B$ in 5 hours if the distance between $A$ and $B$ is 20 km?

Problem solutions

1. Let $O$ be the center of the ball; $B, D$ and $C$ are the points of intersection with the surface of the ball of the lines drawn from the point $A; O_1$ - the center of the section of the ball by the plane $\Delta BCD$. Considering,

*) The exams were allotted: at the Faculty of Psychology and the Department of Political Economy of the Faculty of Economics - 4 hours, at the Faculty of Philology and the Department of Economics of the Faculty of Economics - 5 hours (one hour is equal to sixty minutes).

that the plane $\Delta BCD$ is perpendicular to the diameter $AO_1$ (except for the case $l = r$), we find
$$OO_1 = \sqrt{r^2 - \frac{4}{3}l^2 \sin^2 \frac{\alpha}{2}}.$$
$$AO_1 = l \sqrt{1 - \frac{4}{3} \sin^2 \frac{\alpha}{2}}.$$

Then
$$AO = OO_1 + AO_1,$$ if points $A$ and $O$ lie on opposite sides of the plane $\Delta BCD$;
$$AO = OO_1 - AO_1,$$ if points $A$ and $O$ lie on the same side of the plane $\Delta BCD$, and points $O$ lie outside the parmid $ABCD$ ($OO_1 > AO_1$);
$$AO = AO_1 - OO_1,$$ if points $A$ and $O$ lie on the same side of the plane $\Delta BCD$, and point $O$ lies inside the parmida $ABCD$.

2. Using the difference of cosines formula
   $$2 \sin [(7a + 4)x] \cdot \sin [(2a + 13)x] = 0.$$

Since $a > 0$, the solutions of the equation have the form
$$\frac{\pi k}{7a + 4}, \frac{\pi l}{2a + 13}; k, l - целые.$$

It is clear that the non-negative solutions (corresponding to the non-negative $k$ and $l$) form two arithmetic progressions with the first term 0 and the differences of the corresponding integer factored into the other, i.e. when either $\frac{d_1}{d_2}$ is natural or $\frac{d_1}{d_2}$ the analogous result is: $\frac{d_2}{d_1}$ is natural.

Conversely, if, for example, $\frac{d_2}{d_1}$ is natural, then in connection with the arithmetic progression with the first term 0 and the difference $\frac{d_2}{d_1}$ is also a member of the arithmetic progression with the first term 0 and the difference $\frac{d_1}{d_1}$, therefore the terms of the two progressions together form an arithmetic progression with the first term 0 and the difference $\frac{d_1}{d_1}$.

So, $a > 0$ satisfies the conditions if and only if either $\frac{d_1}{d_2}$ is natural, or $\frac{d_2}{d_1}$ is natural, or $\frac{d_1}{d_2}$ is also a member of an arithmetic progression with the first term 0 and the difference $\frac{d_1}{d_1}$, so the terms of the two progressions together form an arithmetic progression with the first term 0 and the difference $\frac{d_1}{d_1}$.

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Therefore we have to consider two cases:

1. $\frac{d_1}{d_2}$ natural. $\frac{\pi}{7a+4} = n \cdot \frac{\pi}{2a+13}$, $n$ natural. $a = \frac{13-4n}{7n-2}$. $7n-2 > 0$ for natural $n$. Therefore, $a > 0$ if and only if $13-4n > 0$, i.e., when $n=1, 2, 3$, respectively $a = \frac{9}{5}, \frac{5}{12}, \frac{1}{19}$.
2. $\frac{d_2}{d_1}$ natural. Similarly follows 1) we get $a = \frac{9}{5}, \frac{22}{3}, 35$.
   $a = \frac{1}{19}, \frac{5}{12}, \frac{9}{5}, \frac{22}{3}, 35$.

3. Since $\log_2 5 \cdot \log_5 32 = \log_2 32 = 5$, it is enough for us to show that $\log_2 5 > \sqrt{5}$. Since $81 > 80$, i.e. $\frac{81}{16} > 5$, then $\frac{9}{4} > \sqrt{5}$, and to prove the assumed inequality it is enough to verify that $\log_2 5 > \frac{9}{4}$ or $5 > 2^{\frac{9}{4}}$. But $5^4 = 625 > 512 = 2^9$; Taking the fourth root, we obtain the required inequality.
4. The speed of the second pedestrian is higher.
5. $8x^2 + y^2 + 11z^2 + 4xy - 12xz - 5yz = 2\left(2x + \frac{y}{2} - \frac{3}{2}z\right)^2 + \frac{1}{2}(y - 2z)^2 + \frac{9}{2}z^2$.

Consequently, the original expression is non-negative, as reducible to a sum of squares, and if it vanishes, then $x=y=z=0$, i.e. $x^2+y^2+z^2=0$.
Pozny, if $x^2+y^2+z^2>0$, then
$8x^2+y^2+11z^2+4xy-12xz-5yz>0$.
a) The required volume is $\frac{1}{\sqrt{3}}(2r)^3\left(3-4\sin^2\frac{\alpha}{2}\right)\sin^2\frac{\alpha}{2}$.
6) Let $x$ be the speed of the river, $y$ be the speed of the boat in still water. According to the conditions we have: $\frac{20}{(y-x)+x} \geq 2$, $\frac{20}{y-x} < 3\frac{1}{3}$, whence $y \leq 10$, $x < y-6$; $x<4$, $\frac{20}{x} > 5$. Consequently, the raft will not have time to cover the path from $A$ to $B$ in 5 hours.

Department of Political Economy, Faculty of Economics

1. The Nauchsky single-track railway, 28.5 km long, needs to be laid. There are 30 m and 15 m long rails available for laying. If all 30 m long rails are laid, then 15 m long rails will need to be added to 40% of the total number. If you lay all the rails 15 m long, then 60% of the total number of rails 30 m long must be added. Determine the number of rails of both types.
2. Solve inequality $|x^2-1|-2x<0$.
3. In the acute triangle $ABC$ from the vertex $B$ a medana $BD$ equal to $m$ is drawn. Angle $A$ is equal to $a$, angle $C$ is equal to $y$. Calculate the area of ​​triangle $ABC$.
4. Solve the equation
   $$\sin x \cdot \mathrm{ctg}^2 \frac{x}{2} + 4(1+\cos x) - 5\sin x = 0$$
5. Solve the equation
   $$\sqrt{x} + \sqrt{x(x+2)} - \sqrt{(x+1)^2} = 0$$
   3 tasks from other options
   a) In a bookstore, an antique collection of works costs 350 rubles. discounted twice for the same number of projects. Find this number if it is known that after a double price reduction, the collected works cost 283 rubles. 50 kopecks

6. At what real value $x$ does the expression $\frac{2x-1}{2x-x^2-4}$ take its smallest value?

Problem solutions

1. When drawing up the equation, it should be taken into account that a single-track railway of 28.5 km has a track of 57 km. It is necessary to lay rails: 1500 30 m long and 2000 15 m long.
2. $-1 + \sqrt{2} < x < 1 + \sqrt{2}$.
3. $m^2 \frac{\sin \alpha \cdot \sin \gamma \cdot \sin (\alpha+\gamma)}{4\sin^2 \alpha \cdot \sin^2 \gamma + \sin^2 (\alpha-\gamma)}$.
4. $x_1 = (2k+1)\pi$, $x_2 = \frac{\pi}{2} + 2k\pi$.
5. $x = \frac{\sqrt{5}-1}{2}$.
   a) 10%.

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6. Putting $y = \frac{2x - 1}{2x - x^2 - 4}$, we obtain a quadratic (with respect to $x$) equation $yx^2 + 2(1-y)x + 4y - 1 = 0$, the discriminant of which must be non-negative: $(1-y)^2 + y(4y - 1) \geq 0$, since $x$ is a real number, whence
   $$-\frac{1 + \sqrt{13}}{6} \leq y \leq \frac{-1 + \sqrt{13}}{6}.$$
   Therefore, the smallest value of $y$ is $-\frac{1 + \sqrt{13}}{6}$; the corresponding value of $x$ is, as can be easily calculated, $\frac{1 + \sqrt{13}}{2}$.

Department of Economic Cybernetics, Faculty of Economics

1. Solve the equation
   $$\left( \sqrt{\cos x} - 4 \sqrt[4]{2} \sqrt[4]{\cos \left( x - \frac{\pi}{4} \right)} + 5 \sqrt[4]{2}\right)^2 - 2 \sqrt{\cos x} \times \left( \sqrt{\cos x} - 4 \sqrt[4]{2} \sqrt[4]{\cos \left( x - \frac{\pi}{4} \right)} + 5 \sqrt[4]{2}\right) - \sin x = 0.$$
2. Four groups of tourists set off on a Sunday approach along four routes of different lengths. The sum of the distances covered by the first and fourth groups is 6 km greater than the sum of the distances covered by the second and third groups. The second group walked 2 km less than the first. The number of tourists of the third group is equal to the distance traveled by the first group, and the number of tourists of the second group is equal to the distance traveled by the fourth group. The sum of the squares of the distance traveled by each group is 494. How much cybernet did each group ask for?
3. In a circle of radius $R$ there is a hexagon ABCDEF. It is known that $\angle A = \angle C = \angle E$, side $AB = a$, CD = b, EF = c. Calculate the area of ​​the hexagon ABCDEF.
4. Solve inequality
   $$\left( 4x - x^2 - 3 \right) \log_2 \left( \cos^2 \pi x + 1 \right) \geq 1.$$
   3 task
   from other options
   a) The segment $AB = a$ is perpendicular to the segment $BC = 3a$. In the plane containing both of these segments, two circles of radius $a$ with centers at points $A$ and $B$ and a circle of radius $3a$ with center at point $C$ are constructed. Calculate the area of ​​part of the three constructed circles.

6. Solve inequality
   $$\left| \sqrt{2} | x | - 1 \right| \cdot \log_2 \left( 2 - 2x^2 \right) \geq 1.$$

Problem solutions

1. Range of acceptable values ​​(OD3):
   $$-\frac{\pi}{4} + 2k\pi \leq x \leq \frac{\pi}{2} + 2k\pi,$$
   $k = 0, \pm 1, \dots$

Let’s put
$$y = \sqrt{\cos x} - 4 \sqrt[4]{2} \sqrt[4]{\cos \left( x - \frac{\pi}{4} \right)} + 5 \sqrt[4]{2},$$
Then
$$y^2 - 2 \sqrt{\cos x} y - \sin x = 0,$$
where
$$y = \sqrt{\cos x} \pm \sqrt{\sqrt{2} \cos \left( x - \frac{\pi}{4} \right)},$$
i.e. we have
$$\sqrt{\cos x} - 4 \sqrt[4]{2} \sqrt[4]{\cos \left( x - \frac{\pi}{4} \right)} + 5 \sqrt[4]{2} = \sqrt{\cos x} \pm \sqrt[4]{2} \sqrt{\cos \left( x - \frac{\pi}{4} \right)}.$$

Putting $z = \sqrt[4]{\cos \left( x - \frac{\pi}{4} \right)}$, we get $-4z + 5 = \pm z^2$. After the equation has roots $z_1 = 1, z_2 = -5, z_3 = 2 \pm i$. Roots $z_2, z_3, z_4$, very/no, construction.
From $z_1 = 1$ follows $\cos \left( x - \frac{\pi}{4} \right) = 1$. Thus, $x = \frac{\pi}{4} + 2k\pi, k = 0, \pm 1$.

2. Let $x_i$ be the distance traveled by the $i$-group, $i = 1, 2, 3, 4$. From the problem condition we have $x_1 + x_4 - 6 = x_2 + x_3, x_1 - x_2 = 2, x_1^2 + x_2^2 + x_3^2 + x_4^2 = 494$, whence
   $$(x_1 - 1)^2 + (x_4 - 2)^2 = 242,$$
   $$x_1 = 1 + \sqrt{242 - (x_4 - 2)^2}.$$

According to the conditions, $x_1$ and $x_4$ are natural numbers ($x_1$ is the number of tourists in the 3rd group, and $x_4$ is the number of light tourists by selection based on the last correspondence (and replacing that by virtue of the same correspondence $x_4 < 18$). Sort through sequentially
$x_4 = 1, 2, 3, \dots, 17,$ we find the only solution:
$x_4 = 13, x_1 = 12.$
In this case $x_2 = 10, x_3 = 9.$

3. $$\frac{3 \sqrt{3}}{4} R^2 + \frac{\sqrt{3}}{8} a \left( \sqrt{12R^2 - 3a^2} - a \right) + \frac{\sqrt{3}}{8} b \left( \sqrt{12R^2 - 3b^2} - b \right) + \frac{\sqrt{3}}{8} C \left( \sqrt{12R^2 - 3C^2} - C \right).$$

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$$\frac{\cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ - 1}{\sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ} = 2 + \sqrt{2} - \sqrt{3} - \sqrt{6}.$$

In conclusion, we suggest solving one option for each table on your own.

Faculty of Philology

1. There are three pipes installed into the pool through which water flows into it. The first and second pipes together fill the pool 35/18 hours faster than the first and third pipes together fill it in 10 hours. What part of the pool does the second pipe fill per hour if it alone takes $\frac{7}{5}$ times longer to fill than the first pipe alone?
2. Solve the equation:
   $$\log_{\sin x} \left( \sin x - \frac{1}{4} \cos x \right) = 3.$$
3. Solve inequality:
   $$2\log_3 x - \log_1/5 (4 - x) \leq \log_3 (x - 1)^2 + 2\log_2 (10 - x).$$
4. N people exchanged handshakes. How many handshakes were there?
5. Indicate at least one pair of positive integers $k_1$ and $k_2$ such that $36 \cdot k_1 - 25 \cdot k_2 = 1$.
6. Prove, without using tables, that $\log_4 9 > \log_5 11$.
7. Does a triangle have an obtuse angle whose sides are 10, 14 and 17 respectively?

Faculty of Psychology

1. A sphere contains a diameter $AB$ and two equal chords $AM$ and $AN$, each at an angle $a$ to the diameter. Find the angle between the chords if the segment $MN$ is visible from the center of the ball at an angle $\beta$.
2. Find all positive numbers $a$ for which all distinct non-negative values ​​$x$ satisfying the equation $\cos [(8a-3)x] = \cos [(14a+5)x]$ and arranged in ascending order form an arithmetic progression.
3. Prove, without using decimals, that $\log_3 16 > \log_16 729$.
4. The distance between stations $A$ and $B$ is 360 km. At the same time, two trains leave from $A$ and from $B$ towards each other. Trains left $A$ and arrive at station $B$ no earlier than 5 hours later. If his speed had been 1.5 times higher than it actually was, then he would have met the second train more than 2 hours after he left $A$. Which train is faster?
5. Prove that
   $$2x^2 + 5y^2 + 3z^2 - 6xy - 2xz + 5yz > 0,$$
   if $x^2 + y^2 + z^2 > 0$.

Department of Political Economy

1. Five camping tents and twenty stopley sleeping bags 490 RUR. After a ten percent discount on one tent and a five percent discount on one bag, for the same number of tents and bags you need to pay 457 rubles. How much is one stop for one tent and how much is one stop for one bag?
2. Solve inequality: $x^2 + x - 10 < 2/x - 2$.
3. In an acute triangle $ABC$, angle $A$ is equal to $a$, side $AC$ is equal to $b$. From the vertex $B$ the median $m$ is drawn. Calculate the area of ​​triangle $ABC$.
4. Solve the equation: $(3 - \text{ctg}^2 x)\sin 2x = 2(1 + \cos 2x)$.
5. Find the maximum value of the fraction $\frac{3x - 2}{4x - x^2 - 6}$ if $x$ can take any real value.

Department of Economic Library

1. Solve the equation
   $$\left( 4\sqrt{\cos \frac{x}{2}} - 5 - \frac{\sqrt{2}}{2} \right)^2 + \sqrt{2} \left( 4\sqrt{\cos \frac{x}{2}} - \frac{\sqrt{2}}{2} - 5 \right) - \frac{\cos x}{2} = 0.$$
2. There are four art clubs at the pioneer camp. Give each member of the circle the same set of colors. In the first and fourth circles together, and in the second and fourth circles together there are even more pioneers than in the first and third circles together. There are no more than ten pioneers in the third circle. How many pioneers are in each circle, if the price in rubles of one paint is equal to the number of members of the first circle divided by ten, and 52 rubles were paid for paints? 80 kopecks?
3. A circle is inscribed in the trapezoid $ABCD$. Find the difference in the lengths of the diagonal segments $AC$ located in a circle if the height of the trapezoid is $h$, and the angles at the base $AD$ are $a$ and $\delta$.
4. Solve inequality: $2^{-1}x - 2^1\log_2(4x - x^2 - 2) \geq 1$.

A. S. KUZICHEV
V. S. USPENSKY

41

GEOMETRICAL MEANING OF SOME INEQUALITIES WITH TWO VARIABLES

A. Y. MARGULIS, B. A. RADUNSKY

When studying a mathematics course, a student often comes to solve inequalities and systems of inequalities with one unknown. At the same time, a geometric illustration of the solution (in combination with the method of integrals) helps them find the correct solutions in the shortest way.

Consider, for example, the inequality
$$\log_2 (x^2 - 4x + 3) \leq 3.$$
Its solution coincides with the solution to a system of two inequalities with one unknown
$$\begin{cases} x^2 - 4x + 3 > 0 \\ x^2 - 4x + 3 \leq 8 \end{cases}$$
(necessary for the existence of a logarithm),
(obtained by potentiating a given inequality).

The set of values ​​of $x$ satisfying the first inequality is a set of two integrals
$$x < 1 \text{ и } x > 3,$$
and the second - a segment
$$1 \leq x \leq 5$$
(this follows from the properties of the square triterm).

Having marked these intervals (Fig. 1) on the number axis, we see that the set of values ​​$x$ satisfying the system of inequalities fills two semi-closed integrals, $-1 \leq x < 1$ and $3 < x \leq 5$ (on figure they are given in the paragraph).

For greater clarity, on figure itself it is noted which of the ends of the integrals belong to the solution domain and which do not.

The geometric illustration turned out to be very useful here. Its role increases even more when solving inequalities and systems of inequalities with two unknowns.

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Let us consider inequalities with two unknowns of the form

$$y - f(x) > 0.$$

(1)

Let’s construct a graph of the function $y = f(x)$ and take an arbitrary point $A'_1$ with the abscissa $x = x_0$ (Fig. 2). If, with the same $x = x_0$, its ordinate $y_1$ is greater than the ordinate $y_0$ corresponding to the point $A$ of the curve $y = f(x)$, then the point $A'_1$ lies above the curve. Point $A'_1$ lies under curve $y = f(x)$ if its ordinate $y_2$ is less than ordinate $y_0$. Therefore, the coordinates of all points lying below the curve $y = f(x)$ satisfy the inequality $y < f(x)$, and the coordinates of all points lying above the curve $y = f(x)$ satisfy the inequality $y > f(x)$.

Thus, the graph of the function $y = f(x)$ divides the coordinate plane into two parts, in one of which the inequality $y - f(x) > 0$ is satisfied, and in the other $y - f(x) < 0$.

Let us introduce the following definition. The power set of a plane is called convex if it, together with any two powers $A$ and $B$, contains all the powers of the segment $AB$.

Consider, for example, the interior of any triangle, circle, square. It can be shown that these sets are convex.

It is obvious that those parts into which the curve $y = f(x)$ divides the plane can be either convex or non-convex (see Fig. 2).

Let us now illustrate with a number of examples how using geometric concepts one can solve inequalities and systems of inequalities with two unknowns.

Consider the linear function $f(x) = ax + b$ (2).

The graph of the linear function $y = ax + b$ is a straight line (Fig. 3). This straight line divides the plane into two half-planes: upper and lower. In the first of them the inequality $y - (ax + b) > 0$ is satisfied, and in the second $y - (ax + b) < 0$. Note that both of these half-planes are convex sets.

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Example 1. Find on the geometric coordinate plane the locations of points whose coordinates satisfy a system of two linear inequalities (solve graphical given systems of inequalities):

$$\begin{cases} y > -\frac{1}{2}x + 2, \\ y < -\frac{1}{2}x + 1; \end{cases}$$

$$\begin{cases} y > -\frac{1}{2}x + 1; \\ y < -\frac{1}{2}x + 2, \end{cases}$$

$$\begin{cases} y \geq \frac{1}{3}x + 1, \\ y \leq 2x - 2. \end{cases}$$

Solution. figure 4 shows graphical solutions to these inequalities. If the solution to each of the inequalities is colored yellow or gray, then the solution of the system is the part of the plane colored with both colors. From figure 4 it is clear that system a) has no solutions (Fig. 4,a); the solution of system b) is the half-plane lying above the line $y = -\frac{1}{2}x + 2$ (Fig. 4,b); the solution to system b) is an infinite strip between lines $y = -\frac{1}{2}x + 1$ and $y = -\frac{1}{2}x + 2$ (Fig. 4,c); and the solution to system d) is the part of the plane lying inside the double-shaded angle, including its sides (Fig. 4,d).

Fig. 4 a, b, c, d.

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running_head:
folio:

Note that in all cases, the solution domains of a system of linear inequalities are convex sets. This is no coincidence. We have already noted that any straight line divides the plane into two convex sets, and the intersection (common part) of any number of convex sets is always a convex set.

Example 2. Solve graphically a system of three linear inequalities

$$\begin{cases} y < \frac{1}{3}x + 2, \\ y > -\frac{1}{2}x + 3, \\ y > 2x - 4. \end{cases}$$

Solution. This problem can be solved in the same way as the previous one. But due to the large number of restrictions, it is more convenient to paint not those areas where one or another inequality holds, but, on the contrary, those areas where there cannot be solutions. In this case, there is no need for different colors: after all, where at least one of the inequalities of the system is not satisfied, there cannot be solutions to the entire system. See figure 5 for a graphical solution to this example. The set of solutions is the set of interior points of triangle ABC. For clarity, the triangle is painted green.

To a system of linear property inequalities and some irrational and transcendental inequalities.

Example 3. Solve graphically the inequality $\sqrt{y-(x+1)}<1$.

Solution. The given inequality is equivalent to the following system of inequalities:

$$\begin{cases} y-(x+1) \geq 0 \\ y-(x+1) < 1 \end{cases}$$ or \begin{cases} y-(x+1) \geq 0, \\ y-(x+2) < 0.

The graphical solution of this example is shown on figure 6. Just as in example 3, the set of points satisfying this system is painted green. Primaya

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$y = x + 1$, which belongs to the solution area, is drawn as a solid line, and the straight line $y = x + 2$, which does not belong to the solution area, is drawn as a dotted line.

Example 4. Solve graphical inequalities $\log_{y-1}(x-2) < 1$.

Solved and e. The range of permissible values ​​(APV) of unknowns is obviously determined by a system of inequalities

$$x > 2, y > 1, y \neq 2.$$

In accordance with this system, let us first eliminate the set that cannot have a solution. To do this, on figure 7 we paint half-plane $x < 2$, half-plane $y < 1$ with gray paint, and draw a straight line $y = 2$ with a dotted line.

Let’s now consider two cases:

a) $1 < y < 2$. Potentiating the original inequality, we obtain $x - 2 > y - 1$ or $y < x - 1$.

b) For $y > 2$ we get $x - 2 < y - 1, y > x - 1$.

The solutions corresponding to these cases are shown in green on figure7.

Solve examples 5 and 6 on your own.

Example 5. Solve graphical inequalities $\log_{y-1}(y - x + 2) > 0$ (see Fig. 8).

Example 6. Solve graphical inequalities $\log_x \log_y y > 0$ (see Fig. 9).

Let us now consider the square three-fold

$$f(x) = ax^2 + bx + c.$$

As is known, the graph of the function $y = ax^2 + bx + c$ is a parabola with an axis of symmetry parallel to the ordinat axis and with a vertex at point $B\left(-\frac{b}{2a}, -\frac{D}{4a}\right)$, where $D = b^2 - 4ac$.

The graph of one of the quadratic trinomials (at $a > 0$) is shown schematically in figure 10. This graph divides the plane into two parts, in one of which the inequality $y - (ax^2 + bx + c) > 0$ holds, and in the other $y - (ax^2 + bx + c) < 0$. Reader without difficulty

46

Let him notice that only one of these parts is a convex set.

Example 7. Solve graphical inequalities $\log_y(5x-x^2-6)>1$.

Solution i.e. The range of admissible values ​​is obviously determined by the inequalities

$$\begin{cases} 5x-x^2-6>0, \\ y>0, \\ y\neq 1. \end{cases}$$

The first of these inequalities is equivalent to the inequality $x^2-5x+6<0$. Its solutions fill the interval $2<x<3$.

On figure 11, those parts of the plane where $x<2$ and $x>3$ are shaded with gray, resulting in $y<0$, and the points lying on the dotted line $y=1$ are also excluded.

Let’s consider two cases: $y>1$ and $0<y<1$.

a) $y>1$. Potentiating the original expression, we obtain the inequality $5x-x^2-6>y$, or $y-(-x^2+5x-6)<0$. The last inequality is satisfied by the coordinates of all points located under the parabola $y=-x^2+5x-6$, but none of them corresponds to the conditions $y>1$. Thus, for $y>1$ there are no solutions.

b) $0<y<1$. Similarly, we obtain the inequality $5x-x^2-6<y$, i.e. $y-(-x^2+5x-6)>0$.

Obviously, the solution is the area above the parabola $y=-x^2+5x-6$ shaded green at figure 11.

Solve examples 8 and 9 yourself.

Example 8. Solve graphical inequalities $\sqrt{y}<x-1$ (see Fig. 12).

Example 9. Solve graphical inequalities (see Fig. 13)

$$\left(\frac{1}{3}\right)^{x-1} > \left(\frac{1}{3}\right)^{\sqrt{y-x+3}}$$

Example 10. Solve graphical inequalities

$$\sqrt{1-(x-3)^2-(y-2)^2} < \sqrt{1-(x-4)^2-(y-2)^2}$$

Solution i.e. In this case, the ODZ is determined by the inequalities

47

$$\begin{align*} 1 - (x - 3)^2 - (y - 2)^2 &\geq 0, \\ 1 - (x - 4)^2 - (y - 2)^2 &\geq 0. \end{align*}$$

or

$$\begin{align*} (x - 3)^2 + (y - 2)^2 &\leq 1, \\ (x - 4)^2 + (y - 2)^2 &\leq 1. \end{align*}$$

The equation $(x - 3)^2 + (y - 2)^2 = 1$ is the equation of a circle of unit radius centered at the point $O_1(3;2)$. The inequality $(x - 3)^2 + (y - 2)^2 \leq 1$ is satisfied by all points distant from the point $O_1$ by no more than one, that is, all points of the indicated circle, including its boundary. The set of points whose coordinates satisfy the second unknown is defined similarly and is the common part of two circles (Fig. 14). It remains to find the set of points for which the given inequality holds. After estimated transformations, we obtain inequalities that in this ODZ are equivalent to the original one:

$$\begin{align*} 1 - (x - 3)^2 - (y - 2)^2 &< \\ < 1 - (x - 4)^2 - (y - 2)^2, \\ (x - 4)^2 - (x - 3)^2 &< 0, \\ x &>\frac{7}{2}. \end{align*}$$

Thus, the solution is the area painted on figure 14 with it (the part of the ODZ lying to the right of the straight line $x = \frac{7}{2}$).

Finally, consider the following example.

Example 11. Among the points whose coordinates satisfy the condition $x + 2y \leq 3$, $2x - y \leq 1$, $x \geq 0$ and $y \geq 0$, find a point for which the expression $x^2 + y^2$ will be the most accurate.

48

Solution. Let us find using a graphical solution of the system of inequalities

$$x + 2y \leq 3, \quad 2x - y \leq 1, \quad x \geq 0, \quad y \geq 0$$

a set of points whose coordinates satisfy specified conditions. From figure 15 it is clear that this will be a quadrilateral OABC. (The coordinates of its vertices are easily determined from the equations of the boundary lines and are indicated on the same figure.)

Now our task is formulated as follows. Among the points of the quadrilateral AOBC, find a point for which the expression $x^2 + y^2$ is the most 简便. But we know that the expression $x^2 + y^2$ determines the square of the distance from the point with coordinates $(x, y)$ to the origin. Consequently, the task comes down to finding the point in the OABC quadrilateral that is most convenient from the origin. Obviously, such a point is the point $C\left(0; \frac{3}{2}\right)$. Prove it!

Solve two problems of this type for yourself (these problems were presented in the 1968 written entrance exams in mathematics).

Example 12 (physicist at Leningrad University). Among all pairs of numbers that satisfy the inequality $\log_{x^2 + y^2}(x + y) \geq 1$, find those with $y$ the most.

Example 13 (Mechanics and Mathematics of Novosibirsk University). For each $a$ among all points of the plane whose coordinates $(x, y)$ satisfy the system of inequalities

$$\begin{align*} y - 2x &\geq 0, \\ -x^2 + 2ax - a^2 + a + 1 - y &\geq 0, \end{align*}$$

find the point with the most $y$.

At the beginning of the lecture, the lecturer formulated the famous Cauchy theorem, according to which the arithmetic mean of unequal positive numbers is greater than their geometric mean.

At this moment, the student was called to the dean’s office. When he returned to the lecture, conditions were still outside the door as the lecturer finished formulating the problem:

“The clue is that all the roots of the equation are not only real, but also positive.”

When the student entered the classroom, everything on the board had already been erased except for the note $x^{20} - 20x^{19} \dots +1 = 0$.

“Everything is clear,” said the student and named all the roots of the equation that the lecturer wrote down in his absence.

Can you name them?

49

14 DAYS IN BUCHAREST

Residents of Bucharest are accustomed to tourists of all ages. This must be why hardly anyone paid attention to the multilingual flocks of boys and girls who got acquainted with the Romanian capital last July. And only the initiated knew why schoolchildren from 14 countries came here.

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However, many people later learned about this from the newspapers.

What brought young guests from England, Belgium, Bulgaria, Hungary, East Germany, Holland, Mongolia, Poland, Romania, USSR, France, Sweden, Czechoslovakia, Yugoslavia to Bucharest? Mathematics brought them here: the XI International Mathematical Olympiad was held here.

The path of the schoolchildren to Bucharest was not easy. To do this, they had to take recognized places at all our Olympiads of various “ranks” - school, district, regional, All-Union. Moreover, winning only one year’s competition did not guarantee success. Preference was given to those who had a good “Olympic biography” for several years.

And so we name the names: Vladimir Drinfeld - 10th grade at Kharkov School No. 27, Andrey Zelevinsky - 9th grade at Moscow School No. 2, Arkady Klimov - 9th grade at Moscow School No. 18 (physics and mathematics school at Moscow State University), Elena Neklodova - 10th grade at Moscow School No. 7, Andrey Prasolov - 10th grade and Andrey KhodULEV, 9th grade, both from Moscow school No. 18, Valery Solovev - 10th grade at Kazan school No. 131, Pavel Suvorov - 10th grade at Leningrad school No. 45 (physics and mathematics school at Leningrad State University). These eight-member Olympiad participants had to endure “battles” with the same number of opponents from each foreign team.

The “exam” was taken by a reputable jury, headed by Romanian academicians Moisil and Teodersecu. Its members included prominent scientists and teachers from all countries that sent teams to the Olympiad; From the USSR, the jury included the head of the Soviet team, prof. V. I. Levin.

Having arrived in Bucharest on July 7, our Olympians, V.I. Levin and his deputy - the author of these lines - toured the sights of the city, its historical monuments, museum, parks, and visited the Soviet embassy.

On July 10, two days of perfection began. According to the conditions of the Olympiad, each participant must solve three problems in four hours on both the first and second days; their text is given below. Each task was assessed by establishing a number of points. For the correct answer, the solver received the maximum number of points, and if there were flaws in the solution, the number of points was reduced.

Between July 12 and 19, Olympic participants traveled around the country. Of course, they really liked the Romanian cities, where they felt great, but their hearts and minds were in Bucharest, where the jury summed up the results of the competition. It became known on July 19, when it was licensed. N. Belchesu (lyceums are called high schools there) were named the winner. The jury awarded the first degree diploma to the participants who scored the maximum possible number of points - 40. Three were awarded it: Vladimir Drinfeld (USCP), Simon Norton (England) and Filala Tibor (Hungary). Second degree diplomas of doctors and participants who scored from 30 to 37 points participated in the training. Of our participants, they were received by Andrey Prasolov, who scored 32 points, Andrey Zelevinsky and Andrey KhodULEV, who scored 30 points each. Third degree diplomas were awarded to those who scored from 24 to 29 points. Of the Soviet students, third degree diplomas were achieved by Elena Neklodova and Valery Solovev, who scored 27 points, as well as Arkady Klimov, who scored 24 points. Pavel Suvorov scored 21 points. He received a participant’s diploma.

In terms of the total number of points (231), our team took third place, the Hungarians won first place - 247 points. They were awarded one divolma of the first degree, four - of the second and two - of the third. One Hungarian schoolchild received a participant’s diploma. The GDR team is in second place with 240 points: four second degree diplomas and four third. If we count according to the “Olympic score” 7 points for first place, 4 for second and 3 for third, then the Hungarian team received 29 points, and the team of the GDR and the USSR received 28 points each.

All members of our team who graduated from the 10th grade were enrolled in the mechanical and mathematical faculties of universities without the need for entrance exams.

Here are the tasks given by the Olympiad participants. The complete solution to all problems was published in the journal “Mathematics at School” No. 1 for 1970.

On the first day, perfection was given the following tasks.

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1. Prove that there is an infinite set of natural numbers $a$ with the following properties: the number $z = n^4 + a$ is not prime or for which natural number $n$ (GDR, 5 points).

2. Let $a_1, a_2, \dots, a_n$ be real constants, $x$ be a real variable and
   $$f(x) = \cos(a_1+x) + \frac{\cos(a_2+x)}{2} + \frac{\cos(a_3+x)}{2^2} + \dots + \frac{\cos(a_n+x)}{2^{n-1}}.$$
   Prove that from $f(x_1) = f(x_2) = 0$ it follows that $x_1 - x_2 = m$, where $m$ is an integer (Veneria, 7 points).

3. For each value $k = 1, 2, 3, 4, 5$, find the necessary and sufficient conditions that must be satisfied by the number $a > 0$ in order for there to exist a terraer whose $k$ edges have length $a$, and the remaining $6 - k$ edges have length 1 (Polysha, 7 points).

On the second day the following tasks were proposed:

4. The semicircle $\gamma$ is built on the diameter $AB$. The point $C$ lies on $\gamma$ and distinguishes $A$ and $B$. We denote the orthogonal projection of $C$ onto $AB$ by $D$. Consider three circles $\gamma_1, \gamma_2, \gamma_3$ having $AB$ as a common tangent: of them $\gamma_1$ is inscribed in the triangle $ABC$, $\gamma_2$ and $\gamma_3$ both tangent to the segment $CD$ and $\gamma$. Prove that $\gamma_1, \gamma_2, \gamma_3$ have a second common tangent (Holland, 6 points).

5. There are $n > 4$ points in the data plane, and no three lie on the same straight line. Show that it is possible to find at least $C_n^2 - 3$ convex quadrilaterals with vertices at four of the given points (Mongolia, 7 points).

6. Prove that if $x_1 > 0, x_2 > 0$ and $x_1y_1 - z_1^2 > 0, x_2y_2 - z_2^2 > 0$, then
   $$\frac{8}{(x_1+x_2)(y_1+y_2)-(z_1-z_2)^2} \leq \frac{1}{x_1y_1 - z_1^2} + \frac{1}{x_2y_2 - z_2^2}.$$
   Establish necessary and sufficient conditions under which equality occurs in this inequality (CCCP, 8 points).
   N. S. Petrakov

The boarding school at Leningrad State University accepts those who have successfully completed the intermediate and eighth grade of a general education school and who have demonstrated ability to study natural subjects, living in the Arkhangelsk, Volgodsk, Kaliningrad, Kirov, Leningrad, Murmansk, Novgorod, Pskov regions, Karelian and Komi Autonomous Soviet Socialist Republics, Latvian, Lithuanian and Estonian SSRs.

To enter the eighth grade of a boarding school, you must take a competitive exam in mathematics (written) and pass a test with the admissions committee (with a physicist, mathematician or biologist).

Students who live in rural areas, working-class villages, and small towns in the North-West are admitted to the eighth grade.

Admission to ninth grade will include a written exam in mathematics and an oral exam in physics. Admission to the physico-mathematical specialization is based on mathematics, and to the chemical-biological specialization - either chemistry or biology.

Entrance exams are held during the spring holidays simultaneously with the regional (national) round of the Physics, Mathematics, Chemistry and Biology Olympiad, as well as in June in the regional and national centers among candidates selected by public education organizations.

Desire to participate in competitive examinations of subjects for public education authorities (OBLONO or the Ministry of Education of the Republic):

A recommendation from the teaching council to your school.

Characteristics (approved by the pedagogical council).

Statement from the parent or their surrogate.

Medical certificate (form No. 26).

The issue of admission to a boarding school is decided by the admissions committee in Leningrad. The decision of the selection committee is communicated by July 10.

Our address: Leningrad, P-228, Savushkina Street, 61. Specialized boarding school at Leningrad University.

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running_head:
folio:

12 DAYS IN CZECHOSLOVAKIA

About the III International
Physical Olympiad for schoolchildren,
held in Brno
in June 1969,
members told
Soviet team.

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The initial acquaintance with Prague was very short. A few hours after our arrival, we were seated in the Prague-Brono train station and wished us success. True, with a small caveat: “Just don’t overtake the Czechoslovakian team!”

The day was sunny, and we spent the entire trip at the windows. Beautiful dark groves on the light green slopes of the cold gave way to blue cliffs going up - a very beautiful peaceful country. The journey was pleasant, but short. Just four hours in posed and we are in Brono.

The hosts of the Olympics tried their best to entertain their guests. On the eve of the tour, we walked up and down Brono and showed castles and other sights of the city. On the bus we gradually got to know the guys from other delegations - Poles, Romanians, Germans. We discovered that one girl came to the Olympics - a member of the Romanian room. An expectant conversation was taking place (a mixture of English and Russian).

And in the evening - the grand opening of the III International Olympiad *. The regulations for improvement are announced: there will be two rounds, the first is theoretical, the second is experimental. At the theoretical level, four problems are proposed (in mechanics, thermodynamics, electricity and optics), at the experimental level - one. Each tour lasts five hours. Problems are assessed using a full-point system. The second round takes into account the implementation of both the experimental and theoretical parts of the work. The maximum number of points that can be scored is 48. It is announced in advance how many points you need to have to get first place.

*) Schoolchildren from Bulgaria, Hungary, the GDR, Poland, Czechoslovakia and Yugoslavia took part in it.

first prize, second, third, good review. It is announced that it is mandatory to wear the badge of an Olympiad participant, since the emblem will serve as a pass to the territory of the military academy where the Olympiad will be held.

During the break, we met the Bulgarians: “We just arrived and we want to talk in Russian…” We were very, very happy about this - after all, we were pretty tired of speaking English.

The next morning, June 24th, we got up, had breakfast, ran to the hall of the military academy, concentrated… but it turned out that Zhorny had not finished translating the conditions of the problem into the languages ​​of the Olympiad participants, and everyone spent an hour looking at paintings on a plot from Greek mythology, which occupied one of the halls.

Finally, the first round is over, tomorrow is the second round, and we are looking forward to a completely free evening. But… the hosts of the Olympics felt our craving for art, and we had to go listen to Smetana’s opera “The Eaten Bride.”

Despite the lack of sleep, we completed the experimental tour successfully and went to see Austerlitz. We really liked the masterfully executed figures of “mourning matter” and “consolable bride”, standing at the entrance to the MEMORIAL HOUSE IN MEMORY OF THE AUSTRIAN AND RUSSIAN SOLDIERS who died at Austerlitz. The task has a curious action effect: if you say something quietly while standing in a corner, then nothing can be heard in the center, but everything can be clearly heard in the opposite corner. A minute later, everyone (Vedic physicists!) was pinching something in the corners.

The next morning we were taken to Bratislava. Our guides around the city were two guys - Martin and Ladiislav, who treated us sincerely in a friendly manner. We saw the Liberator Monument and Bratislava Castle. Despite the fact that the day was cloudy and gray, we still had a very bright and bright memory of Bratislava.

The next morning, the head of our delegation, Professor Yantsov, informed us that, having received three first, one second and one third first, we lost the first and second team places to the Czechoslovakians and Hungarians.

In the afternoon we explored the Punkva cave. It is difficult to describe its beauty; you need to see everything with your own eyes. There is one black stone, on which they place their hand and make a wish; according to legend, it should be used. It turned out that members of the Czech team came to the cave before the Olympics and wanted to get first place. Maybe that’s why they won?

In the evening - the closing ceremony of the Olympics. Our team received five sentences: three watches and two illustrated books. Then there was a banquet.

On June 29 we went to Prague by bus. The road was tiring, it was shaking, sometimes they woke us up and showed us castles or took us to lunch. In the evening we arrived in Prague.

After the day, we walked around the city, looked around and thought: “Oh, I’d rather go home!” It was good in Czechoslovakia, of course, but at home, as they say, it was better.

A. KLIMOV,
N. KONDRATIEV,
A. CHERNOUTSAN.

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OBJECTIVES OF THE OLYMPICS

Task 1

The system of bodies, presented on figure, is formed by three carts A, B, C, whose masses are $m_1=300$ e, $m_2=200$ e and $m_3=1500$ e, respectively.

Cart C is acted upon by a horizontal force F of such magnitude that carts A and B are at rest relative to cart C.

1. Determine: a) the tension of the insoluble thread connecting carts A and B; 6) force F.

2. Assume that cart C is stationary and determine: a) the acceleration of carts A and B; 6) threading.

When solving the problem, use air resistance, friction, moments of inertia of the block and wheels, as well as the mass of the thread.

Problem 2

A copper calorimeter, mass $m_1$, contains water of mass $m_2$ at a general temperature $t_2$. Ice is placed in the calorimeter, the mass of which is $m_3$ and the temperature is $t_3<0^\circ$ C.

a) Determine the mass of water and ice and their temperature at equilibrium for the most general values ​​of $m_1, m_2, m_3, t_2, t_3$. Write the heat transfer equations in the indicated system.

6. Determine the temperature and mass of water and ice, if up to $m_1=1,00$ ke$^*$), $m_2=1,00$ ke, $m_3=2,00$ ke, $t_2=10^\circ$ C, $t_3=-20^\circ$ C.

When solving a problem with energy losses, apply. Barometric pressure is considered normal. The specific heat of copper is $c_1=0,094$ kcal ke$^{-1}$ erad$^{-1}$, the specific heat of ice is $c_3=0,492$ kcal ke$^{-1}$ erad$^{-1}$, the specific heat of fusion of ice is $l=70,7$ kcal ke$^{-1}$.

Problem 3

A ball of mass $m$, charged with an electric charge $q$, is attached to one end

*) Two zeros after the charge indicate that the answer must be given with an accuracy of one hundredth.

conductive thread. The other end of the thread is attached to the highest point of a ring with radius $R$, which is in the vertical plane. The stake is made of rigid wire, the diameter of which can be adjusted. The charge $Q$ of the same sign as $q$ is uniformly distributed on the ring. Determine the length $l$ of the thread at which, after deflection, the ball will be on the axis of the ring, perpendicular to its plane.

Solve the problem first in general form, and then for the numerical values ​​$Q=q=9,0\cdot10^{-8}$ k, $R=5,0$ cm, $m=1,0$ e, $\varepsilon_0=8,9\cdot10^{-12}$ fm$^{-1}$. Massey threads pretenbr.

Problem 4

A polished plate is placed on top of a ground glass cube, the edge length of which is 2.00 cm, so that a thin interference layer appears in the space between it and the cube. If the plate is illuminated from above at a right angle with radiation with wavelengths from 400.0 mc to 1150 mc, for which the plate is transparent, then the condition for the maximum intensity of light reflected by the interference layer is satisfied only for two waves from the mentioned interval: for $\lambda=400,0$ mc and for another wavelength. Determine this wavelength. Calculate how much the temperature of the cube needs to be increased so that it touches the plate. The coefficient of linear thermal expansion of the stack is $\alpha=8,0\cdot10^{-6}$ erad$^{-1}$, the refractive index of air is $n=1$. The distance from the base of the cube to the plate does not change during heating.

Task 5 (experimental)

Use: two iron-nickel batteries, one dry element, one rheochord (a homogeneous wire of unknown resistance $X$, stretched along the millimeters of the scale and equipped with a sliding contact), one resistance store of value $R$, one galvanimeter (zero value for finding the scale resistance) and one hardened resistance.

Consider: a) a closed circuit consisting of two series batteries, a resistance store and an unknown resistance $X$ (reohord), 6) a series connection of a dry element and a galvanimeter with a protected resistance.

Suggest and explain such a connection of branch “b” to circuit “a”, which allows us to find such a position of motion of the rheochord $X$ in which no current flows through the galvanimeter. Create a circuit based on the diagram.

By making changes, determine: 1) the relationship between the potential difference at the terminals of two series batteries and the electromotive force of a dry cell (consider the potential difference at the terminals of both batteries constant), 2) unknown resistance $X$.

Find for what resistance $R$ the problem has a solution.