Kvant 1970, No. 11 · Pages

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Editor-in-Chief - Academician I. K. KIKOIN
First Deputy Editor-in-Chief - Academician A. N. KOLMOGOROV

Editorial collection:

L. A. Artsimovich,
M. I. Bashmakov
V. G. Boltyansky,
I. N. Bronstein
N. B. Vasiliev
I. F. Ginzburg
V. G. Zubov,
P. L. Kapitsa,
V. A. Kirillin,
G. I. Kosourov
V. A. Letskovtsev
V. P. Litsevsky
A. I. Markutsevich,
M. D. Millyuonitkov,
N. A. Patrikeeva
N. Kh. Rozov
A. P. Savin
I. SH. SLOBODECHKY
M. L. Smolyansky
Y. A. Smorodinsky,
V. A. Fabrikant,
J. E. Schneider

academician
Corresponding Member of the Academy of Pedagogical Sciences of the USSR
academician
academician
(deputy editor-in-chief)
Academician of the Academy of Pedagogical Sciences of the USSR
academician
(deputy editor-in-chief)
Doctor of Physical and Mathematical Sciences
Academician of the Academy of Pedagogical Sciences of the USSR
(executive secretary)

Head of the editorial office L. V. Chernova.
Chief artist E. P. Leonov.
Technical editor T. M. Makarov.
Proofreader L. S. Somova.
Publishing house “Science”. Main editorial office of physical and mathematical literature
Editorial address:
Moscow, V-71, Leninsky Prospekt, 15.
Phone: 2340811.

Delivered as a set on 7/24/70. Under. for publication 10/13/70
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Electrostatics on the tongue
power lines

Generally, electrostatics is studied this way: they write Coulomb’s law for the force of interaction between two exact charges $q_1$ and $q_2$: $F = \frac{q_1 q_2}{r^2}$ (r is the distance between the charges) - and then, with the help of sometimes simple and sometimes complex mathematical calculations, they obtain all sorts of consequences of this law, considering that the charge of any body can be represented as the perfection of properly distributed exact charges.

But often an accurate mathematical analysis of the electric field and charge distribution turns out to be difficult. Then simple messages come to the rescue - symmetry and patterns of field lines. Of course, they cannot give as complete a picture as mathematical calculations, but they are visual and help to “see” and predict what features the electric field will have and how charges will interact.

What are field lines

Let’s take the exact charge $q$. This charge creates an electric field in the surrounding space, which is completely determined if at each point in space the direction of the field is given - the force that would act on an exact unit positive charge if it were placed at that point. In order to visually represent the electric field throughout space, you can try to represent it by a variety of vectors, for example, at different points in space so that each of them shows the direction of the field (its magnitude and direction) at the point from which it emerges. However, it immediately becomes obvious that this method of depicting the field is unsuccessful: individual arrows, overlapping each other, create a confusing picture, which will need to be understood.

Faraday proposed that the field be represented by lines of force. These are lines, the direction of which at each point coincides with the direction of the field at this point. To accurately charge, drawing field lines is quite simple. These are straight lines (rays) passing through the point at which the charge is located. If the charge is serial, then the lines of force are

Fig. 1.

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is national to the square of the distance to the charge. Since $\frac{n_1}{n_2} = \frac{r_2^2}{r_1^2}$, then $n \sim \frac{1}{r^2}$.

The directivity of the field of the exact charge changes in the same way: $E = \frac{q}{r^2}$.

Thus, the density of field lines is proportional to the direction of the field: $n \sim E$. It is only necessary that the power lines are not interrupted anywhere. For an exact positive charge, the lines of force must begin at the charge and go to infinity. In the case of a negative charge, the lines of force must go from infinity and end at the charge.

It turns out that it is always possible to draw a picture of the field lines so that their density is proportional to the direction of the field. To do this, the lines of force must be driven out on positive charges and end on negative ones (or go to infinity), and the number of lines of force leaving the charge or going to it must be proportional to the magnitude of this charge.

Let’s mentally draw a sphere with the center at the point where we find the charge. If we agree to assume that $4\pi q$ lines of force emerge from each positive charge (and end at each negative charge), then their density on the spherical surface will be equal to $\frac{4\pi q}{4\pi r^2} = \frac{q}{r^2}$, that is, it will be equal to the direction of the field. The total number of lines of force intersecting the surface of the sphere does not depend on its radius and is equal to $4\pi q$. It is clear that the same will be the number of lines of force coming out of any closed surface inside which we find a charge (or going into it), no matter what shape this surface has.

This statement is called Gauss’s theorem. It is true, of course, for any charge distribution. It can be shown that for any system of charges the number of force

lines emerging from a closed surface of arbitrary shape, minus the number of force lines entering it, is equal to $4\pi Q$, where $Q$ is the total charge that is located in the volume limited by this surface. If the charge $Q$ is negative, then the number of lines that leave the volume bounded by the surface is less than the number of lines of force entering it. If there are no charges in the volume bounded by the surface, then the same number of lines of force come out of it as they enter.

Potential and equipotential surfaces

Before we begin to implement our knowledge - solving specific problems, let us remember one more characteristic of the field - potential. The fact is that for it you can find a geometric picture that is closely related to the picture of lines of force. This connection often helps to draw and analyze a picture of field lines.

Let us first recall what an electric field potential is.

If, by moving a charge in the field, we return to the starting point, then the total work that is done when crossing an exact charge from one point in the field to another does not depend on the shape of the path. Otherwise, it would be possible to first follow the path where it is smaller, and return along the path where it is greater. The total work would not be zero. Due to the fact that the work does not depend on the shape of the path, the concept of potential can be introduced. The field potential at a given point is equal to the work done by the field when crossing a unit exact positive charge from a given point in the field to infinity.

Potential can be either positive or negative. The field potential of a positive charge is positive, and the field potential of a negative charge is negative. In the first case, the force acting

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and non-predestination from Coulomb’s law.
Let us take an arbitrary point $A$ (Fig. 7) inside the sphere. Let’s imagine a narrow cone that comes from point $A$, cutting out two small spherical areas $\Delta S_1$ and $\Delta S_2$ on the surface of the sphere. If $r_1$ and $r_2$ are the distances from point $A$ to these sections, then $\frac{\Delta S_1}{\Delta S_2} = \frac{r_1^2}{r_2^2}$.

Since the sphere is charged uniformly, the charges of the areas are proportional to their areas. Therefore, $\frac{q_1}{q_2} = \frac{r_1^2}{r_2^2}$, and the fields created at point $A$ by these sections cancel each other out:

$$\frac{E_1}{E_2} = \frac{\frac{q_1}{r_1^2}}{\frac{q_2}{r_2^2}} = 1.$$

In this way, the entire sphere can be divided into pairs of sections. Therefore, the field at point $A$ is zero.

The idea of ​​plum lines made it possible to obtain the same result easier and faster.

Uniformly charged plane and parallel-plate capacitor

Let us now consider the field of an infinite (more correctly said, very large) uniformly charged plane, per unit area of ​​which there is a charge $σ$.

It is clear that the plum lines of the field created by the plane are perpendicular to the plane and their density is the same (Fig. 8): if the plum lines were inclined in one direction, this would mean that one direction along the plane is different from the others; if the density of the plumbing lines were different at different points of the plane’s approach, then this would mean that one point, which, of course, cannot be, differs in some way from the others.

Let us draw a cylindrical surface with bases parallel to the plane and equal in area to one. Drain field lines do not intersect the side surface

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Definition. The greatest common divisor of numbers or polynomials is their common divisor that is divisible by any of their common divisors without a remainder.

We will denote the gcd of numbers or polynomials $a, b$ as follows: gcd $(a, b)$. For example, GCD $(5, 10) = 5$; GCD $(x^2, 3x-x^2) = x$; GCD $(x^3-x^2,x-1) = x-1$.

Japanese most common divisor

For numbers, there are two known ways to find GCD.

Method 1. We decompose both numbers into prime factors and take the product of those of them that appear in both numbers. For example, reducing the fraction $\frac{3770}{9425}$, you can write: $\frac{3770}{9425} = \frac{2\cdot 5\cdot 13\cdot 29}{5\cdot 5\cdot 13\cdot 29}$;

GCD $(3770, 9425) = 5\cdot 13\cdot 29 = 1885$, $\frac{3770}{9425} = \frac{2}{5}$

What’s wrong with this method? First, you must first prove that any natural number can be factorized into prime factors and, moreover, in a unique way. Secondly, in order to find this decomposition, a lot of work will have to be done, mainly in the direction. After all, we will find all the divisors of each of the numbers, but we only need their most common divisor.

As we will see later, it will be even worse to bypass the matter with polynomials: the expansion always exists, but, generally speaking, it is impossible to find it, unless someone specially arranged it so that the expansion is easy to find (as happens in specific examples).

But you don’t have to decompose numbers into polynomials.

Method 2. Euclidean algorithm. It is based on the following lemmas.

Lemma 1. Any natural number $q$ can be divided with a remainder by another natural number $p$, that is, represent $q$ as $q = kp+p_1$, where $0 \leq p_1 < p$.

Lemma 2. If the number $d$ is the GCD of the numbers $p$ and $q$.

Note that this lemma allows us not only to find, but also to prove the existence of the GCD of two numbers, because it reduces questions about the existence and location of the GCD to smaller numbers.

Prove these lemmas yourself. In Lemma 1, we need to break the number line into pieces by points $0, p, 2p, 3p, \dots$ and find which of them contains the point $q$. In Lemma 2, we must assume that the number $d$ is divisible by any common divisor of the numbers $p$ and $p_1$, and also that both numbers $p, p_1$ are divisible by $d$, and prove that the number $d$ has the same two properties relative to the numbers $p$ and $q$.

Now, using the example discussed in method 1, we will outline the Euclidean algorithm itself - a sequence of actions that allows you to find the gcd of the numbers $q$ and $p$. Next we write the sequence of formulas in general form, and on the right - for example.

Step 1. Using Lemma 1, divide $q$ by $p$. Let’s say:
$q = k_0p+p_1, 0 \leq p_1 < p$; $9425 = 2\cdot 3770+1885, 0 \leq 1885 < 3770$.
Based on Lemma 2, we conclude that
GCD $(q, p) = \text{НОД}(p, p_1)$; $\text{НОД}(9425, 3770) = \text{НОД}(3770, 1885)$.

Step 2. See if $p_1$ is equal to zero.
If $p_1$ is equal to zero, then we end the process and go to step 3.

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In Lemma 1, the number $p_1$ was smaller than $p$ in magnitude, and for polynomials, the “size” is taken to be the degree of the polynomial. An example of dividing polynomials with a remainder is given below, using it you yourself will prove this lemma.

Example. Divide the polynomial $2x^5 + 4x^4 - 2x^3 - x + 1$ by the polynomial $x^2 + 2x + 5$ using a corner. It turns out:

$$\frac{-2x^5 + 4x^4 - 2x^3 - x + 1}{2x^5 + 4x^4 + 10x^3} = \frac{x^2 + 2x + 5}{2x^3 - 12x + 24}$$

$2x^5 + 4x^1 - 2x^3 - x + 1 = (2x^3 - 12x + 24)(x^2 + 2x + 5) + (11x - 119)$. Now we have the required lemmas and we can act according to the Euclidean algorithm, that is, construct a sequence of divisions with remainders (for polynomials!), in which the polynomial is successively decreased:

$$Q(x) = K_0(x) P(x) + P_1(x),$$
$$P(x) = K_1(x) P_1(x) + P_2(x),$$
$$P_1(x) = K_2(x) P_2(x) + P_3(x),$$

$$P_{m-2}(x) = K_{m-1}(x) P_{m-1}(x) + P_m(x),$$
$$P_{m-1}(x) = K_m(x) P_m(x).$$

In this case, GCD $(Q(x), P(x)) = \text{НОД}(P(x), P_1(x)) = \dots = \text{НОД}(P_{m-1}(x), P_m(x)) = \text{НОД}(P_m(x), 0) = P_m(x)$. The last quotient $P_m(x)$ is the gcd of the original polynomials (check all this).

These practical formulas are obtained by dividing with a corner, which in the very first example looks like this:

$$\frac{x^9 - x^8 - x^7 - 3x^6 + x^5 + 4x^4 + 4x^3 + x^2 - 2x - 1}{x^9 - x^8 - 3x^6 + x^5 + x^4 + 2x^3 - x} = \frac{x^8 - x^7 - 3x^5 + x^4 + x^3 + 2x^2 - 1}{x^8 - x^7 - 3x^5 + x^4 + x^3 + 2x^2 - x - 1}$$

Now it remains to reduce the algebraic fraction (2) by the last divisor $-x^7 + 3x^4 + 2x^3 + x^2 - x - 1$, that is, divide the numerator and denominator in (2) by the resulting gcd. Having done all this, we get $\frac{x - 1}{x^2 - x - 1}$.

This is the answer.

If you read N.M. Beskin’s article “Continued Fractions” («Kvant» No. 8), you may notice that the procedure described here is exactly the reduction of polynomials by expansion into whole fractions, carried out in the same way as for rational numbers. We rewrote the fraction $\frac{P(x)}{Q(x)}$ as follows:

$$\frac{P(x)}{Q(x)} = \frac{P(x)}{K_0(x) P(x) + P_1(x)} = \frac{1}{K_0(x) + \frac{P_1(x)}{P(x)}}$$,

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$$\frac{P(x)}{Q(x)} = \frac{1}{K_0(x) + \frac{P_1(x)}{K_1(x) P_1(x) + P_2(x)}} = \frac{1}{K_0(x) + \frac{1}{K_1(x) + \frac{P_2(x)}{P_1(x)}}}$$

and so on. In the end $\frac{P_m(x)}{P_{m-1}(x)} = \frac{1}{K_m(x)}$, all that remains is to turn the virgin fraction from polynomials. In the example under consideration, this is done like this: $\frac{1}{x + \frac{1}{-x+1}} = \frac{x-1}{x^2-x-1}$.

Simplify expressions:

1. $\frac{x^4 - 3x^2 + 1}{(3x^2 + 3x - 3)^2}$.
2. $\frac{x^3 + a^3}{x^4 + a^2x^2 + a^4}$.
3. $\frac{a(x^2 + 2) - 2ax - (x-1)^2 - 1}{x^4 + 4}$.
4. $\frac{x^2 + 6}{2(x^4 + 4x^2 - 12) + \frac{1}{7}(x^4 + 11x^2 + 30)}$.

TYPOS NOTED

The following should appear in the places marked below:

«Kvant» № 6

Page 6, right column. In figure for task 3 the bottom line is redundant; the entire figure is upside down.

Page 21, right column, 1 line from top: …=(1+x^t)[(1+x)^p-(1+x^p)]×…;
7th line from the top: x^{kp+s}…
2nd line blue: …C_{119}^3 ≡ C_{23}^6 C_{4}^3…;

Page 22, left column, 2nd line from top: …C_{23}^6 ≡ C_4^1 C_3^1…;

Page 30, right column. In the solution to task 3, the constraint $b \geq 0$ is redundant; the case $b < 0$ must be considered separately.

Page 31, left column, 6th line:
$$x = \frac{a(2+tg\alpha)}{tg\alpha};$$
Line 3 blue:
$$V = \frac{a^3(2+tg\alpha)^3}{6tg^2\alpha};$$

Page 4, in task 15 of the system has the form:
$$\begin{cases} x+3y=3, \\ ax+4y=6. \end{cases}$$

in task 35 of the system has the form:
$$\begin{cases} 3|x|-5y=-3, \\ 2x+|y|=11. \end{cases}$$

Page 35, in task 3 (varnant 2) the equation is:
$$\log_3(\log_2x-1)^2=\sqrt{5}^{-\log_54}.$$

Page 52, right column, 3rd line blue: … $p^2+13=17$…

Page 59, right column. In task 4, 6) answer: 5(11) 540; There was an error in dividing.

Page 60, right column,
8. $\frac{5}{6} < x < 2.$ 23. $S_\sigma = \frac{1}{2}\pi b^2$.
24. $V = -\frac{2}{3}l^2$.
Page 61, left column, 37. $0 < t \leq 0,1$.
Page 63, right column, 11th line:
$$\begin{cases} 500-80=420=100a+b, \\ 2500-496=2004=100b+a. \end{cases}$$
3 pages cover, crossword facing $90^\circ$.

«Kvant» № 7

Page 43. At the intersection of row 5 and column 8 there should be 256;
12 » » » » 46656;
14 » » » » 823 543.
Page 49, right column, 14th line from top:
$$\frac{4}{9}$$.
Page 52, right column, 15 line from top:
$$\sqrt{4r^2-h^2} = \frac{r}{\sin \frac{\pi}{n}}$$
left column, 14th line blue: … $A_nA_1 = 2r$.
Page 59, task 1234: … using four weighings…

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Fig. 5.

then $\angle AC_2C_0 = \angle A_1C_2C_0$. On the other hand, $\angle C_0C_2A_1 = \angle BC_2N$. Therefore, $\angle AC_2C_0 = \angle BC_2N$, and therefore $\alpha = \beta$.

It is not difficult to report that if the rays $AC_2, BC_2$ and the perpendicular $DC_2$ did not lie in the same plane, then the path from point $A$ to the mirror and then from the mirror to point $B$ would be longer.

Thus, the law of reflection follows entirely from Fermat’s principle.

Interestingly, the law of reflection was first obtained by Heron of Alexandria, who lived in the 1st century. Heron believed that light travels along the shortest path. For a single-sided medium, this coincides with Fermat’s principle of least time. But it turned out to be impossible to explain the law of refraction in this way.

3. Law of refraction. Light from point $A$ falls on the flat interface between two media (Fig. 5) and is refracted, then hitting point $B$. If the speeds of light in the first (upper) and second (lower) media are different (for definiteness, we will assume that $v_1 > v_2$), then it is clear that the straight path $AB$ requires no minimum time. The time will be shorter if the light travels a slightly longer path in the first medium,

Fig. 6.

where it propagates at a higher speed, but over a slightly shorter path in the second medium.

Suppose that light takes the least time if its path passes through the point $C$. Then any other path to the left and right of point $C$ will take a long time. An arbitrary beam trajectory $AFB$ can be set using the variable $x$—the distance between point $C$ and point $F$ on the interface between the media. The curve of light propagation time versus $x$ should have a minimum at point $C$, that is, at $x=0$ (Fig. 6). Near the minimum, time changes very little with $x$. Approaching at very small $x$ this time can generally be considered constant: on a very small segment in height $x=0$ the curve can be seen as a small segment of a straight line parallel to the $x$ axis.

Let us compare the propagation times of light on paths $ACB$ and $AFB$, assuming that $x=FC$ is very short. On the path $ACB$, light in the first medium passes an additional path $MC$ (FM is perpendicular to $AC$; since $FC$ is very small, then $AF=AM$). In the second medium, the additional path is already passed by the ray propagating along the path $AFB$ ($CN \perp FB$). Since the time of light propagation from point $A$ to point $B$ should not change when

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Fig. 9.

speed of light propagation in glass. Let’s find at what distance $b$ from the surface of the sphere a narrow beam of rays searching from the source $S$ will be collected. According to Fermat’s principle, this distance is determined by the condition that the time of light propagation along a rectilinear path $SDS_1$ is equal to (the application of) the time of light propagation along an arbitrarily chosen path $SAS_1$, if only the point $A$ is close to the straight line $SDS_1$, perpendicular to the surface of the sphere.

From the point $A$ we drop the perpendicular $AB$ onto the line $SS_1$. Let’s denote its length by $h$. Since we are considering a paraxial ray of rays, then $h \ll a$, $h \ll b$ and $h \ll R$. We denote the distance $DB$ as $x$. On ray $SA$ we plot the segment $SD_1 = a + x$, and on ray $AS_1$ the segment $S_1B_1 = b - x$. Then, along the trajectory $SAS_1$, the light will travel only a path $\Delta_1 = D_1A$ in the air and $\Delta_2 = AB_1$ in the glass compared to the trajectory along the straight line $SS_1$. But then, on a straight path, the light will travel a path $x - \Delta_2$ longer in the glass than in the case when it follows a path $SAS_1$. The light delay in the glass on path $x - \Delta_2$ must be the same as the delay in air on path $\Delta_1$ plus the delay in the glass on path $\Delta_2$. (At $x \leq \Delta_2$ such compensation is very impossible.)

Let us now find the paths $\Delta_1$, $\Delta_2$ and $x$.

The square of $h$ of the right triangle $SAB$, according to the Pythagorean theorem, is equal to

$$h^2 = (a + x + \Delta_1)^2 - (a + x)^2,$$

or

$$h^2 = (a + x + \Delta_1 + a + x)(a + x + \Delta_1 - a - x),$$

but $a + x + \Delta_1 - a - x = \Delta_1$, and $a + x + \Delta_1 + a + x = 2a + \Delta_1 + 2x \approx 2a$ (since $\Delta_1 \ll a$ and $x \ll a$). That’s why

$$h^2 = 2a\Delta_1$$

And

$$\Delta_1 \approx \frac{h^2}{2a}.$$

Similarly, from the right triangle $ABS_1$ we find that

$$\Delta_2 \approx \frac{h^2}{2b}.$$

Finally, from the right triangle $ABC$ one can obtain the equation to determine $x$:

$$h^2 = R^2 - (R - x)^2.$$

From here

$$x \approx \frac{h^2}{2R}.$$

If the speed of light in air is $v_1$, and the speed of light in glass is $v_2$, then the light travels $\Delta_1$ in time $\tau_1 = \frac{\Delta_1}{v_1} = \frac{h^2}{2av_1}$, and the light travels $\Delta_2$ in glass in $\tau_2 = \frac{\Delta_2}{v_2} = \frac{h^2}{2bv_2}$.

On the segment $x$, the delay of light, caused by the fact that it travels in glass and not in air, is equal to the location of the time it takes light to travel through segment $x$ in glass and in air:

$$\tau = \frac{x}{v_2} - \frac{x}{v_1} = \frac{x}{v_1}\left(\frac{v_1}{v_2} - 1\right) = \frac{x}{v_1}(n - 1) = \frac{h^2(n - 1)}{2Rv_1}.$$

To focus rays, the travel time of light from point $S$ to point $S_1$ along all trajectories must be the same. Hence,

$$\tau_1 + \tau_2 = \tau.$$

Substituting here the expressions for $\tau_1$, $\tau_2$ and $\tau$ and then canceling by the common factor $\frac{h^2}{2}$, we obtain the equation

$$\frac{1}{a} + \frac{n}{b} = \frac{(n - 1)}{R}.$$

This equation allows, using the known assumption exponent $n$ and the radius of the spherical surface $R$, to find the image position (distance $b$) for a given distance from the source to the spherical surface. The fact that the distance $b$ does not depend on $h$ shows that all the rays in a narrow beam, distinguishing the values ​​$h$ from each other, converge at one point.

In the partial case when $a \to \infty$ (the beam consists of parallel rays), the position of the point at which the rays converge (focus) is determined by the formula

$$f = \frac{Rn}{n - 1}.$$

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disguise Thus, there is a chain of inequalities

$$\frac{a_1 + a_2 + \ldots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \ldots a_n} \geq \frac{n}{a_1 + \frac{1}{a_2} + \ldots + \frac{1}{a_n}}.$$

Try to prove the last inequality yourself.

In conclusion, we note that in higher mathematics there is a method for finding the maxima and minima of functions by differentiation. However, the considered inequalities turned out to be more convenient for calculating the largest and smallest values ​​of certain types of expressions on electronic computers. These methods are called geometric programming.

Exercise

1. From the circular sectors of a given perimeter $P$ find the sector with the largest area.

2. In a given circle, find the smallest chord passing through the point $A$ inside the circle.

3. Find the minimum of the expression

$$2x^3 + \frac{3}{\sqrt{x}}.$$

4. Prove that if the sum of positive variables $x_1, x_2, \ldots, x_n$ is constructed, then the product $x_1^{k_1} x_2^{k_2} x_3^{k_3} \ldots x_n^{k_n}$, where $k_1, k_2, \ldots, k_n$ are given positive integers, has the greatest value when the variables are proportional to their exponent: $\frac{x_1}{k_1} = \frac{x_2}{k_2} = \frac{x_3}{k_3} = \ldots = \frac{x_n}{k_n}$.

Prove that the same is true for positive rational numbers $k_1, k_2, \ldots, k_n$.

HOW TO FIND AN ERROR

You have multiplied two long polynomials. How to check if you made a mistake in your calculations?

We present two methods that allow us to find some possibilities of error (but, of course, do not provide a complete guarantee of the correctness of the verified calculations).

Method 1. Take the sum of the coefficients of the first polynomial and multiply it by the sum of the coefficients of the second. Should get the sum of the coefficients of the product. This will be the value of the polynomial at $x=1$.

The above method allows you to find errors in arithmetic (for example, if the “equality” $2 \cdot 3=5$ was used), but not in algebra. Let’s say, if at some stage you wrote $2x^2 \cdot 3x^3=6x^4$, then this method will not produce an error.

Method 2. It differs from the first in that the coefficients for even powers $x$ are taken with a plane sign, and for odd ones with a minus sign. Otherwise, this method is the same as the first. Here we get the value of the polynomial at $x=-1$.

The second method is more complicated than the first; this also allows us to identify some algebraic errors. When checking for “equality” $2x^2 \cdot 3x^3=6x^4$ the result is $2 \cdot (-3)=6$. This equality is absurd, which means there was an error in the calculations.

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masca units of length equal to $\rho_1$, the hoop rolls with speed $v_1$. At what speed will the hoop roll along the second section of the rope?

V. D. Eegorov

F63. A man, in order not to slip on a sticky slide, runs away from it. Why is this appropriate?

F64. Four experiments are carried out with a wooden ball and a tall vessel with water: in the first experiment the ball is weighed floating in the vessel (Fig. 2, a), in the second experiment the ball is weighed while attached to the bottom of the vessel (Fig. 2, b), in the third experiment the ball is held under water using a thin rod (Fig. 2, c), and finally, in the fourth experiment the ball floats up during weighing (Fig. 2, d). In which case will the mass of the weight balancing the vessel with the ball be greater?

F65. The plates of a flat capacitor are charged to potentials +φ and -φ relative to the ground. The capacitance of the capacitor formed by the plates is $C$, and the capacitance of the capacitors that form each plate with the ground is $C_1$. How many times will the electron field voltage between the plates change if one of them is grounded?

L. G. Aslamazov

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CL/DC = $\sqrt{b}/\sqrt{a}$. Try to find out when it is possible to split a trapezoid into two similar quadrangles in a more cunning way, corresponding to the following case 2), as shown in figure 4 (answer: if and only if $\sqrt{a}/\sqrt{b} < \sin \alpha/\sin \beta$, where $\alpha < \beta$ are the angles at the larger base of the trapezoid $b$).

**M13. Prove that if the difference between the largest and smallest $n$ real numbers $a_1, a_2, \ldots, a_n$ is $d$, and the sum of the models of all $\frac{n(n-1)}{2}$ pairwise differences of these numbers

$$\sum_{i<j} |a_i - a_j| \text{ равна } s, \text{ то } (n-1) d \leqslant s \leqslant \frac{n^2}{4} d.$$

Solution: Let’s plot the points $a_1, a_2, \ldots, a_n$ on the number line. Then $d$ is the distance between the outermost of these points, the leftmost and the rightmost, and $s = \sum_{i<j} |a_i - a_j|$ is the sum of all pairwise distances between these numbers.

One can obviously assume that the points are designated by $a_1, a_2, \ldots, a_n$ in ascending order: $a_1 \leqslant a_2 \leqslant \ldots \leqslant a_n$ (Fig. 5). Let us denote the distance between neighboring points $a_k$ and $a_{k+1}$ by $d_k(k=1, 2, \ldots, n-1)$. Obviously $d = d_1 + d_2 + \ldots + d_{n-1}$.

Let us now express $s$ in terms of the quantities $d_k$. To do this, we replace the total $s$ length of each segment $[a_i - a_j]$ with the sum of those $d_k$ of which it consists: $|a_i - a_j| = d_i + d_{i+1} + \ldots + d_{j-1}$. It is clear that $d_k$ is included in those segments whose left end lies at one of the points $a_1, \ldots, a_k$, and the right end lies at one of the points $a_{k+1}, \ldots, a_n$, that is, in total $d_k$ is included in the sum $k(n-k)$ times. That’s why

$$s = \sum_{k=1}^{n-1} k(n-k)d_k.$$

Now the assertions to be proved follow from two very simple inequalities: for all $k=1, \ldots, n-1$

1. $k(n-k) \geqslant n-1 \iff kn-k^2-n+1 \geqslant 0 \iff (k-1)(n-k-1) \geqslant 0$;
2. $k(n-k) \leqslant \frac{n^2}{4} \iff n^2-4nk+4k^2 \geqslant 0 \iff (n-2k)^2 \geqslant 0$.

Let’s use these examples and use what we get, what we get

$$\sum_{k=1}^{n-1} k(n-k)d_k \geqslant \sum_{k=1}^{n-1}(n-1)d_k = (n-1)d,$$

$$\sum_{k=1}^{n-1} k(n-k)d_k \leqslant \sum_{k=1}^{n-1}\frac{n^2}{4}d_k = \frac{n^2}{4}d.$$

The correct solutions to this problem were sent by M. Praeger from Tomsk, tenth-class student I. Protasov from Kyiv, S. Pozdnyakov from Vilnius and other readers.

It is also interesting to find out whether the estimates indicated in the problem statement are exact; is it possible, say, to put large numbers together with $n-1$ in the left inequality? In order to be convinced of the opposite, it is enough to give an example of a case when inequality turns into equality (and both of its parts contain positive numbers). It’s easy to come up with such an example, we’ve figured out our proof: we need to arrange

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points $a_1, a_2, \dots, a_n$ so that all $d_k$, except the first one - $d_1$, are equal to zero, that is $a_1 < a_2 = a_3 = \dots = a_n$. Then $s = (n-1)d_1 = (n-1)d$.

As for the second inequality $s \leqslant \frac{n^2}{4}d$, then for even $n = 2m$ equality can also be achieved in it (it is enough to take $a_1 = a_2 = \dots = a_m < a_{m+1} = \dots = a_{2m}$), and for odd $n = 2m+1$ it can be somewhat refined: it is not difficult to report that for odd $n$ the most of the numbers $k(n-k)$ is equal to $\frac{n-1}{2} \cdot \frac{n+1}{2} = \frac{n^2-1}{4}$ (prove this rigorously!); using this instead of inequality 2), we can prove the stronger inequality $s < \frac{n^2-1}{4}d$ in the same way as above. Equality in it is achieved when $a_1 = \dots = a_m < a_{m+1} = \dots = a_{2m+1}$.

M14. A convex white polyhedron has some faces painted black red so that no two black faces have a common edge. Prove that if at least one of the following conditions is met:
a) more than half of black faces;
b) the area of ​​the black faces is more than half the surface area of ​​the polyhedron, then a ball cannot be inscribed in this polyhedron.

Solution. Let’s start with task 6). Let us assume that a ball can be inscribed in a polyhedron. On each face, mark the point of contact with the ball and draw segments from it to the vertices of this face. Then the face will be divided into triangles. Each edge of the polyhedron is adjacent to two such triangles lying in adjacent faces. They are equal on three sides (since two tangents drawn from one point to the ball are equal). By condition, each edge is adjacent to at most one black face. Thus, each black triangle is adjacent to an equal white one*). Consequently, the sum of the areas of the black faces is not greater than the sum of the areas of the white faces. Problem 6) has been solved.

Problem a) is solved in exactly the same way, which was already discussed in the article “Irrepressible polyhedra” by E. M. Andreev (see «Kvant» No. 8, problem 9 at the end of the article). This article examines a number of questions that arise in connection with our task: are there polyhedra that satisfy similar conditions, what other “signs of indescribability” exist, etc. True, as one can judge from the purpose of the article, it is about whether this or that polyhedron can be inscribed in a sphere, and in our task, whether it can be described near a sphere. It turns out, however, that both theories are completely similar.

More precisely, each convex inscribed polyhedron can be assigned a corresponding description so that for each face, the planes corresponding to them will intersect at one point - the vertex of the new polyhedron, corresponding to the face of the original inscribed one. If two vertices of an inscribed polyhedron are connected by an edge, then the corresponding tangent planes will intersect along the edge of the inscribed polyhedron. Thus, we have constructed a mapping of inscribed polytopes into the set of described ones; it is easy to see that this is a mapping onto the entire set of described polyhedra and that it has an inverse mapping

*) But, of course, white can also be applied to a white triangle - the only important thing is that different black triangles correspond to different white ones.

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(we use the terminology discussed in the article by A. N. Kolmogorov “What is a function” in «Kvant» No. 1)—in other words, that for each described polyhedron one can associate an inscribed one, from which this described one is obtained in the manner indicated above. It is enough to mark the point of contact in each face—the vertex of the future inscribed polygon—and prove that all points of contact of faces that have a common vertex lie in the same plane (Fig. 6). Both of these mappings, connecting the set of convex inscribed and circumscribed polytopes, constitute an excellent example of a situation that mathematicians usually call the word “duality” (this word was mentioned in detail in the article by E.M. Andreev). Think about which described polytopes will be dual to such inscribed ones: a cube, a regular n-gonal prism, a regular n-gonal prism, a regular doctrinal prism, a dodecadron.

We highly recommend that readers return to the article “Indescribable polytopes” and, using the duality we just talked about, translate its content into “indescribable” polytopes.

M15. The square table $n \times n$ is filled with non-negative numbers so that the sum of the numbers in each row and the sum of the numbers in each column is equal to 1. Prove that from the table you can choose $n$ to put the calf of $x$ numbers, no two of which are in any column or row.

Solved and e. Let us first note that if we rearrange the rows (or columns) of our table, this will not affect either the condition of the problem or what we need to prove. Suppose that we collect the permutations of rows and columns into a regular angle of size $m \times k$ one zero (Fig. 7). Let us prove that $m + k$ is always no greater than $n$. Let’s calculate the sum of the numbers in the piece marked on figure by the letter $C$. It is clear that the sum of the numbers in the piece $B$ is equal to $m$ (m columns, in each of which the sum is 1). Similarly, the sum of the numbers in the piece $A$ is equal to $k$ (k ​​lines). Since the total sum of all numbers in the table is $n$, then the share of the piece $C$ remains $n - m - k$. This number is non-negative, since by condition in the table all numbers are non-negative. So, $n - m - k \ge 0$, that is $m + k \le n$, which is what we argued.

Now we can reformulate the problem as follows: there are numbers in the table $n \times n$, and it is known that by lower rearranging the rows and columns it is impossible to

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Divide in it an angle of zeros of size $m \times k$ such that $m + k > n$; prove that it is possible to choose $n$ non-zero numbers, no two of which appear in any column or row. (It doesn’t matter to us now that the numbers are positive and that the sums in the rows and columns are equal to 1. We have already used these conditions. In what follows, we distinguish only between “zeros” and “non-zeros.”)

If you read the article by M.I. Bashmakov “Matchings and transport networks” in No. 4 of our magazine, then you will easily be informed that our new formulation of the problem is simply a record in the language of “tables” of the matchmaking theorem. Indeed, let us write boys in the rows of the table $n \times n$, and girls in the columns (both $n$ are empty). At the intersection of the $s$th row and the $t$th column we put “non-zero” (an asterisk) if the $s$th boy is familiar with the $t$th girl, and “zero” if he is unfamiliar. Let’s get a table (as mathematicians say, a “matrix”) of dating.

Let’s take an arbitrary group of $k$ young men. Let’s put the corresponding lines first. If none of these boys is familiar with a girl, then according to the column in the first $k$ lines there are zeros. If we rearrange all the columns corresponding to all such girls last (their number is $m$ empty), then we get an angle of zeros of size $m \times k$ (Fig. 8). Each of the first $m - n$ columns has at least one “non-zero” in the first rows, since the corresponding girl has another of our chosen $k$ boys. Since $m + k \leq n$, then $m - n \geq k$, that is, there are at least $k$ such girls. We see that these problems completely coincide with the data of the “matchmaking theorem”. The choice of $n$ brides for all the young men is the choice of $n$ “non-zeros”, standing in different rows and columns.

Now you can read a simple (by induction) proof of the “matchmaking theorem”, which gives a complete solution to the problem. Of course, the properties of the problem to this theorem are not at all necessary. Inductive reasoning can be carried out immediately - by rearranging rows and columns (this is what, for example, reader D. Grigoriev from Leningrad did). In this case, a new proof of the “matchmaking theorem” is obtained*).

*) But, of course, by simply crossing out the last column and the last row in a table that satisfies the conditions of the problem, we can get a table that no longer satisfies this condition - indexing cannot be carried out “roughly” in this way.

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F18. If you look at the light through two combs with different frequencies of teeth, superimposed on each other, then the luminous areas will intersect with the dark ones. At what speed will these sections move if one of the combs is moved relative to the second at a speed of 1 cm/sec? A stationary comb has 5 teeth per 1 centimeter, while a moving comb has 6.

Solved and e. Let’s combine some comb teeth (Fig. 9). Then the next teeth will be at a distance of $1/5$ cm - $1/6$ cm = $1/30$ cm from each other. Let us now move the comb by this distance. Then the pattern of light and dark stripes will shift by $1/5$ cm—the distance between the teeth of the stationary comb. If the comb is shifted by 1 cm, then the pattern of light and dark stripes will shift by $1/5$ cm·30 = 6 cm.

This means that when the comb with a higher frequency of teeth moves at a speed of 1 cm/sec, the light and dark stripes move at a speed of $v = 6$ cm/sec in the same direction in which the moving comb moves.

The picture of alternating light and dark stripes moving at a constant speed is a very convenient model for the movement of waves of any physical nature. If you look at the moving stripes through the slit (Fig. 10), then the slit will periodically be covered with either rear or black stripes:

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we will see fluctuations. The black and yellow colors of the stripes correspond to two opposite phases of oscillations (Fig. 11). The wavelength is equal to the total width of the light and dark bands. By counting the number of black or yellow stripes passing through the slits in one second, the frequency of oscillations and their period can be determined.

This simple model allows us to understand many physical phenomena and interesting process experiments.

Using black and yellow stripes, you can simulate not only linear waves, but also any others, such as ring waves (Fig. 12). The movement of such ring waves can be shown, for example, the black and yellow Archimedean spiral (Fig. 13). When the spiral rotates through the slit, it will be visible how the waves run towards the center or away from the center, depending on the direction of rotation of the spiral.

If the stripes do not mix, we will get a picture of motionless waves. Such pictures are called rasters. Stationary raster waves should not be confused with standing waves. The standing wave pattern is an alternating black and yellow stripe that periodically changes its color to the opposite.

When several sine waves are superimposed on each other, new waves are created that are no longer sine waves. They are called group waves. Group waves differ from folded sine waves in length, propagation speed, and other parameters. The simplest group waves are beat waves. They are obtained when two waves with unequal, but close to each other, lengths are added.

In order to build a group wave model, two systems of bands need to be superimposed on each other. If these systems are the same, they can be superimposed so that the stripes of one system coincide with the stripes of the other. Such waves will be superimposed by the corresponding ones. If the systems of stripes differ slightly from each other in the distance between the stripes, then we get a picture of beats (Fig. 14). You got a similar picture in your experiments with combs.

It is not difficult to find the wavelength of the beats generated by waves with lengths $\lambda_1$ and $\lambda_2$ ($\lambda_1 > \lambda_2$). If two black bands coincide, then the system of bands with wavelength $\lambda_1$ and $(n+1)$ are close to the system of bands with wavelength $\lambda_2$. This means that the wavelength of the beats is

$$L = n\lambda_1 = (n+1)\lambda_2.$$

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From the equality $n\lambda_1 = (n+1)\lambda_2$ we find that $n = \frac{\lambda_2}{\lambda_1 - \lambda_2}$.

Because $L = \frac{\lambda_1\lambda_2}{\lambda_1 - \lambda_2}$.

The smaller the differences between the superimposed waves, the greater the length of the group waves. This is used when changing the wavelength of high frequency vibrations. To change, the length of the second wave is selected so that when the waves are added, beats with a long wavelength are obtained. Such beats are easy to see and change using an oscilloscope. Knowing the beat length and the wavelength of some vibrations, it is not difficult to determine the wavelength of the vibrations under study.

The beat waves, which are shown in figure 14, are stationary waves. If one or both wave systems are set in motion, the group waves will also mix. The speed $v$ of group waves can be found in the same way as we find the speed of movement of group waves resulting from the application of combs. It is equal

$$v = \frac{\lambda_1 v_2 - \lambda_2 v_1}{\lambda_1 - \lambda_2},$$

where $v_1$ is the speed of waves with length $\lambda_1$, and $v_2$ is the speed of waves with length $\lambda_2$.

This formula is a generalization of the formula we derived for combs.

In this way, this problem was solved by Vyacheslav Mileshin (Balti, Moldova) and Alexander Karpenko (Lisichansk, Voroshilgovgrad region). Lyutsman Petraschik (Lida, Grodno region) and Sergey Yushman (Vilany, Latvia) solved it in a similar way. Some of the readers, while solving the problem correctly, incorrectly calculated the distance between the teeth of the combs. If a comb contains $n$ teeth per 1 cm, then per 1 cm there are also $n$ spaces between teeth (not $n-1$). Of the two outer teeth on one centimeter, the first belongs to this centimeter, and the last belongs to the next one.

F19. The shape of the communicating vessels is shown in figure 15. Where will the water flow through the tube connecting the vessels if you heat the water in one of the vessels?

Solved and e. Regardless of which vessel is heated, the water warms

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Even right. Let the liquid in the right vessel heat up. Then it will expand and occupy more volume. If the vessel were cylindrical (Fig. 16), then the pressure of the liquid at the bottom would not change: the reduced density of water would be exactly compensated by the increase in the height of the liquid column. This follows from the fact that the pressure force on the bottom of a cylindrical vessel, on the one hand, is equal to the gravitational force acting on the water in the vessel, and on the other hand, is equal to the pressure of the liquid on the bottom, multiplied by the bottom area: $P = pS$. Since neither the force of gravity nor the area of ​​the bottom changes when the liquid is heated, the pressure on the bottom of the cylindrical vessel does not change either.

In a finite expanding vessel, with the same decrease in liquid density, the increase in column height is less than in a cylindrical vessel. This happens for two reasons: firstly, less liquid expands, which means there is less change in its volume; secondly, when expanding, the water must fill the volume protected at figure. Therefore, the pressure at the bottom of the vessel decreases when the liquid is heated. Since when the liquid in the tube connecting the vessel is in equilibrium, the pressure at its end in the left vessel must be equal to the pressure at its end in the right vessel, the liquid will flow from the left vessel to the right.

Considering in a similar way the heating and expansion of the liquid in the left vessel, we will find that when the left vessel is heated, the pressure on its bottom increases. This means that in this case the liquid will flow from the left vessel to the right.

Vladimir Metlitsky (Angarsk, Irkutsk region) and Alexaンドr Karpenko solved this problem correctly. 8th grade student Efim Kaplun (Vinnitsa, Ukraine) noticed that in a heated vessel, water evaporates more intensely than in one whose temperature does not change. This should affect the flow of water. However, if the vessel temperatures are low, then this effect can be neglected.

F20. A round hole is drilled parallel to the axis of the cylinder of radius $R$ at a distance $\frac{R}{2}$ from its center. The response radius is $\frac{R}{2}$. The cylinder lies on a plate, which is slowly lifted at one end (Fig. 17). Find the maximum angle of inclination of the swing arm at which the cylinder will still be in right balance. The friction coefficient of the cylinder on the plate is 0.2.

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the cylinder will lose stability at the angle of inclination of the plate

$$\alpha = \alpha_1 = \arcsin \frac{1}{6}.$$

Lyutsman Petrasik, Vladimir Metlitsky, Mikhail Kuznetsov (village of Golchevskaya, Volgoda region), Vladimir Korneev (Moscow), AlexaンドR Karpenko, Alexey Efimov (Moscow) and others correctly solved this problem.

F21. The armature axes of two identical DC motors are also connected to each other. If identical sources with e.g. are connected to the armature windings. d.s. E, then the angular speed of rotation of the anchors without load is equal to $\omega_0$; if the motors are slowed down so that they do not rotate, then current $I_0$ flows through the armature windings. One of the current sources was switched so that the torques of the armatures were opposite. What external moment must be applied to the axis of the armatures in order for them to rotate at a given angular velocity $\omega$? There are few frictional motor claims; The magnetic field of the stat is coefficientd by a permanent magnet.

Solved and e. When the armature is completely inhibited, a current $I_0 = \frac{E}{R}$ flows through its winding. This means that the armature winding resistance is $R = \frac{E}{I_0}$.

When the armature of an electric motor rotates, e.m. occurs in its winding. d.s. induction $E_n$, proportional to the angular velocity of rotation of the armature:
$$E_n = k\omega (k - \text{козффициент пропорциональнойности}).$$
Let us write the law of conservation of energy for an electric motor:
$$EIt = I^2Rt + Nt.$$
Or, reducing the engine operation time $t$, we get
$$EI = I^2R + N;$$
here $N$ is the power consumed by the electric motor from the network, $I^2R$ is the heat loss in the armature winding, $N$ is the power supplied to the load, $R$ is the resistance of the armature winding and $I$ is the current in the winding.

From this equation it follows that in the absence of load ($N = 0$) the current through the armature winding is zero. But according to Ohm’s law, the current passing through the winding must be equal to $I = \frac{E - E_n}{R}$. This means that in the absence of load e. d.s. the induction arising in the armature winding is equal to e. d.s. source: $E = E_n$.

After switching one of the sources in the armature windings, e.m. will be excited. d.s. $E'_n = k\omega = E_n \frac{\omega}{\omega_0} = E \frac{\omega}{\omega_0}$. In this case, currents will flow through the windings
$$I_1 = \frac{E - E'_n}{R} = \frac{\left(E - E \frac{\omega}{\omega_0}\right)}{E/I_0} = I_0 \left(1 - \frac{\omega}{\omega_0}\right),$$
$$I_2 = \frac{E + E'_n}{R} = I_0 \left(1 + \frac{\omega}{\omega_0}\right).$$

Moreover, both the work of the sources and the work of the external force will go towards heating the windings. Therefore $EI_1 + EI_2 + N' = I_1^2R + I_2^2R$.

Since $N' = F \cdot v = F \cdot r\omega = M\omega$ (F is the external force applied to the anchor, M is its moment relative to the axis and v = $\omega r$ is the angular velocity of points on the surface of the anchor axis), then, replacing $N'$, $I_1$, $I_2$ and R with their expressions in terms of M, $\omega$, $\omega_0$ and E, we obtain
$$M\omega + EI_0 \left[ \left(1 + \frac{\omega}{\omega_0}\right) + \left(1 - \frac{\omega}{\omega_0}\right) \right] = I_0^2 \frac{E}{I_0} \left[ \left(1 + \frac{\omega}{\omega_0}\right)^2 + \left(1 - \frac{\omega}{\omega_0}\right)^2 \right],$$

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from where $M\omega + 2EI_0 =$
$= EI_0 \left[ 2 + 2 \left( \frac{\omega}{\omega_0} \right)^2 \right]$
and $M = \frac{2EI_0\omega}{\omega_0^2}$.

F22. Two blacksmiths form a piece of iron. First, they place it on an anvil and hammer it one at a time, then hang it from the ceiling and hit it from different sides at the same time. The force of each blacksmith’s blow is the same in both cases. In what case does a piece of iron heat up more in one blow?

Solution. In the second. It is clear that both when a blacksmith strikes a piece of iron lying on an anvil, and when a blacksmith strikes a hanging piece of iron, the same work is performed: the role of the second blacksmith, striking a piece of iron, in the first case is taken on by the anvil. This work goes to heating a piece of iron and, in addition, to performing mechanical work during the expansion of this iron. When iron lies on an anvil, its center of gravity rises as it expands—work is done against gravity. But in the case when the piece is suspended, its center of gravity drops as it expands. This means that in the second case, when not only the work done by the blacksmiths is converted into heat, but also part of the potential energy of a piece of iron (the work done by iron during expansion is negative), it will heat up in one blow more than in the first.

F23. On a horizontal plane there are two identical thin-walled pipes of mass $m$ each. Their axes are parallel, and their radii are equal to $R$. First, one of the pipes will pile up, and the second will roll without slipping towards the first until it collides. The translational speed of the pipe is $v$. How do the speeds of translational motion and the angular speeds of rotation of the pipes depend on time (write graphs)?

The friction coefficient of sliding pipes on a horizontal surface is $k$, the friction between pipes during a collision is negligible, the impact is absolutely elastic.

Solution. Since the friction between the pipes is negligible, we can assume that when the pipes hit, the rotation of one of them is not transmitted to the second. Therefore, when considering the collision of pipes, we can not take into account the rotation of the first (initial moving) pipe.

Let us write down the laws of conservation of energy and momentum for the alignment of pipes:

$$mv_0 = mv_1 + mv_2, \frac{mv_0^2}{2} = \frac{mv_1^2}{2} + \frac{mv_2^2}{2}$$

($v_1$ and $v_2$ are the translational velocities of the corresponding first and second pipes after the collision).

Solving these equations together, we find that $v_1 = 0$ and $v_2 = v_0$, that is, when the pipes hit, they exchange translational velocities - just like when two identical balls hit.

Let us now consider what will happen to the first, initially moving pipe after the impact. In the coordinate system associated with a pipe rolling without slipping along a plane with a speed of $v_0$, the speed of a point touching the plane is equal to - $v_0$. This means that such a pipe rotates around its axis so that the linear rotation speed

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points on its surface is equal in magnitude to the speed of translational movement of the pipe axis. Therefore, after the collision, the first pipe rotates around its axis with an angular velocity $\omega = \frac{v_0}{R}$.

The friction force $F_{\text{тр}} = kmg$ (Fig. 20) acting on this pipe replaces its rotation and at the same time imparts to it an acceleration $a = \frac{F_{\text{тр}}}{m} = kg$ in the direction of the initial movement of the pipe. By the moment $t$ this pipe will have a translational speed $u'_1 = at = kgt$ and will rotate around the axis with an angular velocity $\omega_1 = \frac{v_0 - kgt}{R}$. The speed of translational movement of the pipe increases, and the speed of rotation of the pipe decreases in proportion to time. By the moment $t_0$, when the speed of translational movement of the pipe axis becomes equal to the linear speed of rotation of the pipe around the axis, the slipping of the pipe relative to the plane will stop, and after that the speed of rotation of the pipe $\omega'_1$, nor the speed of translational movement of the pipe axis $u'_1$ will no longer change. From condition $kgt_0 = v_0 - kgt_0$ we find that $t_0 = \frac{v_0}{2kg}$. At this moment

$$u_1 = \frac{v_0}{2} \text{ и } \omega_1 = \frac{v_0}{2}.$$

Considering the movement of the second pipe in a similar way, we find that the friction force acting on it decreases in its translational motion, $u_2 = v_0 - kgt$, and the angular velocity of rotation $\omega_2 = \frac{kgt}{R}$ increases.

By moment $= \frac{v_0}{2kg}$, the slipping of the pipe relative to the plane will stop. At this moment, the pipe will have a translational speed $u_2 = \frac{v_0}{2}$ and an angular rotation speed around the axis $\omega_2 = \frac{v_0}{2R}$ that will not change in the future.

Graphs of the dependence of the speeds of translational movement of pipes and their angular speeds of rotation on time are shown in figure 21.

The correct solution to this problem was provided by Mikhail Kuznetsov and Sergey Yushmanov.

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QUANTUM LABORATORY

G. I. Kosourov

ORANGE SKY

In this article we present several experiments on color vision. All of them are associated with various color illusions caused by unusual operating conditions of the eye or its color fatigue.

The mechanism of color perception by the eye is very complex and has not yet been fully studied. It is known that the retina of the eye contains two kinds of light-sensitive cells, which are called rods and cones. The rods contain phototympeschi floral pigment - visual purple, or rhodopisin. When exposed to light, rhodopisin supplies and acts on nerve fibers that transmit stimulation to the brain. In bright light it ensures that the sticks become completely blind. Walking in the dark

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the reverse process is the restoration of the visual cortex. Rod, twilight vision has very high sensitivity, but it is achromatic. Rods do not sense color. At sufficiently high brightness, the cone mechanism of vision comes into force. Cone vision is color sensitive. There are many convincing experiments leading to the conclusion that cones contain three types of photochemical sensitive pigments with maximum sensitivity in the red, green and blue regions of the spectrum. Varying degrees of expression of all three pigments give our consciousness the sensation of color and allow us to see the world in all its diversity of colors, shades, halftones and transitions.

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The three-color nature of color vision is the basis for color reproduction in film, television, color photography, and printing. Quantitative color measurement methods are also based on the three-color principle.

However, color errors can be caused not only by paints, but also, for example, by excessive lighting. Draw the black and white disks shown in Figures 1 (a, 6, c, d) using ink on drawing paper. It is enough for the disks to have a diameter of 8-12 cm. Cut the disks and rotate them not very quickly, about 1-3 revolutions per second. To do this, you can put them on the winder axis of a film projector, player, tape recorder, or simply make tops out of disks. Instead of black arcs, you will see colored circles. The color depends on the speed of rotation, illumination and the nature of figure. For example, for a disk at figure 1,a, the arcs following the black sector (in the direction of rotation) appear red in low light, and yellow in bright light. At certain rotation speeds and bright lighting, black sectors appear blue. There is no complete explanation for this phenomenon yet.

Paints differ not only in color, but also in saturation. If you gradually add white to red paint, the color will become more and more pink. In paintings, and even more so in printed copies from paintings, it is difficult to obtain rich tones and a greater gradation of brightness. The ratio of the rage of the whitest paint to the rage of the blackest is barely one hundred, but in nature we are dealing with a difference in rage of many thousands of times. Therefore, copies from paintings seem too pale or, conversely, too dark. This violates one of the elements of illumination reproduction - the impact respectrum. The image of a landscape or genre scene appears flat. This means a wider range of transmitted rages when projected onto the screen. This is why photographs on color reversible film—slides—are so expressive and three-dimensional.

The perception of a painting can be significantly improved by lighting it in a special way. Take a color inclusion from Ogonyok, shape it and print it using contact method from film to film of contrasting black and white ranges. Insert the slides into a projector with a large luminous flux (a “Light” type projector is suitable) and project the slides onto the original so that the contours of the projection and the original exactly match. The result will not make you regret the work expended. The painting will become animate, acquire volume and special expressiveness. Turn off the projector, and you will see how dull and inexpressive the picture is under uniform lighting.

The following experiments relate to the so-called sequential color image. This phenomenon is due to the fact that complete restoration of color-sensitive pigmenta is a relatively slow process. If you look at a single-color figure for a long time, and then turn your gaze to white paper, a wall or a ceiling, then the white color will be perceived as if it lacks color, which tires the eye. On a white surface the same figure will be visible, but rendered in a color that is called complementary. Cut out red, orange, yellow, green, blue and purple squares measuring 2x2 cm from colored paper. Place one of the colored squares in front of you on a sheet of white paper and look at it without straining your eyes for about thirty seconds. You need to look, as they say, at one point, so that the image of the square does not move across the retina. Turn your gaze to the white field, and after a second or two you see on the paper a clear image of a square in a complementary color. So are you

47

Fig. 2. The complementary colors of squares in one row are green, blue is orange, and yellow is violet (Fig. 2). Each pair of complementary colors in the mixture should produce a white or gray achromatic color.

To mix additional colors, place two additional squares next to each other and place a glass plate between them (Fig. 3). Position your eye so that one square is visible through the glass, and the other is reflected in the glass. By changing the inclination of the glass and thereby the correspondence between the light fluxes coming from the squares, it is possible to ensure that the coloring in the place where the image overlaps almost disappears. To completely achromatize an image, you need precise selection of colors. Most often the result is a brown color. But if you take two clearly non-complementary colors, for example green and yellow or red and violet, any mixture of them will produce a bright color. Even more vibrant color images arise if the squares are placed not on white paper, but on an additional background.

After you have made observations on simple figures, try making an applique from colored paper similar to the one shown at figure in front of the statue, that is, replacing the colors in it with additional ones.

Fig. 3.

body By looking at the pictures in the same way as you looked at the squares, you will very soon learn to see landscapes in natural colors on a white background. For success, only bright enough lighting is needed.

The most striking and most incomprehensible color iliosis is provided by our latest experience. We have already said that the basis for color reproduction is a three-color system. If you repeatedly photograph the same scene through three light filters: red, green and blue - and, having printed positive copies, project them with three projectors onto the same place on the screen through the same light filters, then a color image with correct color reproduction will appear on the screen. In this case, the light filters must be selected so that in total they give a white color. You can limit yourself to two photographs through two additional light filters, for example, red and green. This also produces good color rendition, although not as perfect as in the three-color method. But it turns out that during projection you can limit yourself to just one color filter.

Shoot the same color scene, without moving the camera, twice on panchromatic film through red and green filters.

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Fig. 4.

There is no need for precise selection of filters. The filters from the school kit are quite suitable. Contact print positive copies onto film. Insert the resulting slides into two projectors and project the slides onto the screen so that the depicted outline exactly matches. In front of the projector lens in which the slide is inserted, remove the red filter, place a red filter, and leave the second image black and white (Fig. 4). You will see a color image in all its richness, despite the fact that you are only designing red and black and white images with different distributions of light and shadow. Any objective study of light reflected from different places on the screen will only give red color of varying degrees of purity or saturation. The feeling of colors in this case is purely subjective. It is better to turn on projectors through authorization formats (Fig. 4) so ​​that you can independently adjust the illumination given to the screen by each of them. As a rule, the most correct color reproduction occurs when the screen illumination is low.

Thus, simple and very full of mystery and mystery.

Ice slipperiness

One English physicist admitted in print that we are so accustomed to the slipperiness of ice that if he had been asked casually - before he himself thought about this question - why ice is slippery, he would have answered: “because it is slippery,” and only after thinking about it would have come to the correct explanation.

In places where ice or snow is pressed, they float under the influence of pressure caused by the weight of the body - and between the foot and the soil a layer of liquid lubricant is obtained, which significantly reduces the friction force and makes the snow and ice slippery, just like a slippery panel is covered with a layer of liquid mud. When the Hora leaves its place, this lubricant freezes again, forming an icy trail on the snow.

That this is not a matter of degree of smoothness can be seen from the fact that the roughest and most lumpy ice is almost more slippery than mirror-smooth ice, because the latter has a larger area of ​​contact with the foot, and therefore the pressure of the foot on the ice is less.

In general, the smaller the surface where the pressure of the body on the ice is concentrated and the lower the temperature of the ice, the more slippery it should be. Therefore, you can skate further down the mountain than without skates, and skating is easier if the temperature is close to 0°. When the air temperature is above 0°, the ice becomes loose, at least in the upper layers, and skates slip into the ice over time. In severe frosts, the skates seem to stick to the ice: very little liquid lubricant is formed, and it freezes before the hora has time to move from its place.

Prof. B. Weinberg.
“Snow, ice, hail, ice and glaciers”, 1919

49

How many
polyosov
at the magnet

N. A. Mints

Of course, everyone knows that a magnet has two poles - north and south. It is quite easy to verify this. If you take a metal iron needle or ruler and stroke it with a permanent magnet all the time in the same direction, the needle will become magnetized. Using a compass, it is easy to find out which end of the spoke has become the north pole and which has become the south pole. If the spoke is broken, then each of its pieces will be a separate magnet, and each will, of course, have two poles - north and south. No matter how many pieces we break a knitting needle, each piece will always have two poles - north and south.

An experiment with magnetizing an iron strip was carried out by Igor Gofman, an 8th grade student from the city of Artemovsk, Donetsk region, but he ended up with a magnet with amazing properties. Here is what he writes: “I used a steel magnet ruler as a strip magnet, and metal filings to determine the lines of force. The result is a picture in which 5 poles are visible (Fig. 1). Driving the compass along the ruler, I noticed that the arrow turns first one end, then the other, which proves the presence of opposite poles.”

What’s the matter? We have just said with confidence that a magnet always has two poles, but the magnet made by Igor has several. The picture of magnetic field lines looks like there are several

50

It seems that it has three poles - two northern ones at the edges and a southern one in the middle.

Igor does not write how he managed to obtain a magnet with several poles. It can be done like this. Take a knitting needle and stroke it simultaneously with the same poles of two magnets from the middle to the right and left. You will get a magnet with three poles. A large number of stripes can be obtained by magnetizing only the ends of the spokes, or using coils. It is necessary to turn on several coils so that the magnetic fields created by neighboring coils are opposite. If you now place a rod made of high quality magnetic steel inside the coils and then turn off the current, the rod will have several stripes. In this case, the coils must be long enough, since short magnets are very easily demagnetized. This is connected with this. Any magnet can be demagnetized by placing it in an external magnetic field directed against the field inside the magnet itself. But the magnetic fields created by the magnet itself inside and outside are directed in the opposite direction (Fig. 3). The external field created by the magnet itself demagnetizes it. The shorter the magnet, the stronger the external demagnetizing

Fig. 3.

Fig. 4.

the field it creates at the same magnetization of the sample. Therefore, long magnets are preserved better than short ones, and very short magnets cannot be obtained.

The ability to retain magnetization depends not only on the distances between the poles, but also on the properties of the materials from which the magnet is made. For example, a tape on which music is recorded consists of a large number of short transverse magnets. These magnets differ from each other in the magnitude of the field they create. You can estimate the number of such magnets per 1 mm of tape. Let’s assume that the sound frequency is 1000 Hz and the tape moves at a speed of 10 cm/sec. Then the number of magnets in one centimeter will be $\frac{1000}{10} = 100$. That is, 10 magnets in every millimeter!

figure 4 shows the alternation of regions with different magnetic orientations in the crystal. Light stripes - the northern stripes are directed towards the viewer, dark stripes - vice versa. Simok was obtained under special conditions under a microscope. Magnified 500 times.

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ALL-UNION CORRESPONDENCE OLYMPINAD

Olympiad conditions

We publish the texts of the problems of the All-Union Correspondence Olympiad in mathematics, physics and chemistry for students of schools and other secondary educational institutions. This Olympid, which has already become traditional, is an integral part of the All-Union Physics, Mathematics and Chemistry Olympids.

Those who successfully cope with the tasks of the correspondence Olympiad will be invited to the regional (territorial, resource) Olympiads on an equal basis with the winner of the regional Olympiads. Invitations will be sent to public education organizations at home addresses. The best of the winners of regional and resource competitions will take part in the final type of the Weight-Union Olympiad, and will also be invited to the summer school of the Siberian branch of the USSR Academy of Sciences.

Before each problem, the classes for whose students it is offered are indicated in brackets, but you can also try your hand at problems for older classes. To become an Olympiad winner, you don’t have to solve many problems. Sometimes a witty solution to one is enough. But solving a problem doesn’t just mean giving an answer; it is necessary to justify it, to give precise evidence. The assignment does not require any knowledge beyond the high school curriculum. Solutions to the problem for each subject must be sent in a separate standard format envelope. On the envelope indicate the name of the item. The letter must include a double stamped envelope with a return home address and a stamp (in this envelope you will be sent an extract about the results of the work). Solutions must be written very legibly. The conditions of the tasks are not rewritten, but their numbers are indicated.

At the beginning of the letter, do not forget to indicate (legibly):

1. Your last name, first name and patronymic.
2. Postal home address (detailed).
3. Class, school (school, technical, etc.), address.
4. Last name, first name and patronymic of your teacher in this subject.

Aftereffect in Siberia, Kazakhstan, the Far East and Central Asia: Novosibirsk, 90, Olympida;

• living in the Karal and Komi Autonomous Soviet Socialist Republics, in the Arkhangelsk, Vologda, Kaliningrad, Kirov, Leningrad, Murmansk, Novgorod and Pskov regions of the RSFSR: Leningrad, V-164, Leningrad State University, Olympida;
• residents of the rest of the territory of the European part of the RSFSR, including the Urals: Moscow, V-234, Moscow State University, Olympida;
• residents of the Estonian SSR: Tartu State University, Olympida;
• those living in other resource areas - to the university of the capital of their resource area, marked “Olympida”.
  The Central Organizing Committee appeals to all parents, teachers of mathematics, physics, chemistry and school directors with a request to familiarize themselves with the conditions of the tasks of high school students.

We wish all Olympiad participants the greatest success.

Central organizing committee of the Weight-Union Olympiad for schoolchildren

52

Mathematics

1. (7-10) 19 teams participate in the football tournament. Prove that at each moment of the tournament a team will find an even number of matches (perhaps none).

2. (7-10) An arbitrary circle of radius $R > 1$ is drawn on the checkered paper from the side of the cell equal to one. Prove that there are at least $l(R-1)^2$ mesh nodes inside this circle.

3. (7-10) In a 5 by 5 square, 16 cells are shaded. Prove that it is possible to design a 2 by 2 square in which at least three cells are shaded.

4. (7-8) Prove that the number 1970 cannot be represented as the difference of the squares of two integers.

5. (7-8) Is it possible to make a closed chain from 99 identical square tiles (so that each two adjacent squares have a common side and do not overlap each other)?

6. (7-10) In the equilateral triangle $ABC$ with base $AC$ there is a bicsetris $CD$ (D is a point on the side $AB$) and a line $DE$, perpendicular $CD$ (E is a point on the line $AC$). Prove that $CE=2AD$.

7. (9-10) Prove that of at least two numbers $\sqrt{m}; \sqrt{n}$ does not exceed $\sqrt{3}$; $m$ and $n$ are any natural numbers greater than 1.

8. (8-10) An interior point of a quadrilateral $ABCD$ is called nice if the sum of the areas of triangles $APB$ and $CPD$ is equal to the sum of the areas of triangles $BPC$ and $DPA$. Which quadrilaterals have three nice points that do not lie on the same line?

9. (8-10) Solve for $x$ the equation

$$\frac{1}{x} = \frac{1}{a-x} + \frac{1}{b-x} + \frac{1}{a+b-x}.$$

10. (7-10) Is it possible to represent a single one as a sum

$$\frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3} + \dots + \frac{1}{n_{1971}}$$

where $n_1, n_2, n_3, \dots, n_{1971}$ are different odd natural numbers?

Physics

1. (8) The heavy rod is aligned in the middle at an angle $90^\circ$ and hung freely at one of the ends. What angle does the attached side make with the vertical?

2. (8) When a neutron flow passed through a 1 mm thick kadmium plate, the number of particles decreased by 15%, but their speed did not change. What fraction of the neutron flux will pass through an 8 mm thick plate of the same material?

3. (8-9) A mathematical mat of length $l$ and mass $m$ is swung as follows: each time the mat passes the equilibrium position, a force $F$, directed parallel to the velocity, acts on it for a short period of time. After how many oscillations will the matnik turn off at $90^\circ$?

4. (8-10) A bullet of mass $m$, flying with an initial speed $v$, pierces one suspended weight of mass $m$ and sticks in the second similar one. Neglecting the time of interaction of the bullet with the weight, find the amount of heat released in the first weight if the amount of heat released in the second is equal to $Q$.

5. (8-10) A single thread hangs freely so that both ends are at the same height. Thread tension at the bottom point $T$. Thread weight $m$. Find the tension at close suspension points.

6. (9-10) $P-V$ diagram of the process over an ideal gas is represented by a parallel gram (Fig. 1). Draw a $T-V$ diagram of this process.

7. (10) Find the charge on the grounded plate of the capacitor (Fig. 2).

8. (10) Trem identical insulated capacitor extropyto $C$ each was given the charges $q_1, q_2$ and $q_3$ corresponding. Then the capacitors were connected as shown in figure 3. What charges remained on the capacitors?

9. (10) The circuit shown in figure 4 is connected to an alternating voltage source with an amplitude of 220 V and a period of 1/50 sec. Find the maximum value of voltage across resistance $R$ and dopudenta, during which the current in the circuit is non-zero. Draw a graph of the voltage across resistance $R$ versus time. Consider that the dependence of the current flowing through the didot on the voltage applied to it (shown as a dotted line on the graph) with sufficient accuracy appears to be broken $ABC$. The internal resistance of the battery can be overcome.

10. (9-10) In an optical communication system, the transmitting laser beam has the form of a cone with an apex angle of $\alpha=10^{-4}$ rad (the angle of beam divergence). In the receiving device, light energy is focused using lenses with a diameter of $D=1$ m onto a photocell. It turned out that when the distance $L$ between the transmitter and receiver changed from 5 to 10 km, the signal at the focusent output decreased by 2 times due to light absorption in the atmosphere. How many times does the signal change when the distance $L$ changes from 10 to 20 km?

Chemistry

1. (8) If we assume that the Tunguska meteorite, which exploded in 1908, was made of iron and stone and its fragments fell into the ground, then can you say what

chemical production processes with these fragments during their stay in the ground?
(The composition of the meteorites includes: iron, magnesium, nickel, silicon; the soils in the area of ​​the explosion are acidic, the depth of the fragments is no more than 0.3 m.)
2. (8) There are three alloys: prina, brass and bronze. What reactions must be carried out to determine each of the alloys?
3. (8) When 12.8 g of a white crystalline substance was heated, it completely decomposed into nitrogen and water. At 0° C, the volume of nitrogen released was 4.48 liters. Determine what substance was heated.
4. (8) When ferrous mineral waters stand, sediment is released from them. How to avoid this phenomenon? Explain the processes that occur during this process.
5. (9) When a few drops of hydrochloric acid solution are added to a saturated solution of silver chloride, the solution becomes cloudy, but if a significant amount of HCl solution is added, the solution becomes clear again. Explain these phenomena.
6. (9) 31.6 g of potassium permanganate was treated with concentrated hydrochloric acid. The gaseous product of the reaction is completely consumed by interaction with 625 ml of a solution of sulfurous acid with a concentration of 0.8 M. Determine the concentration of hydrogen in the solution obtained at

Fig. 1.
Fig. 2.
Fig. 3.
Fig. 4.

54

DEVELOP
SPATIAL
IMAGINATION

N. G. Fedin

Exams often offer stereometric problems, for which the most important thing is to imagine how the figures are located in space. Foreign messages suggest ideas for solving planimetric problems. In this article we will talk about some questions and tasks that should, to some extent, interest high school students and help them expand the horizons of their spatial imagination.

Crossbreeding primo

1. Let’s draw two crossing primaries $a$ and $b$. On figure 1 their image is given.
   Are you surprised that the primes are shown parallel? In vain. This is a correct drawing, although not visual. It is known that crossing primes $a$ and $b$ can be placed in parallel planes $a$ and $b$ so that the prime $a$ lies in the plane $\alpha$, and the straight line $b$ lies in $\beta$, and then project them onto the plane $\gamma$, intersecting the planes $\alpha$ and $\beta$ (Fig. 2). In this plane we will obtain images $a_1$ and $b_1$ of straight lines $a$ and $b$.

2. Let the given crossing primes $a$ and $b$ and points $A_1, A_2$ on the line $a$ and $B_1, B_2$ on the line $b$ (Fig. 3). Let’s carry out receptions $A_1B_1$ and $A_2B_2$.

Fig. 1
Fig. 2

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Can these lines intersect?

Of course, the drawing can derail us. But if we assume that the lines $A_1B_1$ and $A_2B_2$ intersect, then it turns out that the four points $A_1$, $A_2$, $B_1$, $B_2$ lie in the same plane, and therefore the original lines will also lie in the same plane, which cannot be the case. We come to the conclusion that the lines $A_1B_1$ and $A_2B_2$ cannot intersect.

3. Does a figure consisting of two intersecting lines have a center of symmetry?

It may seem that the center of symmetry of two crossing lines exists and coincides with the midpoint $M$ of their common perpendicular (Fig. 4). However, this is not true. In fact, if the center of the symmetry of these two were straight, then the straight line $a$ with the central symmetry would turn into a parallel straight line $b$, but $b$ is not parallel to $a$. Consequently, intersecting lines cannot have a symmetrine center.

4. Do two intersecting straight lines have an axis of symmetry?

This question is much more difficult than the question about the center of symmetry of intersecting lines. It turns out that two crossing lines have three axes of symmetry: $l_1$, $l_2$ and $l_3$. However, one of them is a common perpendicular to the original lines $a$ and $b$. The other two are mutually perpendicular, intersect in the middle $M$ of the segment $KP$ and are bisectors of adjacent angles formed by straight lines $a_1$ and $b_1$, corresponding to parallels $a$ and $b$ (Fig. 4). Try to prove this fact yourself.

Solving planetary problems using a stereometric method

Obviously planimetric problems are solved on a plane, and stereometric problems are associated with the image of a spatial figure on a flat drawing. Such direct solutions to problems or proofs of the theorem are not surprising; they are generally acceptable. But solving planimetric problems in a stereometric, spatial, three-dimensional way is already unusual and interesting. Of course, not all problems can be solved by this method, but only problems of a certain type.

5. There are three rays on the plane, emanating from one point (Fig. 5, $a$) and dividing the plane into three angles, each of which is less than $180^\circ$. In each of the three angles, one point is given: $A_1$, $B_1$ and $C_1$ - so that these points do not lie on the same straight line. Required to construct a triangle

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Fig. 5, a.

ABC, the vertices of which lie on the given rays, and the sides pass through the given points.

It is planned to solve this problem in a difficult way, but stereometrically it can be solved quite easily. Let us imagine (Fig. 5, b) that the rays form in space the edges of a trihedral angle with the vertex at point O. Let us imagine the points A₁, B₁, C₁ located in the faces of the trihedral angle. These points define a certain plane. Let us denote this plane by β; it will intersect the faces of the trihedral angle along the required sides BC, CA and AB. It remains to construct the line of intersection of the planes of the faces of the trihedral angle with the plane β.

To do this, we project these points onto the xOy plane parallel to Oz. Points A₁, B₁, C₁ will move to points A₀, B₀, C₁ respectively (point C₁ will remain in place).

Thus, we have reduced the problem to a projection one, which in school is often called the problem of cutting a polyhedral angle with a plane. We find the point x₀ of intersection of the line A₁B₁ with the plane xOy, this will be the point of intersection of the lines A₁B₁ and A₀B₀. Points x₀ and C₁ lie in both the β plane and the xOy plane, therefore, the cutting plane β intersects the xOy plane along the straight line x₀C₁.

Then we find the intersection points

Fig. 5, b.

straight line x₀C₁ with rays (edges) Ox, Oy, we get points A and B. Wires BA₁ and AB₁, we get the third point at the intersection with the axis (ray Oz) - vertex C of the desired triangle ABC. The spatial representation helped us find an idea for solving the planimetric problem.

6. Two triangles ABC and A₁B₁C₁ are given on the plane; also, lines AA₁, BB₁ and CC₁ intersect at one point S, and lines AB and A₁B₁, BC and B₁C₁, AC and A₁C₁ intersect at points x, y and z. Prove that points x, y and z lie on the same line l.

This problem is known in geometry under the name Desargues’ theorem*).

Here is the idea of ​​its proof. Let’s imagine that point S is the vertex of a paramid with base ABC and cutting plane A₁B₁C₁ (Fig. 6). Then the lines AB, BC and AC lie in the plane ABC, and the lines A₁B₁, B₁C₁ and A₁C₁ lie in the plane A₁B₁C₁. Therefore, points x, y, z lie both in the plane of triangle ABC and in the plane of triangle A₁B₁C₁, that is, they are located on the line of intersection of these planes, on straight line l.

Conduct a rigorous proof of Desargues’ theorem on your own.

*) Gerard Desargues - French mathematician and engineer (1593-1662).