Kvant 1971, No. 6 · Pages

Popular science physics and mathematics
quantum
6
1971
magazine
Academy
Sciences of the USSR
and the Academy of Pedagogical
Sciences of the USSR

To ensure in the new five-year period:
further development of problems of theoretical and applied mathematics and cybernetics…
development of research in nuclear physics, solid state and semiconductor physics, quantum electronics, plasma physics, low temperature physics…

Directive XXIV Congress of the CPSU on the five-year plan for the development of the national economy of the USSR for 1971-1975

In the room:
On the best approximations 1 D. V. Fuks,
M. V. Fuks

Semiconductor diodes and triodes 8 M. A. Fedorov

Crystal growth 16 R. Fullman

Snow drifts 23 L. G. Aslamazov

Math circle
Solutions to the problems of the introductory test in ZMSH 25 A. L. Toom

Laboratory “Quenta”
Strange pendulum 30 N. A. Mints

Problem “Quenta”
Problems 33

Problem solutions
M34—M38, F52—F58 35 A. K. Tolpew,
A. G. Kushinirenko,
N.V. Vasiliev,
L. M. Lopovok,
I. Sh. Slobodetsky

Applicant’s workshop
Logarithmic equations 46 V. V. Goldber

Geometry problems 52 Yu. V. Sidorov

Written exam in mathematics
at the Faculty of Chemistry of Moscow State University
in 1970 56 N. N. Kolesnikov,
V. M. Tikhomirov

Written exam in physics at MIPT
in 1970 60 V. E. Belonuchkin,
S. M. Kozel

Answers, directions, solutions 63

Collector’s Corner
Stamps associated with Galileo
3rd cover page

Escott’s game
4th cover page

relies on ideas that are interesting in themselves.

We will prove this theorem using so-called lattices.

§ 2. Lattices

Let us assume that two intersecting lines are given on the plane: $l_0$ and $m_0$ and, in addition, two scale segments are fixed: $a$ and $b$. On both sides of line $l_0$ we draw parallel lines $l_{\pm 1}, l_{\pm 2}$… at a distance $a, 2a$, … from it.

Similarly, on both sides of $m_0$ at a distance $b, 2b$, … draw lines $m_{\pm 1}, m_{\pm 2}$, … Mark all points of intersection of lines $l_i$ with lines $m_j$. The set of these points is called a lattice (Fig. 2).

We emphasize that we call a lattice the set of points of intersection of lines, without referring to the most straight lines. This is important because the same lattice can be obtained from completely different families of lines. Thus, the lattice shown in figure 3 is obtained both by the intersection of blue lines and by the intersection of red lines.

There is no need to look for examples of daleko grilles. The intersection points of the lines in the “checkered” terraces make up a lattice. And also, the points in the terrades in which the first classicists write do not form a grid - the distances between adjacent horizontal lines are not the same (Fig. 4).

Let us fix a family of straight lines $l_i, m_j$ on the plane and denote the lattice constructed with their help by $R$.

Theorem 2 (parallelogram property). If ABCD is a parallelogram and points A, B, C belong to the lattice, then each of them lies on one of the lines $l_i$ and on one of the lines $m_j$. Both statements can be proven in exactly the same way, and we will limit ourselves to proving the first.

Since the points A, B, C belong to the lattice, then each of them lies on one of the lines $l_i$, say, A lies on $l_i$, B on $l_j$, C on $l_k$ (see Fig. 5). From point A we drop the perpendicular AE to the line $l_j$, from point D - the perpendicular DF ​​to the line $l_k$. Since BA || CD, $l_j || l_k$ and AE || DF (as perpendiculars to parallel lines), then triangles BAE and CDF are similar. Since, in addition, BA=CD, then these triangles are congruent. Therefore AE=DF. The line of the segment AE is the distance between

$$2 \times 2 = 4$$

lattice

not a grille

Fig. 4.

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straight lines $l_i$ and $l_j$, that is, $AE=a|i-j|$. Consequently, a line drawn through the point $D$ parallel to lines $l$ is located from $l_k$ at a distance $DF=AE=a|i-j|$, a multiple of $a$, and this line is one of the lines $l$ (more precisely, line $l_{k+i-j}$).

The theorem is proven.

We will not need other properties of lattices. We will present some of them without proof - for your own reflection.

1. If the vertices are parallel to the lines $l$, is located from $l_k$ at a distance $DF=AE=a|i-j|$, a multiple of $a$, and this line is one of the lines $l$ (more precisely, line $l_{k+i-j}$).

For example, the main ones are all parallel lines into which the plane is divided by straight lines $l_i$, $m_j$.

2. In order for a parallel, the vertices of which are parallel to the lattice, to be the main one, it is necessary and sufficient that within this parallelum and on its edges there is not a single lattice point, except for the vertices of the parallelum.

3. Every parallel on the plane serves as the basis for one and only one lattice.

§ 3. Relationship between lattices and rational approximations

Let us now turn to our original goal: finding approximations for the irrational number $\alpha$. First, we associate some special lattice with the number $\alpha$.

Let us introduce a rectangular coordinate system on the plane. As straight lines $m_0, m_{\pm 1}, m_{\pm 2}, \dots$ we take straight lines parallel to the axis $Ox$ and passing through points $(0,0), (0,\pm 1), (0,\pm 2), \dots$. (equation of straight line $m_q: y=q$). For the line $l_0$ we take the line passing through the points $(0,0)$ and $(\alpha, 1)$; its equation is $x=\alpha y$. Finally, as lines $l_{\pm 1}, l_{\pm 2}, \dots$ we take lines parallel to line $l_0$ and passing through points $(\pm 1,0), (\pm 2,0), \dots$. (line equation $l_p: x=\alpha y-p$). This lattice is shown on figure 6. Let us attach the designation “$R$” to this lattice.

It turns out that the question of how good an approximation for the number $\alpha$ is $\frac{p}{q}$ can be reduced to the question of where on the plane the point of intersection of the lines $l_p$ and $m_q$ is located. Indeed, let us find the coordination

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dinata of this point. This can be done both by elementary geometric means and by jointly solving the equations of these lines:

$$\begin{cases} x = \alpha y - p \\ y = q \end{cases}$$

(lₚ), (mₙ).

We find that the coordinates of the intersection point are: αq—p, q. We see that the ordinata of this point is equal to the value of the fraction $\frac{p}{q}$. Its abscissa is equal to αq—p = q($\alpha - \frac{p}{q}$). Thus, the abscissa model coincides with the quality factor of the approximation, and it is positive if α > $\frac{p}{q}$ (or $\frac{p}{q}$ is an approximation “with a disadvantage”), and negative if α < $\frac{p}{q}$ (or $\frac{p}{q}$ is an approximation “with excess”). Note that since α is irrational, we have α ≠ $\frac{p}{q}$, and, therefore, not a single lattice point, except the point (0,0), falls on the basis of Oy.

So, to which point of our lattice, lying above the Ox axis (we limit ourselves to the case q > 0), does a certain rational approximation of the number α correspond; in this case, considering this point from the Ox axis is a significant approximation; considering it from the Oy axis is the quality factor of the approximation; if a point lies to the right (left) of the Oy axis, then its corresponding approximation is an approximation with a deficiency (with an excess).

§ 4. Proof of the main theorem

We will prove a stronger proposition than the main theorem.

Theorem 3. For any natural numbers N, k there are at least k distinct fractions $\frac{p}{q}$ maxims that

a) $q \leqslant Nk$,

b) $q \left| \alpha - \frac{p}{q} \right| < \frac{1}{N}$.

(The fact that Theorem 1 follows from Theorem 3 is obvious: k can be arbitrarily large, so there is an infinite number of approximations with a quality coefficient less than 1/N).

Proof. Let’s see what our statement means for the lattice R constructed in § 3. As we have seen, the points of this lattice correspond to the approximation, and the denominator is equal to the ordinates, and the quality coefficients are the abscissa model of these points. So, we need to prove that there are k points of the lattice R contained within the direct line P given by the inequalities: $-\frac{1}{N} < x < \frac{1}{N}, 0 < y < kN + 1$ (see Fig. 7).

To prove this, consider the wider straight line $\tilde{P}$: $-\frac{1}{2} < x < \frac{1}{2}, 0 < y < kN + 1$ (Fig. 8). Inside this straight line

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lie $kN$ points of the lattice $R$. Indeed, on each of the lines $y=1, y=2, \dots, y=kN$ the lattice points lie with intervals equal to 1; therefore, exactly one lattice point lies on the segment of each of these straight lines enclosed in the rectangle $\tilde{P}$ (the lattice points do not fall on the vertical sides of the rectangle $\tilde{P}$, since their abscissas are irrational). Divide the rectangle $\tilde{P}$ into $N$ equal rectangles $p_1, p_2, \dots, p_N$ with the base $\frac{1}{N}$.

It is clear that either each of them contains exactly $k$ lattice points, or one of them (say, $p_m$) contains at least $k+1$ lattice points (say, $A_0, A_1, \dots, A_k$).

In the first case, the theorem to be proved is obtained immediately: the rectangle $P$ obviously contains one of the rectangles $p_i$, which means it contains $k$ lattice points.

In the second case, we perform the following construction (Fig. 9). To determine the type, assume that $A_0$ is the lowest point of $A_0, \dots, A_k$. Let’s connect point $A_0$ with point $O$ and draw segments equal and parallel to $A_0O$ from points $A_1, \dots, A_k$.

Let these be segments $A_1B_1, \dots, A_kB_k$.

The points $B_1, \dots, B_k$ belong to the lattice $R$; in fact, the figure $A_0OB_iA_i$ with $1 \leq i \leq k$ is a parallelogram (since $A_0O \parallel A_iB_i, A_0O=A_iB_i$) and the points $A_0, O, A_i$ belong to $R$; This means that by the parallelogram property (Theorem 2) $B_i$ belongs to $R$.

The points $B_1, \dots, B_k$ lie in the rectangle $P$. Indeed, for $1 \leq i \leq k$ the segments $A_0A_i$ and $OB_i$ are equal and parallel, and, therefore, there are projections onto the $Ox$ axis. But the segment $A_0A_i$ lies in the rectangle $p_m$ and its projection onto the axis $Ox$ is less than $\frac{1}{N}$. This means that the abscissa of the point $B_i$ according to the model is less than $\frac{1}{N}$, that is, $B_i$ lies in $P$. So, $B_1, \dots, B_k$ are $k$ lattice points lying in the rectangle $P$. The theorem is proven.

§ 5. Conclusion

From Theorem 3 other interesting consequences can be obtained.

For example, this:

Theorem 4. There is an infinite set of fractions $\frac{p}{q}$ maximum

$$\left| \alpha - \frac{p}{q} \right| < \frac{1}{q^2}.$$

Proof. In Theorem 3 we take $k=1$. Then, according to this theorem, for each $N$ there is an approximation $\frac{p}{q}$ of the number $\alpha$ such that $q \leq N$ and

$$q \left| \alpha - \frac{p}{q} \right| < \frac{1}{N}.$$

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Conductors, semiconductors and dielectrics

Before talking about the semiconductor diode and transistor, we will talk about what semiconductors are, what their internal structure is and what properties they have.

Of the large number of semiconductors known in nature, silicon and germanium are the most studied. These are elements numbered 14 and 32 in the periodic table.

Strictly speaking, semiconductors are dielectrics, only very weak dielectrics. Weak in the sense that they are easy to convert into semiconductors. There are several ways to do this. The main ones are: irradiation, bombardment of semiconductors with fast particles, heating and adding appropriate impurities to the semiconductor.

As you know, dielectrics do not conduct electric current, that is, they are insulators. If a potential difference is created at the ends of a dielectric sample and the current is changed, it will be equal to zero. This is explained by the fact that inside the sample there are no free loaded particles, for example, electrons, which, under the influence of an electric field, would move from one end of the sample to the other and thus create a current. The dielectric atoms firmly hold all their electrons and even if they release them from themselves, they have only exchanged with their neighbor for another electron.

It’s a different matter in metal. The bulk of electrons there behave similarly to electrons in a dielectric. But some of them, those that are located far from the positively charged nuclei and are therefore weakly connected with them, during the formation of a crystal receive such freedom that they can be throughout the entire crystal. These electrons are called free electrons, or conduction electrons. If you place a metal in an electric field, then these

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free electrons (because they have a negative charge) move in the direction opposite to the direction of the electric field. The electric field, like wind, blows them towards the boundary of the sample. A galvanometer connected in this way will register an electric current.

So, a dielectric does not conduct electric current because there are no free charge carriers in it that can move freely along the sample under the influence of an electric field.

But we ourselves can create such charges. One of the ways is to act on the atom of the dielectric with some suitable “tool”, for example, a quantum of light - a photon. We irradiate the dielectric with ordinary light, or even better, RENTen rays, since they have higher energy, and knock out electrons from the atoms that form the crystal. In this way, free electrons can be created in the dielectric and turned into a conductor. It is clear that the energy required to release an electron depends primarily on what atoms the crystal is made of and how these atoms are located in the crystal. In other words, this energy is different for different crystals. The minimum energy required to remove an electron from an atom and convert it into a free one is called the energy gap of a given dielectric. If radiation falls on the crystal, the energy of the quanta of which is lower than the minimum energy required to release an electron, then none of the electrons will be knocked out and the dielectric will remain an insulator.

The difference between dielectrics and semiconductors lies precisely in the size of the energy gap. Dielectrics with a sufficiently small energy gap are called semiconductors. The energy gap of semiconductors is approximately 10 times smaller than that of pronounced dielectrics, such as, for example, diamond. The energy gap width for diamond is 7 38, while for thermanium we have 0.72 38, and for silicon 1.1 38*). Thus, in order to turn a semiconductor into a conductor, it is enough to impart it to electrons relatively small small small small small smallest.

Now imagine that we are heating a dielectric. As the temperature increases, the average energy of the electrons will increase, and many of them will have enough energy to bridge the energy gap. As a result, there are more and more free electrons, and the appearance of free electrons occurs more often the higher the temperature, and, of course, most often in those crystals that have a smaller energy gap, that is, in semiconductors. That is why semiconductors are called semiconductors, because at sufficiently high temperatures (for example, room temperature ~300°K) they are conductors, and when the temperature drops they turn into dielectrics. The transistor example transmits work in freezing cold, since the semiconductors used in its transistors turn into real insulators.

Electrons and holes

Let us somehow knock out an electron from a dielectric atom and it flies away from this atom. In the crystal, in this place, a non-chewing electron or, as they say, a hole is formed. The interesting thing is that the hole does not stand still, but begins to move and generally behaves like a positively charged particle. Here’s how it happens.

When some atom lacks an electron, the electron can transfer

*)1 electron-volt (1 38) - the energy acquired by an electron when passing through a potential difference of 1 volt. 1 38=1.6·10⁻¹⁹ j.

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reck from a neighboring atom that has as many electrons as needed (Fig. 1, a). But then a shortage of electrons is detected on the neighboring atom, that is, the hole moves to the neighboring atom (Fig. 1, b). Of course, in reality it is electrons from filled atoms that jump to neighboring unfilled ones, but it is more convenient to talk about the movement of a hole. There is still only one hole, but it reflects the movement of many electrons. The movement of a hole is similar to the movement of a positively charged particle, because if, under the influence of an electric field, electrons move in one direction, then their “lack” moves in the other.

It will be useful to highlight the difference between the behavior of metals and semiconductors in an electric field. Current in a metal is created only by free electrons. Metal is structured in such a way that it always contains one or another amount of them. In a semiconductor, at sufficiently low temperatures, there are no free electrons. These free electrons

Fig. 1.

can be created. But at the same time, by tearing an electron away from the atom and presenting it with freedom, we also create a hole. Subsequently, these electrons and holes move independently of each other. It turns out that the current in a semiconductor consists of the electron current and the hole current.

It may happen that a free electron and a hole meet, that is, that an electron, while moving through a crystal, encounters an atom that just lacks an electron. It will occupy this free space, and at this moment both the electron and the hole will disappear. In this case, a recommendation is said to have occurred. You have the right to ask, why then create electrons and holes if sooner or later they will simply destroy each other? First, in the presence of an electric field, electrons and holes, as particles with different charges, move in opposite directions, and we can detect the current they create even before they recombine. Secondly, we do not knock out electrons from every atom, but, for example, from every thousandth. In this case, we create a significant current (since there are approximately $10^{23}$ atoms in each cubic centimeter of the crystal) and leave enough freedom for the movement of electrons and holes.

Semiconductor crystals are grown extremely pure. This is also done so that nothing interferes with the movement of electrons and holes.

Donors and Accelerators

There is also a way to create free electrons in a semiconductor. Let us imagine a thermanium crystal in which one of the thermanium atoms is replaced by an arsenic atom. Thermanium atoms have a valency of 4, and arsenic atoms have a valence of 5. When an arsenic atom is surrounded by thermanium atoms, the fifth

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The electron of arsenic is bound to it very weakly (Fig. 2). To tear it off, an energy of 0.1 eV is sufficient, which does not exceed the thermal energy of the crystal. Thus, one of the electrons of arsenic atoms becomes free and flows through the crystal. By introducing impurities of higher valency into the semiconductor, we can always obtain a sufficient number of negative current carriers. In this case, the impurity atoms are called donors, and the semiconductors themselves containing donor-type impurities are called $n$-type semiconductors.

It is possible to introduce trivalent atoms, for example, aluminum, into a thermanium crystal. An aluminum atom, in order to sit firmly in a thermanium crystal, must have two outer electrons, so it captures the missing electron from neighboring thermanium atoms (Fig. 3). A hole is formed in the crystal. Impurity atoms of this type are called $^*$ acceptors, and semiconductors are called $p$-type semiconductors.

$n$- ​​and $p$-type semiconductors have a corresponding excess number of negative or positive current carriers. Beaten - in comparison with those electrons and holes that create thermal excitations. However, the overall mystery of semiconductors is, of course, zero, since their crystals are formed from neutral atoms. The negative supply of electrons in a $n$-type semiconductor is compensated by positively charged donor centers fixed at the sites of the crystal lattice, for example, arsenic atoms. In a $p$-type semiconductor, the positive riddle gives rise to twos of the negative riddle of the accelerator centers.

Transitions between semiconductors

Let’s take two different samples of termanium ($p$- and $n$-types) and apply them to each other $^*$). We are interested in what will happen at the interface (Fig.4).

Until the contact between the samples was created, electrons in the $n$-region and holes in the $p$-region, approaching the sample boundary, were reflected from the surface and returned deep into the semiconductor. How

$\cdot$) From the root “accett” - to accept.

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Fig. 4.

As soon as a transition is formed, the situation changes. Electrons from the $n$ region moving towards the interface have the opportunity to jump to the $p$ region. Similarly, holes from the $p$-area will be counted into the $n$-area. Both of these processes act in the same direction: they tend to charge the $n$-type term positive, and the $p$-type term negative. In fact, the $n$-region, the total charge of which was equal to zero, now becomes positively charged, donating negatively charged electrons to the $p$-region. Positively charged holes coming from the $p$-region only contribute to this. Thermal $p$-type acquires a negative charge both due to the arrival of electrons and the departure of holes.

However, all this process must be left behind or later. When the $n$-type thermal is charged positively, and the $p$-type thermal is charged negatively, then a capacitor is formed, and an electric field appears at the boundary, directed from the $n$-region to the $p$-region (Fig. 5). This means that, for example, electrons from the $n$-region will no longer be able to go to the $p$-region. A strong electric field at the boundary will cause them to turn back. The same can be said for holes. The resulting electric field will reduce their chances of moving into the $n$-region. So, an equilibrium will occur in which the $n$-type thermal will be positively charged, and the $p$-type thermal will be positively charged.

Fig. 5.

negative, and at the boundary there will be a strong electric field directed from the $n$-region to the $p$-region.

If we take the potential of the electric field at the boundary to be zero, then the potential of the $n$-region is greater than zero, and the potential of the $p$-region is less than zero (Fig. 6), as will be the plate of a charged capacitor. The potential changes sharply in the region where the electric field exists, i.e., at the interface between $p$- and $n$-type thermals. The potential difference in the $p$-$n$ transition is equal to the energy that an electron needs to have to move from the $n$ region to the $p$ region, or a hole to transition from the $p$ region to the $n$ region. As for holes in the $n$-region and electrons in the $p$-region, if they are lucky enough to accidentally get to the interface, the electric field will only push them into another region. However, if this happens, then this leads to a decrease in the charge of both the $n$ and $p$ regions, i.e., to a decrease in the transfer

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a drop in potential at the boundary and a decrease in the electric field.
But then it will find electrons in the $n$-region and holes in the $p$-region, which have sufficient energy to overcome the diminished potential bartender $^*$). They will flow through the barrier, and balance will be restored. We see that the equilibrium between the $p$- and $n$-regions is dynamic. There will always be an exchange of electrons and holes between them, but this exchange is equivalent: the average charge of the $p$- and $n$-regions remains unchanged and, therefore, the potential drop and the electric field at the boundary remain unchanged. However, the total current across the boundary is zero. This equilibrium can only be disrupted by creating an external potential difference at the ends of the $p$—$n$ junction. Let us now consider this case.

Semiconductor diode

A sample with a $p$—$n$ junction has one important property: it can work as a diode, that is, a system that passes current in only one direction.

Let’s see what happens if we turn on the $p$—$n$ transition in the electrical target, that is, we create an external potential difference at the ends of the sample. There are two possible cases. Either the potential drop at the $p$–$n$ transition boundary increases or decreases. the balance between the $p$ and $n$ regions, carried out through the exchange of electrons and holes, is disrupted, and the average current across the boundary remains depending on whether the outer direction of the potential increases or decreases, which exists in transition by itself.

If the external direction increases at the boundary, that is, reduce the potential difference if it increases, and vice versa. However, he fails to do this due to the transition to an electrical target. The average charge transferred from one side of the $p$—$n$ junction to the other is not retained in the sample, but goes further along the chain. the balance remains disrupted.

So, when turning on the $p$—$n$-transition under the direction, the exchange of charge

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between the $p$- and $n$-regions changes and current flows through the junction. This current is very large at one level of external voltage and very small at another.
The property of a semiconductor diode to pass current in only one direction is used, for example, to create rectifiers - devices that convert alternating current into direct current.
The simplest circuit of such a rectifier is shown in figure 7. To analyze the operation of this circuit, the only important thing for us is the fact that with one direction symbol the resistance of the diode ($R_D$) is practically equal to zero (when it passes a huge current), and with another direction symbol the diode practically does not conduct current and, therefore, its resistance is very high.

Now suppose that the diode allows current to pass only from left to right, that is, it is open when the input direction is greater than zero, and closed when it is negative.
At $U_{bx} > 0$ total circuit resistance
$$R_D + R \approx R,$$
and the output direction will be practically equal to the input.
At $U_{bx} < 0$ the diode resistance is infinite ($R_D \gg R$), and therefore the direction of $R$ is practically zero. the output direction is zero. Thus, our system converts an alternating direction into a constant direction.

Transistor
Now we can consider a transistor - a system that combines the two transitions just discussed. A $p-n-p$-transistor is a small block made, for example, of thermium, composed of three sections: $p$-region, $n$-region and again $p$-region (Fig. 8). Each transition in the transistor behaves as described above. In particular, in each transition there should be a potential drop - a drop from the $n$-region to each of the $p$-regions, so that the overall potential distribution will be as shown in figure 8.

In a working transistor, each of the three areas is connected to the source

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stick to the surface of the crystal and immediately leave it again. Thus, even in an idealized picture, the growth of crystals is not at all as orderly as the stacking of crystals. The growth of crystals should proceed approximately as shown in figure 1, 6. A cube falling onto the surface of a crystal binds firmly to it only when at least three of its six faces adhere to cubes already sitting on the crystal. Moreover, in order to guarantee the growth of any layer, it must consist of a certain minimum number of cubes.

In the case where too many of the cubes that make it up are located at the corners or on the edges, they will not be able to hold new cubes that fall to the surface.

The rate of transition of molecules from a solid to the surrounding vapor depends on the intensity of molecular movement - this depends on temperature. The rate of return from vapor to a solid depends on the concentration of molecules in the vapor, that is, on the vapor pressure. For each temperature there is a certain “saturated vapor pressure” at which the rate of transition from solid to vapor and back is balanced. Under these conditions the crystal will not grow. It can only if the steam is too long.

This model of crystal growth further assumes that once a layer on the surface is filled, further crystal growth is hindered: a higher degree of oversaturation is needed to begin to fill a new layer.

It was at this point that the classical theory of crystal growth failed. Calculations have shown that the formation of islands of condensation centers large enough for the construction of new words to begin and the crystal to continue to grow requires pressure exceeding the saturated vapor pressure by values ​​of 25 to 50 percent. However, in 1931, physicists M. Volmer and W. Schulze were able to grow crystals with a steam supersaturation of only one percent. The theory predicted that with such a degree of supersaturation, the crystal could grow at a rate 10^1000 times less than the rate at which it actually grows. This is apparently one of the most record-breaking discrepancies between theory and experiment.

How does such a relatively easy construction of new crystalline planes begin and continue? Physicists continue calculations again and again, but within the framework of the old model, they could not come to results that make any sense, as is now clear. This continued until 1949, when the English physicist F. Frank proposed a new model. Frank noted that the growth of a crystal can be quite simply understood if we assume that during the building process its crystal planes are not arranged independently of one another, but form a spiral structure. This would be possible if the crystal had such a deviation from the ideal structure, which is called a “screw dislocation.”

Let us imagine a crystal lattice of two adjacent sections, in which, throughout the entire depth of the crystal, the crystal elephants of one of the parts (the right one at Fig. 2) are located below the corresponding words of the other part. Why do we call such a structural disturbance a screw dislocation? Let the tiny observable walk around the inner edge of the cut along the surface of the crystal, starting its path at its upper end. He will end up at a lower level. And this would happen with each subsequent round, thus moving our observed elephant into the depths of the crystal. The model we described is called a screw dislocation, because the spiral path into the depth of the crystal always makes it turn to the right.

Once Frank understood the picture of a crystal with a dislocation, he immediately realized that the presence of a dislocation constantly creates an open step on the surface. Atoms and molecules can always easily attach themselves to it and continue to build a crystal without starting a new row. This step, as new building units are added, spreads across the surface of the crystal. However, it bends (Fig. 2). To understand the reason for this, consider a skater’s chain, which turns around one of its ends. If they want to maintain formation, then those skaters who are close to the second end of the rotation will have to run faster than those who are close to the center of the rotation. After all, they must run in a circle with a large radius. In our case with a step on the surface of a crystal, the problem is that its outer edge moves no faster than the area in the center. The bricks are laid at a more or less constant speed along the entire length of the step. In this case, as construction proceeds, the rotation of the outer part of the step around the center does not occur quickly enough and the step bends, forming a spiral.

Was the new picture of crystal growth correct? Confirmation did not take long to arrive. At one of the conferences where issues of this theory were discussed, micrographs of the crystal surface were demonstrated, which revealed the presented patterns of spiral steps. This answered the questions quite well. Subsequent intensive microscopic examinations of the crystals revealed a variety of growth patterns that could only be attributed to

Fig. 2. figure $a$ shows the formation of a screw dislocation, which leads to the spiral crystal growth illustrated in subsequent figures $b, c, d$.

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theories of screw dislocations were previously predicted by him.

Perfectly regular crystals rarely occur in nature. They usually contain at least one dislocation. This is a necessary condition for growth with slight supersaturation. In the general case, the crystal has a lot of dislocations. The presence of various dislocations complicates the growth picture. So, for example, two closely spaced dislocations, one of which is right-handed and the other left-handed, can lead to a step fixed at both ends (Fig. 3, a). As new creeps sit on such a step, it takes on a semicircular shape (Fig. 3, b). As a result, we arrive at a quantity-shaped step (Fig. 3, θ). This is a fully built-up crystalline surface on such a crystal. However, in the meantime, a new step is formed at the top of the original step, which serves as the core for building a new row (Fig. 3, e). As this process is repeated, the crystal will develop into a series of close-packed circles rather than spirals, as it did with a single screw dislocation. You can see a photograph of such a crystal on figure 4. The steps here are rectangles, not circles, for the reason that the conditions of this crystal at the periphery and in the center are slightly different.

Fig. 3. Closed loops can arise from steps that begin and end on the surface of the crystal (figure a) and which are caused by the proximity of two dislocations: right and left. Such a step is fixed at both ends and can only be raised by unfurling. Figure b shows a swelling step beginning to turn out. The next figure shows the moment of growth when the left and right protruding parts came together. The last figure shows how the internal parts have separated from the filled layer. A dent is visible in this step, which will be filled as soon as the area formed on the top of the loop grown from the first area begins to fill with molecules.

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Fig. 4. A four-row step, formed due to the presence of four dislocations of the same type located very close to each other. A series of a large number of dislocations produces multi-edge steps.

The height of the crystal step can be changed. It turned out that some crystals have steps many rows of atoms high (Fig. 5). The theory explains the appearance of such multi-row steps as starting from a group of closely spaced dislocations, which, moreover, are either all right-handed or all left-handed. From each dislocation the step receives one row of atoms. figure 6 shows a crystal of podmium kadmium with a step many rows high. A short straight line at the center of the spiral is a collection of screw dislocations.

The theory helped to understand other features of the crystal structure that had long mystified crystallographers. We are talking about the packing of crystal planes of a crystal. To understand this, we need to somewhat clarify the shape of the building units of the crystal, which until now we assumed were simply bricks. These lines

Fig. 5. Microphotography allows one to identify growth steps on the surface of a paraffin crystal.

Fig. 6. An unwinding spiral in four successive positions on the surface of a crystal of subsurface kadmium. Such a step, the height of which is many layers, is generated by a large group of close-packed dislocations.

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Fig. 7. In the diagram, close-packed molecules are represented by circles. The triangles indicate two possible positions for the molecules of the next layer.

Intelligent blocks of a crystal - atomic or molecular cells - are not creaks, but rather look like balls. Certain crystals have a close-packed structure, which means that these spherical blocks are not arranged directly on top of one another, but rather like cannonballs (or biliary balls) in a paramide in which each ball fits into a depression between its nearest neighbors. It is clear that each subsequent layer in such a paramid can fit tightly to the previous one with two possible positions of the balls. On figure 7 these positions of the balls of the upper layer are indicated by the symbols ▲ and ▼. When one layer is stacked on top of another, close packing conditions require that all the beads of that layer be located either at the positions marked ▲ or at the positions marked ▼. The next layer can again be adjusted so that its balls are located in either one or another position. Let the layers be arranged so that their balls are arranged sequentially, now in one position, now in another. This layer sequence is indicated by the symbol ▲. However, crystals are often packaged in much more complex ways. For example, one type of silicon carbide crystal is characterized by the following arrangement:

▲▲▼▼▲▼▼▼ and so on.

Here the group consisting of a heel of layers is repeated. In some types of silicon carbide crystals, a group of 29 layers is repeated.

How can information about the lower location of the balls in the layer be transmitted over such a large distance? How does each ball in the thirtieth layer end up in the same position as the corresponding ball in the first layer? Classical crystallography could not offer any explanation and there were reports of unknown long-range forces between molecules. Crystal growth theory using dislocation theory offered a simple, although incomplete, answer. The presence of a series of screw dislocations can lead to the appearance of a step height in a multilayer. This step becomes the only repetition in the structure of the crystal, which is constantly reproduced as it grows. The theory does not explain only one significant spectrum of implementation: in the case of silicon carbide, only 14 variations of the crystal structure pattern have been found, while the theory leads to the appearance of an almost unlimited number of such structures.

How does the initial screw dislocation occur? The answer to this question is not clear yet. Almost all crystals begin to grow from a small foreign core. Some of these nuclei may have surface inhomogeneities on which a screw dislocation will arise, propagating throughout the growing crystal. Other crystals may acquire dislocations as they grow. Dislocations can also occur on trapped parts of the dirt.

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Sections of railways or highways running in hollows will be covered with snow, even if there is no snowfall. Why is this happening? At first glance, the answer is clear: snow is carried by the wind. However, to understand in detail the mechanism of this process, a whole study was required.

In 1936, the English physicist Bangold studied the transport of sand by wind in a wind tunnel.

It turned out that if the wind speed is less than some speed $v_1$, then the sand does not move. At wind speeds greater than $v_1$, but less than some other speeds $v_2$, sand can also remain at rest. But if at the same time a grain of sand falls from somewhere into a motionless mass of sand, several other grains of sand jump up from its impact. These grains of sand are carried away by the wind, fall, set other grains of sand in motion, and as a result, the sand is carried along the wind. When the wind speed is greater than $v_2$, the grains of sand rise in the wind, forming a sand-air flow that is quite significant, but the density decreases upward. figure 1 gives an idea of ​​the trajectories of grains of sand.

Now we can explain why snow fills the recesses in windy weather. At the point of deepening, the flow expands (this immediately becomes visible from the picture of linear figure 2), and therefore its speed decreases. As a result, the balance between those carried by the wind and those falling -

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Fig. 2.

mi whizz particles are disrupted: more particles fall than rise, and the depression is gradually filled with snow.

Similar processes occur when snow, carried by the wind, encounters some obstacle in its path, such as a tree. In front of the tree trunk, on the side from which the wind is blowing, an upward movement of air occurs. It leads to the formation of a deep depression on the windward side of the trunk and on the sides of the snow surface. In front of this notch and slightly behind the trunk, where the wind speed is lower, on the contrary, an elevation is formed.

The described phenomenon is used to protect sections of roads passing through hollows from snow. At a certain distance in front of the excavation or hollow on the side from which the wind blows, install a fence made of rickety sheets. Behind the shields, a calm zone with a uniform weak wind is formed, in which all the snow carried by the wind is completely deposited.

The movement of sand dunes is also obligated. A strong enough wind blowing onto a sand dune lifts the sand from the windward side. On the leeward side, where the wind speed is lower, the sand falls down. As a result, over time, the dunes gradually move in the direction of the wind - “coaching”.

DON’T BELIEVE

TO YOUR EYES

On the pages of our magazine («Kvant» No. 10, 1970) we have already written about critical illusions.

Here is another interesting example of such an exchange of vision. Concentric circles are drawn on the cover of the magazine. However, due to the background consisting of colored spirals, the lines of circles, sometimes transformed, are perceived by the eye as spirals.

Using a compass, you can easily verify that this is a real image of a circle.

It is curious that if you illuminate figure with short flashes of light, then after a certain flash duration the illusion disappears.

The same illusion will arise when looking at figure placed below. The circles shown here also look like spirals.

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Solutions to the problems of the introductory test in the ZMS

A. L. Toom

In the first issue «Kvant» the problems of the introductory test for the Correspondence Mathematical Scale were published. ZMSH received about 6,000 letters from eighth-graders. About 2,500 of those who sent decisions were accepted into ZMSH and its branches.

In this issue we publish solutions to the most difficult problems.

Problem conditions

5. Four statements have been made about triangle ABC:

1. triangle ABC is right-angled;
2. ∠A = 30°; 3) AB = 2BC; 4) AC = 2BC.
   We know that two of these statements are true and the other two are false. Find the perimeter of triangle ABC if BC = 1.

6. It is known that a+b+c<0 and that the equation ax²+bx+c=0 has no real roots. Determine what sign the number c has.

7. The following is known about points A, B, C: for any point M on the plane, the segment AM is less than at least one of the segments BM and CM. Prove that point A lies on segment BC.

8. Is it possible to wrap a cube with edge 1 in a square piece of paper with edge 3?

9. Is it possible to write down nine numbers 1,2,3,…, 9 in a circle in such an order that the sum of no two adjacent numbers is divisible by 3, 5, or 7?

10. Residents of the city of Pravdin always tell the truth, and residents of the city of Liars always tell lies.

In one of these cities, a football match took place between the teams of these cities. After the match, near the stadium where it was played, the following happened:

A (to B and C): “I wasn’t at the match. Tell me, who won?
B: “G told me that their team lost.”
B: “Our team won.”
Γ (getting into a bus going to another city, and addressing A): “Let’s go together, A, because we are from the same city.”
A: “You are mistaken, G. I live here. This is B from your city."
Determine who is from each city, where the match takes place and which team won.

12. On the highway in one direction at a constant speed at regular intervals, buses travel without stopping. One person walked 4 km along the highway, and during this time he was overtaken by 6 cars. Another time he walked 7 km, and during this time he was overtaken by 8 autovosu. The third time he walked 17 km. How many cars did he overtake? (All three times the person walked at the same speed.)

Problem solutions

5. We are told that out of four statements some two are true, but we are not told which ones are true and which are not. Therefore, let’s consider several cases. In each such case we will assume that some two certain statements are true and the other two are true. In order not to miss any case, we first find out

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Case 1.2. Since $\Rightarrow A=30^\circ$, the right angle is either $B$ or $C$. If $\Rightarrow B$ is direct, then $AC=2BC$, which should not be the case, since statement 3 is false in this case. If $\Rightarrow C$ is straight, then $AB=2BC$, which also contradicts the conditions. So case 1.2 is impossible.

Case 1.3. Angle $A$ cannot be right, since the opposite side does not lie opposite it. If $\Rightarrow C=90^\circ$, then $\Rightarrow A=30^\circ$, which cannot be in this case. So, $\Rightarrow B=90^\circ$. then statement 2 and 4 are false. This means that such a case is possible. Then, by the Pythagorean theorem, $AC=\sqrt{5}$, and the perimeter is $3+\sqrt{5}$.

Case 1.4. It is analyzed similarly to case 1.3. Possible if $\Rightarrow C=90^\circ$. The perimeter is $3+\sqrt{5}$.

Case 2.3. If $\Rightarrow A=30^\circ$ and $AB=2BC$, then $\Rightarrow C=90^\circ$, which contradicts the conditions.

Case 2.4. Similar to case 2.3 is impossible.

Case 3.4. Statement 1.2 is incorrect. So this case is possible. The perimeter is 5.

Answer. a) $3+\sqrt{5}$; b) 5.

6. Since the square triangular $ax^2+bx+c$ has no real roots, then $b^2-4ac<0$, hence $ac>\frac{b^2}{4}$. Let us prove that $c(a+b+c)>0$.

Really,

$$c(a+b+c)=ac+bc+c^2>\frac{b^2}{4}+bc+c^2=\left(\frac{b}{2}+c\right)^2\geq 0.$$

Since in a positive product one factor $a+b+c$ is negative, then the other factor $c$ is negative. So, $c<0$.

Solution option. It is known that if a square triangular $ax^2+bx+c$ is not equal to zero for any $x$, then its sign is the same for all $x$. Note that the number $a+b+c$ is the value of our triangle at $x=1$, and the number $c$ is its value at $x=0$. This means that these two numbers have the same sign. Since $a+b+c<0$, so is $c<0$.

7. Let us prove the required contradiction. Let us assume that the point $A$ does not lie on the segment $BC$, and we will come to a contradiction with the conditions. That is, we will prove that then it is not true that “for any point $M$ on the plane, the segment $AM$ is less than at least one of the segments $BM$ and $CM$.” In other words, let’s find a point $M$ on the plane such that the segment $AM$ is not less than one of the segments $BM$ and $CM$. Let’s consider two cases.

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the columns are numbered from 1 to 9. If two numbers can stand next to each other, then at the intersection of the row and column with these numbers there is a square, and if they cannot, then minutes. You can do otherwise. Let’s draw nine circles and place numbers from 1 to 9 in them. Let’s connect two linear circles, the numbers in which can be next to each other. And if two numbers cannot be adjacent, then we will not connect them. We will redraw the resulting diagram in order to “unravel” it as much as possible. The result is depicted in figure 4. Once such figure is constructed, the question can be sorted as follows: is it possible, and for only along lines, to go around all the circles of this figure once at a time and return them to their original place? We will solve the problem by looking at this drawing, although we could use a table in the same way.

Since 4 is connected only to 7 and 9, it means that 7 and 9 are its neighbors in location on the circle. Similarly, neighboring 2 are the numbers 6 and 9. This means that 2 and 4 are neighboring 9, and 7 and 8 are obviously not 相邻 9. This means that the connections of 9 with 7 and 8 can be crossed out (which is what was done on figure). Similarly, 1 has neighbors 3 and 7. This means that the neighbors of 7 are numbers 1 and 4, and the connection between 7 and 6 can be crossed out. Now 8 is connected only to 3 and 5, which means that the connection between 3 and 5 can be crossed out. The remaining lines will process a closed path, which constitutes the only (up to the direction of traversal) solution.

11. Let’s make a table:

| And from Pravdin? | G from Pravdin? |
| :— | :— |
| B from Pravdin? | Did the match take place in Pravdin? |
| B from Pravdin? | Did Pravdin’s team win? |

Solving a problem means filling in the empty columns of this table with the words “yes” or “no” so as not to conflict with the conditions. More precisely, we need to find all such filling methods, since there may be more than one of them. At worst, you can rework all possible filling methods (there are 64 in total) and check for each one whether it contradicts the conditions or not. But this, of course, is an ugly way to solve it. We will decide otherwise.

1. From statement B it follows that Pravdin’s team won. Let’s prove it. If B is from Pravdin, then what he said is true, and Pravdin’s team won. If B is from Liar, then what he said was not true, and again Pravdin’s team won.

2. From B’s statement it follows that B is a Liar, since his statement is completed incorrectly. Let’s prove it. We already know that Pravdin’s team won.

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Dina. If $\Gamma$ is from Pravdin, then he could not say that their team lost, since that would be a lie. If $\Gamma$ is from Liar, then he also could not say this, since it would be the truth.

3. From the statement $\Gamma$ it follows that $A$ is from Pravdin. Let’s prove it. If $\Gamma$ is from Pravdin, then $A$ is valid from the same city, that is, from Pravdin. If $\Gamma$ is from Lgunov, then $A$ is not from the same city, that is, $A$ is again from Pravdin.

4. Now it’s easy to answer all the questions. From the last statement $A$ it follows that the case is happening in Pravdin, $B$ and $\Gamma$ - from Lgunov.

Answer. $A$ - from Pravdin, $B$, $B$, $\Gamma$ - from Lgunov, the match was hosted in Pravdin, Pravdin’s team won.

12. It is clear that auto heights overtake a person at regular intervals. (Any such interval is equal to $\frac{v_1 t}{v_1 - v_2}$, where $t$ is the time interval between autovises, $v_1$ is the speed of autovises, $v_2$ is the speed of a person. However, we will not formulate this.) Therefore, between those moments when the autovises overtake him, the person has time to travel the same distance each time. Let’s denote this distance by $x$.

On figure 5, the segment $AB$, marked with a curly bracket, represents the 4 km that the person walked for the first time. Let’s mark on it those 6 points $P_1$, $P_2$, $P_3$, $P_4$, $P_5$, $P_6$, where it was cut by the autovis. There are 5 spaces between them, and the length of each is $x$. There remain two pieces of the path at the ends $AP_1$ and $P_6B$, which the person passed before the first autovis overtook him, and after he was overtaken by the last autovis (drawn with a red line at figure). The length of each of these segments is less than $x$. Let’s prove it. Let’s say, for example, that $P_6B \geq x$. Then on the segment $P_6B$ there will be another point $P_7$ such that $P_6P_7 = x$. At this point, of course, a person will also be overtaken by an autovis, and in total over these 4 km a person will be overtaken by more than six autovises, which is contrary to the conditions. The fact that $AP_1 < x$ proves similarly. Taking all this into account, we can write:

$$5x \leq 4 < 7x. \tag{1}$$

Similarly, since on a segment of 7 km there will be 8 points spaced $x$ from each other, and the end segments will again be less than $x$:

$$7x \leq 7 < 9x. \tag{2}$$

From (1) we obtain that $x \leq 4/5$, and from (2) we obtain that $x > 7/9$.

Now we must determine how many points, spaced $x$ km apart, can be located on a segment of 17 km.

Since $\frac{17}{4/5} = 21 \frac{1}{4}$ and $\frac{17}{7/9} = 21 \frac{6}{7}$, then $21 \frac{1}{4} \leq \frac{17}{x} < 21 \frac{6}{7}$ or

$$21 \frac{1}{4}x \leq 17 < 21 \frac{6}{7}x. \tag{3}$$

Let us prove that on a segment of 17 km a person will be overtaken by 21 or 22 cars. Indeed, if a person is passed by 20 autovises (or less), then there will be 19 intervals of length $x$ between them (or less), hence $17 < 21x$, which contradicts (3). If a person is overtaken by 23 autovises (or more), then $17 > 22x$, which also contradicts (3).

Answer. 21 or 22 autovis.

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STRANGE PENDULUM

N.A. MINC

An ordinary mathematical mat, familiar to you, is no less than the plane of its oscillations.

The well-known demonstration of the rotation of the Earth, Foucault’s experiment, is based on this property of the matnik. The matnik on the long pendant oscillates. Below it is a circle that fills the dial. Due to the fact that the plane of vibration of the matnik relative to the fixed stars does not change, and the Earth rotates around its own axis, over time the matnik passes sequentially over all marks of the circle. At the pole, within 24 hours the circle under the mat will complete a full revolution. For the first time such an experiment was carried out by the French physicist L. Foucault in 1851 under the dome of the Pantheon in Paris with a 67 m long mat.

But does every motherboard necessarily maintain the plane of oscillations? After all, the suspension thread allows it to oscillate in any vertical plane.

Try to make a mat like that shown in figure 1. To do this, take a thread, fold it in half, and tie another thread to the middle. Attach a spoon, strings or some other object to the other end of this second thread - and the mat is ready. (The vertical pendant thread should be long enough. At least not less than the inclined pendant thread.)

Hang the matnik at both ends of a thread folded in half on horses or nails (for example, in a doorway). If you now open the mat from the equilibrium position and then lower it, you will see an interesting picture. The matnik will move along an ellipse, and this ellipse will constantly change, stretching in one direction or the other. Why is this happening?

For a matnik with one suspension point (Fig. 2), the oscillation plane is not highlighted. Whatever it is

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

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the initial deflection of the matrix, all the forces acting on it lie in the same plane. It is only necessary, when releasing the matrix, not only his choice.

Indeed, let us draw a plane through the initial and deflected positions of the matrix. Obviously, both the gravitational force $mg$ and the tension force of the thread $T$ will lie in this plane. Consequently, the acting force of these forces, which, in fact, causes the matrix to oscillate, is equal; for any deflected position, the matrix also lies in the same plane. This means that there are no forces, which can be quite complex, especially at close frequencies of longitudinal and transverse vibrations. If the frequencies are the same, the motion trajectories will be an ellipse. The figure shown in Fig. 4 describes a matrix whose equation of motion is as follows: $x=\sin 3t$, $y=\sin 5t$. Figure 5 corresponds to vibrations $x=\sin 3t$, $y=\sin 4t$.

The frequency ratio can be adjusted by changing the length ratio

$\cdot$ Of course, if the matrix is ​​deflected in line with the primary plane of the suspension, it will oscillate in one plane, the primary plane of the suspension. But in practice there is always a deviation from the initial point and a speed that does not lie in the plane of the initial deviation.

Fig. 4.

Fig. 5.

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Fig. 6.

vertical and inclined suspension threads. At the same time, it is quite difficult to calculate the vibration frequencies of a magtin, but it is much easier to see the figures drawn by it.

In order to see the Lissajous figures, you can attach a bucket filled with sand to the suspension of our magtin, and put a piece of cardboard painted in a dark color on the floor. Then the magtink will “draw” a clear trajectory of its movement.

Fig. 7.

You can get a photograph of the magtin’s movement trajectories.

Take a sinker or heavy small ball and paint it white and red. Make the same pendant from dark threads as the magtinik shown in figure 1. Place a sheet of black matte paper under the magtinik. Smooth paper can cut off, and this will interfere with what is being observed. You need to take photos from above. Position the camera so that the object plane is horizontal. With a sufficiently long shutter speed, you can see clear trajectories on the photos.

Figures 6 and 7 show a photograph of the trajectories obtained in this way. It can be seen how the magnetite changed the direction of vibrations. In figure figure 6 this change will occur more sharply. The exposure time of the two photographs is different. This can be seen at least from the fact that the length of the trajectories is not the same.

The resulting curves seem to be inscribed in a parallelogram. In fact, they should be entered into a straight line. The fact that we did not get a straight line can be explained simply: the plane of the object was not strictly horizontal.

In experiments with magnetism, it should be taken into account that a more or less correct trajectory is obtained only in the case when there are no strong attenuations. The oscillations of a magnet with a small load mass and a sufficiently large volume will be damped. So magtinik qualitatively several times, quickly reducing the amplitude. Naturally, when moving with strong attenuation, it is difficult to see and photograph the change in the direction of oscillations of the magnet.

You often encounter Lissajous figures in cases where the oscillations are mutually perpendicular. So, they inevitably appear when setting up an oscilloscope.

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TASKS

M86. The bottom of the rectangular box was lined with tiles of sizes $2 \times 2$ and $1 \times 4$. The tiles spilled out of the box and in the process lost one tile $2 \times 2$. Instead, we managed to get tile $1 \times 4$. Prove that it is now impossible to tile up to the box.

L. G. Limanov

M87. Prove that if three circles of the same radius pass through one point, then three other points of pairwise intersection of these circles lie on a circle of the same radius (Fig. 1).

M88. What condition must the coefficients $a, b, c$ of the equation $x^3 + ax^2 + bx + c = 0$ satisfy in order for its three roots to form an arithmetic progression?

M. F. Bezborodnikov

M89. Prove that in any convex polygon except pa-

Fig. 1.

rappelogram, you can select these three sides, when continuing

Fig. 2.

which form a triangle enclosing the given polygon. (For example, at figure 2, where the polygon is outlined with a black line, the three red lines satisfy the required condition, but the three blue lines do not.)

I. M. Yaelom

M90. Prove that if $x_1 < x_2 < x_3 < \dots$ are natural numbers,

$$\frac{\sqrt{x_2-x_1}}{x_2} + \frac{\sqrt{x_3-x_2}}{x_3} + \dots$$

$$\dots + \frac{\sqrt{x_n-x_{n-1}}}{x_n} < 1$$

$$+\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n^2-1} + \frac{1}{n^2}$$

G. I. Natanson

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F98. What is the maximum difference in potential power to obtain if you have a source with e.g. d.s. $E$ and $n$ identical capacitors with a capacity of $c$ each?

L. G. Aslamazov

F99. figure 3 shows $V$—$T$ diagrams for two circular processes. In which of them does the gas do more work: in process $1 \rightarrow 2 \rightarrow 3 \rightarrow 1$ or in process $1 \rightarrow 3 \rightarrow 4 \rightarrow 1$?

I. Sh. Slobodetsky

F100. At an altitude of 200 km, the atmospheric density is $1,6 \cdot 10^{-10} \text{ ке/м}^3$. Estimate the drag force experienced by a satellite of cross-section $0,5 \text{ m}^2$ and mass $10 \text{ ке}$ flying at this altitude.

F101. A ball is suspended from a hanging light spring with stiffness $k$. At first the spring does not grow. Then the ball is released. What is the maximum speed the ball will reach while moving? The mass of the ball is $m$.

S. A. Belleev

F102. Oxygen and helium were mixed equally in a hermetically sealed vessel. Then small responsible ones were made in the wall of the vessel. What is the composition of the molecular beam emerging from it?

I. Sh. Averbukh

In this issue we feature several problems from Scientific American magazine in the Mathematical Games section. The permanent editor of Mathematical Games is Martin Gardner. The problem on the fourth page of the cover is also taken from this section. (Translation by I. I. Moroz).

PLACE THE COINS

This little challenge was created by Kobon Fujimura from Japan. Place the ten coins as shown at figure. What is the maximum number of coins you must remove so that there is not a single equilateral triangle (of any size) with vertices in the centers of the remaining coins?

STRANGE NUMBER

Find a tenth number in which its first digit indicates the total number of zeros in the number, the second - the total number of ones, and so on until the last digit indicates the number of nines in the number (of course, zero is a digit). There is only one solution.

35

SOLUTIONS

In this issue we publish solutions to problems M34—M38.

M34

Prove that if a natural number is divisible by 10 101 010 101, then in its decimal notation at least six digits are different from zero.

The problem can be solved very easily if the number $A$ is divisible by 10 101 010 101 and has no more than 12 digits; then it necessarily has the form $ababababab$, and the problem is solved.

It remains to come up with some way to reduce the problem to this case.

We denote the number 10 101 010 101 by $M$. Let us prove that if the number $A$ is divisible by $M$, then we will find a twelve-(or less) digit number $C$, which is also divisible by $M$ and has no more non-zero digits than $A$.

Lemma. If $A$ is divisible by $M$ and has more than 12 digits (counting zero), it will find the number $B$, also divisible by $M$, less than $A$ and having no more non-zero digits than $A$.

Let’s prove it. Let $A = a_1a_2a_3\dots a_n$; cross out its first digit $a_1$ and add it 12 digits lower, for example:

$$123\ 456\ 789\ 123\ 456 \rightarrow 23\ 456\ 789\ 123\ 456$$
$$+\frac{1}{23\ 456\ 789\ 123\ 556}$$

The resulting number will certainly be smaller than the original: it is easy to check that we are simply calculating $a_1 \cdot (10^{n-1} - 10^{n-13})$. In parenthesis

36

kah is the product $10^{n-13} \times (10^{12}-1)$; the second factor is obviously divisible by $A$. It remains to replace that the resulting number of significant (non-zero) digits is no more than the original one. In our example, their number even decreased by one; This could, however, not be accidental if there had been a 0 in the 13th place or if there had been a sufficiently large number there and a 0 in the previous place, as in the example

$$888 888 888 880 421 \rightarrow 88 888 888 880 421$$
$$+\frac{8}{88 888 888 881 221}$$

(you have not yet considered the case when the transfer occurs in the 12th digit, but it is easy to see that then the number of significant digits will even decrease). The lemma is proven.

Now if $A$ is divisible by $M$, subtract from $A$ its first digit multiplied by $10^{n-13} \cdot (10^{12}-1)$. With the resulting number, we sell the same operation and will do this until we add to the 12-digit number. Since it has at least six significant digits, $A$ also has at least six significant digits.

Most solvers tried to consider part of the division $A$. on $M$;

but solving it in this way is very difficult, since when this quotient is multiplied back by 10 101 010 101, its digits “stick together” with each other, a lot of transfer of units in digits can occur, and it is almost impossible to figure out what happens. The idea behind the solution is to look at these transfers one at a time.

A. K. Tolpieo

M35

A sharp 19-sided face is described near a sphere of radius 10. Prove that there are two points on its surface the distance between which is greater than 21.

First decision. Let us assume the opposite, that is, that the distance between any two points on the surface of our 19-hedron is no more than 21. Then this polyhedron lies inside a sphere of radius 11, concentric to a sphere of radius 10, and each of its faces lies between the spheres. It is clear that the area of ​​each face is not too large, namely, it does not exceed the area of ​​\u200b\u200bthe circle shown at figure 1, the radius of which is $\sqrt{21}$. Our polyhedron has 19 faces, so the area $S$ of its surface does not exceed $19\left[\pi\left(\sqrt{21}\right)^2\right]=\pi\left(20^2-1^2\right)=399\pi$.

No polyhedron is described around a sphere of radius 10. Hence, its surface area is greater than the area of ​​this sphere $4\pi\left(10\right)^2$*. So, on the one hand, $S<399\pi$, on the other hand, $S>400\pi$. The resulting opposite solves the problem. (A similar solution was sent to us by D. Grisoriev from Leningrad.)

In this (non-rigorous) solution, we missed the proofs of three statements that begin with the three words highlighted above: moeda, pozmotu, hence. We leave these simple proofs to the reader, but we want to warn that although the required statement is easy to prove for a convex polyhedron

*) Recall that for a ball of radius $R$ the volume is $\frac{4}{3}\pi R^3$, the area of ​​a segmented surface with height $h$ is $2\pi Rh$ and, in particular, the area of ​​the sphere is $4\pi R^2$.

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nik by comparing its volume with the volume of a sphere *), nevertheless, the intuitively clear and correct statement that our polyhedron is convex is difficult to prove strictly, since the strict definition of a polyhedron itself is very complex. (Look, for example, at the book by I. Lakatok “Proofs and Refutations”, M., “Science”, 1967).

Second solution. Let us pose a more general question: what is the maximum number of faces that a polyhedron can have that is described around a sphere of radius $r$ and lies entirely in a concentric sphere of radius $R > r$. (Here is an everyday situation that prompted the author to solve this problem: with what maximum number of straight strokes can you use the top layer of the husk of an orange without destroying a single piece of the pith? Obviously, after weeding the entire top layer of the kokura, the remainder will be a polyhedron, since there will not be a single rounded section on its surface, so this question is equivalent to the previous one.) We do not know the exact answer to this more general question, but we will prove it for a number faces of some inequalities, which for $r = 10$, $R = 11$ shows that $N > 22$. Thus, we will prove that if under conditions problem M35 we take a 22-hedron together with a 19-hedron, then the statement of the problem remains valid.

So, let the $N$-hedron be described near a sphere of radius $r$ and lie entirely inside a sphere of radius $R$. Some edge of it at a distance.

A plane passing through it cuts off a cap (segmental surface) of height $R - r$ from the sphere. It is clear that if we build caps for all the faces of our polyhedron, then their union will cover the entire outer sphere. Each of $N$ caps

*) Indeed, the volume of a polygon is equal to $\frac{R \cdot S}{3}$, where $R$ is the radius of the inscribed sphere, and $S$ is its surface area.

A natural surface has height $R - r$, and therefore has area $2\pi R (R - r)$. The sum of the areas of all caps is greater than the area of ​​the sphere. Therefore, $N \cdot 2\pi R (R - r) > 4\pi R^2$, hence $N > \frac{2R}{R - r}$, in particular, for $R = 11$, $r = 10$ we obtain $N > 22$.

It is interesting that for any set of caps that completely cover the outer sphere, it is possible to construct a polyhedron described around the inner sphere. (Prove it!) Therefore, our question about the minimum number of faces is completely equivalent to the following question. What is the minimum number $N = N(h)$ of caps of height $h$ that completely cover spheres of radius 1? (In problem $M35 h = \frac{1}{11}$).

Obviously $N(h) > \frac{2}{h}$, but this incorrectness simply reflects the fact that the sum of the areas of the caps is larger than the area of ​​the sphere, while intuitively it is clear that for $h < 1$ the caps should overlap quite a lot. Indeed, it can be proven that for sufficiently small $h$ $N(h) > 1,2 \frac{2}{h}$. Try to prove for yourself, for example, that for $h < 1$

$N(h) > 1,001 \frac{2}{h}$.

(The solutions to this problem and other similar problems can be found in the book by L. F. Toth. “Location on a plane, on a sphere and in space.” M., Fizmatgiz, 1958).

A. G. Kushirienko

M36

Prove that it is impossible to arrange seven lines and seven points on a plane so that three lines pass through each point and three points lie on each line.

Let us assume that such an arrangement of seven points and seven straight lines

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Fig. 2. The convex hull of a set of a finite number of points is a convex polygon with vertices at some of these points (or a segment if all points lie on the same line).

marches. First of all, we prove that every two of these points lie on one of the given lines. Indeed, if $A$ is one of these points, then three lines pass through $A$, and each of them contains two of these points (not counting $A$); thus $A$ and any of the six points other than $A$ lie on one of the given lines. In the same way, it is proved that every two of the given lines intersect at one of the given points: if $a$ is one of the lines, then through each of the three points lying on it there are two straight lines (not counting $a$), and therefore each of these lines intersects with $a$ at one of the given points.

Let us now consider the convex hull of the set of our seven points - the smallest convex figure containing these seven points.

Fig. 3. This configuration almost completely satisfies the requirements of problem M36, only one “straight” had to be bent.

This will be some convex $n$-gon, where $n \leqslant 7$ (it can be obtained like this: be at the given seven points of nails and then stretch a rubber thread covering all these nails; the nitka will stretch along the contour of a polygon with vertices at some $n$ of the given points, and all other ($7-n$) points will lie inside or on the border of this polygon; see Fig. 2). It is clear that the $n \geqslant 4$ case is impossible - after all, on each side of the $n$-gon there must be one more third point, in addition to the vertices, and there are seven points in total. The case $n=3$, when the convex hull is some triangle $ABC$, the same is impossible. Indeed, if $E$ and $F$ are points (from among the given seven) that lie on the sides $AB$ and $BC$, then $EF$ and $AC$ - two of the given lines - must intersect at one of the given seven points, and they will intersect on the continuation of side $AC$, that is, outside the triangle $ABC$. We got a contradiction.

Figure 3 will help find the error for those who sent us the wrong solution to this problem.

M37

A certain number is inscribed in each cell of an infinite sheet of checkered paper so that the sum of the numbers in any square (the sides of which follow the grid lines) does not exceed one in absolute value. Prove that then there exists a number $c$ such that the sum of the numbers in any rectangle (whose sides follow the grid lines) is not greater than $c$; in other words, that the sums of numbers in all rectangles are limited. Prove that this is true for $c=4$. You may be able to improve this estimate (for example, prove that our statement is true for $c=3$ or even for $c=2$).

The solution we present is close to proposal $A$. Klimov (Moscow).

Let’s prove that $c$ equal to 4 is suitable.

Suppose that in some rectangle with sides $a$ and $b$ the modulo sum is $4+\epsilon$, where $\epsilon > 0$ ($a < b$). Let’s build four squares, each of which has three sides along certain three hundred

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Fig. 6. Let $k_n = \frac{a_n}{b_n}$. Then $k_{n+1} = f(k_n)$, where $f(k) = \frac{|1-2k|}{2-k}$. Here is a graph of this function on the segment $0 \leq k \leq 1$. When solving the problem, we obtain that for any point $M$ on the right half of the graph $MQ/MP > 2$; if $0 \leq k \leq \frac{1}{2}$, then $0 \leq f(k) \leq \frac{1}{2}$, and $f(f(k)) = k$, that is, the function $f$ on the segment $0 \leq k \leq \frac{1}{2}$ coincides with its inverse function, and its graph is symmetrical with respect to the bisector of the angle between the coordinates.

Fig. 7. All unfilled cells contain zeros.

Secondly, if $\frac{a_n}{b_n} > \frac{1}{2}$, then

$$\frac{1 - \frac{a_{n+1}}{b_{n+1}}}{1 - \frac{a_n}{b_n}} = \frac{1 - \frac{2a_n - b_n}{2b_n - a_n}}{1 - \frac{a_n}{b_n}} = \frac{3b_n}{2b_n - a_n} > \frac{3b_n}{2b_n - \frac{b_n}{2}} = 2;$$

Thus, the value of $1 - \frac{a_n}{b_n}$ in the transition from $n$ to $n+1$ is mutilated by at least two times until we arrive at the rectangle with $\frac{a_n}{b_n} \leq \frac{1}{2}$. Therefore, how could we consider that it is already $\frac{a_1}{b_1} < \frac{1}{2}$.

At the same time

$$a_3 = b_2 - 2a_2 = (2b_1 - a_1) - 2(b_1 - 2a_1) = 3a_1,$$

$$b_3 = 2b_2 - a_2 = 2(2b_1 - a_1) - (b_1 - 2a_1) = 3b_1.$$

Therefore, in general, $a_{2k+1} = 3^k a_1$ and $b_{2k+1} = 3^k b_1$, and the sum of the numbers in this rectangle $3^k a_1 \times 3^k b_1$ is greater than $4 + 9^k e$, that is, by choosing $k$ large enough, we can make it angle-wise large. Now it is not difficult to obtain a contradiction: it is clear that any rectangle $Na_1 \times Nb_1$, where $a_1$, $b_1$ and $N$ are integers, can be divided into $a_1b_1$ squares (from the side $N$), and therefore the sum of the numbers in it cannot exceed the absolute value of the number $a_1b_1$.

Figure 6 shows the second glaze of the proof. figure 7 shows an example for which the sum of the numbers in some rectangles is equal to those. Thus, for $c = 2$ the statement of the problem is false. It is very plausible that the exact estimate is $c = 4$, but we do not know of examples showing that $c > 3$.

N. B. Vasiliev

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M38

A circle constructed at the height AD of a right triangle ABC as a diameter intersects the leg AB at point K, and the leg AC at point M. The segment KM intersects the height AD at point L. It is known that the segments AK, AL and AM form a geometric progression (that is, $\frac{AK}{AL} = \frac{AL}{AM}$).

It is not difficult to prove that AKDM is a rectangle and L is its center (Fig. 8). Let us lower the redistribution of AP to KM.

Then
$$AL^2 = AK \cdot AM = KM \cdot AP = 2AL \cdot AP,$$
$$AL = 2AP, \angle PLA = 30^\circ,$$

and since $AL = KL = ML$, it follows that the acute angles of triangle AKM are $15^\circ$ and $75^\circ$. Triangle ABC has the same acute angles: $\angle C = \angle BAD = \angle AKM$.

Fig. 8.

This decision was added by S. Komanov from Tbilisi and V. Krivitsky from the village of Kletnoye, Minsk region.

L. M. Lopovok

In this issue we publish solutions to problems F52—F58

F52

The ball is thrown vertically upward. What is greater: the rise time or the fall time?

The kinetic energy of the ball at any height during the bet is greater than when the ball is dropped. Indeed, if there were no air resistance, then these energies would be equal. Due to air resistance during descent, the kinetic energy of the ball is less than during the rebound by the amount of work spent on overcoming air resistance. This means that at any height, the speed of the ball on the upswing is greater than on the downswing. Obviously, the average speed of the ball is also high. Therefore, the time it takes to pick up the ball is less than the time it takes to fall.

F53

The vessel contains two immiscible liquids with densities $\rho_1$ and $\rho_2$ and thicknesses $h_1$ and $h_2$, respectively. A small streamlined body is lowered from the surface of the liquid into a vessel, which reaches the bottom just at the moment when its speed becomes zero. What is the density of the materials from which the body follows?

Let us take advantage of the fact that the potential energy possessed by the body on the surface of the upper liquid is spent when the body moves to overcome resistance. That’s why
$$mg(h_1 + h_2) = A_1 + A_2,$$
(*)

where $m$ is the mass of the body, $g$ is the acceleration of free fall, $A_1$ is the work of resistance forces in the upper fluid and $A_2$ is the work of resistance forces when the body moves in the lower fluid. Since the body is streamlined, the drag force is the Archimedean buoyant force: $F_1 = \rho_1Vg$ in the upper fluid and $F_2 = \rho_2Vg$ in the lower fluid (V is the volume of the body). That’s why
$$A_1 = \rho_1Vgh_1$$ and $A_2 = \rho_2Vgh_2$.

Substituting these expressions for $A_1$ and $A_2$ into equation (*) and taking into account that $m = V \cdot \rho$ (ρ is the density of the body), we obtain $\rho (h_1 + h_2) = \rho_1h_1 + \rho_2h_2$.

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From here
$$\rho = \frac{\rho_1 h_1 + \rho_2 h_2}{h_1 + h_2}.$$

The correct solution to add V. Kalintsev from Moscow, Yu. Polonsky from Zelenyodolsk TASSR and other readers.

However, a great reader solved the problem in a more complex way.

A body moving in a fluid is acted upon by two forces: the gravity force, equal to $mg = \rho Vg$, and the Archimedean buoyant force (equal to $F_1 = \rho_1 Vg$, when the body moves in a fluid with a density of $\rho_1$, and $F_2 = \rho_2 Vg$, when it moves in a fluid with a density of $\rho_2$).

Since the body is streamlined, the forces acting on it while moving in each fluid do not change. Consequently, the body moves uniformly accelerated. Let’s write down the equation of motion of the body. In the upper fluid: $mg - F_1 = ma_1$ (acceleration is directed downward), in the lower fluid: $F_2 - mg = ma_2$ (acceleration is directed upward). From these equations we find that in a fluid with density $\rho_1$ the body moves with acceleration $a_1 = g \frac{\rho - \rho_1}{\rho}$, and in fluid with density $\rho_2 - c$ with acceleration $a_2 = g \frac{\rho_2 - \rho}{\rho}$.

Having passed the first liquid, the body will acquire speed $V_0 = a_1 t_1$, where $t_1$ is the time of movement. This time can be found from the kinematic equation $h_1 = \frac{a_1 t_1^2}{2}$. Therefore $V_0 = \sqrt{2 a_1 h_1}$. At this speed the body begins to move in the lower liquid. Since the speed of the body at the end of the movement should become equal to zero, then $V_0 - at_2 = 0$, where $t_2$ is the time of movement of the body in the second fluid.

Hence $t_2 = \frac{V_0}{a_2}$. During this time, the body will travel a distance $h_2$ in the second liquid. Therefore $h_2 = V_0 t_2 - \frac{a_2 t_2^2}{2}$.

Substituting the expressions for $V_0$, $a_2$ and $t_2$ into the last equation, we obtain
$$h_2 = \frac{V_0^2}{a_2} - \frac{1}{2} \frac{V_0^2}{a_2} = \frac{1}{2} \frac{V_0^2}{a_2} = \frac{a_1 h_1}{a_2}.$$

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It is equal to 7.5 ohms. In Pozny, current $I_0 = \frac{30 \, \theta}{7,5 \, \omega} = 4 \, a$ flows through the source.

The current going through the source is the sum of the current going through the ammeter and the current $I_1$ going through the resistance $R_1$. Let’s find the current $I_1$.

At point $A$ the current $I_0$ is divided equally between the upper and lower branches of the circuit. Posny through resistance $R_2$ current $I_2$, equal to 2$a$. At point $B$ this current is again divided equally between equal resistances $R_1$ and $R_3$. To Posny, current 1$a$ flows through resistance $R_1$. Now you can find the current flowing through the ammeter. It is equal to $I_0-I_1=3a$.

F55

Estimate the height $H$ of an arrow shot vertically upward from a child’s bow. Arrow weight $m=20 \, \text{г}$, string length $AB=1 \, \text{м}$. The bowstring is pulled back to $h_0=5 \, \text{см}$. The bowstring tension is considered constant and equal to 250 $\text{н}$.

The energy acquired by the arrow when fired is equal to the work done by the force acting on the arrow from the bowstring. This force is obviously equal to the equal $F$ string tension force $T$ (Fig. 11). If the angle formed by the bowstring at points $A$ and $B$ with lines $AB$ is $\alpha$, then it is not difficult to find that $F=2T\sin \alpha$. So, the force acting on the arrow depends on how much the bowstring is pulled back. Since the bowstring is pulled back to a distance that is small compared to its length, the angle $\alpha$ is small, that is, $\sin \alpha \approx tg\alpha \approx \alpha$. Pozny $F=2T\alpha$.

Since $tg\alpha = \frac{h}{AB}$ ($h$ is the height of the bowstring deflection), then

$$F = 4T \frac{h}{AB}.$$

So, the force acting on the arrow is proportional to $h$ (Fig. 12).

It is not difficult to find the work of this force. It is equal to the area of ​​the figure between

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Fig. 13.

container theory of flight and waves on water" in «Kvant» No. 5 for 1970. The behavior of the flag is explained in the same way as the formation of a wave on water.

Let’s assume that the flag is slightly curved at some point (Fig. 13). In this case, at the top, when flowing around the protrusion, the wind speed is greater, and at the bottom, in the area of ​​concavity of the flag, the wind speed is less than away from the flag. From Bernoulli’s law it follows that the air pressure at point A is greater than the pressure at point B. Therefore, the formation of a protrusion should increase. Also, since the formation behind the bump causes the pressure behind it to be less than the pressure in front of the bump, the bump will get mixed up towards the end of the flag. So, by chance the formation of the concavity of the flag will increase. If we also take into account that, due to the formation of an outlet when flowing around even a flat flag with a uniform wind, the pressure on different sides of the flag may be different and therefore small bends in the surface will easily form, it becomes clear why the flag “flashes” in the wind.

F57

Two identical balls, connected by a weightless spring, move along a smooth horizontal floor at the same speed, perpendicular to the vertical wall. Describe how the collision of the system with the wall will occur. How will the balls move after the impact? The impact of the ball on the wall is absolutely elastic, the time of impact is short compared to the period of oscillation of the ball outside.

Since the speeds of the balls are initially identical and constant, the torus is not deformed. Indeed, otherwise, a force would act on each of the balls from the side of the spring, imparting acceleration to it, and the speed of the ball would change with time. After the first ball hits the wall, its speed is reversed. The balls begin to move towards each other with

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equal in speed (see Fig. 14), the spring is released. The kinetic energy of the balls turns into potential energy of deformation of the spring. Then the spring begins to straighten, imparting speeds of equal magnitude and opposite direction to the balls. In this case, just as during the roll of the spring, the center of mass of the system is motionless relative to the wall - no horizontal forces act on the system. From the law of conservation of energy it follows that at the moment when the ball hits the wall correctly, the speeds of the balls are equal to the initial speed of the system. After the ball collides with the wall, its speed again changes to the opposite and the balls move away from the wall at the same speeds. Since the spring is not deformed at the moment the ball hits the wall, it does not deform during further movement of the system.

F58

In a magnetic field, a ring with a diameter of $d$ and a resistance of $R$ falls from a great height. The plane of the ring is always horizontal. Find the steady speed of the fall of the ring if the field induction varies with height according to the law

$$B = B_0 (1 + \alpha H).$$

The force acting on the ring in a magnetic field is proportional to the current inducting through the ring, the current is proportional to e. d.s. induction excited in the ring during its movement, and e. d.s. induction, in turn, is proportional to the rate of change of the magnetic flux, proportional to the plane of the ring, and therefore proportional to the speed of the ring. Consequently, at a certain speed of movement of the ring, the force acting on it in the magnetic field becomes a cell. This speed is the speed of steady motion of the ring.

Since the kinetic energy of the ring does not change when the ring moves, the change in potential energy must be equal to the heat losses in the ring.

Let the speed of the ring be $v$. E.m.f. the induction excited in the ring during its movement is equal in magnitude

$$E = k \frac{\Delta \Phi}{\Delta t},$$

where $\Phi$ is the magnetic flux, an arbitrary ring, and $k$ is the proportionality coefficient.

$$\Phi = \frac{\pi d^2}{4} B = \frac{\pi d^2}{4} B_0 (1 + \alpha H),$$

$$\frac{\Delta \Phi}{\Delta t} = \frac{\pi d^2}{4} B_0 \alpha \frac{\Delta H}{\Delta t}.$$

But $\frac{\Delta H}{\Delta t} = v$, therefore

$$E = k \frac{\pi d^2}{4} B_0 \alpha v$$

and current flows through the ring

$$I = \frac{E}{R} = \frac{k \pi d^2 B_0 \alpha v}{4 R}.$$

Let us now write down the law of conservation of energy. Let the mass of the ring be $m$. During time $t$, heat equal to $I^2 Rt$ is released in the ring. If during this time the ring drops to a height of $h$, then

$$mgh = I^2 Rt.$$

Since $\frac{h}{t} = v$, then

$$mgv = I^2 R,$$

$$mgv = \frac{k^2 \pi^2 d^4 B_0^2 \alpha^2 v^2}{16 R}.$$

From here

$$v = \frac{16 Rmg}{k^2 \pi^2 d^4 B_0^2 \alpha^2}.$$

I. Sh. Slovodetsky

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Logarithmic equations
V. V. Goldberg

Logarithmic equations are usually called equations containing an unknown quantity under the logarithm sign.
Solving the simplest logarithmic equation

$$\log_a x = b$$

(1)
is based on the following property of logarithms: if the logarithms of two (positive) numbers to the same base $a(a > 0, a \neq 1)$ are equal, then these numbers themselves are equal.

Having written equation (1) in the form $\log_a x = \log_a a^b$ and using the indicated property, we obtain the answer: $x = a^b$.

This technique is also applicable to solving more complex logarithmic equations of the form

$$\log_a f(x) = b.$$

(2)
For equation (2) we obtain $f(x) = a^b$.

Example 1: Solve the equation

$$\log_5 [2 + \log_3(3+x)] = 0.$$

Solution e. First we find that

$$2 + \log_3(3+x) = 5^0, \quad 2 + \log_3(3+x) = 1, \quad \log_3(3+x) = -1.$$

Next we find: $3 + x = 3^{-1}, x = \frac{1}{3} - 3, x = -\frac{8}{3}$.

Some logarithmic equations are easily properties of the simplest. Thus, equations of the form

$$f(\log_a x) = 0$$

(3)
by replacing $\log_a x = y$ they are added to the equation $f(y) = 0$. If $y_1, \dots, y_k$ are the roots of the last equation, then we obtain a set of simple logarithmic equations:

$$\log_a x = y_1, \dots, \log_a x = y_k.$$

Example 2. Solve equation $\lg^3 x - \lg^2 x - 6 \lg x = 0$.

Solution e. Let us denote $\lg x$ by $y$. Then we obtain a cubic equation for $y$:

$$y^3 - y^2 - 6y = 0.$$

Let’s solve it: $y(y^2 - y - 6) = 0, y_1 = 0, y_2 = 3, y_3 = -2$. We consider three possible cases:

a). $\lg x = 0, x_1 = 1$; 6) $\lg x = 3, x_2 = 1000$; b) $\lg x = -2, x_3 = 0,01$.

Exponential-logarithmic equations are similar to equations of the form (3), if they are first polygonalized.

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Technique 3. Solve the equation $x^2 + \log_3 x = 3^8$.

Solution. Since both sides of the equation are positive (ODZ: $x > 0$), we equate their logarithms to base 3: $\log_3 x^2 + \log_3 x = \log_3 3^8$, which means that $(2 + \log_3 x) \log_3 x = 8$. Denoting $\log_3 x$ by $y$, we arrive at the quadratic equation $y^2 + 2y - 8 = 0$ with roots $y_1 = -4, y_2 = 2$. Let’s break down two cases:

a) $\log_3 x = -4, x_1 = 1/81$ 6) $\log_3 x = 2, x_2 = 9$.

Technique 4. Solve the equation $9x^{\lg x} + 91x^{-\lg x} = 60$.

Solution. Let us introduce the notation $x^{\lg x} = y$; then $x^{-\lg x} = \frac{1}{y}$, otherwise the property equation to the following equation with respect to $y$: $9y^2 - 60y + 91 = 0$. The roots of the latter are $y_1 = \frac{13}{3}, y_2 = \frac{7}{3}$. We are considering both possibilities:

a) $x^{\lg x} = \frac{13}{3}$; taking logarithm, we get $\lg x^{\lg x} = \lg \frac{13}{3}, \lg^2 x = \lg \frac{13}{3}$, whence $\lg x = \pm \sqrt{\lg \frac{13}{3}}, x_1 = 10^{\sqrt{\lg \frac{13}{3}}}, x_2 = 10^{-\sqrt{\lg \frac{13}{3}}}$;

b) $x^{\lg x} = \frac{7}{3}$; similarly to how it was done in the first case, we find $x_3 = 10^{\sqrt{\lg \frac{7}{3}}}, x_4 = 10^{-\sqrt{\lg \frac{7}{3}}}$.

When solving logarithmic equations, it is often a matter of properties to solve an equation of the type

$$\log_a f(x) = \log_a g(x).$$

Here, using potentiation, we need to go to the equation

$$f(x) = g(x).$$

But it should be borne in mind that with the maximer we are expanding the ODZ equation, and therefore the appearance of extraneous roots is possible. Therefore, it is necessary to check whether each of the roots of equation (5) is included in the ODZ of equation (4).

Technique 5. Solve the equation $\log_2(2x^2 - 2) = \log_2(5x - 4)$.

Solution. Finding ODZ:

$$\begin{cases} 2x^2 - 2 > 0, \\ 5x - 4 > 0, \end{cases}$$

that is, $$\begin{cases} x < -1, x > 1, \\ x > \frac{4}{5}. \end{cases}$$

Therefore, ODZ: $x > 1$. The potency of this equation, we get $2x^2 - 2 = 5x - 4$, or $2x^2 - 5x + 2 = 0$, from where we find $x_1 = 2, x_2 = 1/2$.

Since $x_2 = 1/2$ is not included in the ODZ, the equation has only one solution: $x = 2$.

The following properties of logarithms of numbers are well known:

$$\log_a(AB) = \log_a A + \log_a B \quad (A > 0, B > 0, a > 0, a \neq 1),$$

$$\log_a \frac{A}{B} = \log_a A - \log_a B \quad (A > 0, B > 0, a > 0, a \neq 1),$$

$$\log_a A^p = p \log_a A \quad (A > 0, p - \text{любое действительное число}),$$

$$\log_a q^p = \frac{p}{q} \log_a A \quad (A > 0, q \neq 0).$$