Kvant 1971, No. 1 · Pages

Popular science physics and mathematics
quantum
1
1971
magazine
Academy
Sciences of the USSR
And
Academy of Pedagogical
Sciences of the USSR

4

In the room:

Rotational motion of bodies 1 A. K. Kikoin
Combinatorics 13 N. Ya. Vilenkin
Movement of planets 20 Ya. A. Smorodinsky
Math circle
Auxiliary circle 28 E. G. Gotman
Laboratory «Kvant»
Waves in a shallow plate 32 A. G. Kosourov
Problem book «Kvant»
38
Applicant’s workshop
Substitution method for solving equations and systems of equations 41 M. I. Skanavi
Entrance exams in mathematics at Moscow State Pedagogical Institute named after. V. I. Lenina 45 E. A. Shegolkov
Written exam in physics at NSU 51 G. V. Meledin
Information
Zoological mathematical school 56 J. M. Rabot
Answers, directions, solutions 58
Collector’s Corner
Stamps associated with N. Copernicus 62
Review, bibliography
Tasks, games, experiments 64 V. P. Litsevsky
«Kvant» for younger schoolchildren, 3rd cover page
Briefly about exhibite 4th cover page

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It is well known that any body can perform two types of movement - translational and healing. The first of them is studied in sufficient detail in the school physics course. But the medical movement was “unlucky.” Due to lack of time, very little attention is paid to it. Meanwhile, the healing movement is found in both nature and technology no less often than the progressive one. In many cases, the actual motion of bodies is a combination of both types of motion. In other cases, translational movement or even resting bodies have healing details. This primarily includes all devices with wheels - from wheelbarrows and carts to airplanes and helicopters. Just try to find a machine that doesn’t have anything in it! Such a common device as a watch is, one might say, “chocked” with wheels interlocking with each other.

Rotational motion is not uncommon in nature. It is enough to recall the complex healing movements of the planets, their satellites, the healing of the Sun and, probably, other stars. Let us note by the way that when applied to planets, the word “rotation” is used in relation to two different types of their movements. They say, for example, that the Earth is healed around the Sun, but they also talk about the Earth being healed around its axis. So, are these the same movements? Can it be said that both are caused by the same principles? It is known that the Earth moves around the Sun along a trajectory close to a circle because it is affected by the force of gravity towards the Sun. But can it be said that the same or a similar force forces the Earth to make healing movements around its axis? Obviously not. In fact, the Earth’s movements can be considered as a combination of translational and linear motion - translational motion around the Sun and linear motion around an axis passing through both earth’s poles.

This article talks about the medical movement of bodies, what distinguishes it from translational movement, and how it is similar to it.

Cinematika of the medical movement

First of all, let’s fill the reader with what they already know about the medical movement from the school physics course. Let’s start with a definition.

Rotational movement of the body is the type of movement in which all points of the body move in circles, the centers of which lie on the same straight line - the axis of healing.

If the axis of healing is fixed, then the body as a whole does not move, and the only thing that changes over time is the angle of rotation, which here plays the same role as the movement of some point of the body during its forward movement.

When a body is rotated through an angle φ, each point passes through a certain arc of a circle. If we denote the length of the arc by S, then there is a simple relationship between the values ​​of S and φ

$$S = \varphi r,$$

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where $r$ is the distance of the point from the rotation axis. Points located at different distances $r$ from the axis of rotation describe arcs of different lengths, but the angle $\varphi$ is the same for the entire body.

The task of the mechanics of rotational movement is to be able to achieve a certain angle of rotation of the body at any time, if the initial angle of rotation is known. To do this, you need to know the rapid change in the angle of rotation, that is, the corner speed, which is usually denoted by the letter $\omega$. For a body rotating at a constant angular velocity,

$$\omega = \frac{\varphi - \varphi_0}{t},$$

where $\varphi - \varphi_0$ is the change in the rotation angle over a period of time of $t$ seconds.

If the angular velocity is constant, then the value of the angle of rotation at any time can be easily found:

$$\varphi = \varphi_0 + \omega t;$$

if the initial rotation angle $\varphi_0$ is zero, then

$$\varphi = \omega t.$$

This expression is similar to the formula $S = vt$ for uniform translational motion. Consequently, the angular velocity in rotational motion plays the same role as linear velocity $v$ for translational motion.

Angular velocity, like linear velocity, may not be constant. Then, in order to find the speed at any time, you need to know the angular acceleration, which, like linear acceleration $a$, is defined as a rapid change in angular velocity. In uniform rotation

$$\alpha = \frac{\omega - \omega_0}{t},$$

where $\omega$ and $\omega_0$ are the values ​​of angular velocity at the end and at the beginning of a long time period of $t$ seconds.

It is easy to find the connection between angular velocity and acceleration and their linear “twins” $v$ and $a$. Since $\omega = \frac{\varphi}{t}$ and $\varphi = \frac{S}{r}$, then $\omega = \frac{S}{rt}$.

But $\frac{S}{t} = v$ therefore

$$\omega = \frac{v}{r} \text{ и } v = \omega r.$$

In the same way, since $\alpha = \frac{\omega - \omega_0}{t}$, and $\omega = \frac{v}{r}$ and $\omega_0 = \frac{v_0}{r}$, then $\alpha = \frac{v - v_0}{rt}$. But $\frac{v - v_0}{t}$ is equal to the linear acceleration of $a$, so

$$\alpha = \frac{a}{r} \text{ и } a = \alpha r.$$

Thus, the numerical values ​​of the linear quantities $S$, $v$ and $a$ for each point of the rotating body are equal to the corresponding values ​​of the angular quantities (they relate not to individual points, but to the entire body), multiplied by the radius $r$ of rotation of this point.

If you use the quantities $\varphi$, $\omega$ and $\alpha$, then you can solve problems related to the rotation of bodies, using well-known formulas of kine-

matics of translational motion, but replacing linear quantities with angular ones. Let us present some formulas for the kinematics of translational motion and an analysis of these formulas for rotational motion:

| Forward movement | Rotational movement |
| :— | :— |
| $S = vt$ | uniform $\varphi = \omega t$ |
| $v = v_0 + at$ | uniformly accelerated $\omega = \omega_0 + \alpha t$ |
| $S = v_0 t + \frac{at^2}{2}$ | $\varphi = \omega_0 t + \frac{\alpha t^2}{2}$ |

Recall also that the rotation angle $\varphi$ varies in radians. Then the angular velocity changes in radians per second ($\dot{p}_{\text{rad}} / \text{сек}$). The unit of angular acceleration is radian divided by second squared ($\dot{p}_{\text{rad}} / \text{сек}^2$).

The description of the rotational motion, however, is incomplete if we know only the numerical values ​​of the above angular quantities. To fully describe rotation, you also need to know around which axis and in which direction (clockwise or counterclockwise) the body rotates. Both will be determined if we consider both the angle of rotation and angular velocity to be a kind of vector quantities. It is generally accepted that the vector $\varphi$, as well as the vector $\omega$, is directed along the rotation axis, so if a vector of one of these quantities is given, then the rotation axis is also given. For example, rotation is determined by the so-called gimlet-corkscrew rule or the screw rule (this is far from the only case in physics when one has to resort to the “services” of such a device). It is as follows: if you mentally rotate the hand of a gimlet in the same way as the body is rotated, then the direction of the translational movement of the gimlet will show the direction of the angular velocity vector $\omega$. If, for example, the red arrow on figure 1 represents the vector $\omega$ for a rotating disk, then this means that the disk is rotating as shown by the yellow arrow. As usual, the length of the vector on the depicted scale gives us the numerical value of the corresponding angular magnitude.

Angular acceleration is also a vector quantity. If the direction of the rotation axis in space does not change with time, then the vector $\alpha$, like the vector $\omega$, is directed along the rotation axis. In this case, if the angular velocity returns with time, then

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The direction of vector $\vec{\alpha}$ coincides with the direction of vector $\omega$, but if the angular velocity decreases, then the directions of vectors $\vec{\alpha}$ and $\omega$ are opposite.

Dynamics of rotational motion

The dynamics of translational motion studies, as is known, the reasons for the appearance of accelerations in bodies and makes it possible to determine their magnitudes and directions. Rotational dynamics do the same for angular acceleration. If the acceleration of the translational motion of a body is caused by a force applied to it, then the reason for the acceleration of the angular acceleration of the body is the torque of the force applied to it. Readers are familiar with it from the 8th or 9th grade physics course. Let us recall that the numerical value of the moment of force about any axis is equal to the product of the force applied to the body by the length of the perpendicular drawn from the axis to the line of action of the force. For example, if a force $F$ is applied to the rim of a flat disk, shown in figure 2, the vector of which lies in the plane of the disk, then the moment of this force relative to the axis perpendicular to the plane of the disk and passing through the point $O$ is equal to $Fr$. If the axis passed not through the point $O$, but through the point $O_1$, then the moment of the same force relative to the new axis would be different: it would be equal to $Fr_1$. If the same force is applied as shown in figure 3, then its moment relative to the axis passing through point $O$ is equal to zero, since the line of action of the force passes through this point. Such a force cannot cause rotation around an axis and cannot change angular acceleration. But the same force can cause rotation of the disk around an axis passing through point $O_1$.

The moment of force is also a vector quantity. And the direction of the force moment vector is determined using the same expression. Namely, we agreed to consider the direction of the moment of force to be that in which the gimlet (screw) moves along the axis if its hand rotates in the same way as a body would rotate under the influence of a force applied to it. For example, the vector moment of force $F$ relative to the axis passing through the point $O$, at figure 2 is directed perpendicular to the drawing plane “from us”.

As already indicated, Newton’s laws of mechanics are also valid for rotational motion. From what has just been said about the moment of force, it follows, for example, that Newton’s law, when applied to rotational motion, states that a body rotates with a constant angular velocity or does not rotate at all if the moment of the force acting on it is

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was neglected, and the ball is so small compared to the length of the rod that it can be considered a material point.

Now all our difficulties will immediately disappear. In the formula

$$M = mr^2\alpha$$

$m$ is the mass of the ball, and $r$ is the distance from the ball to the axis of rotation. Since for a given ball and rod neither $m$ nor $r$ change, the product $mr^2$ is a constant quantity and can be denoted by one letter. Let us choose the letter $I$ for this value. Then the formula of Newton’s second law takes the form

$$M = I\alpha.$$

The similarity between this formula and the formula is striking

$$F = ma.$$

Only the force $F$ is replaced by the moment of force, and together with the linear acceleration $a$ the formula includes the angular acceleration $\alpha$, as it should be, since the furnace is not about translational, but about rotational motion. The role of mass in this formula is played by a completely new, unusual quantity, equal to the product of the mass by the square of its distance to the axis of rotation.

Just as the acceleration of a translationally moving body depends, in addition to force, also on the mass of the body being accelerated, so the acceleration of a rotating body depends not only on the moment of force, but also on the value $mr^2$, which is, so to speak, the rotational “double” of the mass. This quantity is called the moment of inertia of the body relative to the axis. For a rotational movement, it is not only mass that is important. It is also important how far this mass is removed from the axis of rotation. It is clear that a small mass at a large distance from the axis of rotation can receive the same angular acceleration at a given moment of force as a large mass located near the axis.

All this applies as if to a ball rotating on a thin rod. But in reality, the formula turns out to be true even when a disk, wheel, cylinder, and in general any “real” solid body, which also has a certain moment of inertia, rotates. To obtain the moment of inertia of a body relative to each axis, you need to mentally divide it into many small parts (in the limit of infinitely small), for each of them determine the product of its mass by the square of the distance from this axis and add all the products. This will be the moment of inertia of the whole body. It is clear that the moment of inertia of a body is greater, the greater the mass of those parts of it that are located away from the axis. In other words, the moment of inertia of a body depends on how the mass of the body is distributed over its volume. For different axes the moment of inertia of the same body will be different. For example, the moment of inertia of a rod of length $l$ and mass $m$, relative to the axis perpendicular to the rod and passing through its middle (Fig. 6), is equal to $$\frac{ml^2}{12}$$. And the moment of inertia of the same rod is

, relative to the axis passing through the end (Fig. 7), is 4 times larger and equal to $\frac{ml^2}{3}$: it is more difficult to rotate the rod around the end than around the middle. “More difficult” means that to obtain the same velocity acceleration, a larger moment of force will be required.

Calculating the moment of inertia is usually a difficult task. But for bodies that are homogeneous and have a regular geometric shape, such a calculation with respect to some axes of rotation is possible, although not always in simple ways. For one special simple case we will also present the calculation.

Let us calculate the moment of inertia of the wheel, in which almost all the mass is concentrated in its rim, relative to the axis of rotation, by redistributing the plane of the wheel and passing through its center. An example is a weight distribution wheel with thin spokes.

Let us divide the rim into separate small parts with masses $m_1, m_2, m_3$, etc. Each of them can be considered a material point located at a distance $R$ from the rotation axis, where $R$ is the radius of the wheel. The moment of inertia of each point is equal to $m_1R^2, m_2R^2, m_3R^2$, etc. The total moment of inertia of the entire wheel is equal to the sum of the moments of inertia of its part:

$$I = (m_1 + m_2 + m_3 + \dots)R^2.$$

But the sum $m_1 + m_2 + m_3 + \dots$ is obviously equal to the mass of the entire wheel $m$. So the moment of inertia of the wheel about the axis passing through its center and redistributing its plane is equal to

$$I = mR^2.$$

For arbitrary shapes or heterogeneous in composition, or for relative axes of rotation that are not as “convenient” as in the case under consideration, the moments of inertia are determined experimentally, for example, by changing the angular acceleration at a known moment of force *). As we see, op-

*) The table shows formulas for the moments of inertia of some bodies.

| body | how does the axis go | moment of inertia |
| :— | :— | :— |
| homogeneous ball | through the center of the ball | $\frac{2}{5}mr^2$ |
| spherical shell | through the center of the sphere | $\frac{2}{3}mr^2$ |
| kernel | through the center of gravity | $\frac{1}{12}ml^2$ |
| kernel | through the end | $\frac{1}{3}ml^2$ |

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The determined masses of bodies are a much simpler task than determining its rotational “double”—the moment of inertia. If the mass is determined by simple weighing, then “weighing” for rotational motion is a rather complex operation, which, moreover, for the same body, but for different axes of rotation, must be carried out separately, and “scales” for such “weighing” do not exist.

Momentum, angular momentum

So, for rotational motion, Newton’s second law is written as $M = I\alpha$, where $M$ is the moment of force, $I$ is the moment of inertia and $\alpha$ is the angular acceleration of the body.

No angular acceleration is expressed by the formula

$$\alpha = \frac{\omega - \omega_0}{t}.$$

Therefore, the formula of Newton’s second law can be written as

$$M = \frac{I\omega - I\omega_0}{t}.$$

In this form, this formula means that the moment of force is equal to the change in value $I\omega$ unit time. But, as is known, Newton’s second law can be written in the form

$$F = \frac{mv - mv_0}{t},$$

where $mv$ is a quantity called the momentum or impullas of the body. It is clear that implulys $mv$ also has an analogue in the rotational motion diaphragm and this analogue is the value $I\omega$. This follows from the fact that the moment of inertia $I$ plays the same role in rotational motion as mass does in translational motion, and the angular velocity $\omega$ replaces the linear velocity $v$. The quantity $L = I\omega$ is therefore called the angular momentum of the body or the angular momentum. This is also a vector quantity (the same as $\omega$, $F$, $\alpha$, $mv$, $M$), directed in the same way as the angular velocity vector. Similar to the other values ​​we have given that relate to rotational motion, the momentum of the impullas is different relative to different axes of rotation.

If the body has such a shape or size or such a radius of rotation that we can talk about the rotation of a point (for example, a ball on a thin rod, a planet as it rotates around the Sun), then the moment of impullas is easy to associate with impullas. Indeed, the moment of inertia $I$ is equal to $mr^2$, and the angular velocity $\omega$ is equal to $\frac{v}{r}$. That’s why

$$L = I\omega = \frac{mr^2v}{r} = rmv.$$

The moment of impuls of a point, therefore, is equal in magnitude to the product of the radius $r$ of the circle along which the point moves, and the impuls $mv$. It is clear that for a body of arbitrary shape such a simple correspondence is impossible.

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moment of inertia of the body relative to the vertical axis of rotation (the mass of the arms is close to this axis). Since this decrease in the moment of inertia $I$ is achieved not by external, but by internal forces, the moment of momentum $I\omega$ cannot change. Therefore, the angular velocity $\omega$ increases by the same number of times as the figure skater managed to reduce her moment of inertia. On the contrary, before completing the rotation, when the speed needs to be reduced, the skater spreads her arms wide, increasing the moment of inertia of her body. Dancers do the same when performing pruets, acrobats when they perform somersaults, etc. Interesting experiments of this kind can be carried out using the so-called Zhukovsky skama. It is a platform that can rotate around a vertical axis with little effort (Fig. 9). By standing on such a rotating platform, you can, by manipulating your hands, noticeably change the angular speed of rotation.

Conservation of angular momentum means conservation of not only the numerical value of the quantity $I\omega$. We must not forget that angular momentum is a vector quantity. Therefore, conservation of angular momentum also means conservation of its direction. But the moment of impulse is directed, as we know, along the axis of rotation. Therefore, any rotating body that is not affected by external moments of forces cannot change the position of the rotation axis. This is the “secret” of the stability of a rolling wheel, tops, etc. This is why discus throwers, throwing their projectile, cause it to rotate. This allows the disc to fly without tumbling. The ability of a rotating body to maintain the position of its axis of rotation (in the absence of moments of external forces) is widely used in technology. The use of so-called gyroscopic compasses, various stabilizers, etc. is based on this property. A gyroscopic compass is essentially a rotating body. If, at launch, the entrance was installed so that the weld was oriented to the north, then it will maintain this direction of the best magnetic compass, which is influenced by both the iron hull of the ship and changes in the Earth’s magnetic field. Gyroscopic stabilizers are sometimes installed on the hulls, which are huge multi-ton wheels driven by powerful engines. In an effort to maintain the position of the rotation axis, such a stabilizer changes the quality of the ship. Gyroscopic stabilizers find important applications in space flight technology and in automatic devices for various purposes.

Kinetic energy of rotating bodies

A rotating body is a moving body, since all its parts, except the point located on the rotation axis, move. Therefore it has kinetic energy. To express the kinetic energy, let’s start again with the rotation of a small ball on a thin rod, that is, with the rotation of a point. In this case, the kinetic energy can be written in the general form

$$K = \frac{mv^2}{2},$$

where $m$ is the mass of the ball, and $v$ is its linear speed. But since $v = \omega r$, the expression for $K$ can be written in the form

$$K = \frac{m\omega^2r^2}{2}.$$

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In conclusion, we again present for comparison some of the mechanics of translational motion, and similar formulas relating to rotational motion.

| Forward movement | Rotational movement |
| :— | :— |
| $\vec{F} = ma$ | $\vec{M} = I\alpha$ |
| $\frac{\Delta(mv)}{\Delta t} = \vec{F}$ | $\frac{\Delta(I\omega)}{\Delta t} = \vec{M}$ |
| $\Delta(mv) = 0$ at $\vec{F} = 0$ | $\Delta(I\omega) = 0$ at $\vec{M} = 0$ |
| $A = FS$ | $A = M\varphi$ |
| $K = \frac{mv^2}{2}$ | $K = \frac{I\omega^2}{2}$ |

Assignment and questions

1. In what units is the moment of inertia measured?
2. Two cylinders of identical diameters and masses roll down an inclined plane without piercing. One of them is lead and has a cork stop, the other is cork with a lead shell. Which cylinder will roll off the plane in less time?
3. A wheel whose moment of inertia is 0.01 SI units is expressed with an angular velocity of 4 rps. What must be the friction torque for the wheel to stop rotating after 5 revolutions?
4. Tides in the seas and oceans act on the rotating Earth like brake pads and create a braking friction force. Because of this action of the tides, in 100 years the length of a day on Earth will be 1 second longer than the next day. Calculate the tidal friction force.
5. A ship’s gyroscopic stabilizer with a radius of $R = 2$ m and a mass of $M = 120$ m is driven by a motor. What power is required to bring the stabilizer rotation speed to 13 rpm within 10 hours? Assume that the mass of the stabilizer is concentrated on its rim (in this case, the moment of inertia is equal to $MR^2$).
6. A homogeneous ball of radius $R$ and mass $M$ is initially launched along a plane so that it slides along it without rolling with a speed $v_0$. There is friction between the ball and the plane, the coefficient of which is $k$. How far will the ball travel before it stops piercing relative to the plane? What will be the speed of the ball at this moment?
7. Two windows of the spherical satellite are located at diametrically opposed points of the shell. The astronaut walks from the center of the ship along a radial ladder to one of the windows, then walks along the side of the ship to the second window, and finally returns along the radial ladder to his place in the center of the ship. At what angle does the ship rotate relative to the stars, if the mass of the ship is $M$, its radius is $R$, the mass of the astronaut is $m$ and we can assume that the entire mass of the ship is concentrated in its shell.
8. To replace the rotation of satellites, it is proposed to use the following devices. Two weights of mass $m$ each are attached to the side surface of the satellite on threads of length $l$. The loads are released simultaneously. They fly away from the satellite and at the moment when the threads are stretched, they follow from the hooks to which they are attached. At what length of the thread will the rotation speed of the satellite after this operation decrease by $n$ times?
   Assume that the satellite is a cylinder of mass $M$ and radius $R$; all the mass of the ship is concentrated in the side shell.

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Probably every reader «Kvant» is familiar with the great excitement before a major sporting competition - be it the World Cup in football or the “match of the century” in chess. Further, from those who do not quite accurately remember the castling rules, you can hear thoughtful substitutions like “I would put Porttish, not Larsen, on the first board of the world team.” Chess experts and fans argue about the chances of victory between Smyslov and Keres, and football fans almost come to blows when discussing the possible outcome of the match between the national teams of Brazil and England.

Very often, before such competitions, newspapers announce a knowledge competition, offering to guess both the outcome of the competition as a whole (predict who will get the “Golden Goddess”, or the score with which the USSR chess team will defeat the world team), and the outcomes of individual compositions. At first glance, there is a very simple way to win such a competition: send as many letters to the editor as there are theoretically possible outcomes of the competition, and in each letter give one of the options. At least one letter will hit the target!

But the matter is not at all that simple - there are too many options. Of course, if you only need to predict the score with which the match of the century will end, then it is enough to send 81 letters: 40 games were played, and the USSR national team could score any number of points from 0 (brrr…) to 40 (that would be great!). But if you also need to indicate the outcome of the match on each board (at least predict who will win it), then the number of possible options grows κατάstrophically.

Problems in which it is necessary to find the number of possible options for a particular operation, a particular event, arise in a variety of areas of human activity. These problems are called combinatorics now in school and are not taught, it is studied as an elective).

1. Sum and product rules. The basis for solving combinatorial problems are two simple rules - the correct sum and the correct product.

Sum rule. If we have $m$ ways to choose element $a$ and (independently) $n$ ways to choose element $b$, then the choice “either $a$ or $b$” can be made in $m+n$ ways.

For example, if there are 5 apples and 9 pears on a plate, then the choice “apple or pear” can be made in 14 ways - choose either one of 5 apples or one of 9 pears.

The correct work is somewhat more difficult. It deals with the selection of a pair of elements $(a, b)^*$.

Product rule. If we have at our disposal $m$ ways to select element $a$ and $n$ ways to select element $b$, then the pair $(a,b)$ can be selected in $mn$ ways.

Thus, if there are 5 apples and 9 pears on a plate, then the pair (apple, pear) can be selected in $5\cdot 9=45$ ways.

*) Two pairs obtained from each other by rearranging elements are considered different, for example, the pair (1, 4) is considered not to coincide with the pair (4,1).

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To verify the validity of the product rule, we denote all ways of choosing the element $a$ by $a_1, \dots, a_m$, and all ways of choosing the element $b$ by $b_1, \dots, b_n$. Then all the pairs that can be made from these elements are arranged in the following table of $m$ rows and $n$ columns:

$$(a_1, b_1), \dots, (a_1, b_n),$$
$$(a_m, b_1), \dots, (a_m, b_n).$$

In this table there are just $mn$ pars.

Note that sometimes the choices are more complex - you must first select the element $a$, and then, depending on this choice, the element $b$. But if the element $a$ can be chosen in $m$ ways, and after each can be chosen in $n$ ways, then the number of distinct pairs $(a, b)$ is again $mn$. Prove it yourself.

Of course, it may be that we need to make not two, but several choices. In this case, you need to run through the number of options for each choice. In other words, it follows the rules correctly.

If the element $a_1$ can be chosen in $n_1$ ways, the element $a_2$ in $n_2$ ways, the element $a_k$ in $n_k$ ways, then the set $(a_1, \dots, a_k)$ can be chosen in $n_1 \dots n_k$ ways (in this case, sets that are distinct from each other only by the order of the elements are considered different).

Now we can already solve the problem of the number of possible outcomes of the match of the century. Each board could have 3 outcomes: win, draw and defeat. Since the number of the board is 10, there could be $3^{10} = 59049$ different outcomes. And if you need to imagine not only the outcome of each winner, but also his score, then the answer will be $9^{10}$.

Using product rules, you can also solve quite complex problems in which additional conditions are imposed on the choice of options.

Example. There are 33 different letters in the Russian alphabet. How many words containing 5 letters can you form if you do not allow words in which two identical letters appear in a row?

Of course, “words” here have a different meaning than in philological terms - such “words” as abbed or ababa are allowed. But words like quarrel, press, aachen are not allowed - they contain two identical letters in a row.

To solve this example, we will select the letters one by one - first the first, then the second, the fourth, the fifth, and then choose in 33 ways - after all, we have at our disposal all 33 letters of the Russian alphabet. But the second letter can be chosen in only 32 ways - after all, it must be different from the first. You can choose the third letter in 32 ways - it must be different from the second; The fourth and fifth letters can also be chosen in 32 ways. And in total we get $33 \cdot 32 \cdot 32 \cdot 32 = 34603008$ different words that satisfy the given condition.

2. k-subsets. Using the sum and product rules, you can solve various combinatorics problems. But this is not always convenient, just as solving any geometric problem with axioms is not always convenient. Along with the axioms, we use the geometric theorem, which allows us to shorten the solution to the problem. In the same way, in combinatorics there are several simple, “standard” problems, to which it is often possible to reduce solutions to other problems.

These problems are formulated in the language of set theory. If the reader is not familiar with what a “set”, “element of a set”, “intersection of a set”, “union of a set” is, we refer him to the literature, a list of which is given at the end of the article. However, we will use only the simplest concepts of set theory, which will now be studied in the 4th and 5th grades of high school, and as an example we will take a set of four elements and select its subsets and words made up of the elements of this set (Fig. 1-5).

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Fig. 1. Let the set $N$ consist of four elements: knight, bishop, queen and rook.

Let’s solve the following problem. 5 students take the swimming test. The test is passed if the student swims 100 meters (any time). If the student comes to catch him, then the test is not passed. In how many ways can a swim end?

Various cases are possible. It may be assumed that all students will be good swimmers, or it may be that all five will have to be rescued. The total number of outcomes can be calculated like this. Let’s arrange all the students in some order, say alphabetically: Andreev, Borisov, Voldin, Grigoriev, Dmitriev. For each of them there are two possibilities - either he passes the test or he doesn’t. In the first case, we will assign the number 1 to this student, and in the second case, the number 0. Then the outcome of the game will be expressed as a sequence of length 5 from zeros and ones. For example, the sequence 0, 1, 1, 0, 1 means that only Borisov, Voldin and Dmitriev swim well enough. And if only Grigoriev takes the test, then the report on the outcome of the competition will look like this: 0, 0, 0, 1, 0. The sequence 0, 0, 0, 0 means that not a single student reached the finish line.

Thus, the problem of the number of outcomes of a game has been reduced to the following: how many sequences of length 5 can be counted from the numbers 0 and 1? The answer to this problem is given by the correct product: since at each place in the sequence we have a choice of two possibilities, then the total number of outcomes is $2 \cdot 2 \cdot 2 \cdot 2 = 32$. So, there are 32 possible outcomes of the swim.

In the analyzed problem, we needed to find out how many subsets can be counted from the set $N$: {Andreev, Borisov, Voldin, Grigoriev, Dmitriev}.

In this case, an empty subset was allowed, as well as a subset coinciding with the entire set $N$. The answer turned out to be $32 = 2^5$. In the same way it is proved that if a set $N$ contains $n$ elements, then it has $2^n$ subsets. The same result can be proven using mathematical induction.

In the Olympic Games they do it this way: the top two qualify for the finals. Let’s find out how many outcomes the competition can have in this case. Mathematically, the problem is formulated as follows: in how many ways can you select 2 elements from a set containing 5 elements? It is easy to check that this can be done in 10 ways (write down all these methods yourself). But this “direct search” method is only suitable for a small number of participants. I would like to have a formula that allows me to immediately answer the question. So, we need to solve the following problem.

How many $k$-subsets are there in the set $N$ of $n$ elements?

Here we say “$k$-subset” together with “a subset containing $k$ elements” for brevity.

The choice of a subset from a given set $N$ can be found in the following way. Let’s put the elements of the set $N$ into a bag, and then put our hand into this bag and pull out a subset. Of course, all $k$ elements of the subset are pulled out at once and no reasonable order can be established for them.

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Fig. 5. There are $P_4=4!=24$ permutations of elements of the set $N$.

It is necessary to pull out from it one by one the elements of this set. But this time we will not return these elements to the bag, but will place them in a row. After $k$ selections we will again have the $k$-word $(a_1, \dots, a_n)$, but this time there are no duplicates among the elements $a_1, \dots, a_n$ *). In another way, we can say that $k$-words without repetitions are ordered $k$-subsets of the set $N$.

The number of $k$-words without repetitions of $n$ elements is denoted by $A_n^k$. It immediately follows from the product rule that

$$A_n^k=n(n-1)\dots(n-k+1).$$

In fact, the first element can be selected in $n$ ways, the second - only in $n-1$ ways, because you cannot take an already selected element. The third element can be selected in $n-2$ ways, …, $k-1-(n-k+1)$ ways.

*) Such $k$-words are called placements without repetitions.

This means that the total number of real ways is expressed by formula (3).

For example, if out of 10 commission members it is necessary to appoint a representative, his substitute and a secretary, then this can be done in $10\cdot 9\cdot 8\cdot 720$ ways.

5. Rearrangements. If the elements of the set $N$ are pulled out from the bag and laid out in a row one after another (and not a single element is returned back), then in the end all the elements of the set $N$ turn out to be located in some order. In this case we speak of permutations without repetitions of $n$ elements. Thus, a permutation of $n$ elements is a $k$-word without repetitions of $n$ elements. Polyag in formula (3) $k=n$, we find that the number of permutations of $n$ elements is $n!$. It is represented by the symbol $P_n$. So,

$$P_n=n!$$

6. Permutations with repetitions. We already know that from $n$ “letters” we can

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This article talks about some of the properties of the ellipse and the different laws of planetary motion, the most important of which are called Kepler’s laws.

Gravity
The law of inertia now seems to be the most elementary and natural law - the study of mechanics will begin with it. However, in order to understand it, it was necessary to distract ourselves from what was happening directly before our eyes. Under normal conditions, the movement of the body in the horizontal plane is hampered by friction and air resistance; When moving in a vertical plane, the forces of air resistance and gravity act on the body. It was necessary to get used to their influence, at least mentally. It took many years.

Only after the law of inertia—the law of body motion under idealized conditions—was understood, could it be possible to consider the movement of bodies under the action of forces.

The law of falling bodies on Earth was established by Galileo (1564-1642). He was the first to set himself the task of understanding how the speed of a falling body changes in successive periods of time. Thus, he introduced acceleration into mechanics and established that the accelerations of all bodies falling to the ground are identical and constant. But the fact that the cause of acceleration is gravity was only later understood by Newton.

Almost this years after the publication of Galileo’s work, Newton discovered the law of universal gravitation (1687), according to which the force of gravity is not constant, but decreases with increasing distance between bodies as the square of this distance.

If physics had historically developed in the order in which it is presented in textbooks, then only after Newton’s discovery would it be possible to move on to the study of the laws of planetary motion. But the development of science does not follow the logic of the textbook. Much earlier than Newton, these laws were established by Galileo’s contemporary Johannes Kepler (1571-1630).

Kepler’s meanings in mechanics were extremely meager in comparison with those of a modern schoolchild. Not only did he not know Newton’s laws (they had not yet been discovered), but he also had a very vague idea of ​​what force was. All that Kepler had in his hands were very neat ancient observations of the movements of celestial bodies, which his teacher Tycho Brahe had conducted for many years *) and he himself continued to conduct. Only extraordinary hard work and confidence in the existence of simple laws governing the movement of the planets allowed him to discover the laws that have since bear his name.

Kepler’s laws are not so strong even now. This is mainly due to the fact that ellipses - the curves along which the planets move - are poorly understood by the average reader. In reality, obtaining Kepler’s laws is not so difficult. To do this, you need to use two laws of mechanics: the law of conservation of energy (which was not known to either Kepler or Newton) and the law of conservation of angular momentum. The latter, as we will see, is equivalent to Kepler’s second law: the areas slowed down by the planet’s radius vector over equal periods of time are the same.

*) Tycho Brahe was one of the last astronomers to observe the movement of celestial bodies with the naked eye.

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Law of Conservation of Energy

On the surface of the Earth the force of gravity is $mg$. If we introduce the potential energy of a body into the gravitational field $U = mgh$ (h is the distance from the body to the Earth), then we can say that when the body moves, the sum of its kinetic and potential energy $E = \frac{mv^2}{2} + mgh$ does not change.

However, the expression for $U$ that we wrote is valid only at small distances from the body to the Earth, when $h$ is small compared to the radius of the Earth and gravity can be considered constant. If the height becomes greater, then the force of gravity decreases with distance. According to Newton’s law

$$|F| = \gamma \frac{mM}{R^2},$$

where $\gamma$ is the gravitational constant, $M$ is the mass of the Earth and $R$ is the distance from the body to the center of the Earth.

If we use the exact expression for gravity, then the potential energy must be replaced by $U = -\gamma \frac{mM}{R}$. Let’s check it out. Let us remember that the work done by a body during movement is equal to the decrease.

Fig. 2.

Fig. 3.

reducing its potential energy. We calculate the work done by the body feeding from the height $R_2$ to the height $R_1$ from the center of the Earth (Fig. 2). To do this, you need to calculate the difference in potential energies of the body:

$$\text{работа} = U_2 - U_1 =$$

$$= \gamma m M \left( -\frac{1}{R_2} + \frac{1}{R_1} \right) =$$

$$= \gamma m M \left( \frac{R_2 - R_1}{R_1 R_2} \right).$$

If $R_2 - R_1 \ll R_1$ ($R_2 \approx R_1$), then the product $R_1 R_2$ can be replaced by $R^2$ - the square of the geometric mean $R_1$ and $R_2$ (for $R_2 \approx R_1$: $R \approx R_2 \approx R_1$), and $R_2 - R_1$ can be replaced by $l$ - the path traversed by the body. Then the last formula takes the form

$$\text{работа} = \text{сила} \times \text{путь} = \gamma \frac{mM}{R^2} l.$$

force path of gravity

This means that the expression for $U$, which we wrote above, is the real potential energy of the body.

Now we need to write the formula for the total energy of the body differently:

$$E = \frac{mv^2}{2} - \gamma \frac{mM}{R} = m \left( \frac{v^2}{2} - \gamma \frac{M}{R} \right).$$

Ellipse

Let’s start with the general definition of an ellipse (Fig. 3). An ellipse is the geometric locus of lobes, the sum of the distances of which to two lobes (the foci of the ellipse) is constructed.

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Fig. 4.

From this definition we can deduce an interesting “optical” property of the ellipse: if a light source is placed at one of the focuses of the mirror ellipse ), then all the rays coming out of it will be collected at another focus of the ellipse*). This means that the angles between the straight lines going to the point of the ellipse from its foci and the tangent to the ellipse are equal to each other (Fig. 4).

Let us prove that the ellipse has one more property, which is even convenient to take as the definition of an ellipse. To do this, we draw a tangent $t$ to the ellipse at point $A$ and lower the redistributions from both foci of the ellipse to this tangent (Fig. 5). We extend one of the redistributions beyond the straight line $t$ to a distance $F'_2 N_2$ equal to $N_2 F_2$. Let us now connect the points $F'_2$ and $F_1$ with the point $A$. Let us prove that $F'_2 AF_1$ is a straight line.

This is immediately clear from the fact that $\alpha_1 = \alpha_2$ (by construction) and $\alpha_1 = \alpha_3$ (according to the optical property of the ellipse). Therefore, $\alpha_2 = \alpha_3$, and therefore $F'_2 AF_1$, is a straight line.

*) An ellipside is the geometric locus of a point, the sum of the distances from which to two fixed points is constructed. Such a surface is obtained by growing an ellipse around its axis.

**) The easiest way to prove this property of an ellipse is to use Fermat’s principle (see «Kvant» No. 11, 1970).

Fig. 5.

Further, since $AF'_2 = AF_2$, then $F_1 F'_2 = F_1 A + AF'_2 = F_1 A + AF_2$, that is, the segment $F_1 F'_2$ is equal to the sum of the distances from the ellipse point to the foci. Let’s denote this sum $2a$. Let us now lower the redistribution $F_1 B$ onto the line $F'_2 F'_2$, calculate the length of the leg $F_1 B$ from two right triangles - brown and yellow - and equate both expressions for $F_1 B$ to each other:

$(2a)^2 - (\rho_1 + \rho_2)^2 = (2c)^2 - (\rho_2 - \rho_1)^2$.

Opening the brackets, we get

$\rho_1 \rho_2 = a^2 - c^2$.

So, the product $\rho_1 \rho_2$ does not depend on the point at which the tangent is drawn. Let’s draw it at the end of the semi-minor axis (Fig. 6). In this case $a^2 - b^2 = c^2$. Therefore $\rho_1 \rho_2 = b^2$.

Thus, we got the main theorem:

The product of the distances from two foci to any tangent of the ellipse is a constant value equal to the square of the semi-minor axis.

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Kepler’s first law

Now we can consider the movement of the planet. For simplicity, we will not prove that the planet’s orbit is an ellipse, but, assuming this and using the laws of conservation of the moment of motion and energy, we will find it and make sure that the movement along ellipses in a gravitational field does not contradict the laws of conservation.

Let us note two positions of the planet $A$ and $A'$, symmetrical relative to the center of the ellipse (Fig. 7). We can say it another way: point $A$ is the true position of the planet at some point in time, and $A'$ is obtained from the first by transferring the Sun to another focus, while the planet remains in the same place.

Let’s write down the law of conservation of energy. Marking the quantities with index 1 for point $A$ and index 2 for point $A'$, we obtain $(E_1 = E_2 = E)^*$.

$$-E = \frac{mv_1^2}{2} - \gamma \frac{mM}{r_1} = \frac{mv_2^2}{2} - \gamma \frac{mM}{r_2}.$$

The angular momentum of the planet is equal to the product of the amount of motion $mv_1$ (or $mv_2$) and the “play” - the impact parameter $\rho_1$ (or $\rho_2$), while the total energy is $T + U = -E < 0$.

How is the angular momentum for both positions of the planet the same:
$$L = mv_1\rho_1 = mv_2\rho_2.$$
Let us calculate the product of the kinetic energies of the planet at points $A$ and $A'$:
$$T_1T_2 = \frac{mv_1^2}{2} \cdot \frac{mv_2^2}{2} = \left(-E + \frac{\alpha}{r_1}\right) \times \left(-E + \frac{\alpha}{r_2}\right).$$
(Here $\alpha = \gamma Mm$.)
Replacing the velocities in this equation with their expressions through the moment of quantity of motion, after simple transformations we obtain:
$$\frac{1}{4} \frac{L^4}{m^2\rho_1^2\rho_2^2} - E^2 = \frac{-\alpha(r_1 + r_2)E + \alpha^2}{r_1r_2}.$$

It is easy to see that this equality holds if the planet’s orbit is an ellipse. Indeed, for this you need to put
$$r_1 + r_2 = \frac{\alpha}{E} = 2a$$
And
$$\rho_1\rho_2 = \frac{L^2}{2mE} = b^2.$$

Both of these equalities are nothing more than the two definitions of an ellipse given above.

Thus, our trajectory is an ellipse with semi-axes: large
$$a = \gamma \frac{mM}{2E} \text{ и малой } b = \frac{L}{\sqrt{2mE}}.$$

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Fig. 8.

Then the speed of the planet will be dimensionless. Speed ​​is a number that shows how many times $v_0$ is contained in the speed of the planet:

$$\beta_1 = \frac{v_1}{v_0} \quad \text{и} \quad \beta_2 = \frac{v_2}{v_0}.$$

According to the previous speeds $\beta_1$ and $\beta_2$ are related by the relation

$$\beta_1 = \frac{1}{\beta_2}.$$

What does it mean? If a particle moves at some point in the orbit with a speed of $\beta$, then at a point symmetric relative to the center of the orbit, its speed will be equal to $\frac{1}{\beta}$

We described the trajectory of the planet, specifying its position (coordinates) at each moment in time. You can do it differently - set the speed of the planet. The curve that we get in this case is called the velocity graph (we can say that this is a trajectory in velocity space). Since the product of the segments $\beta_1$ and $\beta_2$ has been constructed, the speed graph of the planet is a circle (Fig. 8) (remember the theorem: “A point is given inside a circle. Then for all chords passing through this point, the product of the lengths of the segments on which this point divides the chords are the same”).

If we replace the speed $\beta$ with $\frac{1}{\beta}$ at each point of the godgraph, then the godgraph will not change, but simply the surface speed at $180^\circ$. This property is called relative surface motion invariance.

If the planet’s trajectory is circular, then $\beta = 1$ (prove it!). In this case, the godgraph is a circle with its center at the origin.

Circulation period

In order to calculate the orbital period - the time of a complete revolution of a planet around Coln or an artificial satellite around the Earth, it is necessary to calculate the time of passage of each section of the trajectory (separately, since the speed changes all the time) and then sum it over all sections. Of course, this cannot be done simply, but all difficulties can be circumvented.

From Kepler’s second law we know that in time $\Delta t$ the radius vector of the planet “sweeps” an area equal to $\frac{L}{2m} \Delta t$ (recall that $L = \text{const}$). This means that the period of revolution can be calculated by dividing the area of ​​the ellipse by the constant speed of “sweeping”

$$T = \frac{S_{\text{эллипса}}}{L/2m}.$$

Since the ellipse is obtained from a circle sloping in one direction $\frac{b}{a}$ times$^*$), then the area of ​​the ellipse is equal to

$$S_{\text{эллипса}} = S_{\text{крута}} \cdot \frac{b}{a} = \pi a^2 \frac{b}{a} = \pi ab.$$

From these formulas we get

$$T = \frac{2\pi ab}{L}.$$

But $b = \frac{L}{\sqrt{2Em}}$, therefore

$$T = \pi a \sqrt{\frac{2m}{E}}$$

$^*$) See the article by I. N. Bronstein “Ellipse” («Kvant» No. 9, 1970).

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or, substituting here the expressions for $a \left( a = \gamma \frac{Mm}{2E} \right)$ and replacing $\left( \frac{2E}{m} \right)^{3/2}$ with $v_0^3$, we get
$$T = 2\gamma M \frac{m}{2E} \sqrt{\frac{m}{2E}} = 2 \frac{\pi \gamma M}{v_0^3}.$$
From this formula it is clear that the period of revolution depends only on the value $v_0$. If several identical satellites are released from the same point above the Earth’s surface in different directions, but with the same speed $v_0$, then all of them will simultaneously (after a time $T$) converge at the same point.

Kepler’s third law
If now in the formula (*) for the period of revolution of the planet together the energy $E$ substitute her expression with a larger one $\left( E = \gamma \frac{mM}{2a} \right)$, then we get
$$T = 2\pi a^{3/2} \frac{1}{\sqrt{\gamma M}}.$$
Therefore, $T^2$ is proportional to $a^3$, that is, the square of the planet’s orbital period is proportional to the cube of the large half-ellipse. And this is Kepler’s third law.

Fig. 9.
An ellipse is the geometric locus of the centers of a circle passing through the focus and touching a given circle (guide circle).

In this article we talked a little about geometry, optics and mechanics. Now solve the problems.

Optical problem
In one of the focuses of the ellipson, the inner surface of which is mirrored, there is an exact light source. Construct an image of this source for an observer located at the second focus of the ellipson.

Geometric problem
Show that the ellipse is the locus of the centers of a circle passing through the focus and touching a given circle (guide circle) (Fig. 9).

At figure it is clear that the bases of the perpendiculars $\rho_1$ and $\rho_2$ lie on the “principal circle”. Are you listening to this?

Space challenge
The orbit of the Earth’s satellite has the following parameters: greatest distance $R$, shortest distance $r$. Calculate the total energy and angular momentum of the satellite. Find the speed of the satellite at the ends of the major and minor axes.

Beautiful formation
The drawing on the cover of the magazine is constructed like this: a point is selected inside a circle, and a circle smaller than it is drawn from it, as from the center. Rays are drawn from the center of this second circle until they meet the larger circle. Through a point on each ray lying in the middle between two circles, a chord is drawn perpendicular to this ray. The spot inside has the shape of an ellipse.

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AUXILIARY CIRCLE

The solution to many geometric problems begins with drawing auxiliary lines that help establish a connection between data and unknown elements of the figure. Recalling a successful auxiliary formation is often difficult. Therefore, you should approach each solved problem and try to find out why certain auxiliary lines lead to the goal. Is it possible to use the found example to solve some other problems?

At the next meeting of the mathematical circle, we will consider one of the interesting examples of solving geometric problems, which consists of introducing an auxiliary circle into the drawing.

Example 1. The altitudes AP, BQ and CR are drawn in an acute triangle. Prove that $\angle ABQ = \angle APR$).

Solved and e. Let H be the point of intersection of the altitudes of triangle ABC (Fig. 1). Since $\angle APB$ and $\angle CRB$ are straight lines, a circle can be described around the quadrilateral BPHR by taking BH as the diameter. Having constructed it, we replace that $\angle ABQ = \angle APR$ (as inscribed angles based on the same arc).

*) See «Kvant» No. 1, 1970, p. 59.

Fig. 1.

Thus, constructing an auxiliary circle allows you to use the inscribed angles theorem and thereby establish a connection between the angles specified in the task.

Find out how the result changes if $\angle B$ of triangle ABC is obtuse.

Solve several similar problems yourself.

1. From an arbitrary point M ka-tega BC of right triangle ABC, a perpendicular MN is dropped onto the hypotenuse AB. Prove that $\angle MAN = \angle MCN$.

2. Prove that the line connecting the vertex of a right angle of a right triangle with the center of a square externally constructed on the hypotenuse bisects the right angle of the triangle.

3. From an arbitrary point M inside a given angle, perpendiculars MP and MQ are dropped to the sides of the angle. From the vertex of angle A a perpendicular AK’ is dropped onto the segment PQ. Prove that $\angle KAP = \angle MAQ$.

Example 2. Prove that a segment connecting the bases of two altitudes of an acute triangle cuts off from it a triangle similar to the given one.

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Fig. 2.

Solved and e. Let $AA_1$ and $BB_1$ be the altitudes of the acute triangle $ABC$ (Fig. 2). $\angle AA_1B$ and $\angle AB_1B$ are straight, so a circle constructed on the $AB$ side of the triangle as a diameter will pass through the points $A_1$ and $B_1$. Further, $\angle ABC = 180^\circ$ - $\angle AB_1A_1$, $\angle A_1B_1C = 180^\circ$ - $\angle AB_1A_1$, therefore $\angle ABC = \angle A_1B_1C$, and triangles $ABC$ and $A_1B_1C$ are similar.

This problem can be solved without auxiliary constructions, if we notice that the triangles $AA_1C$ and $BB_1C$ are similar and write down the proportion

$$\frac{A_1C}{AC} = \frac{B_1C}{BC}.$$

Solve each of the following two problems in ways to work around them.

4. From the vertex $A$ of the parallelogram $ABCD$ the heights $AM$ and $AN$ are drawn. Prove that triangles $AMN$ and $ABC$ are similar.

5. From the base $H$ the heights $CH$ of the triangle $ABC$ are lowered to the sides $AC$ and $BC$ perpendiculares $HM$ and $HN$. Prove that triangle $CMN$ is similar to triangle $ABC$.

Sometimes it is advantageous to describe a circle and a triangle.

Example 3. Prove that the square bicsectrix of a triangle is equal to the difference between the product of its enclosing sides and the product

segments of the third side into which it is divided by a bicector.

Solved and e. We describe a circle around the triangle $ABC$ and extend the bicsectrix $CD$ of the triangle until it meets the circle at the point $E$ (Fig. 3). Let $BC = a$, $AC = b$, $AD = m$, $BD = n$, $CD = l$, $DE = x$.

According to the conditions, $\angle ACE = \angle BCE$, in addition, $\angle AEC = \angle ABC$, are like inscribed angles based on the same arc. Consequently, the triangles $ACE$ and $BCD$ are similar and the equality $$\frac{l + x}{b} = \frac{a}{l},$$ is true, hence $l^2 = ab - lx$. Chords $AB$ and $CE$ intersect at point $D$. Therefore the equality $lx = mn$ holds. Therefore, $l^2 = ab - mn$.

Example 4. The altitude and median of a triangle drawn from one vertex inside it are different and form equal angles with the sides emerging from the same vertex. Prove that the triangle is right-angled.

Solved and e. Let the height $CH$ and the median $CM$ of the triangle $ABC$ oblige the sides $AC$ and $BC$ to be equal angles (Fig. 4). Let us describe a circle around the triangle $ABC$ and extend the median $CM$ until it meets the circle at point $D$. Consider triangles $ACH$ and $BCD$. Since $\angle ACH = \angle BCM$ is by conditions and $\angle A = \angle D$,

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Fig. 4.

as inscribed angles based on the same arc, then $\angle AHC = \angle CBD = 90^\circ$. Therefore, $CD$ is the diameter of the circle.

The center of the circle lies on the diameter $CD$ and on the perpendicular $m$ to the side $AB$ at its midpoint $M$. Since the median $CM$ is not a height, the lines $CD$ and $m$ have only one common point $M$, which is the center of the circumscribed circle. Therefore, $AB$ is the diameter of the circle and $\angle ACB = 90^\circ$.

Use the result to solve the next two problems.

6. Determine the angles of the triangle in which
   a) the median and height drawn from one vertex divide the angle into three equal parts;
   b) the median, bisector and altitude drawn from one vertex divide the angle at this vertex into four equal parts.
7. Construct a triangle, knowing the angles formed by the median, bisector and altitude emerging from one vertex.
   Example 5. Construct an equidistant triangle so that its vertices lie respectively on three given parallel lines.

Fig. 5.

There are several known solutions to this interesting problem using expressions around a point, similarity and algebraic methods. A very short solution is obtained by constructing an auxiliary circle.

Solution. Let $ABC$ be the original equidistant triangle whose vertices lie respectively on the parallel lines $a, b$ and $c$ (Fig. 5). The circle circumscribed around it intersects the line $c$, in addition to the point $C$, at some point $D$. Then, based on the property of inscribed angles, we have $\angle ADC = \angle ABC = 60^\circ$.

The following construction follows from this. From an arbitrary point $D$ of line $c$ we draw rays that form angles $60^\circ$ and $120^\circ$ with line $c$. Let one of the rays intersect line $a$ at point $A$ and line $b$ at point $B_1$, and the other ray intersect line $b$ at point $B$. Then the segment $AB$ will be the side of the triangle $ABC$.

Let us plot on the line $c$ the segment $DC$ equal to the segment $AB_1$, so that the points $C$ and $A$ lie on the same side of the line $BD$. Then the point $C$ is a triangle of triangle $ABC$, since triangles $ABB_1$ and $BCD$ are equal, which means that $AB = BC$ and $\angle ABC = 60^\circ$.

Thus, the construction technique

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Fig. 6.

It is advisable to use the construction of an auxiliary circle when solving some construction problems.

Let’s consider another difficult problem, in which the most difficult thing to solve is to guess how to describe a circle around the squares.

Example 6. On sides AC and BC of triangle ABC, squares ACA₁A₂ and BCB₁B₂ are constructed outside it. Prove that lines AB₁, A₁B and A₂B₂ intersect at one point, and determine the angles between these lines.

Solution. If ∠ACB = 90°, then the validity of the theorem is obvious. Lines AB₁ and A₁B are redefined and each of them intersects line A₂B₂ at point C at an angle of 45° (Fig. 6).

If ∠ACB ≠ 90°, then the circles circumscribed about the squares have, in addition to the point C, another common point C₁, which turns out to be the intersection point of the lines AB₁, A₁B and A₂B₂ (Fig. 7). Indeed, ∠A₂C₁C = 90°, like a hanging one, resting on a semicircle. Similarly ∠CC₁B₂ = 90°. Therefore, ∠A₂C₁C + ∠CC₁B₂ = 180°, i.e. rays C₁A₂ and C₁B₂ form a straight line.

In the same way, it is proved that the point C₁ lies on the lines AB₁ and A₁B.

Considering the obtained inscribed angles, we find that the straight lines AB₁ and A₁B are mutually redefined and each of these straight lines makes an angle of 45° with the straight line A₂B₂.

Fig. 7.

The next task has a lot in common with the previous one.

8. On the sides AB, BC and CA of an arbitrary triangle ABC, equilateral triangles ABC₁, BCA₁ and CAB₁ are constructed outside it. Prove that lines AA₁, BB₁, CC₁ intersect at one point and each of them forms an angle of 60° with the other.

9. Prove that if two opposite angles of a quadrilateral are obtuse, then the diagonal connecting the vertices of these angles is shorter than the second diagonal.

Let us present two more problems involving maximum and minimum, for solving which it is also advisable to construct an auxiliary circle in a certain way.

10. On one side of the angle with the vertex at point S are the given points A and B. Find on the other side the point C from which the segment AB is visible at the most angle.

11. From an arbitrary point M lying on side AB of triangle ABC (or on its extension), perpendiculars MP and MQ are drawn to sides AC and BC. Under what conditions will the segment PQ be the longest?

Thus, various geometric problems are solved using the same technique. The construction of an auxiliary circle allows you to increase the number of theorems that can be used when solving a problem, and thanks to this, you can find dependencies between the elements of the figure.

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There is no more universal phenomenon in nature than the propagation of waves. Waves on the surface of water, sound, light, radio, transfer of deformation from one part of a solid body to another - all these are wave processes. Quantum mechanics shows that the movement of microparticles is controlled by wave laws. All these waves have different physical natures, different propagation speeds, different frequencies and wavelengths. But regardless of this, the movement of any waves has a lot in common. Having studied the laws of propagation of waves of one nature, it is possible to practically use a large number of them without changing them to waves of another type. It is most convenient to study waves on the surface of the wave.

What is a wave? Throw a stone into the ore. On the calm horizontal surface of labor a running circle will appear. The points on the water surface that the wave should reach begin to oscillate relative to their equilibrium position. This position corresponds to a horizontal surface. The farther a point is from the place where the stone fell, the more late it “learns” about the fall of the stone. Disturbance spreads at a certain speed. The points to which the disturbance must simultaneously be located are in the same stage of oscillatory motion (in the same phase). The speed of movement of this phase is the speed of the wave. It changes in the direction normal to the wave front.

For waves of any nature, it is always possible to indicate physical objects that, under the influence of the wave, will experience a disturbance, that is, a deviation from their equilibrium values. For sound, these are periodic increases and decreases in pressure. For radio waves and light - rapid changes in the direction of the electric and magnetic fields.

The properties of all media without exception are such that a disturbance that arises in a certain area spreads, being transmitted from point to point with a finite speed, which depends on the nature of the disturbance and the properties of the medium.

For a wave to occur, a source of disturbance is required, that is, an external cause that causes a disturbance in propagation in a certain region of the medium. A small source, such as, for example, a stone thrown into water, emits in a homogeneous medium (that is, in a medium in which the speed of the wave does not depend on the direction of its propagation) spherical waves (circular on the surface of the wave), propagating along radii. Such wave sources are called point sources.

One of the basic principles of elementary wave theory is the principle of wave independence or superposition principle. He argues that the disturbance caused by a wave at an observed point does not depend on the fact that other waves simultaneously pass through this point. The principle of superposition gives a simple rule for finding the total action of waves from several sources: the total amount of work is equal to the sum of the oscillations caused by each source separately.

A characteristic feature of wave processes is wave interference. Interference is called a set of phenomena that arise in a medium during the propagation of waves from two or more sources that oscillate in accordance (synchronously). It turns out that at some points the oscillations caused by the simultaneous action of two sources will be more confluent or weaker than the oscillations caused by each source separately. It may happen that coordinated waves will cancel each other out.

At first glance, the phenomenon of interference contradicts the principle of superposition. Before discussing this, we decide to look at the interference with our own eyes. An experienced eye can easily see interference when waves from two stones thrown into ore intersect. However, to study interference

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this method is not suitable. We will obtain a stable interference pattern of water waves on the LAB table.

First of all, you need a vessel with water. So that the waves reflected from its walls do not mask the waves coming from the sources, a vessel with flat walls is needed. In this regard, it is well treated almost to the very top. Running onto the walls, the waves quickly attenuate and are almost not reflected. An electric bell with an unscrewed cap can serve as a wave generator. Attach a piece of wire to the bell hammer and place a cork ball on its end, which will be the exact source of the waves. When testing wires, make sure that the wires are well insulated.

The bell must be secured above the plate so that by turning the bell the cork ball can be immersed in the water at the edge of the plate. It is best to power the call through an author-former. This allows you to adjust the amplitude of vibrations. The author-former from the children’s railway is very convenient. A transformer from a burning device will also work. Turning on the device, we will see circular waves on the surface of the water with a distance between adjacent humps—the wavelength—of about 1 cm (Fig. 1 on page 36).

The best way to observe waves is by looking at the shadow on the bottom of the plate in direct sunlight or the light of a bright lamp. Each wave, acting like a cylindrical lens, produces a light stripe on the bottom of the plate, repeating the shape of the wave front. However, waves traveling at a speed of about 10 cm/sec merge for the gaze fixed on a stationary plate. They are visible only near the source, where their amplitude is large. To increase them over the entire surface of the water, you need to quickly turn your head. In the same way, with rapid head movements, you can see the spokes of a rolling wheel. It is very effective to observe waves on the frosted glass of a camera, especially large-format ones. Holding the camera in your hands and showing it slowly, it is easy to ensure that the waves are visible over the entire surface. At the same time, it seems that they are running very slowly. You can look at the reflection of the wave surface in a mirror. Slight movements of the mirror also make waves visible across the entire surface. But it is most convenient to observe waves under stroboscopic lighting. If you illuminate the installation with short flashes of light with a frequency equal to the frequency of the wave source, then from one flash to another the waves will move by one wavelength, and the wave pattern of the waves will seem motionless. This kind of lighting is extremely easy to achieve. It is enough to turn on a small light bulb parallel to the windings of the bell magnet. From a distance of 0.5-1 m it will illuminate the installation well and evenly, and the motionless shadow pattern of the waves will be clearly visible. The best way to take photographs is to use direct sunlight.

Now attach a fork with pieces of cork at the ends to the bell hammer according to the wiring. The distance between the ends of the fork is 2-3 cm. If the ends are lowered into the water at the same time, you will get two sources of waves that oscillate not only synchronously, that is, in time, but also in phase, that is, waves from both sources will arise at the same moment in time. The picture will be approximately the same as in figure 2. From the wave sources, areas of large amplitudes diverge upward, separated by areas of “silence”. The central region of large amplitudes is located perpendicular to the line connecting the sources. Both areas of silence and areas of large amplitudes pass between the sources.

A study of the interference pattern with a ruler will show that the distance between the two maxima on the line connecting the sources is equal to half the distance between the two waves, i.e., half the length of the wave.

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Fig. 2. Interference of waves from two sources (2d/λ=4).

us. If you change the distance between the sources, then the number of large amplitude bands (Fig. 3) will also change. The greater the distance between the sources, the more “feathers” there are in our fan. But the distance between the maxima on the line connecting the sources is always equal to half the wavelength. Therefore, the total number of bands with maximum amplitude will be twice the number of wavelengths indicated by the distance between the sources. It follows that if the distance between the sources is less than half the wavelength, then there will be no interference at all. Such sources act as one, producing one system of circular waves. This can be seen by gradually decreasing the distance between the sources. Please also note that, if we continue the wave front passing through the hump in some region of large amplitudes, we will encounter a vladina in the neighboring region. In other words: when crossing the zero region, the phase of the wave changes to half a full oscillation.

Now realize that we do not have two hammers, but two light sources emitting light waves, and we place a screen perpendicular to the surface of the wave. We will see illuminated places corresponding to the transformation

Fig. 3. Interference of waves from two sources (2d/λ=2)

cross section of the screen and areas with large amplitude, and dark, unlit areas. We will see dark and light interference fringes. We’ll talk about how to carry out interference experiments in optics some other time, but now let’s try to explain what we saw.

Let’s draw a system of waves on a piece of paper as if they froze at some point (Fig. 4). We mark the humps of the waves in red, and the peaks in blue. I will renumber the waves, marking with the same numbers those that were embroidered from the sources at the same time. It can be seen from the drawing that waves with the same numbers simultaneously arrive at a line equidistant from the sources. This is understandable, since the same paths were taken to the point of the spicy wave. Based on the principle of superposition, we can conclude that on this line both the heights of the humps and the depths of the valley will double. To the right and to the left of this line lie the points at which the humps of one wave system coincide with the humps of another. While waves from one source will cause an upward deflection at these points, waves from another at the same moment will cause a downward deflection at these points. The total deviation at these points will be close to zero. Let’s connect everything

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Fig. 4. Addition of oscillations during the interference of waves from two precise sources, $2d/\lambda=3,6$. Red lines are humps, blue lines are depressions. The thick lines show the resulting waves.

some points of a continuous line. If you follow the numbering of the humps and troughs, you can easily conclude that to all points on the right line, the left waves travel a path half the wavelength longer than the right ones. For the left line, the right waves lag behind by half the wavelength.

To the right and left of the zero lines are the intersection points of the first hump with the second, the second with the third, etc. It is easy to understand that these are also areas of maxima. By connecting these points, we get a line, up to the points of which one wave system lags behind the other by one wavelength.

Continuing the analysis of the drawing, you can find all the zero lines and all the lines

37

maximums. The Linneys we obtained are called hyperbolas by mathematicians *).

Now it is completely clear why the distance between the maxima on the line connecting the sources is equal to half the wavelength. Indeed, at the midpoint of this wavelength, the waves arrive in the same phase and reinforce each other. If you move half a wavelength from this point, then the path of one wave will increase by half the wavelength, and the path of the other will decrease by the same amount. The path difference between the waves at this point (the difference in the path traveled by the waves from the sources to the point) will be equal to one wavelength, and the waves will again be mutually amplified, etc., through each wavelength.

The maximum corresponding to zero path difference is called zero maximum or zero order interference. Maxima with a path difference of one wavelength are maxima of the first order of interference, etc. The maximum order of interference is determined by an integer close to 2d/λ, where d is the distance between the sources and λ is the wavelength. Now try to analyze it yourself using a drawing, and maybe check experimentally how the picture will change if one of the sources emits waves delayed by half the period or some other part of it. What if the shift between oscillations changes randomly? To experimentally implement waves with a phase shift, it is enough to make the ends of the fork of different lengths.

*) A hyperbola is the locus of points, the difference in distances from which to two points (foci) is constructed.

From the examiner’s notebook

An X-ray tube consists of a glass balloon filled with vacuum.

Light sources are divided into two camps: natural and artificial.

The unit of frequency, the hertz, is named after the scientist who discovered this unit.

Inductance is measured in hens.

The further the object is from the lens, the more real the image will be; the closer, the smaller.

Lenses can be double-convex or single-convex.

The atom is complex and it fooled scientists for a long time.

All objects around us: tables, chairs, etc. press on us with a force equal to their weight.

Normal people have good vision, both near and at a distance.

Gamma rays are highly penetrating.

Saturated vapors differ from unsaturated ones in that they have already been saturated.

The red beam is the longest, and the violet beam is the shortest.

Angular velocity is expressed in omegas.

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This section of the magazine is maintained from issue to issue. It publishes problems in physics and mathematics, some of which are quite difficult. Further, specialization in physics and mathematicians, solving them is not easy. Therefore, do not check in if you cannot cope with this or that task. Try to bring the words back to her over time.

Solving a problem means understanding what mathematical apparatus, what equations (or theorems) need to be used for this, what factors to take into account, and what factors to take into account; this means also trying to establish where and how the phenomena indicated in the problem are applied in practice. But a solution has been found. You are satisfied. One should not, however, rest on this. Think about how you can generalize the problem.

In the second part of the section (about six months after the start of publication of the problems themselves), we will post their solutions, taking into account the letters sent by readers. We need this time because it takes more than four months to prepare for printing the next issue.

Schoolchildren who regularly submit particularly original and successful solutions will receive the right to participate in the areas of three-weight optimization for schoolchildren on an equal basis with the winners of the Weighted Correspondence Optimization. In addition, the editors of the magazine established several special prizes for the authors of the most interesting solutions.

And one last thing. Coming up with a problem is usually more difficult than solving it. It’s more difficult, but it’s also more interesting. Try to come up with problems and send them to us. We will publish the best of them in the Quantum Problem Book.

Send your solutions and new problems to the following address: Moscow, V-71, Leninsky Prospekt, 15, editorial office of the journal «Kvant», “Problem Book «Kvant»”, mathematics (physics). The solution to each problem (if you send several attempts at once on a separate piece of paper (sheets), the page is readable. At the end, indicate your last name, first name, patronymic, as well as class and school. Solutions are considered only if they are received no later than a month and a half after the publication of the corresponding issue of the journal.

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Fig. 1.

M61. Two wise men play a new game, which consists of the following. The numbers 0, 1, 2,…, 1024 are written out. The first sage crosses out 512 numbers of his choice, the second crosses out 256 of the remaining numbers, then again the first crosses out another 128, then the second – another 64 numbers, etc. With his last fifth move, the second crosses out one number. There will be two numbers left, and the second one pays the first one the difference between these numbers. How should you play the first game to get as much as possible? How can the second one lose as little as possible? How much will the second one pay the first one if she plays better?

XXXII Moscow Mathematical Olympiad

M62. Prove that for any odd number $a$ there is a natural $b$ such that $2^b - 1$ is divisible by $a$.

M63. Is it possible to lay out a square out of 18 tiles of size $1 \times 2$?

Fig. 2.

So that there is not a single straight “seam” connecting the opposite sides of the square and running along the edges of the tiles? (For example, a tile arrangement like figure 1 would not work because there is a “seam” $AB$).

A. A. Kirillov

M64. On the data plane there is a straight line $l$ and two points $P$ and $Q$ lying on one side of it. Find on the line $l$ also the points $M$ for which the distance between the bases of the altitudes of the triangle $PQM$, dropped to the sides $PM$ and $QM$, is called.

G. A. Palatnik

M65. a) Let $E, F, G$ be points on sides $AB, BC$ and $CA$ of triangle $ABC$ for which $\frac{AE}{EB} = \frac{BF}{FC} = \frac{CG}{GA} = k$, where $0 < k < 1$. Find the ratio of the area of ​​the triangle formed by lines $AF, BG$ and $CE$ to the area of ​​triangle $ABC$ (Fig. 2).

6. Cut the triangle into six straight lines into pieces that can be used to form seven equal triangles.

A. L. Soifer

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F72. Why does the length of the day change faster in March than in December?

G. L. Komkin

F73. Lint the sugar solution inside the glass tube. The lower end of the tube is covered with a film that is permeable to water but impermeable to sugar molecules. The tube is immersed in clean water until the sugar dissolves and secured in this position. At the same time, the liquid level in the tube begins to rise and, after a sufficient period of time, is established at a certain height above the water level (Fig. 3).

What explains this accumulation of water inside the cylinder? Where does the energy to lift liquid come from?

F74. Two capacitors are connected in series. The first has a capacity of $C_1$ and is designed for a maximum voltage of $U_1$, the second has a capacity of $C_2$ and is designed for a voltage of $U_2$. To what maximum zero does

Can you connect this bank of capacitors?

A. R. Zilberman

F75. A piston with an area $S$ hinged with a washer $C$ sliding without friction along the lever $AB$. Lever length $L$. What is the minimum force that must be applied to lever $AB$ in order to increase the pressure in the fluid by $\Delta p$ (Fig.4)?

B. B. Bukhovtsev, Moscow State University Physics Olympiad, 1970

F76. A heavy hoop with weightless knitting needles is located in a vertical plane and can rotate around a horizontal axis passing through the center. A material point is fixed in the thickness of its rim, having the same mass as the hoop itself. What will be the period of oscillation of the hoop around its axis? How will it change if: a) the motherboard is moved to the Moon, b) placed in a liquid in which it will move without friction?

SUBSTITUTION METHOD FOR SOLVING EQUATIONS AND SYSTEMS OF EQUATIONS

M. I. Scanavi

The substitution method involves moving from the original value to some other value. It allows you to reduce solutions to equations to more simply constructed ones or to equations of studied types. There is a great variety of cases when one has to use this method. It is impossible to cover them with any general recommendation. Therefore, success in choosing an appropriate substitution significantly depends on such advantages of the decision as his experience, ingenuity, intuition… The examples under consideration demonstrate the substitution method in action.

Example 1: Solve the equation
$$\frac{1}{x(x+2)} - \frac{1}{(x+1)^2} = \frac{1}{12}.$$

Solution. The difference of the fractions located on the left side of this equation is a fraction whose denominator is a polynomial of four degrees with respect to $x$. In this case, template processing of the equation will lead it to an equation of four degrees, which the solver cannot do. After thinking about it, we will begin to understand that it is useful to choose a substitution
$$x^2 + 2x = z,$$
(*)

because it transforms the given equation into a much simpler one:
$$\frac{1}{z} - \frac{1}{z+1} = \frac{1}{12}$$

With respect to $z$ we obtain a quadratic equation with roots $z_1=3$ and $z_2=-4$. Substitution (*) results in two quadratic equations for $x$:
$$x^2 + 2x = 3$$
And
$$x^2 + 2x = -4.$$

From them we find all four roots of this equation:
$$x_1 = 1, x_2 = -3, x_3 = -1 + \sqrt{3}i,$$
$$x_4 = -1 - \sqrt{3}i$$

and by direct testing we are convinced of the suitability of each of them.

Example 2: Solve the equation
$$4x^4 + 12x^3 - 47x^2 + 12x + 4 = 0.$$

Solution. A so-called reciprocal equation has been proposed for solution - the coefficients of the terms equally spaced from the ends of its left side are equal. For reciprocal equations, a standard method has been developed for reducing them to an equation of lower degree. An idea of ​​this method can be gained by observing how it is applied to the equation that is being solved. Noticing that $x \neq 0$, we divide the equation by $x^2$; we get
$$4x^2 + 12x - 47 + \frac{12}{x} + \frac{4}{x^2} = 0.$$

Now you should apply the grappirum members:
$$4\left(x^2 + \frac{1}{x^2}\right) + 12\left(x + \frac{1}{x}\right) - 47 = 0.$$

With this writing, the equations are already

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not containing radicals of the second degree!): let’s compose the difference of the subdecimal expressions of the radicals located on the left, and we get that $(t+3) - t=3$.

Now we factorize this difference (using the formula for the difference of squares) and find that $(\sqrt{t+3} + \sqrt{t})(\sqrt{t+3} - \sqrt{t}) = 3$.

From here, by dividing by the original equation (with respect to $t$), we obtain that

$$\sqrt{t+3} - \sqrt{t} = \frac{3}{\sqrt{2t+7}}.$$

Subtracting from the original equation what we now have, we find that

$$2\sqrt{t} = \sqrt{2t+7} - \frac{3}{\sqrt{2t+7}},$$

$$2\sqrt{t} = \frac{2t+4}{\sqrt{2t+7}}, \sqrt{t} = \frac{t+2}{\sqrt{2t+7}}.$$

The procedure we have just carried out allows us to get rid of radicals by simply squaring the parts of the equation. Let’s do this:

$$t = \frac{t^2+4t+4}{2t+7}, t^2+3t-4=0.$$

Hence $t_1=1, t_2=-4$. Next, note that we are only interested in positive values ​​of $t$, since $t = x^2+x+1 > 0$

for all values ​​of $x$. Therefore, we discard the unsuitable root $t_2$ and obtain one quadratic equation for finding $x$

$$x^2+x+1=1, \quad x^2+x=0$$

with roots $x_1=-1$ and $x_2=0$. It is easy to check that both of them are roots of this irrational equation.

Example 5. Solve a system of equations

$$\begin{cases} \frac{x^3}{y} + xy = 40, \\ \frac{y^3}{x} + xy = 10. \end{cases}$$

(Limit yourself to finding real solutions).

Solving a system of equations students practice a significant solution. At the same time, in these cases it often turns out to be very effective. Before going to the solution of the proposed system, I will allow myself to express some general messages about the substitution method when solving a system of two equations with two exceptions. If these latter are designated by the letters $x$ and $y$, then it is often useful to move on to the new quantities $\rho$ and $\varphi$ in accordance with the following equalities:

$$x = \rho \cos \varphi,$$

$$y = \rho \sin \varphi.$$

The reader who is sufficiently familiar with the method of coordinates on the plane, almost that if $x$ and $y$ serve as the Cartesian coordinates of some point, then $\rho$ and $\varphi$ are the so-called polar coordinates of the same point. The geometric meaning of polar coordinates and their connection with Cartesian ones will become clear from figure, which we now recommend everyone to CAREFULLY consider:

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We note that the origin $O$ has $\rho=0$, and the angle $\varphi$ has no defined value for it. All other points of the plane will be subordinated to a single pair of polar coordinates ($\rho, \varphi$), if we require that there be

$$0 \leqslant \rho < +\infty,$$
$$0 \leqslant \varphi < 2\pi.$$

We use equalities (!) to solve this system of equations, for which instead of $x$ and $y$ in the domain of the system equation we substitute their expressions through $\rho$ and $\varphi$; we get that

$$\begin{cases} \frac{\rho^3 \cos^3 \varphi}{\rho \sin \varphi} + \rho \cos \varphi \rho \sin \varphi = 40, \\ \frac{\rho^3 \sin^3 \varphi}{\rho \cos \varphi} + \rho \cos \varphi \rho \sin \varphi = 10. \end{cases}$$

or after processing that

$$\begin{cases} \rho^2 \text{ctg} \varphi = 40, \\ \rho^2 \text{tg} \varphi = 10. \end{cases}$$

By dividing the first equation by the second we find that

$$\text{ctg}^2 \varphi = 4.$$

Hence $\text{ctg} \varphi = 2$ or $\text{ctg} \varphi = -2$. The second (negative) value $\text{ctg} \varphi$ is immediately discarded, since we have $\text{ctg} \varphi$ and $\rho^2$ having the same sign, and $\rho^2$ cannot be negative. Now, using the known cotangent, we need to find the sine and cosine of the angle $\varphi$. I follow this without words:

$$\cos^2 \varphi = 1 + \text{ctg}^2 \varphi = 5,$$
$$\cos \varphi = \sqrt{5},$$
$$\cos \varphi = -\sqrt{5},$$

$$\sin \varphi = \frac{1}{\sqrt{5}}, \quad \sin \varphi = -\frac{1}{\sqrt{5}},$$
$$\cos \varphi = \frac{2}{\sqrt{5}}, \quad \cos \varphi = -\frac{2}{\sqrt{5}}.$$

Next we find $\rho$ from the fact that

$$\rho^2 \text{ctg} \varphi = \rho^2 \cdot 2 = 40, \quad \rho^2 = 20.$$

We get that $\rho = 2\sqrt{5}$. Using equalities (!) using the known $\rho$ and $\varphi$, we find the following two pairs of numbers as a solution to this system of equations:

$$x_1 = 2\sqrt{5} \cdot \frac{2}{\sqrt{5}} = 4,$$
$$y_1 = 2\sqrt{5} \cdot \frac{1}{\sqrt{5}} = 2;$$
$$x_2 = 2\sqrt{5} \left(-\frac{2}{\sqrt{5}}\right) = -4,$$
$$y_2 = 2\sqrt{5} \left(-\frac{1}{\sqrt{5}}\right) = -2.$$

It is incorrect to assume that this system can be solved only by passing to polar coordinates (it can be solved, for example, also this way: transfer $xy$ from the left sides of the equation to the right, then recombine the results and obtain a quadratic equation for $xy$; the rest is obvious). Meanwhile, this technique should be adopted. It allows us, in a certain sense, to regulate our transformations and makes it possible to use the entire apparatus of trigonometry. In what cases is it beneficial to use this technique, and in what cases is it not? Your experience helps answer this question, which you can enrich with your own work!

While there is time before the entrance exams, act!

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ENTRANCE EXAMINATIONS
IN MATHEMATICS IN MATHEMATICS
FACULTY OF MSPI
them. V. I. LENIN IN 1970
E. A. Shegolkov

The written examination in mathematics at our department in 1969 (the report on it can be read in «Kvant» No. 3 for 1970) was somewhat unusual. The unusualness of the state is in some imitation of competitive exams at Moscow State University. As a result, out of 475 applicants, 224, that is, about half, wrote the paper with a bad mark and dropped out of the further “struggle.” This method of exchanging applicants does not seem suitable for a pedagogical institute; too many candidates ended up without personal acquaintance with the examiners.

In 1970, the selection committee of the Faculty of Mathematics decided on almost different paths. The written work included four tasks to get to know the student well.

1. Stereometric problem with application of trigonometry.
2. The task of composing a quadratic equation with physical or other content.
3. Algebraic equations or inequality.
4. Trigonometric equation.

All tasks did not require any special knowledge or training. But at the same time, they assumed a thorough knowledge of school mathematics and known abilities of applicants. 4 hours were allotted to complete the written work.

When assessing written work, “five” was given only if all tasks were completed; “four” - if all tasks were completed, but with minor shortcomings. “Two” was given, for example, if the first tasks were not solved due to a group of errors. At the same time, the rating of “three” acquired a greater share and did not exclude the applicant from participation in the competition.

In 1970, 508 people entered the Faculty of Mathematics, and there were 175 places; the competition was about three people per place.

Of the 508 applicants for the written work, 120 wrote it as “four” and “five”, 233 as “three” and 155 as “two”.
As an example, below we will analyze the written work option and point out the typical mistakes made by applicants when solving it.

Option

1. The length of the generatrix of the cone is equal to l, the angle of the generatrix with the plane of the base is equal to α. Find the volume of a primid circumscribed around a cone whose base is a rhombus with an acute angle β.
2. In n hours, the tractor plows p hectors more than the horse. How many hectors can a horse plow in n hours and how many hectors can a tractor plow in n hours, if a tractor plows one hector in t hours faster than a horse.
3. Solve inequality
   $$\sqrt{1+2x}-\sqrt{x+4}<1.$$
4. Solve the equation
   $$\frac{3(\cos 2x+\mathrm{ctg} 2x)}{\mathrm{ctg} 2x-\cos 2x}-2(\sin 2x+1)=0.$$

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As you already noticed, when solving this problem I had to solve right-angled triangles three times. The applicants made mistakes here too. Solid ability to solve right-angled triangles is a prerequisite for successfully solving most stereometric problems.

2. Let the tractor turn $x$ ($2a$) in $n$ hours, therefore $1$ ($2a$) in $\frac{n}{x}$ hours.

A horse turns in $n$ hours $(x-p)$ ($2a$), and $1$ ($2a$) in $\frac{n}{x-p}$ hours.

According to the conditions of the problem, $\frac{n}{x}$ is less than $\frac{n}{x-p}$ by $t$ hours, that is, $\frac{n}{x} + t = \frac{n}{x-p}$ ($x \neq 0, x \neq p$). After the transformation we obtain the equation $tx^2 - ptx - np = 0$. Solving it, we find $x_{1,2} = \frac{pt \pm \sqrt{(pt)^2 + 4ptn}}{2t}$.

Which of these two roots satisfies the conditions of the problem? By condition, $n > 0, t > 0, p > 0$, so $\sqrt{(pt)^2 + 4ptn}$ has a real value. $(pt)^2 + 4ptn > (pt)^2$ and therefore $\sqrt{(pt)^2 + 4ptn} > pt$.

So the root

$$x_1 = \frac{pt + \sqrt{(pt)^2 + 4ptn}}{2t}$$

gives the answer to the question of the problem, exactly how many hectares does a tractor turn in $n$ hours, and a horse turns in $n$ hours

$$x - p = \frac{-pt + \sqrt{(pt)^2 + 4ptn}}{2t}$$.

The second root of the equation is negative and disappears according to the meaning of the problem.

When compiling the initial equation from the value $\frac{n}{x}, \frac{n}{x-p}, t$, some of the applicants for some reason subtracted more from less and put the equation $\frac{n}{x-p} + t = \frac{n}{x}$.

This is the result of not paying attention.

When solving a problem with specific physical or other content, you should be especially careful and always check your calculations. Formal statements must correspond to the actual state of affairs. In addition, it should be remembered that the whole solution to this type of problem is not the solution of the compiled equation, but obtaining an answer to the question posed in the task.

3. Let’s find the range of acceptable values ​​for $x$ (OD3):

$$\begin{cases} 1 + 2x \geq 0 \\ x + 4 \geq 0 \end{cases}$$ or $$\begin{cases} x \geq -\frac{1}{2} \\ x \geq -4. \end{cases}$$

Consequently, OD3 is determined by the condition

$$x \geq -\frac{1}{2}.$$ (1)

Let us transform the original inequality as follows:

$$\sqrt{1 + 2x} < 1 + \sqrt{x + 4}.$$

In marked OD3, both sides of the last inequality are non-negative; squaring them we get

$$\sqrt{x + 4} > \frac{x}{2} - 2.$$ (2)

The left side of inequality (2) is positive, the right side is negative, and the left side is positive.

1. $\frac{x}{2} - 2 < 0$ therefore

$$-\frac{1}{2} \leq x < 4.$$

For such values ​​of $x$ inequality (2) depends negatively, and the left one is positive.

2. $\frac{x}{2} - 2 \geq 0$ and by condition

$$\sqrt{x + 4} > \frac{x}{2} - 2.$$

Solving the first of these inequalities, we obtain $x \geq 4$. To solve the second inequality, we square both sides (this can be done,