Kvant 1971, No. 3 · Pages

Popular science physics and mathematics
quantum
3
1971
magazine
Academy
Sciences of the USSR
And
Academy of Pedagogical
Sciences of the USSR

In the room:

Economics and linear inequalities 1
A. B. Skating rink

Converting electrical circuits 10
A. R. Zilberman

Do you like messing around with goal numbers! 15
M. I. Batsimakov

Amazing skating rinks 21
B. Yu. Kodan

Math club
Semiregular polygons 25
L. M. Lopovok

Problem «Kvant»
Problems 30

Solutions to problems M26—M28; F35—F40 32
N. B. Vasiliev,
A. N. Vilenkin,
I. Sh. Slobodetsky

Applicant’s workshop
Exponential Equations 40
V. V. Goldberd

Written exam in mathematics at mexmate Moscow State University in 1970 45
N. N. Kolesnikov

Written exam in physics at the Moscow Automobile and Road Institute 51
A. N. Ageev,
Yu. G. Rudoy

Physics News 57

Reviews, bibliography
If you need to calculate something…
57
R. S. Guter

The Sorcerer from Plutalova Street 58
G. I. Mishchevich

Answers, directions, solutions 62

Collector’s Corner
Stamps dedicated to N. E. Zhukovsky 64
A. V. Altykis

«Kvant» for younger students
3rd cover page
Crossword
4th cover page

2

that together with two road bicycles the plant can produce one racing bicycle, and the racing bicycle brings 1.5 times more profit. The factory can produce 700 road bikes per day, but the warehouse can only accept 500 bikes per day. How many racing bikes and how many road bikes do you need to produce per day in order for the plant to get the maximum approximation?

2. Problem about quarries. At points A and B there are brick factories, and at points C and D there are quarries that supply them with sand. The need of factories for sand is no more than the productivity of quarries. It is known how much sand each plant needs and how much is added to each quarry. In addition, it is known how much it costs to transport a ton of sand from each quarry to the factories. It is necessary to organize the supply of sand to factories so that costs are minimal. (Such tasks are called transortins.)

3. Problem about chicken feed. The poultry farmer is engaged in open-casting of chickens. Let us assume for simplicity that two products can serve as food - wheat and peas. In order for a chicken to grow normally, its daily diet must contain at least a certain amount of calories, as well as proteins, fats and carbohydrates. These quantities are established by esophageal specialists, and we should consider them as given. Let it be known how much fat, proteins, fats and carbohydrates are contained in one gram of wheat and one gram of peas, and the price of each type of food is also known. How to create the cheapest but most complete diet?

In each of the above, it is necessary to make the optimal choice, although in each problem it is understood differently.

In order to see what mathematical questions arose when solving such problems, let us analyze in detail the problem of chicken feed.

To analyze this problem, you will receive all the information that is included in the school course, and all calculations can be easily checked using pencil and paper.

Let one gram of wheat contain $a_1$ calories, one gram of peas contain $b_1$ calories, and the minimum required number of calories in the daily diet is $c_1$. We will depict menus containing $x$ grams of wheat and $y$ grams of peas by point $(x, y)$ on the plane. Let’s try to highlight those points on the plane that correspond to diets containing at least $c_1$ calories. Since a menu consisting of $x$ grams of wheat and $y$ grams of peas contains $a_1x+b_1y$ calories, the inequality $a_1x+b_1y \geq c_1$ holds for such points.

From your school algebra course you know that the points for which $a_1x+b_1y=c_1$ lie on the line $l_1$ passing through the points $(0, \frac{c_1}{b_1})$ and $(\frac{c_1}{a_1}, 0)$. This straight line divides the plane into two parts. For the point $(x, y)$, lying on one side of the line $l_1$, the inequality $a_1x+b_1y > c_1$ is satisfied, and for the point lying on the other side, the inequality $a_1x+b_1y < c_1$ is satisfied. Each of these parts, together with the straight line $l_1$, is called a half-plane. Thus, one half-plane consists of all points $(x, y)$ for which $a_1x+b_1y \geq c_1$,

and the other from the point for which $a_1x+b_1y \leq c_1$.

Since in our case $a_1$ and $b_1$ are positive numbers, then the first half-plane is the half-plane $L_1$,

Which is highlighted at figure 1. In addition, from the conditions of our problem it is clear that $x$ and $y$ are non-negative numbers (you MUST NOT give the chicken a negative amount of food). The condition $x \geq 0$ is satisfied by the points of the right half-plane (limited by the ordinat axis), the condition $y \geq 0$ is satisfied by the points of the upper half-plane (limited by the abscissa axis), and the portions containing at least $c_1$ calories are satisfied by the points $(x, y)$, for which three inequalities are simultaneously satisfied:

$$\begin{cases} x \geq 0, y \geq 0, \\ a_1 x + b_1 y \geq c_1, \end{cases}$$

that is, points lying in the intersection of the right half-plane, the upper half-plane $L_1$. On figure 2 a set of points corresponding to the solution of the system of inequalities (1) is highlighted. Let’s move on to other limited ones. If one gram of solutions contains $a_2$ grams of protein, one gram of peas contains $b_2$ grams of protein, and the daily diet of a chicken should contain at least $c_2$ grams of protein, then, obviously, those diets containing $x$ grams of solutions and $y$ grams of peas contain sufficient protein for which

$$a_2 x + b_2 y \geq c_2.$$

Corresponding to these ratios, the points lie in the half-plane $L_2$, bounded by the straight line $l_2$, which is given by the equation $a_2 x + b_2 y = c_2$. Similarly, the diet must satisfy the inequalities

$$\begin{cases} a_3 x + b_3 y \geq c_3, \\ a_4 x + b_4 y \geq c_4, \end{cases}$$

where $a_3$ and $a_4$ are the amount of fats and carbohydrates in one gram of wheat, $b_3$ and $b_4$ are the amount of fats and carbohydrates in one gram of peas, $c_3$ and $c_4$ are the minimum amount of fats and carbohydrates in the daily diet of a chicken. (All quantities are measured in grams.) Points $(x, y)$ that satisfy the first of these two inequalities lie in the half-plane $L_3$, and points satisfying the second inequality lie in the half-plane $L_4$. We will call a diet containing a sufficient amount of all components (calories, proteins, fats and carbohydrates) acceptable. Thus, acceptable diets consisting of $x$ grams of wheat and $y$ grams of peas will be those that satisfy the system of six linear inequalities

$$\begin{cases} x \geq 0, y \geq 0, \\ a_1 x + b_1 y \geq c_1, \\ a_2 x + b_2 y \geq c_2, \\ a_3 x + b_3 y \geq c_3, \\ a_4 x + b_4 y \geq c_4. \end{cases}$$

Correspondence to the admissible ration of the half-plane points lies at the intersection of the right half-plane, the upper half-plane $L_1$, $L_2$, $L_3$, $L_4$, that is, inside or on the boundary of the figure (infinite set-

4

Fig. 3.

sovereign) $A$, allocated to figure 3.

This figure depicts the case when each inequality in system (2) exists, that is, for each of the six inequalities there is a solution to the system of the remaining five inequalities that does not satisfy this inequality. Another possible case is depicted at figure 4.

Geometrically, this means that one of the lines (on Fig. 4 is the line $l_4$) lies outside the set $A$, and algebraically, that system (2) is equivalent to a system from which one of the inequalities is excluded (in our case, the last inequality). Throughout what follows, in order not to make special reservations, we will assume that the case depicted in figure 3 occurs.

A set of points on a plane is called convex if, together with any two properties of points, it entirely contains a segment connecting these points. On figure 5, $a$ the selected non-convex sets are convex, and on figure 5, $b$ the selected non-convex sets.

A half-plane is a convex set, and since the intersection of any number of a convex set is also a convex set, then the set $A$ is convex.

Fig. 4.

Now we will try to choose the cheapest one from all acceptable razions. Let 1 gram of wheat cost $p$ kopecks, and 1 gram of peas cost $q$ kopecks. then a diet of $x$ grams of wheat and $y$ grams of peas costs $px+qy$ kopecks. Thus, the price is a linear function of $x$ and $y$, and the task of finding the cheapest ratio is to find among all pairs of numbers $(x, y)$ that satisfy the system of inequalities (2), such pairs for which the value of this linear function reaches a minimum. In other words, you need to find those from the point of the set $A$ at which the function $px+qy$ reaches the smallest value of all the values ​​​​accepted by this function at the points of the set $A$.

The set of points $(x, y)$ of the plane for which the expression $px+qy$ is equal to some number $k$ is a pr-

Fig. 5, $a$.

Fig. 5, $b$.

My. For different $k$ such lines are parallel. For negative and sufficiently small positive values ​​of $k$ this line does not intersect the set $A$ and, therefore, there is no admissible ration worth $k$ kopecks. If we increase the values ​​of $k$, starting, for example, with $k=0$, then the straight line given by the equation

$$px + qy = k,$$

will blend to the right (see Fig. 6). Finally, for some value of $k=k_0$ the line $px + qy=k_0$ will intersect the set $A$ for the first time, “stumbling upon” one of its vertices (point $M$ on Fig. 6) or immediately on the whole side (this can happen if the lines $px + qy=k$ are parallel to one of the lines $l_1, l_2, l_3, l_4$ or, which is the same, if the relation $\frac{p}{q}$ is equal to one of the ratios

$$\frac{a_1}{b_1}, \frac{a_2}{b_2}, \frac{a_3}{b_3}, \frac{a_4}{b_4};$$

see further, Fig. 8).

A prinya is called an operon to a convex set on a plane if this set lies entirely in one of the half-planes into which this line divides the plane, and at the same time at least one point of the set lies on the line. Thus, the equation $px + qy=k_0$ defines the operatic line to the set $A$.

Let us now examine separately the two cases mentioned above. Suppose first that the relation $\frac{p}{q}$ is not equal to any of the relations

$$\frac{a_1}{b_1}, \frac{a_2}{b_2}, \frac{a_3}{b_3}, \frac{a_4}{b_4}.$$

Then point $M=(x_0, y_0)$ is the only solution to the problem of the cheapest admissible ration. Indeed, any other point of the set $A$ lies in the half-plane $K$ given by the inequality $px + qy \geq k_0$, and therefore corresponds to a more expensive diet. The lines of the infinite polygon $A$ are the rays of the coordinate axes and the line segments $l_1, l_2, l_3, l_4$. Therefore, on each of these sides, one of the six inequalities of system (2) turns into equality, and at each vertex of the polygon $A$ two inequalities of system (2) turn into equalities. In the case chosen at figure 6, the point $M$ lies at the intersection of the lines $l_1$ and $l_3$ (see Fig. 3); hence,

$$\begin{cases} a_1x_0 + b_1y_0 = c_1, \\ a_3x_0 + b_3y_0 = c_3, \end{cases}$$

where

$$x_0 = \frac{b_1c_3 - b_3c_1}{a_3b_1 - a_1b_3}; \quad y_0 = \frac{c_1a_3 - a_1c_3}{a_3b_1 - a_1b_3}.$$

But the remaining four inequalities at this point are satisfied in the strict sense, that is, with the sign “greater than” and not “equal”. This means that a diet consisting of $x_0$ grams of solutions and $y_0$ grams of peas contains the minimum required amount of protein and fat, but an excessive amount of protein and carbohydrates (if point $M$ lay at the intersection of other lines, then the conclusion would change accordingly).

If food scientists had decided that chickens needed a little more than $c_2$ grams of protein or a little more than $c_4$ grams of carbohydrates per day, then the cheapest diet we found would meet this new requirement (indeed, this means that the straight $l_2$ or $l_4$ would shift slightly to the right, but if such a shift is small enough, then the point $M$ still remains the vertex of the reduced polygon $A$). However, if you need to give the chickens a little more than $c_3$

6

Fig. 7.

grams of fat per day, then the problem of the cheapest diet must be solved anew, since the diet corresponding to point $M$ will cease to be acceptable, and the cheapest diet becomes expensive (see Figure 7, where, with the conditional requirement $a_3x+b_3y \geq c_3+\delta$, the cheapest diet corresponds to point $M'$, lying inside the half-plane $K$).

Let us now consider the case where the ratio $\frac{p}{q}$ is equal to one of the four ratios, for example $\frac{a_1}{b_1}$ (see Fig. 8).

In this case, line $px+qy=k_0$ coincides with line $l_1$. All points of the set $A$, with the exception of the point of the segment $M_1M_2$, lie inside the half-plane $K$, and therefore the ration corresponding to any point of the segment $M_1M_2$ is the cheapest admissible ration. Thus, among the points corresponding to the solution to the problem of the cheapest ration, there are always the vertices of the polygon $A$.

A little theory

To find a solution, it is enough to compare the values ​​of the function $px+qy$ at all vertices. If there is one of these values ​​that is smaller than all the others, then the corresponding vertex is the only solution to the problem; if the values ​​at two vertices are equal and less than the values ​​at the other vertices, then these vertices are necessarily adjacent, and all points on the side connecting these vertices serve as solutions to the problem.

For brevity we will call

Fig. 8.

an acute angle formed by a straight line with the abscissa axis, the angle of inclination of the straight line. If dwir lies along the boundary of a set of permissible points, starting from the abscissa ray towards the ray of the ordinat axis, then each subsequent side lies on a straight line with a greater angle of inclination than the previous one. The tangents of these angles will also increase. The tangent of the angle of inclination $\alpha$ of the straight line $ax+by=c$ ($a>0, b>0$) is $\frac{a}{b}$. (If $b=0$, $\alpha=\frac{\pi}{2}$ and it is generally accepted that $\tg\alpha$ is equal to infinity.)

Let the reference line pass through the vertex $M$ of the set of admissible points and the inclination angle of this line is equal to $\beta$, and the inclination angles of the straight lines on which the sides adjacent to the vertex $M$ lie are equal to $\alpha_1$ and $\alpha_2$. Then the angle $\beta$ is greater than one of these angles and less than the other (see Fig. 9).

7

Thus, in order to find the vertex through which a reference line of a given direction passes, you need to compare the slope angle of the lines in a given direction with the slope angles of the lines on which the sides of the set of admissible points lie, and select two adjacent sides between whose slope angles the angle β lies. The intersection point of these sides will be the desired vertex. If the angle β coincides with the angle of inclination of one of the sides, then the desired reference line coincides with the line on which this side lies. Instead of comparing slope angles, you can compare their tangents, which is much more convenient, since the tangent of the slope angle is simply expressed through the coefficients of the equation of the straight line.

Numerical examples

1. Among the solutions to the system of linear inequalities $2x+y \geq 10$, $x \geq 0, y \geq 0$, $x+y \geq 8$, $3x+7y \geq 42$, $x+5y \geq 20$, find those for which the linear function $4x+3y$ reaches a minimum*).

Table 1

| Equation of a line | A point on the coordinate axis through which a straight line passes | Tangent of the angle of inclination of a straight line |
| :— | :— | :— |
| $y=0$ | $(0,4), (20,0)$ | $\frac{1}{5}$ |
| $x+5y=20$ | $(0,6), (14,0)$ | $\frac{3}{7}$ |
| $3x+7y=42$ | $(0,8), (8,0)$ | $\frac{1}{2}$ |
| $x+y=8$ | $(0,10), (5,0)$ | $\infty$ |
| $2x+y=10$ | $(0, \frac{3}{c}), (\frac{c}{4}, 0)$ | $\frac{4}{3}$ |

The set of feasible points is depicted on figure 10, from which

*) The numerical data of this example are in no way related to the values ​​​​that are given in the problem of chicken feed.

Fig. 10.

it is clear that the sides of this set are the rays of the coordinate axes and the segments of all four straight lines: $x+5y=20$, $3x+7y=42$, $x+y=8$, $2x+y=10$.

A supporting straight line with a slope angle $\frac{4}{3}$ must pass through a vertex lying at the intersection of a pair of neighboring sides, one of which has a slope angle less than $\frac{4}{3}$, and the other greater than $\frac{4}{3}$. Such a pair is formed by sides lying on straight lines

$x+y=8$ and $2x+y=10$,

with slope angles of 1 and 2, respectively. Thus, the coordinates of the desired vertex are determined from the system

$$\begin{cases} x+y=8, \\ 2x+y=10, \end{cases}$$

whence $x=2$, $y=6$. The value of the function $4x+3y$ at this point is $4 \cdot 2 + 3 \cdot 6 = 26$, and the reference line is determined by the equation $4x+3y=26$.

For comparison, let’s calculate the values ​​of the function $4x+3y$ at all other vertices of the set of admissible points (see Table 2 on page 8).

2. Among the solutions to the system of linear inequalities $x \geq 0, y \geq 0, 3x+y \geq 15, 3x+4y \geq 36, x+2y \geq 12, 2x+8y \geq 32$, find those for which the function $3x+12y$ reaches a minimum.

8

Table 2

Equation of lines, the vertex lies at the intersection of coordinates
| vertex coordinates | |
| :— | :— |
| $x$ | $y$ |
| $x = 0, 2x + y = 10$ | $0, 10$ |
| $2x + y = 10, x + y = 8$ | $2, 6$ |
| $x + y = 8, 3x + 7y = 42$ | $3, 5, 4, 5$ |
| $3x + 7y = 42, x + 5y = 20$ | $8, 75, 2, 25$ |
| $x + 5y = 20, y = 0$ | $20, 0$ |

Function values ​​$4x + 3y$ at vertex
| $x$ | $y$ |
| :— | :— |
| $0$ | $10$ |
| $2$ | $6$ |
| $3,5$ | $4,5$ |
| $8,75$ | $2,25$ |
| $20$ | $0$ |

Table 3

Equation of a line
$y = 0$
$2x + 8y = 32$
$x + 2y = 12$
$3x + 4y = 36$
$3x + y = 15$
$x = 0$
$3x + 12y = c$

Points on the coordinate axes through which a straight line passes
$(0, 4), (16, 0)$
$(0, 6), (12, 0)$
$(0, 9), (12, 0)$
$(0, 15), (5, 0)$
$(0, \frac{c}{12}), (\frac{c}{3}, 0)$

Tangent of the angle of inclination of a straight line
$0 \frac{1}{4}$
$\frac{1}{2}$
$\frac{3}{4}$
$3 \frac{\infty}{1/4}$

The set of admissible points is depicted in figure 11, from which it is clear that the straight line $x + 2y = 12$ lies entirely outside this set. Thus, the sides of this set are the rays of the coordinate axes and the segments.

The solutions to the problem are all points on the side lying on this line. The ends of this side are the vertices lying at the points of intersection of the line $2x + 8y = 32$ with the neighboring lines $y = 0$ and $3x + 4y = 36$ (recall that we excluded the line $x + 2y = 12$ from consideration).

So, the coordinates $(x_1, y_1)$ of the first vertex $M_1$ are determined from the system of equations
$$\begin{cases} 2x_1 + 8y_1 = 32, \\ y_1 = 0, \end{cases}$$

9

Fig. 12.

whence $x_1=16, y_1=0$, and the coordinates $(x_2, y_2)$ of the second vertex $M_2$ are determined from the system of equations

$$\begin{cases} 2x_2 + 8y_2 = 32, \\ 3x_2 + 4y_2 = 36, \end{cases}$$

whence $x_2=10, y_2=1,5$.

The solutions to the problem are all points of the segment $M_1M_2$, in other words, points of the form $(16\lambda + 10(1-\lambda), 1,5\cdot(1-\lambda)) = (10+6\lambda, 1,5-1,5\lambda)$, where $\lambda$ is any number between 0 and 1. The value of the function $3x+12y$ at all points of the segment $M_1M_2$ is 48.

Exercises *)

1. Solve the bicycle problem.
2. Solve the “carber problem” with the data given in figure 12.
3. How will the set of admissible points in the problem of the cheapest ration change if we additionally require that all rationals do not exceed s gram? Could such a limitation influence the decision?
4. Prove that any convex polygon on a plane can be represented as a polygon of a system of linear inequalities.
5. Let $\frac{p}{q} = \frac{a_1}{b_1}$ (see Fig. 3 and 8). How does the solution to the problem of the cheapest diet change in this situation if we replace in system (2)

$$\begin{cases} a) c_1 на c_1 + \delta, \\ b) c_n на c_n + \delta; \end{cases}$$

Is δ a positive but very small number?

*) Answers to these exercises will be published in the next issue.

2, Quantum No. 3

QUARTZ HAIRY

There is a stone known in nature that is called “hairy quartz” in Russia. This is a rich quartz crystal produced with shiny gold threads (see figure on the cover).

Pliny called the threads “Hairs of Venus”, and in the east they are called “Hairs of Mohammed” and the stone was considered a binding. Currently, these “Hairs”, or whiskers as they are now called, are grown in laboratory conditions. Studies using an electron microscope showed that the “Hair of Venus” is nothing more than very thin and long crystals of titanium oxide $(\text{TiO}_2 - \text{рубин})$, which are tubes of rectangular cross-section with a wall thickness of 1–2 µm (see photograph).

Whiskers have a number of very interesting properties, one of which—high strength—makes it possible to create new, so-called composite materials based on them.

M. A. Golosova

10

This article describes a method that allows you to simplify complex problems in calculating electrical circuits.

What do we mean by “chain conversion”? Let’s assume that we have a complex circuit of resistances that has many pins and connections to sources. Let’s replace this circuit with another one, with the same number of terminals, and so that the resistance between any two terminals of the new circuit is the same as that of the old one. It is clear that the sources “will not know anything” about this replacement, and the points consumed by the circuit will remain the same. But finding these points may be easier.

Therefore, if we want to count currents in a complex circuit, it can be replaced by a simpler equivalent circuit. In this case, the currents inside the replaced part change. Therefore, we can do this only with that part of the scheme that does not directly interest us.

You have already encountered similar substitutions.

Let, for example, in the circuit two resistances $r_1$ and $r_2$ are connected in series, we can replace them with one, equal in value to the sum $r_1 + r_2$. If two resistances are connected in parallel, then they can also be replaced with one, the value of which is $R = \frac{r_1 r_2}{r_1 + r_2}$

These are the simplest examples of circuit conversion. We will focus on more complex schemes.

Let’s see how circuits that have three terminals each are converted into each other - “star” (Fig. 1, a) and “triangle” (Fig. 1, b).

The slightly unusual designations (Fig. 1, b) are very convenient, the index showed between which points the resistance is connected. For example, resistance $R_{13}$ is connected between points 1 and 3, etc.

If we want to replace one of these circuits with another, we need to obtain such correspondences between $r$ and $R$ that the resistances between any points are the same for both circuits.

In the “star” circuit (Fig. 1, a) the resistor

12

Fig. 4.

“triangle” ACD “star” as shown in Fig. 4.

Now it is clear that the resistance between points A and B will be equal to
$$\frac{1}{3} + \frac{28}{33} = \frac{13}{11} \text{ ома}.$$

We replaced the “triangle” ACD with a “star”, but it was possible to solve the problem differently - replacing the “star” ADB with a “triangle”.

Let now a battery with negligible small internal resistance and e.m. be connected to points A and B. With. E=1 in. We need to find the current through section CB.

The circuit must be transformed so that the resistance CB of interest to us is not affected. The transformation we did earlier (Fig. 4) will do.

Using that $R_{AB} = \frac{13}{11} \text{ ома}$ we get $I = \frac{E}{R_{AB}} = \frac{11}{13} a$. After branching, the points are added in the ratio inverse to the resistances of the branches:

$$\frac{I_1}{I_2} = \frac{7}{4}.$$

At the same time $I_1 + I_2 = I$. (I is the current before the branching, $I_1$ is the current in the upper branch, $I_2$ is the current in the lower branch).

From here

$$I_1 = \frac{7}{13} a$$

It would be a little more difficult to find the current flowing through the CD section. To do this we would still have to find a current,

13

Fig. 6.

potentiometer. Moreover, the sum of resistance $5r+R_1$ and $7r+R_2$ remains constant. Let’s see what they should be so that the total resistance is maximum. We denote $5r+R_1$ by $r_1$ and $7r+R_2$ by $r_2$. Then the total resistance of the parts of the circuit connected in parallel is $r_0 = \frac{r_1 r_2}{r_1 + r_2}$. If we take into account that $r_1+r_2 = const = c$, then the resistance $r_0$ will be expressed as follows: $\frac{r_1 (c-r_1)}{c}$. This expression is maximum when the number of slipels is maximum. But $y = (cr_1-r_1^2)$ is the equation of a parabola (Fig. 7), the branches of which intersect the x-axis at points $O$ and $c$. Therefore, the numerical fraction is greatest at $r_1 = \frac{c}{2}$. Since $r_1+r_2 = c$, this means that the resistance between points $A$ and $B$ is maximum if $r_1=r_2$, that is $5r+R_1=7r+R_2$, or $R_1-R_2=2r$.

It is clear that this is only possible if the resistance of the entire potentiometer $R=R_1+R_2$ is not less than $2r$. Otherwise, the maximum resistance between points $A$ and $B$ is achieved when the potentiometer slider is in its extreme position.

So the answer is: $R \leq 2r$.

The method we have described is very convenient for sequentially transforming a complex diagram into a simple form. It allows you to calculate practically any complex target consisting of resistances.

This method can be applied to circuits containing more than just resistances.

Please note that we did not talk anywhere about physical processes in the circuit, but only used the formal expression for Ohm’s law: $U=r \cdot I$. It follows from it that when connected in series, their values ​​are added, and when connected in parallel, the values ​​reciprocal to resistance $\left(\frac{1}{r_{06\pi}} = \frac{1}{r_1} + \frac{1}{r_2}\right)$ are added. It is clear that if any other physical quantities are related by a law similar to Ohm’s law, then all our conclusions are valid for them.

As an example, consider a circuit with a capacitor (Fig. 8). We know that $Q=C \cdot U$, or $U = \frac{1}{C} Q$.

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Fig. 9.

Let’s compare this with the expression for Ohm’s law $U=r \cdot I$.

It can be seen that the laws of positivity, only then together there is a charge, and together with resistance there are quantities, the inverse capacitance. This means that in order to find, say, charges on capacitors, you can do this. Instead of a circuit containing capacitors, draw a target containing resistances, and replace the capacitor with a capacitance of $x$ farads with a resistance $r = \frac{1}{x} \text{ om}$. After we calculate the current in a circuit of resistances, we can immediately write what the charges are on the capacitors: if a current of $I=y$ amers flows through the resistance, then the corresponding capacitor will have a charge of $Q=y$ coulombs.

E.m.f. The batteries remain unchanged during this transformation. But, of course, in a circuit with capacitors, the internal resistance of the batteries does not affect the result. Therefore, the goal changes, we will have to deprive the batteries of their internal resistances.

Let, for example, you need to find the charge on a capacitor with capacity $10 \text{ мкф}$ (Fig. 9). To the capacitor

15

DO YOU LIKE FUCKING?
WITH WHOLE NUMBERS?

M. I. Bashmakov

Several schoolchildren were asked to answer the following question:
Find out whether the equality $3^{100} + 7^{100} = 8^{100}$ is true or false.
After a few minutes they all replied that this equation was false.
Here are the most typical obligations.

1. Let’s look at the last digit of the numbers on the right and left. 3 squared is 9, 3 cubed ends in 7, in the fourth power ends in 1, in the fifth power ends in 3, in the sixth power ends in 9, etc. After four powers, the last digit is repeated, so the number $3^{100}$ has the same last digit as $3^4$, that is, 1. In the same way, it is easy to assume that the last digit of $7^{100}$ is also 1, and $8^{100}$ is 6. So, the last digit of the number is 2, and the last digit on the right is 6. Therefore, the numbers on the left and right are different.

2. $3 + 7$ is greater than 8. But $3^2 + 7^2 = 58$ is less than $8^2 = 64$. Next, $3^{100} + 7^{100} = (3^2)^{50} + (7^2)^{50} < (3^2 + 7^2)^{50} = 58^{50} < 64^{50} = 8^{100}$. We found that the number on the left is less than on the right.

3. The number on the left is divisible by 2, but not divisible by 4. Indeed, any odd number squared gives a remainder of one when divided by 4: $(2n + 1)^2 = 4n^2 + 4n + 1$. The sum of two such numbers $(3^{50})^2 + (7^{50})^2$ gives a remainder of 2. On the right is a number divisible by 4.

Is the evidence given similar to each other? It is easy to notice the similarity of the first and third: in both, the remainders from division by some integer are compared, in one - by 10 (the last digit), and in the other - by 4. But finding an analogy between this proof and the second is not at all so easy. We will try to calculate it at the end of this article. For now, let’s learn how to apply the first method of reasoning to some other problems.

Computations “modulo”

Problem a 1. Prove that the equation $x^3 = 2 + 3 y^2$ has no solutions in integers.

Let us consider this equation “modulo 9”, that is, up to terms that are multiples of division *). Since each integer $x$ has the form $3n$, $3n - 1$ or $3n + 1$, where $n$ is an integer, $a(3n \pm 1)^3 = 27n^3 \pm 27n^2 + 9n \pm 1 = 9k \pm 1$, where $k$ is an integer, then any number cubed is either divisible by 9, or leaves a remainder of 1 or 8 when divided by 9.

*) In general, two integers $a$ and $b$ are said to be comparable modulo $k$ if the difference $a - b$ is divisible by $k$. For those who are not familiar with this concept, we advise you to look at the article by A. A. Egorov “Arithmetic of remainders” in «Kvant» No. 5, 1970.

Likewise, since $3(3n \pm 1)^2 = 27n^2 \pm 18n + 3$,

$3y^2$ gives a remainder of 0 or 3 when divided by 9. So, the right side of our equation modulo 9 can only equal 2 or 5, and the left side can only equal 0, 1 or 8, so they cannot coincide for any purpose. The modulo calculations themselves are easy to do. Let us formulate the basic rule of these calculations: if, when adding (or multiplying), we replace numbers comparable to them in terms of

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modulo $k$, then we obtain a number comparable to the sum (or product) modulo $k$. In other words, in an equality in which the operations of addition and multiplication function, we can replace numbers with modulo $k$ comparable to them whenever in the answer we are interested only in the remainder of division by $k$.

The fact that $a$ and $b$ are comparable modulo $k$ is written as follows: $a \equiv b \pmod{k}$. The rules for operations on comparisons can be written as follows:

$$\text{если } \begin{cases} a \equiv b \pmod{k}, \\ c \equiv d \pmod{k}, \end{cases}$$

$$\text{то } \begin{cases} a + c \equiv b + d \pmod{k}, \\ ac \equiv bd \pmod{k}. \end{cases}$$

Recall that by definition, $a \equiv b \pmod{k}$ means that $a - b = k \cdot q^*$.

Will engage in such a rather artificial task.

Task 2. Prove that the equality $1153^{60} + 211^{60} = 27027 \cdot 100^x + 2$ is impossible for any integer $x$.

It is clear that if for some modulus we do not obtain equality, then it cannot exist in the general sense. Modulo 2 there is no contradiction - there are even numbers on the right (zeros modulo 2). Will work on module 3. $1153 \equiv 1 \pmod{3} = 1153^{60} \equiv 1 \pmod{3}$; $211 \equiv 1 \pmod{3} = 211^{60} \equiv 1 \pmod{3}$. On the right is $27027 \equiv 0 \pmod{3}$. Therefore, the entire right-hand side gives a remainder of 2. Again, we do not obtain a contradiction. If you have enough patience, make sure that modulo all numbers up to 16 inclusive, the written equality is true. (You’ll notice that $27027 \cdot 100^x, x \geq 2$ is divided by each of these numbers, leaving a remainder of 2.)

*) Be careful with division comparisons. If $ac \equiv bc \pmod{k}$, then it is not necessarily true that $a \equiv b \pmod{k}$. For example, $27 \equiv 3 \pmod{6}$ but $9 \neq 1 \pmod{6}$. However, if $c$ is coprime with the modulus $k$, then it is possible to cancel by $c$. This is a consequence of such a gentle and not simply provable fact: if $c$ is coprime with $k$, then there is a number $x$ such that $cx \equiv 1 \pmod{k}$.

If you are consistent, then modulo 17 you will get the following equalities: $(-3)(-2)^x + 2 \equiv 0 \pmod{17}$. By going through the values ​​of $x$ in a row (the cycle here will be of length 8), you will never get the desired equality. A contradiction will be found.

The calculations done will immediately raise many questions:

(1) If the equality is false, can this be detected by any modulus?

(2) Is it possible to somehow indicate those modules for which contradictions are more likely to be obtained?

(3) Is the behavior of equality across different modules related?

If we are talking about some specific numerical equality (as was the case in our very first example), then, of course, there is always a number comparable in modulus to which we find a contradiction (for example, we can take a number greater than the common parts of the equality). Equalities in which there are variables are much more interesting (this was the case in the last two examples). For equations in integers, each of the three questions posed is interesting and not always fully explored.

A little unproven

The first equations in integers that pose serious difficulties in solving are quadratic equations*. Mathematicians have long been interested in questions of this kind: can a given number be represented as the sum of two squares? It turned out that the solvability of equations of the form $x^2 + y^2 = a$ and other more general ones, given by polynomials of the second degree, is evidence for checking the solvability of comparison (positive

*) The theory of linear integer equations is quite simple. At the same time, it contains a number of interesting tasks that are close to elementary, and deserves a separate stat.

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researching questions concerning objective numbers: pose these questions “modulo $p$”, and then draw general conclusions from the behavior of all $p$. This method of strangeness was called the “Minkowski-Hasse principle.” There have been and continue to be problems for which it gives the result: a problem has a solution in terms of numbers if and only if it is solvable in any modulus. It was all the more disappointing to find examples where this principle does not work. One of the first examples of this kind (but not the very first chronologically) was an example found by the Norwegian mathematician Selmer. The equation $3x^3+4y^3+5z^3=0$ turned out to be solvable in any modulus (and, of course, in real numbers), but unsolvable in terms of numbers (other than zero). This example served as the starting point for much research on Diophantine equations (that is, equations in numerical terms) over the past 15 years. It should only be said that during this time the number of unclear questions and hypotheses has greatly increased, and the true roots of what will actually happen have not become clear.

Distances between rational numbers

The most important modern development in number theory is the new understanding of “modulo calculations”. For simplicity, let’s move on from whole numbers to rational ones. (This transition is often made painlessly - for example, the solutions to the equation $x^n+y^n=z^n$ in terms of numbers are very closely related to the solutions to the equation $x^n+y^n=1$ in rational numbers.) It is clear that if we want to solve an equation in rational numbers, we must find out whether it can be solved in real numbers. This solvability is much easier to check: you can extract various roots, take logarithms, write inequalities, etc.

But let’s remember how we got real numbers from rational numbers? One of the points of view is this: we can consider sequences of rational numbers, the distance between neighboring terms of which decreases without limit. For example, 1; 1.4; 1.41; 1.4142; … Such a sequence (which can be imagined as a sequence of changes with age-related accuracy) determines a real number. How does the deviation of neighboring members of the sequence from each other change? If the sequence is written like this: $a_1; a_2; a_3; ...$, then the distance between the terms is written as $|a_m-a_n|$. Thus, the basis for obtaining real numbers from rational numbers is the distance, defined as the absolute value of the difference. Let’s denote the distance between two numbers as follows: $d(a, b) = |b-a|$. Try displaying the following distance properties*:

1. $d(a, b) \geq 0$, with $d(a, b) = 0 \Rightarrow a=b$;
2. $d(a, b) = d(b, a)$;
3. $d(a, c) \leq d(a, b) + d(b, c)$ (triangle inequality);
4. $d(a+c, b+c) = d(a, b)$;
5. $d(ab, 0) = d(a, 0) \cdot d(b, 0)$.

It turned out that, in addition to the general absolute value, rational numbers have many other distances that have the 1-5 property. Let us fix an arbitrary prime number $p$. Any rational number $a$ (except zero) can be represented as $a=p^v \cdot r$, where $v=v(a)$ is an integer, and the rational number $r=r(a)$ is such that when written in

*) The first three of these properties are the main ones that satisfy the general distance on the plane and in space; these are aximons that any “metric” must satisfy (see the article by N.B. Vasiliev “Metric spaces” in «Kvant» No. 10, 1970). The fourth and fifth properties of real distance with the operations of addition and multiplication of numbers. A metric that satisfies them is said to be a field metric (for example, $d(b, a) = |b-a|$ is the metric of the field of real numbers).

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in the form of an irreducible fraction, $p$ is not included in either the numerator or the denominator. (Note that $v(a)$ can be either zero or a negative integer. For example, if $p=3$, $a=\frac{73}{324}=3^{-4}\cdot\frac{73}{4}$, then $v(a)=-4$.) Now take an arbitrary positive number $\lambda<1$. Let’s define the distance $d_p$ as follows: $d_p(a,b)=\lambda v(a-b)$, if $a\neq b$, and $d_p(a,a)=0$. It is easy to check (try it yourself) that all distance properties will be satisfied.

Since there is a new distance, we can try to obtain new numbers by constructing sequences of rational numbers $a_1, a_2, \ldots$, so that the distances between neighboring ones decrease. This means that $\lambda v(a_{n+1}-a_n)$ as $n$ increases should become smaller and smaller and tend to zero, and then $v(a_{n+1}-a_n)$ should increase.

As a result, we obtain new numbers, which are called $p$-adic $^*$. They are very similar to real ones. Just as easily as for real numbers, questions can be explored when it is possible to take the root, logarithm, and perform other operations on $p$-adic numbers. The problem of the solvability of an equation according to the power model of a given prime number will be replaced by the problem of its solvability in $p$-adic numbers.

The question arises whether we have found all the distances between rational numbers. It turns out that yes. There is Ostrovsky’s denominator theorem (you can find it in the book on number theory cited above), which shows that all distances are exhausted by a common absolute value and the $p$-adic distances introduced by us for each prime $p$.

We have, as it were, equalized the behavior of an equation based on the model of a single prime number and its power, on the one hand, and its behavior in ordinary real numbers

$^*)$ It is easy to prove that the choice of $\lambda$ does not affect the forgiveness and properties of $p$-adic numbers. One can, if one likes, assume everywhere that $\lambda=\frac{1}{2}$ or $\lambda=\frac{1}{p}$.

numbers - on the other. In both cases, numbers are compared by their distance from each other, but these distances are chosen differently. (Then we see how the second and third solutions to the problem with which our article began are similar.)

Understanding of this fact greatly enriched number theory, again brought into it powerful methods of function theory and geometry, previously developed for general real numbers, but successfully used for $p$-adic numbers, making this sum the ancient branch of modern mathematics one of the most flourishing.)

Exercises

1. Prove that none of the equations a) $x^2+y^2=1971$, b) $19x^3-17y^3=50$, b) $3x^2-y^2=5^z$, r) $x^3+y^3+7z=3$ has integer solutions.

2. Prove that from the numbers 1, 2, 3, 4, 5, 6, 7 it is impossible to form two different seven-digit numbers, one of which is divisible by the other. (All digits must be used in each digit.)

3. Prove that a number whose decimal notation consists of thirty ones, some zeros, and some nines is not a perfect square.

4. Find the 5-adic distance (that is, the distance $d_5$) between the numbers: a) 0 and $\frac{2}{5}$; b) $1\frac{31}{45}$ and $6\frac{2}{15}$; b) 6 and 631.

5. Prove that for 3-adic numbers the equality is true

$$1+3+9+27+\ldots=-\frac{1}{2}$$

(in other words, a sequence of distances

$$\alpha_n=d_3\left(1+3+\ldots+3^n,-\frac{1}{2}\right)$$

tends to zero).

What is the sum of $1+3^2+3^4+3^6+\ldots$?

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The board moves on round rollers. At the same time, it maintains a horizontal position and remains at the same height. Why is this happening? Yes, because the circle has the same width no matter how it turns.

Will the dock move the same way if you use non-circular rollers? At the first moment it seems that this is impossible. Let’s not rush, however. It turns out that there are curves that, like a circle, have the same width in all directions. They are called lines of the same width.

One of the simplest curves of this kind is shown in figure 1. It consists of two $A_1B_1, A_2B_2, A_3B_3$ of radius $r$ and two $B_1A_3, B_3A_2, B_2A_1$ of radius $R$, the centers of which are located at the vertices of a regular triangle $O_1O_2O_3$. It is easy to see that the width of this curve is the same in all directions and is equal to $R+r$. Consequently, rollers whose profiles are sequential along such curves will work no worse than round ones (Fig. 2).

The curve shown at figure 1 can be obtained as follows. Let’s take the segment $A_1B_3$ and rotate it by $60^\circ$ around the point $O_1$ and move it to position $B_1A_2$. Then, rotate the segment $B_1A_2$ by $60^\circ$ around the point $O_2$, we get the segment $A_3B_2$ and, finally, rotate the segment $A_3B_2$ by $60^\circ$ around the point $O_3$, again we arrive at the segment $A_1B_3$. In the process of these turns, one end of the segment is considered, arcs $A_1B_1A_3B_3$, and the other - arcs $B_3A_2B_2A_1$. Therefore, the curve shown in figure 1 can be considered as a trajectory described by the ends of some

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Fig. 3.

The width of all these curves is constructed and is equal to the length of the rotated segment. You can rotate a segment around one of its ends. This way we get the simplest curve of equal width, different from a circle (Fig. 4). In this case, the segment $AB$ rotates first around the point $A$, moving to the position $AB'$, then around the point $B'$, passing into the segment $B'B$, and, finally, around the point $B$, closing the curve.

Let us now consider the non-actor segment $AB$. Let it roll without sliding along the SS curve. Then in every

Fig. 4.

In a separate moment, such movement can be considered as rotation of a segment relative to the point of contact. The tangent point about which the rotation is currently being made is called the instantaneous center of rotation (for example, point $C$ on figure 5).

The idea of ​​an instantaneous center of rotation makes it easy to calculate the velocities of various points on the segment $AB$. For example, for the speed of points $A$ and $B$ we have

$$v_A = \omega \cdot CA,$$
$$v_B = \omega \cdot CB,$$

where $\omega$ is the angular velocity of the segment $AB$ in its “instantaneous” rotation around the center $C$. In this case, the velocities of the points under consideration will be perpendicular to the corresponding radius of rotation, that is, the segments $CA$ and $CB$. In this way, you can calculate the speed of any point lying on this segment or rigidly connected with this segment. If, for example, a non-actor figure is rigidly connected to the segment $AB$, then the speed of any point $P$ is equal to the product $\omega \cdot CP$.

Let us now return to the constructed width curves. Let the segment $AB$ roll along the curve SS. Since the velocities $v_A$ and $v_B$ are perpendicular to the radii $CA$ and $CB$, they are parallel to each other. But the speed $v_A$ is directional

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Fig. 6.

line is tangent to the trajectory of point $A$, and speed $v_B$ is tangent to the trajectory of point $B$. Therefore, these tangents are parallel to each other, and the distance between them remains equal to $AB$ all the time. But this means that the trajectories $A_1A_2$ and $B_1B_2$ can be considered as parts of some curve of constant cyrin.

However, we need to construct not two sections, but just this curve. If we want to do this using the described operation, we must choose the line $SS$ so that, by drawing it, the segment $AB$ rotates to $180^\circ$, that is, so that the point $A$ comes to the position $B$, and the point $B$ to the position $A$. However, of such lines - $PQRP$ - is shown on figure 6. Let us trace in more detail how the segment $AB$ describes it.

First it rolls along area $PQ$ and comes to position $A'B'$. Moreover, its ends describe the arcs $AA'$ and $BB'$. Then it rolls along the section $QR$, and its ends describe the arcs $A'A''$ and $B'B''$. Finally, it marks the part $RP$, and its ends move along the arcs $A''B$ and $B''A$. As a result, point $A$ comes to position $B$, and point $B$ to position $A$, and the curve of the constructed cirina is closed. In this case, arcs $PQ$, $QR$

and $RP$ can have any shape and do not have to be the same. The only thing that is required of them is that they be convex and touch each other as shown in figure. As for the segments $AP$ and $PB$, their length cannot be arbitrary, or then the arc described by the point $A$ will not “join” the arc described by the point $B$. Here you can build as follows. Let us draw a straight line through the point $P$, tangent to the arc $PQ$, and place the point $A$ on it. Let us now force this straight line to line $PQRP$. Then the end of the arc, which will cover the point $A$, determines the position of the point $B$.

Line $PQRP$, shown in figure 6, consists of three arcs. But the number of arcs may be greater. For example, on figure 7 the curve of the constructed tsirina is shown, obtained by drawing the line $PQRSTP$, consisting of the heel of the arcs.

The curves of the constructed cirina have one more property that makes them similar to the circle. They are expressed by the following theorem:

The perimeter of the curve of the constructed cirina is $l=\pi D$, where $D$ is the width of this curve.

Let’s prove this theorem.

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Fig. 8.

Let the horizontal straight line $PQ$ be moved using rollers of constant width (Fig. 8). In the distance is one of these skating rinks, for example the left one. Let its width be $D$ and its perimeter be $l$. When the straight line $PQ$ moves so much that this roller makes one revolution, it moves relative to the straight line $MN$ by a distance $l$ to the right. But since it rolls not only along the straight line $MN$, but also along the straight line $PQ$, then after one revolution it turns out to be folded relative to the straight line $PQ$ by a distance $l$ to the left. Consequently, the displacement of straight line $PQ$ relative to straight line $MN$ will be equal to $2l$.

But since this same mixing is equal to $vt$, where $t$ is the time during which the roller makes one revolution, then

$$2l = vt. \tag{1}$$

(We assume that straight line $PQ$ moves at a constant speed.) The instantaneous center of rotation of the roller is at point $A$, which means that the speed of movement

$$v = \omega \cdot AB = \omega D,$$

where $\omega$ is the angular speed of the roller. Substituting this expression into equality (1), we obtain

$$2l = \omega Dt. \tag{2}$$

But $\omega t$ is the angle of rotation of the roller per revolution. Hence,

$$\omega t = 2\pi,$$

and equality (2) takes the form

$$2l = 2\pi D,$$

where

$$l = \pi D.$$

Thus, the perimeter of a constant width curve is calculated in the same way as the perimeter of a circle. What is the area enclosed by a curve of constant width? Can it be calculated in the same way as the area of ​​a circle? It turns out not. However, the area of ​​a scapula of constant width can be calculated as the area of ​​the scapula between two concentric circles. More precisely this means the following.

Consider a curve of constant width (Fig. 9). Through an arbitrary point $C$ we draw a normal on it (a straight line perpendicular to the tangent is normally called) and plot a segment $CC'$ of a given length $d$ along it. By setting aside such segments from each point of a given curve, we obtain a new curve, which is the geometric locus of the ends of the delayed segments. Obviously, it will also have a constant width. If we now denote by $S$ the area enclosed between these curves, then the equality will be true

$$S = \frac{\pi D_2^2}{4} - \frac{\pi D_1^2}{4},$$

where $D_1$ is the width of the inner curve, and $D_2$ is the width of the outer one.

Try to prove it yourself.

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SEMI-CORRECT

L. M. Lopovok POLYGONS

By definition, a convex polygon is called regular if it is equiangular and equilateral. However, there is an interesting class of figures that do not possess these two properties at the same time. We will call half-right and the following polygons:

a) equiangular, in which the sides are equal through one;
b) equilateral, in which the angles are equal through one.

An example of a semi-right polygon of the first kind is a rectangle, and an example of the second is a rhombus. From the description of half-right polygons it is clear that each of them has an even number of vertices; let’s denote it by $2n$.

Properties of half-right equiangular polygons

1. Opposite sides of an equiangular polygon are parallel.

Indeed, when traversing the polygon, we replace that each side is rotated relative to the previous one by an external angle, that is, by $\frac{180^\circ}{n}$ (Fig. 1). Therefore, side $A_{n+1}A_{n+2}$ is rotated relative to side $A_1A_2$ by the amount of $n$ external angles, that is, by $180^\circ$.

2. If you connect the vertices of a half-right equiangular polygon through one (all odd or all even vertices), you get a regular polygon (Fig. 1).

The proof is not difficult: the triangles $A_1A_2A_3$, $A_3A_4A_5$, $A_5A_6A_7$, … are equal according to the first criterion, which means $A_1A_3=A_3A_5=\dots$. On the other hand, from the equality of triangles it follows that their corresponding angles are equal. Let us denote the interior angle of an equal $2n$-gon by $\alpha_{2n}$. We find the interior angles of our best polygon: they are equal

$$\alpha_{2n} - (\beta + \gamma) = \alpha_{2n} - (180^\circ - \alpha_{2n}) =$$

$$= 2\alpha_{2n} - 180^\circ = 2 \cdot 180^\circ \left( 1 - \frac{1}{n} \right) -$$

$$- 180^\circ = 180^\circ \left( 1 - \frac{2}{n} \right),$$

Fig. 1.

4. Quantum No. 3

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as it should be for a regular $n$-gon.

3. A circle can be described around a semi-regular equiangular polygon.

Let us describe a circle around the regular $n$-gon $A_1A_3A_5\dots$ (Fig. 2). The interior angle of an equiangular $2n$-gon is $180^\circ \left(1 - \frac{1}{n}\right)$. It is easy to calculate that $\Rightarrow A_1A_2A_3$ is half the degree value of the arc $A_3A_5A_1$, which means that point $A_2$ lies on the constructed circle. This also applies to other vertices: $A_4$, $A_6$,…

Property 3 is based on two methods for constructing semiregular equiangular polygons.

a) Construct a regular $n$-gon $A_1A_3A_5\dots A_{2n-1}$ and describe a circle around it. Then we find (using a compass) on the circle the points $A_2$, $A_4$, $A_6$,…, $A_{2n}$, located at the same distance from the corresponding vertices of the $n$-gon: $A_1A_2=A_3A_4=..$

b) We construct a triangle $A_1A_2A_3$ using two sides (of arbitrary length) and a known angle between them, and describe a circle around this triangle. Then, knowing the sizes of all sides, we successively find the remaining vertices using the notches of a compass.

4. If the number of vertices of a semiregular equiangular polygon is a multiple of 4, then more diagonals intersect at one point and bisect it.

To prove use 2: these diagonals are also the diagonals of a regular polygon with side four, so they intersect at one point—the center of the polygon’s circumscribed circle—and are bisected by that point.

If the number of vertices of a semiregular equiangular polygon is not a multiple of 4, then it is not difficult to construct an example (Fig. 3), when more diagonals intersect at several points.

5. The center of a circle circumscribed about a semiregular equiangular $4n$-gon is its symmetrical center.

From property 4 it follows that with central symmetry with respect to the circle, the vertices of the polygon transform into vertices; therefore, the parties also change sides.

6. The straight line passing through the midpoints of two parallel sides of a semiregular equiangular polygon is its symmetrical axis.

According to the property of parallel chords of a circle, this straight line passes through the center of the circumscribed circle. From the equality of the corresponding chords on both sides of the line, the symmetry of the corresponding vertices follows

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A polygon with respect to a drawn line (see further Fig. 6).

7. Let a semiregular equiangular polygon have 2n vertices, then the center of the circle circumscribed around it is its symmetric center of order n.

This means that when rotated around the center of the circle by an angle that is a multiple of $\frac{360^\circ}{n}$, the polygon aligns with itself. This follows from properties 2 and 3.

8. The sum of the distances from the interior point to the sides (or their extensions) of a semi-regular equiangular polygon is constructed.

As is known, a semiregular polygon has this property.

If you connect the interior point of the equigon with the vertices, then the sum of the areas of the resulting triangles is equal to the area of ​​the polygon. Denoting the side of the polygon by $a$, and the distance to the side by $x_0, x_1, x_2, x_3, \dots$, we obtain

$$S = \frac{ax_0}{2} + \frac{ax_1}{2} + \frac{ax_2}{2} + \frac{ax_3}{2} + \dots;$$

$$x_0 + x_1 + x_2 + x_3 + \dots = \frac{2S}{a} = \text{const.}$$

Let the side $A_1A_2$ be larger than the side $A_2A_3$ in a given polygon. Let us construct a regular 2n-gon (Fig. 4) on $A_1A_2$. Then all the sides of the original polygon (except three) are respectively parallel to the sides of the regular polygon, and three ($A_1A_2$ and funny with it) lie on the sides of the regular polygon. Therefore, the distances from the internal point $M$ to the side of the correct polygon are as follows:

$$x_0 = y_0, \ x_1 = y_1, \ x_2 = y_2, \ x_3 = y_3 + b_1,$$

$$x_4 = y_4 + b_2, \ x_5 = y_5 + b_3, \dots,$$

where $b_1, b_2, b_3$ are constant, and the required sum

$$y_0 + y_1 + y_2 + y_3 + y_4 + \dots = (x_0 + x_1 + x_2 + x_3 + x_4 + \dots) - (b_1 + b_2 + \dots) = \text{const.}$$

9. If you connect the sides of an equiangular equilateral polygon in series, you get a semi-regular equilateral polygon.

Indeed, by Property 2, all small diagonals of a semiregular equiangular polygon are equal to each other. The sides of the equigon (Fig. 5) are half the size of these diagonals, since it is the average of the linear triangles $A_1A_2A_3, A_2A_3A_4, A_3A_4A_5, \dots$. Using property 7, it is easy to install