I hope you have already read about the number circle and know why it is called a number circle, where is the origin of coordinates on it and in which direction is the positive direction. If not, then run! Unless, of course, you are going to find points on a number circle.
Denote the numbers \(2π\), \(π\), \(\frac(π)(2)\), \(-\frac(π)(2)\), \(\frac(3π)(2 )\)
As you know from the last article, the radius of a number circle is \(1\). This means that the circumference is \(2π\) (calculated using the formula \(l=2πR\)). With this in mind, note \(2π\) on the number circle. To mark this number, you need to go from \ (0 \) along the numerical circle, the distance is \ (2π \) in the positive direction, and since the circumference is \ (2π \), it turns out that we will make a full turn. That is, the number \(2π\) and \(0\) correspond to the same point. Don't worry, multiple values for one point is normal for a number circle.
Now let's denote the number \(π\) on the number circle. \(π\) is half of \(2π\). Thus, to mark this number and its corresponding point, you need to go from \ (0 \) in the positive direction half a circle.
Note the point \(\frac(π)(2)\) . \(\frac(π)(2)\) is half of \(π\), therefore, to mark this number, you need to go in the positive direction from \(0\) equal to half of \(π\), i.e. quarter circle.
Let us denote the points \(-\)\(\frac(π)(2)\) on the circle. We move the same distance as last time, but in a negative direction.
Let's put \(-π\). To do this, let's go a distance equal to half the circle in the negative direction.
Now let's look at a more complicated example. Mark the number \(\frac(3π)(2)\) on the circle. To do this, we translate the fraction \(\frac(3)(2)\) into \(\frac(3)(2)\) \(=1\)\(\frac(1)(2)\) , i.e. e. \(\frac(3π)(2)\) \(=π+\)\(\frac(π)(2)\) . So, it is necessary from \ (0 \) in the positive direction to go the distance to half the circle and another quarter.
Exercise 1. Mark the points \(-2π\),\(-\)\(\frac(3π)(2)\) on the number circle.
Denote the numbers \(\frac(π)(4)\), \(\frac(π)(3)\), \(\frac(π)(6)\)
Above, we found the values at the points of intersection of the number circle with the axes \(x\) and \(y\). Now let's determine the position of the intermediate points. First, we plot the points \(\frac(π)(4)\) , \(\frac(π)(3)\) and \(\frac(π)(6)\) .
\(\frac(π)(4)\) is half of \(\frac(π)(2)\) (that is, \(\frac(π)(4)\) \(=\)\ (\frac(π)(2)\) \(:2)\) , so the distance \(\frac(π)(4)\) is half a quarter circle.
\(\frac(π)(4)\) is a third of \(π\) (in other words, \(\frac(π)(3)\) \(=π:3\)), so the distance \ (\frac(π)(3)\) is a third of a semicircle.
\(\frac(π)(6)\) is half of \(\frac(π)(3)\) (because \(\frac(π)(6)\) \(=\)\(\frac (π)(3)\) \(:2\)) so the distance \(\frac(π)(6)\) is half the distance \(\frac(π)(3)\) .
This is how they are located relative to each other:
Comment: Location of points with value \(0\), \(\frac(π)(2)\) ,\(π\), \(\frac(3π)(2)\) , \(\frac(π)( 4)\) , \(\frac(π)(3)\) , \(\frac(π)(6)\) is better to just remember. Without them, the number circle, like a computer without a monitor, seems to be a useful thing, but it is extremely inconvenient to use.
Different distances on the circle clearly:
Denote the numbers \(\frac(7π)(6)\), \(-\frac(4π)(3)\), \(\frac(7π)(4)\)
Denote a point on the circle \(\frac(7π)(6)\) , to do this, perform the following transformations: \(\frac(7π)(6)\) \(=\)\(\frac(6π + π)( 6)\) \(=\)\(\frac(6π)(6)\) \(+\)\(\frac(π)(6)\) \(=π+\)\(\frac( π)(6)\) . This shows that from zero to the positive side, you need to go the distance \(π\), and then another \(\frac(π)(6)\) .
Mark a point \(-\)\(\frac(4π)(3)\) on the circle. Transform: \(-\)\(\frac(4π)(3)\) \(=-\)\(\frac(3π)(3)\) \(-\)\(\frac(π)( 3)\) \(=-π-\)\(\frac(π)(3)\) . So it is necessary to go from \(0\) in the negative direction the distance \(π\) and also \(\frac(π)(3)\) .
Draw a point \(\frac(7π)(4)\) , for this we transform \(\frac(7π)(4)\) \(=\)\(\frac(8π-π)(4)\) \ (=\)\(\frac(8π)(4)\) \(-\)\(\frac(π)(4)\) \(=2π-\)\(\frac(π)(4) \) . So, to put a point with the value \(\frac(7π)(4)\) , it is necessary to go from the point with the value \(2π\) in the negative direction the distance \(\frac(π)(4)\) .
Task 2. Mark the points \(-\)\(\frac(π)(6)\) ,\(-\)\(\frac(π)(4)\) ,\(-\)\(\frac (π)(3)\) ,\(\frac(5π)(4)\) ,\(-\)\(\frac(7π)(6)\) ,\(\frac(11π)(6) \) , \(\frac(2π)(3)\) ,\(-\)\(\frac(3π)(4)\) .
We denote the numbers \(10π\), \(-3π\), \(\frac(7π)(2)\) ,\(\frac(16π)(3)\), \(-\frac(21π)( 2)\),\(-\frac(29π)(6)\)
We write \(10π\) as \(5 \cdot 2π\). We recall that \(2π\) is a distance equal to the circumference, so to mark the point \(10π\), you need to go from zero to a distance equal to \(5\) circles. It is easy to guess that we will be back at the point \(0\), just make five turns.
From this example, we can conclude:
Numbers with a difference of \(2πn\), where \(n∈Z\) (that is, \(n\) is any integer) correspond to the same point.
That is, to put a number with a value greater than \(2π\) (or less than \(-2π\)), you need to select from it an even number \(π\) (\(2π\), \(8π\), \(-10π\)…) and discard. Thus, we will remove from the number “empty turns” that do not affect the position of the point.
Another conclusion:
The point corresponding to \(0\) also corresponds to all even numbers \(π\) (\(±2π\),\(±4π\),\(±6π\)…).
Now let's put \(-3π\) on the circle. \(-3π=-π-2π\), so \(-3π\) and \(–π\) are in the same place on the circle (because they differ by an “empty turn” in \(-2π\)).
By the way, all odd \(π\) will also be there.
The point corresponding to \(π\) also corresponds to all odd numbers \(π\) (\(±π\),\(±3π\),\(±5π\)…).
Now let's denote the number \(\frac(7π)(2)\) . As usual, transform: \(\frac(7π)(2)\) \(=\)\(\frac(6π)(2)\) \(+\)\(\frac(π)(2)\ ) \(=3π+\)\(\frac(π)(2)\) \(=2π+π+\)\(\frac(π)(2)\) . We discard two pi, and it turns out that, to denote the number \(\frac(7π)(2)\), you need to go from zero in the positive direction to a distance equal to \(π+\)\(\frac(π)(2)\ ) (i.e. half a circle and another quarter).
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Lesson and presentation on the topic: "Number circle on the coordinate plane"
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Manuals and simulators in the online store "Integral" for grade 10 from 1C
Algebraic problems with parameters, grades 9–11
We solve problems in geometry. Interactive construction tasks for grades 7-10
What will we study:
1. Definition.
2. Important coordinates of the numerical circle.
3. How to find the coordinate of a numerical circle?
4. Table of the main coordinates of the numerical circle.
5. Examples of problem solving.
Definition of a number circle on the coordinate plane
Let's place the number circle in the coordinate plane so that the center of the circle is aligned with the origin, and its radius is taken as a unit segment. The starting point of the numerical circle A is aligned with the point (1;0).Each point of the number circle has its x and y coordinates in the coordinate plane, and:
1) for $x > 0$, $y > 0$ - in the first quarter;
2) with $x 0$ - in the second quarter;
3) for $x 4) for $x > 0$, $y
For any point $M(x; y)$ of the numerical circle, the following inequalities hold: $-1
Remember the equation of the number circle: $x^2 + y^2 = 1$.
It is important for us to learn how to find the coordinates of the points of the numerical circle shown in the figure. 
Find the coordinate of the point $\frac(π)(4)$
The point $M(\frac(π)(4))$ is the middle of the first quarter. Let us drop the perpendicular MP from the point M to the line OA and consider the triangle OMP. Since the arc AM is half of the arc AB, then $∠MOP=45°$. Hence triangle OMP is an isosceles right triangle and $OP=MP$, i.e. point M has abscissa and ordinate equal: $x = y$.
Since the coordinates of the point $M(x;y)$ satisfy the equation of the number circle, then to find them, you need to solve the system of equations:
$\begin (cases) x^2 + y^2 = 1, \\ x = y. \end(cases)$
Solving this system, we get: $y = x =\frac(\sqrt(2))(2)$.
Hence, the coordinates of the point M corresponding to the number $\frac(π)(4)$ will be $M(\frac(π)(4))=M(\frac(\sqrt(2))(2);\frac (\sqrt(2))(2))$.
The coordinates of the points presented in the previous figure are calculated in a similar way.
Number circle point coordinates
Consider examples
Example 1Find the coordinate of a point on the number circle: $P(45\frac(π)(4))$.
Decision:
$45\frac(π)(4) = (10 + \frac(5)(4)) * π = 10π +5\frac(π)(4) = 5\frac(π)(4) + 2π*5 $.
Hence, the number $45\frac(π)(4)$ corresponds to the same point of the number circle as the number $\frac(5π)(4)$. Looking at the value of the point $\frac(5π)(4)$ in the table, we get: $P(\frac(45π)(4))=P(-\frac(\sqrt(2))(2);-\frac (\sqrt(2))(2))$.
Example 2
Find the coordinate of a point on a number circle: $P(-\frac(37π)(3))$.
Decision:
Because the numbers $t$ and $t+2π*k$, where k is an integer, correspond to the same point of the numerical circle, then:
$-\frac(37π)(3) = -(12 + \frac(1)(3))*π = -12π –\frac(π)(3) = -\frac(π)(3) + 2π *(-6)$.
Hence, the number $-\frac(37π)(3)$ corresponds to the same point of the number circle as the number $–\frac(π)(3)$, and the number –$\frac(π)(3)$ corresponds to the same point as $\frac(5π)(3)$. Looking at the value of the point $\frac(5π)(3)$ in the table, we get:
$P(-\frac(37π)(3))=P(\frac((1))(2);-\frac(\sqrt(3))(2))$.
Example 3
Find points on the number circle with ordinate $y =\frac(1)(2)$ and write down what numbers $t$ do they correspond to?
Decision: 
The line $y =\frac(1)(2)$ intersects the number circle at the points M and P. The point M corresponds to the number $\frac(π)(6)$ (from the data in the table). Hence, any number of the form: $\frac(π)(6)+2π*k$. The point P corresponds to the number $\frac(5π)(6)$, and hence to any number of the form $\frac(5π)(6) +2 π*k$.
We got, as they often say in such cases, two series of values:
$\frac(π)(6) +2 π*k$ and $\frac(5π)(6) +2π*k$.
Answer: $t=\frac(π)(6) +2 π*k$ and $t=\frac(5π)(6) +2π*k$.
Example 4
Find points on the number circle with abscissa $x≥-\frac(\sqrt(2))(2)$ and write down which numbers $t$ they correspond to.
Decision:
The line $x =-\frac(\sqrt(2))(2)$ intersects the number circle at the points M and P. The inequality $x≥-\frac(\sqrt(2))(2)$ corresponds to the points of the arc PM. The point M corresponds to the number $3\frac(π)(4)$ (from the data in the table). Hence, any number of the form $-\frac(3π)(4) +2π*k$. The point P corresponds to the number $-\frac(3π)(4)$, and hence to any number of the form $-\frac(3π)(4) +2π*k$.
Then we get $-\frac(3π)(4) +2 π*k ≤t≤\frac(3π)(4) +2πk$.
Answer: $-\frac(3π)(4) +2 π*k ≤t≤\frac(3π)(4) +2πk$.
Tasks for independent solution
1) Find the coordinate of a point on the number circle: $P(\frac(61π)(6))$.2) Find the coordinate of a point on the number circle: $P(-\frac(52π)(3))$.
3) Find points on the number circle with ordinate $y = -\frac(1)(2)$ and write down which numbers $t$ they correspond to.
4) Find points on the number circle with ordinate $y ≥ -\frac(1)(2)$ and write down which numbers $t$ they correspond to.
5) Find points on the number circle with abscissa $x≥-\frac(\sqrt(3))(2)$ and write down which numbers $t$ they correspond to.
>> Number circle
While studying the algebra course of grades 7-9, we have so far dealt with algebraic functions, i.e. functions given analytically by expressions, in the notation of which algebraic operations on numbers and a variable were used (addition, subtraction, multiplication, division, exponentiation, square root extraction). But mathematical models of real situations are often associated with functions of a different type, not algebraic. With the first representatives of the class of non-algebraic functions - trigonometric functions - we will get acquainted in this chapter. You will study trigonometric functions and other types of non-algebraic functions (exponential and logarithmic) in more detail in high school.
To introduce trigonometric functions, we need a new mathematical model- a number circle, which you have not yet met, but are well acquainted with the number line. Recall that a number line is a line on which the starting point O, the scale (single segment) and the positive direction are given. We can associate any real number with a point on a straight line and vice versa.
How to find the corresponding point M on the line given the number x? The number 0 corresponds to the starting point O. If x > 0, then, moving in a straight line from the point 0 in the positive direction, you need to go n^th length x; the end of this path will be the desired point M(x). If x< 0, то, двигаясь по прямой из точки О в отрицательном направлении, нужно пройти путь 1*1; конец этого пути и будет искомой точкой М(х). Число х - координата точки М.
And how did we solve the inverse problem, i.e. how did you find the x-coordinate of a given point M on the number line? We found the length of the segment OM and took it with the sign "+" or * - "depending on which side of the point O the point M is located on the straight line.
But in real life, you have to move not only in a straight line. Quite often, movement is considered circles. Here is a concrete example. We will consider the stadium treadmill as a circle (in fact, it is, of course, not a circle, but remember how sports commentators usually say: “the runner ran a circle”, “there is half a circle left to run to the finish line”, etc.), its length is 400 m. The start is marked - point A (Fig. 97). The runner from point A moves in a counterclockwise circle. Where will he be in 200 meters? after 400 m? after 800 m? after 1500 m? And where to draw the finish line if he runs a marathon distance of 42 km 195 m?
After 200 m, he will be at point C, diametrically opposite point A (200 m is the length of half the treadmill, i.e. the length of half the circle). After running 400 m (i.e. “one lap”, as the athletes say), he will return to point A. After running 800 m (i.e. “two laps”), he will again be at point A. And what is 1500 m ? This is "three circles" (1200 m) plus another 300 m, i.e. 3
Treadmill - the finish of this distance will be at point 2) (Fig. 97).
We have to deal with the marathon. After running 105 laps, the athlete will overcome the distance 105-400 = 42,000 m, i.e. 42 km. There are 195 m left to the finish line, which is 5 m less than half the circumference. This means that the finish of the marathon distance will be at point M, located near point C (Fig. 97).
Comment. Of course, you understand the convention of the last example. Nobody runs the marathon distance around the stadium, the maximum is 10,000 m, i.e. 25 circles.
You can run or walk a path of any length along the stadium's running track. This means that any positive number corresponds to some point - the “finish of the distance”. Moreover, any negative number can be associated with a circle point: you just need to make the athlete run in the opposite direction, i.e. start from point A not in the opposite direction, but in the clockwise direction. Then the stadium running track can be considered as a numerical circle.
In principle, any circle can be considered as a numerical one, but in mathematics it was agreed to use a unit circle for this purpose - a circle with a radius of 1. This will be our "treadmill". The length b of a circle with radius K is calculated by the formula The length of a half circle is n, and the length of a quarter circle is AB, BC, SB, DA in Fig. 98 - equal We agree to call the arc AB the first quarter of a unit circle, the arc BC - the second quarter, the arc CB - the third quarter, the arc DA - the fourth quarter (Fig. 98). In this case, we are usually talking about an open arc, i.e. about an arc without its ends (something like an interval on a number line).
Definition. A unit circle is given, the starting point A is marked on it - the right end of the horizontal diameter (Fig. 98). Associate each real number I with a circle point according to the following rule:
1) if x > 0, then, moving from point A in a counterclockwise direction (the positive direction of going around the circle), we describe a path along the circle with a length and the end point M of this path will be the desired point: M = M (x);
2) if x< 0, то, двигаясь из точки А в направлении по часовой стрелке (отрицательное направление обхода окружности), опишем по окружности путь длиной и |; конечная точка М этого пути и будет искомой точкой: М = М(1);
0 we assign point A: A = A(0).
A unit circle with an established correspondence (between real numbers and points of the circle) will be called a number circle.
Example 1 Find on the number circle 
Since the first six of the given seven numbers are positive, then in order to find the corresponding points on the circle, you need to go along the circle a path of a given length, moving from point A in a positive direction. At the same time, we take into account that
Point A corresponds to the number 2, since, having passed a path of length 2 along the circle, i.e. exactly one circle, we again get to the starting point A So, A \u003d A (2).
What 
So, moving from point A in a positive direction, you need to go through a whole circle.
Comment. When we're in 7th or 8th grade worked with the number line, we agreed, for the sake of brevity, not to say "the point of the line corresponding to the number x", but to say "the point x". We will adhere to exactly the same agreement when working with a numerical circle: "point f" - this means that we are talking about a circle point that corresponds to the number
Example 2
Dividing the first quarter AB into three equal parts by points K and P, we get: 
Example 3 Find points on the number circle that correspond to numbers 
We will make constructions using Fig. 99. Postponing the arc AM (its length is equal to -) from the point A five times in the negative direction, we get the point!, - the middle of the arc BC. So,
Comment. Notice some liberties we take in using mathematical language. It is clear that the arc AK and the length of the arc AK are different things (the first concept is a geometric figure, and the second concept is a number). But both are denoted the same way: AK. Moreover, if points A and K are connected by a segment, then both the resulting segment and its length are denoted in the same way: AK. It is usually clear from the context what meaning is attached to the designation (arc, arc length, segment or segment length).
Therefore, two layouts of the number circle will be very useful to us.
FIRST LAYOUT
Each of the four quarters of the numerical circle is divided into two equal parts, and their “names” are written near each of the eight available points (Fig. 100).
SECOND LAYOUT Each of the four quarters of the numerical circle is divided into three equal parts, and their “names” are written near each of the twelve points available (Fig. 101).
Please note that on both layouts, we could assign other “names” to the given points.
Have you noticed that in all the examples analyzed, the lengths of the arcs
expressed by some fractions of the number n? This is not surprising: after all, the length of a unit circle is 2n, and if we divide the circle or its quarter into equal parts, then we get arcs whose lengths are expressed as fractions of the number and. And what do you think, is it possible to find such a point E on the unit circle that the length of the arc AE will be equal to 1? Let's guess:
Arguing in a similar way, we conclude that on the unit circle one can find both the point Eg, for which AE, = 1, and the point E2, for which AEg = 2, and the point E3, for which AE3 = 3, and the point E4, for which AE4 = 4, and point Eb, for which AEb = 5, and point E6, for which AE6 = 6. In fig. 102 (approximately) the corresponding points are marked (moreover, for orientation, each of the quarters of the unit circle is divided by dashes into three equal parts).
Example 4 Find on the number circle the point corresponding to the number -7.
We need, starting from the point A (0) and moving in a negative direction (in a clockwise direction), go around the circle path of length 7. If we go through one circle, we get (approximately) 6.28, which means we still need to go ( in the same direction) a path of length 0.72. What is this arc? Slightly less than half a quarter of a circle, i.e. its length is less than number -. 
So, a numerical circle, like a numerical straight line, each real number corresponds to one point (only, of course, it is easier to find it on a straight line than on a circle). But for a straight line, the opposite is also true: each point corresponds to a single number. For a numerical circle, such a statement is not true, we have repeatedly convinced ourselves of this above. For a number circle, the following statement is true.
If the point M of the numerical circle corresponds to the number I, then it also corresponds to the number of the form I + 2k, where k is any integer (k e 2).
Indeed, 2n is the length of the numerical (unit) circle, and the integer |d| can be considered as the number of complete rounds of the circle in one direction or another. If, for example, k = 3, then this means that we make three rounds of the circle in the positive direction; if k \u003d -7, then this means that we make seven (| k | \u003d | -71 \u003d 7) rounds of the circle in the negative direction. But if we are at the point M(1), then by doing more | to | full circles, we will again find ourselves at the point M.
A.G. Mordkovich Algebra Grade 10
Lesson content lesson summary support frame lesson presentation accelerative methods interactive technologies Practice tasks and exercises self-examination workshops, trainings, cases, quests homework discussion questions rhetorical questions from students Illustrations audio, video clips and multimedia photographs, pictures graphics, tables, schemes humor, anecdotes, jokes, comics parables, sayings, crossword puzzles, quotes Add-ons abstracts articles chips for inquisitive cheat sheets textbooks basic and additional glossary of terms other Improving textbooks and lessonscorrecting errors in the textbook updating a fragment in the textbook elements of innovation in the lesson replacing obsolete knowledge with new ones Only for teachers perfect lessons calendar plan for the year methodological recommendations of the discussion program Integrated LessonsWhen studying trigonometry at school, each student is faced with a very interesting concept of "numerical circle". It depends on the ability of the school teacher to explain what it is and why it is needed, how well the student will go about trigonometry later. Unfortunately, not every teacher can explain this material in an accessible way. As a result, many students get confused even with how to celebrate points on the number circle. If you read this article to the end, you will learn how to do it without problems.
So let's get started. Let's draw a circle with a radius of 1. Let's denote the "right" point of this circle by the letter O:
Congratulations, you have just drawn a unit circle. Since the radius of this circle is 1, then its length is .
Each real number can be associated with the length of the trajectory along the number circle from the point O. The direction of movement is counterclockwise as the positive direction. For negative - clockwise:
Arrangement of points on a number circle
As we have already noted, the length of the numerical circle (unit circle) is equal to. Where then will the number be located on this circle? Obviously from the point O counterclockwise, you need to go half the length of the circle, and we will find ourselves at the desired point. Let's denote it with a letter B:
Note that the same point could be reached by passing the semicircle in the negative direction. Then we would put the number on the unit circle. That is, the numbers and correspond to the same point.
Moreover, the same point also corresponds to the numbers , , , and, in general, an infinite set of numbers that can be written in the form , where , that is, belongs to the set of integers. All this is because from the point B you can make a "round the world" trip in any direction (add or subtract the circumference) and get to the same point. We get an important conclusion that needs to be understood and remembered.
Each number corresponds to a single point on the number circle. But each point on the number circle corresponds to infinitely many numbers.
Let us now divide the upper semicircle of the numerical circle into arcs of equal length with a point C. It is easy to see that the arc length OC is equal to . Let's put aside now from the point C an arc of the same length in a counterclockwise direction. As a result, we get to the point B. The result is quite expected, since . Let's postpone this arc in the same direction again, but now from the point B. As a result, we get to the point D, which will already match the number :
Note again that this point corresponds not only to the number , but also, for example, to the number , because this point can be reached by setting aside from the point O quarter circle in the clockwise direction (in the negative direction).
And, in general, we note again that this point corresponds to an infinite number of numbers that can be written in the form 
. But they can also be written as . Or, if you like, in the form of . All these records are absolutely equivalent, and they can be obtained from one another.
Let us now break the arc into OC halved dot M. Think now what is the length of the arc OM? That's right, half the arc OC. I.e . What numbers does the dot correspond to M on a number circle? I am sure that now you will realize that these numbers can be written in the form.
But it is possible otherwise. Let's take in the presented formula. Then we get that 
. That is, these numbers can be written as 
. The same result could be obtained using a number circle. As I said, both entries are equivalent, and they can be obtained from one another.
Now you can easily give an example of numbers that correspond to points N, P and K on the number circle. For example, numbers , and :
Often it is the minimum positive numbers that are taken to denote the corresponding points on the number circle. Although this is not at all necessary, and the point N, as you already know, corresponds to an infinite number of other numbers. Including, for example, the number .
If you break the arc OC into three equal arcs with dots S and L, so the point S will lie between the points O and L, then the arc length OS will be equal to , and the length of the arc OL will be equal to . Using the knowledge that you received in the previous part of the lesson, you can easily figure out how the rest of the points on the number circle turned out:
Numbers that are not multiples of π on the number circle
Let us now ask ourselves the question, where on the number line to mark the point corresponding to the number 1? To do this, it is necessary from the most "right" point of the unit circle O postpone an arc, the length of which would be equal to 1. We can only approximately indicate the location of the desired point. Let's proceed as follows.