Construct a tangent to the tangent circles. Lesson “Constructing tangents to circles. Properties of two tangent circles

State budgetary educational institution

Gymnasium No. 000

Design work in geometry.

Eight ways to construct a tangent to a circle.

9 biological-chemical class

Scientific director: ,

Deputy Director for Academic Affairs,

mathematics teacher.

Moscow 2012

Introduction

Chapter 1. …………………………………………………………………………………4

Conclusion

Introduction

The highest manifestation of the spirit is the mind.

The highest manifestation of reason is geometry.

The geometry cell is a triangle. He also

inexhaustible, like the universe. The circle is the soul of geometry.

Know the circle and you not only know the soul

geometry, but also elevate your soul.

Claudius Ptolemy
Task.

Construct a tangent to a circle with center O and radius R passing through point A lying outside the circle

Chapter 1.

Construction of a tangent to a circle that does not require justification based on the theory of parallel lines.

https://pandia.ru/text/78/156/images/image003_18.gif" width="17" height="16 src=">ABO = 90°. For a circle (O; r) OB - radius. OB AB, therefore, AB is a tangent according to the tangent property.

Similarly, AC is a tangent to a circle.

Construction No. 1 is based on the fact that the tangent of a circle is perpendicular to the radius drawn to the point of contact.

For a straight line there is only one point of contact with a circle.

Only one perpendicular line can be drawn through a given point on a line.

Construction No. 2.

https://pandia.ru/text/78/156/images/image003_18.gif" width="17" height="16"> ABO = 90°

5. OB – radius, ABO = 90°, therefore, AB – tangent by attribute.

6. Similarly, in the isosceles triangle AON AC is the tangent (ACO = 90°, OS is the radius)

7. So, AB and AC are tangents

Formation No. 3

https://pandia.ru/text/78/156/images/image003_18.gif" width="17" height="16">ORM = OVA = 90° (as corresponding angles in equal triangles), therefore, AB – tangent based on tangent.

4. Similarly, AC is a tangent

Construction №4

https://pandia.ru/text/78/156/images/image008_9.jpg" align="left" width="330" height="743 src=">

Construction No. 6.

Construction:

2. I will draw an arbitrary straight line through point A intersecting the circle (O, r) at points M and N.

6. AB and BC are the required tangents.

Proof:

1. Since triangles PQN and PQM are inscribed in a circle and side PQ is the diameter of the circle, then these triangles are right-angled.

2. In triangle PQL, segments PM and QN are heights intersecting at point K, therefore KL is the third height..gif" width="17" height="16 src=">.gif" width="17" height="16 src =">AQS =AMS = 180° - https://pandia.ru/text/78/156/images/image003_18.gif" width="17" height="16">PQN = β, then |AQ| |AS|ctg β. Therefore |PA| : |AQ| = ctg α: ctg β (2).

5. Comparing (1) and (2) I get |PD| : |PA| = |DQ| : |AQ|, or

(|OD| + R)(|OA| - R)=(R -|OD|)(|OA| + R).

After opening the brackets and simplifying, I find that |OD|·|OA|=R².

5. From the relation |OD|·|OA|=R² it follows that |OD|:R=R: |OA|, that is, triangles ODB and OBA are similar..gif" width="17" height="16"> OBA = 90°. Therefore, straight line AB is the desired tangent, which was what needed to be proven.

Construction No. 6.

Construction:

1. I will construct a circle (A; |OA|).

2. I will find a compass opening equal to 2R, for which I will select point S on the circle (O; R) and plot three arcs containing 60º each: SP=PQ=QT=60°. Points S and T are diametrically opposed.

3. I build a circle (O; ST) intersecting w 1What kind of circle is this? at points M and N.

4. Now I’ll build the middle of the MO. To do this, I construct circles (O; OM) and (M; MO), and then for points M and O we find diametrically opposite points U and V on them.

6. Finally, I will construct a circle (K; KM) and (L; LM), intersecting at the desired point B - the middle of the MO.

Proof:

Triangles KMV and UMK are isosceles and similar. Therefore, from the fact that KM = 0.5 MU, it follows that MB = 0.5 MK = 0.5 R. So, point B is the desired point of contact. Similarly, you can find the point of tangency C.

Chapter 3.

Construction of a tangent to a circle based on the properties of secants and bisectors.

Formation No. 7

https://pandia.ru/text/78/156/images/image011_7.jpg" align="left" width="440" height="514 src="> Formation No. 8

Construction:

1. I will construct a circle (A;AP) intersecting straight line AP at point D.

2. Construct a circle w on the diameter QD

3. I will intersect it with a perpendicular to the straight line AP at point A and get points M and N.

Proof:

It is obvious that AM²=AN²=AD·AQ=AP·AQ. Then the circle (A;AM) intersects (O;R) at the tangent points B and C. AB and AC are the required tangents.

A straight line tangent to a circle makes an angle of 90  with the radius drawn to the point of contact. Thus, to construct a line tangent to a circle at a given point, it is necessary to draw the desired line perpendicular to the radius.

Let's look at some examples of constructing tangents and mates.

Example 1

Through point A draw a straight line tangent to the circle with center O 1

To solve the problem, we perform the following constructions:

1) connect points O 1 and A with a straight line;

2) from point O 2 - the middle of the segment O 1 A - draw an auxiliary circle with radius O 2 A until it intersects with the given circle at point B.

The latter is a point of contact, since angle ABO 1 is equal to 90  (it rests

by the diameter AO 1), therefore, the radius O 1 B is the common normal to the straight line and the circular arc at point B.

Example 2

Construct a common tangent to two circles with radii R 1 and R 2 (Fig. 3.4).

To solve the problem, we perform the following constructions:

1) from the center O 1 of the large circle we draw an auxiliary circle with a radius equal to the difference between R 1 and R 2, i.e. R 1 – R 2;

2) to this circle from point O 2 we draw a tangent O 2 K as we did in example 1;

3) we continue the line O 1 K to the intersection with the given great circle, we get point B, which is the point of tangency. From point O 2 we draw a straight line parallel to O 1 B until the line intersects the circle at point A, which is the second point of tangency of the tangent AB.

Rice. 3.3. Construction of tangent-

no straight line to a circle

Rice. 3.4. Construction of a tangent

to two circles

3.3. Pairing two straight lines

Example 3

Construct a conjugation of two intersecting lines m and n with radius

pairing R c (Fig. 3.5).

Rice. 3.5. Constructing a conjugation of two intersecting lines

drop the perpendiculars onto the given lines and get the connecting points A and B; From point O with radius R c we draw a conjugate arc between points A and B.

3.4. Conjugation of a straight line with a circle (internal and external)

Example 4

Construct the external and internal conjugation of a circle with radius R c

with center O 1 with a straight arc t of a given conjugation radius.

D

Rice. 3.6. Construction of the external

conjugation of a circle and a line

Rice. 3.7. Constructing an internal conjugation of a circle and a line

To build an external interface, perform the following steps

1) draw a straight line m parallel to a straight line t at a distance R c and an auxiliary circle from the center O 1 with radius (R 1 + R c); the point of intersection of straight line m and the auxiliary circle - point O - is the center of the conjugation arc; 2) connect the centers O 1 and O with a straight line, its intersection with a given circle will give the first conjugation point - point A; 3) lower the perpendicular from point O to the given straight line t and get the second conjugation point - point B; 4) from point O we draw a conjugation arc AB with radius R c. The construction of an internal conjugation of a circle with a straight line (Fig. 3.7) is carried out similarly to the construction of an external conjugation. The difference is that the radius of the auxiliary circle is not equal to the sum of the radii, but to their difference (R 1 – R c).

Usually in such a problem you are given a circle and a point. It is required to construct a tangent to a circle, and the tangent must pass through a given point.

If the location of the point is not specified, then three possible cases of location of the point should be separately specified.

1. If a point lies inside a circle bounded by a given circle, then a tangent cannot be constructed through it.
2. If a point lies on a circle, then the tangent is constructed by constructing a line perpendicular to the radius drawn to the given point.
3. If a point lies outside the circle bounded by a circle, then before constructing a tangent, a point on the circle is sought through which it must pass.

To solve the second case, on the straight line on which the radius lies, a segment is constructed that is equal to the radius and lies on the other side of the point on the circle. Thus, a point on a circle turns out to be the middle of a segment equal to twice the radius. Next, two circles are constructed whose radii are equal to twice the radius of the original circle, with centers at the ends of the segment equal to twice the radius. A straight line is drawn through any point of intersection of these circles and a point specified by the conditions of the problem. It will be the median perpendicular to the radius of the original circle, that is, perpendicular to it, and therefore tangent to the circle.

You can solve the third case, when the point lies outside the circle bounded by the circle, as follows. It is necessary to construct a segment connecting the center of a given circle and a given point. Next, find its middle by constructing a median perpendicular (described in the previous paragraph). After this, draw a circle (or part of it). The intersection point of the constructed circle and the one specified by the problem conditions is the point through which the tangent passes, which also passes through the point specified by the problem conditions. A tangent line is drawn through two known points.

To prove that the constructed straight line is a tangent, one should consider the angle formed by the radius of the circle given by the conditions of the problem and the segment connecting the point of intersection of the circles with the point given by the conditions of the problem. This angle rests on a semicircle (the diameter of the constructed circle), which means it is straight. That is, the radius is perpendicular to the constructed line. Therefore, the constructed line is tangent.

Secant, tangent - all this could be heard hundreds of times in geometry lessons. But graduation from school is behind us, years pass, and all this knowledge is forgotten. What should you remember?

Essence

The term “tangent to a circle” is probably familiar to everyone. But it’s unlikely that everyone will be able to quickly formulate its definition. Meanwhile, a tangent is a straight line lying in the same plane as a circle that intersects it only at one point. There may be a huge number of them, but they all have the same properties, which will be discussed below. As you might guess, the point of tangency is the place where the circle and the straight line intersect. In each specific case there is only one, but if there are more of them, then it will be a secant.

History of discovery and study

The concept of a tangent appeared in ancient times. The construction of these straight lines, first to a circle, and then to ellipses, parabolas and hyperbolas using a ruler and compass, was carried out at the initial stages of the development of geometry. Of course, history has not preserved the name of the discoverer, but it is obvious that even at that time people were quite familiar with the properties of a tangent to a circle.

In modern times, interest in this phenomenon flared up again - a new round of study of this concept began in combination with the discovery of new curves. Thus, Galileo introduced the concept of a cycloid, and Fermat and Descartes constructed a tangent to it. As for circles, it seems that there are no secrets left for the ancients in this area.

Properties

The radius drawn to the intersection point will be This

the main, but not the only property that a tangent to a circle has. Another important feature includes two straight lines. So, through one point lying outside the circle, two tangents can be drawn, and their segments will be equal. There is another theorem on this topic, but it is rarely taught as part of a standard school course, although it is extremely convenient for solving some problems. It sounds like this. From one point located outside the circle, a tangent and a secant are drawn to it. The segments AB, AC and AD are formed. A is the intersection of lines, B is the point of tangency, C and D are intersections. In this case, the following equality will be valid: the length of the tangent to the circle, squared, will be equal to the product of the segments AC and AD.

There is an important corollary to the above. For each point on the circle you can construct a tangent, but only one. The proof of this is quite simple: theoretically dropping a perpendicular from the radius onto it, we find out that the formed triangle cannot exist. And this means that the tangent is the only one.

Construction

Among other problems in geometry there is a special category, as a rule, not

loved by pupils and students. To solve problems in this category, you only need a compass and a ruler. These are construction tasks. There are also ones for constructing a tangent.

So, given a circle and a point lying outside its boundaries. And it is necessary to draw a tangent through them. How to do this? First of all, you need to draw a segment between the center of the circle O and a given point. Then, using a compass, divide it in half. To do this, you need to set a radius - a little more than half the distance between the center of the original circle and this point. After this, you need to build two intersecting arcs. Moreover, the radius of the compass does not need to be changed, and the center of each part of the circle will be the original point and O, respectively. The intersections of the arcs need to be connected, which will divide the segment in half. Set a radius on the compass equal to this distance. Next, construct another circle with the center at the intersection point. Both the original point and O will lie on it. In this case, there will be two more intersections with the circle given in the problem. They will be the points of contact for the initially specified point.

It was the construction of tangents to the circle that led to the birth

differential calculus. The first work on this topic was published by the famous German mathematician Leibniz. It provided for the possibility of finding maxima, minima and tangents regardless of fractional and irrational quantities. Well, now it is used for many other calculations.

In addition, the tangent to a circle is related to the geometric meaning of tangent. This is where its name comes from. Translated from Latin tangens means “tangent”. Thus, this concept is associated not only with geometry and differential calculus, but also with trigonometry.

Two circles

The tangent does not always affect only one figure. If a huge number of straight lines can be drawn to one circle, then why not vice versa? Can. But the task in this case becomes seriously complicated, because the tangent to two circles may not pass through any points, and the relative position of all these figures can be very

different.

Types and varieties

When we are talking about two circles and one or more straight lines, even if it is known that these are tangents, it is not immediately clear how all these figures are located in relation to each other. Based on this, several varieties are distinguished. Thus, circles may have one or two common points or not have them at all. In the first case they will intersect, and in the second they will touch. And here two varieties are distinguished. If one circle is, as it were, embedded in the second, then the tangency is called internal, if not, then external. You can understand the relative position of the figures not only based on the drawing, but also having information about the sum of their radii and the distance between their centers. If these two quantities are equal, then the circles touch. If the first one is greater, they intersect, and if it is less, then they do not have common points.

The same goes for straight lines. For any two circles that do not have common points, you can

construct four tangents. Two of them will intersect between the figures, they are called internal. A couple of others are external.

If we are talking about circles that have one common point, then the problem is greatly simplified. The fact is that, regardless of their relative position, in this case they will have only one tangent. And it will pass through the point of their intersection. So construction will not be difficult.

If the figures have two points of intersection, then a straight line can be constructed for them, tangent to the circle of both one and the other, but only external. The solution to this problem is similar to what will be discussed below.

Problem solving

Both the internal and external tangent to two circles are not so simple to construct, although this problem can be solved. The fact is that an auxiliary figure is used for this, so you have to come up with this method yourself

quite problematic. So, two circles with different radii and centers O1 and O2 are given. For them you need to construct two pairs of tangents.

First of all, you need to build an auxiliary one near the center of the larger circle. In this case, the difference between the radii of the two initial figures should be established on the compass. Tangents to the auxiliary circle are constructed from the center of the smaller circle. After this, perpendiculars are drawn from O1 and O2 to these lines until they intersect with the original figures. As follows from the basic property of the tangent, the required points on both circles are found. The problem is solved, at least the first part.

In order to construct internal tangents, you will have to solve practically

similar task. Again you will need an auxiliary figure, but this time its radius will be equal to the sum of the original ones. Tangents are constructed to it from the center of one of these circles. The further course of the solution can be understood from the previous example.

Tangent to a circle or even two or more is not such a difficult task. Of course, mathematicians have long stopped solving such problems manually and entrust the calculations to special programs. But you shouldn’t think that now you don’t have to be able to do it yourself, because to correctly formulate a task for a computer you need to do and understand a lot. Unfortunately, there are concerns that after the final transition to a test form of knowledge control, construction tasks will cause students more and more difficulties.

As for finding common tangents for a larger number of circles, this is not always possible, even if they lie in the same plane. But in some cases you can find such a straight line.

Examples from life

A common tangent to two circles often occurs in practice, although this is not always noticeable. Conveyors, block systems, pulley transmission belts, thread tension in a sewing machine, and even just a bicycle chain - all these are real-life examples. So you shouldn’t think that geometric problems remain only in theory: in engineering, physics, construction and many other fields they find practical application.

Lesson Objectives

• Educational – repetition, generalization and testing of knowledge on the topic: “Tangent to a circle”; development of basic skills.
• Developmental – to develop students’ attention, perseverance, perseverance, logical thinking, mathematical speech.
• Educational - through the lesson, cultivate an attentive attitude towards each other, instill the ability to listen to comrades, mutual assistance, and independence.
• Introduce the concept of a tangent, a point of contact.
• Consider the property of a tangent and its sign and show their application in solving problems in nature and technology.

Lesson Objectives

• Develop skills in constructing tangents using a scale ruler, protractor and drawing triangle.
• Test students' problem-solving skills.
• Ensure mastery of the basic algorithmic techniques for constructing a tangent to a circle.
• Develop the ability to apply theoretical knowledge to problem solving.
• Develop students' thinking and speech.
• Work on developing the skills to observe, notice patterns, generalize, and reason by analogy.
• Instilling an interest in mathematics.

Lesson Plan

1. The emergence of the concept of tangent.
2. The history of the appearance of the tangent.
3. Geometric definitions.
4. Basic theorems.
5. Constructing a tangent to a circle.
6. Consolidation.

The emergence of the concept of tangent

The concept of a tangent is one of the oldest in mathematics. In geometry, a tangent to a circle is defined as a line that has exactly one point of intersection with that circle. The ancients, using compasses and rulers, were able to draw tangents to a circle, and later to conic sections: ellipses, hyperbolas and parabolas.

The history of the tangent

Interest in tangents was revived in modern times. Then curves were discovered that were unknown to ancient scientists. For example, Galileo introduced the cycloid, and Descartes and Fermat constructed a tangent to it. In the first third of the 17th century. We began to understand that a tangent is a straight line, “most closely adjacent” to a curve in a small neighborhood of a given point. It is easy to imagine a situation where it is impossible to construct a tangent to the curve at a given point (figure).

Geometric definitions

Circle- the geometric locus of points on the plane equidistant from a given point, called its center.

circle.

Related definitions

• A segment connecting the center of a circle with any point on it (as well as the length of this segment) is called radius circles.
• The part of the plane bounded by a circle is called all around.
• A segment connecting two points on a circle is called its chord. A chord passing through the center of a circle is called diameter.
• Any two divergent points on a circle divide it into two parts. Each of these parts is called arc circles. The measure of an arc can be the measure of its corresponding central angle. An arc is called a semicircle if the segment connecting its ends is a diameter.
• A straight line that has exactly one common point with a circle is called tangent to a circle, and their common point is called the tangency point of the line and the circle.
• A straight line passing through two points on a circle is called secant.
• A central angle in a circle is a plane angle with a vertex at its center.
• An angle whose vertex lies on a circle and whose sides intersect this circle is called inscribed angle.
• Two circles having a common center are called concentric.

Tangent line- a straight line passing through a point on a curve and coinciding with it at this point up to first order.

Tangent to a circle is a straight line that has one common point with a circle.

A straight line passing through a point on a circle in the same plane perpendicular to the radius drawn to this point called tangent. In this case, this point on the circle is called the point of tangency.

Where in our cases “a” is a straight line which is tangent to a given circle, point “A” is the point of tangency. In this case, a⊥OA (straight line a is perpendicular to the radius OA).

They say that two circles touch, if they have a single common point. This point is called point of contact of the circles. Through the point of contact, you can draw a tangent to one of the circles, which is also a tangent to the other circle. Touching circles can be internal or external.

A tangency is called internal if the centers of the circles lie on the same side of the tangent.

A tangency is called external if the centers of the circles lie on opposite sides of the tangent

a is the common tangent to the two circles, K is the point of tangency.

Basic theorems

Theorem about tangent and secant

If a tangent and a secant are drawn from a point lying outside the circle, then the square of the length of the tangent is equal to the product of the secant and its outer part: MC 2 = MA MB.

Theorem. The radius drawn to the point of tangency of the circle is perpendicular to the tangent.

Theorem. If the radius is perpendicular to a line at the point where it intersects a circle, then this line is tangent to this circle.

Proof.

To prove these theorems, we need to remember what a perpendicular from a point to a line is. This is the shortest distance from this point to this line. Let us assume that OA is not perpendicular to the tangent, but there is a straight line OS perpendicular to the tangent. The length OS includes the length of the radius and a certain segment BC, which is certainly greater than the radius. Thus, one can prove it for any line. We conclude that the radius, the radius drawn to the point of contact, is the shortest distance to the tangent from point O, i.e. OS is perpendicular to the tangent. In the proof of the converse theorem, we will proceed from the fact that the tangent has only one common point with the circle. Let this straight line have one more common point B with the circle. Triangle AOB is rectangular and its two sides are equal as radii, which cannot be the case. Thus, we find that this straight line has no more points in common with the circle except point A, i.e. is tangent.

Theorem. The tangent segments drawn from one point to the circle are equal, and the straight line connecting this point with the center of the circle divides the angle between the tangents.

Proof.

The proof is very simple. Using the previous theorem, we assert that OB is perpendicular to AB, and OS is perpendicular to AC. Right triangles ABO and ACO are equal in leg and hypotenuse (OB=OS - radii, AO - total). Therefore, their sides AB=AC and angles OAC and OAB are equal.

Theorem. The magnitude of the angle formed by a tangent and a chord having a common point on a circle is equal to half the angular magnitude of the arc enclosed between its sides.

Proof.

Consider the angle NAB formed by a tangent and a chord. Let's draw the diameter of AC. The tangent is perpendicular to the diameter drawn to the point of contact, therefore, ∠CAN=90 o. Knowing the theorem, we see that angle alpha (a) is equal to half the angular value of the arc BC or half the angle BOS. ∠NAB=90 o -a, from here we get ∠NAB=1/2(180 o -∠BOC)=1/2∠AOB or = half the angular value of the arc BA. etc.

Theorem. If a tangent and a secant are drawn from a point to a circle, then the square of the tangent segment from a given point to the point of tangency is equal to the product of the lengths of the secant segments from a given point to the points of its intersection with the circle.

Proof.

In the figure, this theorem looks like this: MA 2 = MV * MC. Let's prove it. According to the previous theorem, the angle MAC is equal to half the angular value of the arc AC, but also the angle ABC is equal to half the angular value of the arc AC according to the theorem, therefore, these angles are equal to each other. Taking into account the fact that triangles AMC and BMA have a common angle at the vertex M, we state the similarity of these triangles in two angles (second sign). From the similarity we have: MA/MB=MC/MA, from which we get MA 2 =MB*MC

Constructing tangents to a circle

Now let's try to figure it out and find out what needs to be done to construct a tangent to a circle.

In this case, as a rule, the problem gives a circle and a point. And you and I need to construct a tangent to the circle so that this tangent passes through a given point.

In the event that we do not know the location of a point, then let's consider cases of possible locations of points.

Firstly, a point may be inside a circle, which is limited by a given circle. In this case, it is not possible to construct a tangent through this circle.

In the second case, the point is located on a circle, and we can construct a tangent by drawing a perpendicular line to the radius, which is drawn to the point known to us.

Thirdly, let’s assume that the point is located outside the circle, which is limited by the circle. In this case, before constructing a tangent, it is necessary to find a point on the circle through which the tangent must pass.

With the first case, I hope everything is clear to you, but to solve the second option we need to construct a segment on the straight line on which the radius lies. This segment must be equal to the radius and the segment that lies on the circle on the opposite side.

Here we see that a point on a circle is the middle of a segment that is equal to twice the radius. The next step will be to construct two circles. The radii of these circles will be equal to twice the radius of the original circle, with centers at the ends of the segment, which is equal to twice the radius. Now we can draw a straight line through any point of intersection of these circles and a given point. Such a straight line is the median perpendicular to the radius of the circle that was drawn initially. Thus, we see that this line is perpendicular to the circle and it follows from this that it is tangent to the circle.

In the third option, we have a point lying outside the circle, which is limited by a circle. In this case, we first construct a segment that will connect the center of the provided circle and the given point. And then we find its middle. But for this it is necessary to construct a perpendicular bisector. And you already know how to build it. Then we need to draw a circle, or at least part of it. Now we see that the point of intersection of the given circle and the newly constructed one is the point through which the tangent passes. It also passes through the point that was specified according to the conditions of the problem. And finally, through the two points you know, you can draw a tangent line.

And finally, in order to prove that the straight line we constructed is a tangent, we need to pay attention to the angle that was formed by the radius of the circle and the segment known by the condition and connecting the point of intersection of the circles with the point given by the condition of the problem. Now we see that the resulting angle rests on a semicircle. And from this it follows that this angle is right. Consequently, the radius will be perpendicular to the newly constructed line, and this line is the tangent.

Construction of a tangent.

The construction of tangent lines is one of those problems that led to the birth of differential calculus. The first published work related to differential calculus, written by Leibniz, was entitled “A new method of maxima and minima, as well as tangents, for which neither fractional nor irrational quantities, nor a special type of calculus, are an obstacle.”

Geometric knowledge of the ancient Egyptians.

If we do not take into account the very modest contribution of the ancient inhabitants of the valley between the Tigris and Euphrates and Asia Minor, then geometry originated in Ancient Egypt before 1700 BC. During the tropical rainy season, the Nile replenished its water reserves and overflowed. Water covered areas of cultivated land, and for tax purposes it was necessary to determine how much land was lost. Surveyors used a tightly stretched rope as a measuring tool. Another incentive for the accumulation of geometric knowledge by the Egyptians was their activities such as the construction of pyramids and fine arts.

The level of geometric knowledge can be judged from ancient manuscripts, which are specifically devoted to mathematics and are something like textbooks, or rather, problem books, where solutions to various practical problems are given.

The oldest mathematical manuscript of the Egyptians was copied by a certain student between 1800 - 1600. BC. from an older text. The papyrus was found by the Russian Egyptologist Vladimir Semenovich Golenishchev. It is kept in Moscow - in the Museum of Fine Arts named after A.S. Pushkin, and is called the Moscow papyrus.

Another mathematical papyrus, written two to three hundred years later than Moscow’s, is kept in London. It is called: “Instruction on how to achieve knowledge of all dark things, all the secrets that things hide in themselves... According to old monuments, the scribe Ahmes wrote this.” The manuscript is called the “Ahmes papyrus”, or the Rhind papyrus - after the name of the Englishman who found and bought this papyrus in Egypt. The Ahmes papyrus provides solutions to 84 problems involving various calculations that may be needed in practice.