Municipal budgetary educational institution

secondary school No. 34 with in-depth study of individual subjects

MAN, Physics and Mathematics Section

“Geometric constructions using compass and ruler”

Completed: student 7 "A" class

Batishcheva Victoria

Head: Koltovskaya V.V.

Voronezh, 2013

3. The construction of an angle equal to this.

P we draw an arbitrary circle centered at the vertex A of a given angle (Fig. 3). Let B and C be the points of intersection of the circle with the sides of the angle. With radius AB we draw a circle centered at the point O-starting point of the given half-line. The intersection point of this circle with this half-line is denoted by C 1 . We describe a circle with center C   1 and Fig. 3

the radius of the sun. Point B 1 the intersection of the constructed circles in the indicated half-plane lies on the side of the desired angle.

6. Construction of perpendicular lines.

Draw a circle with an arbitrary radius r centered at point O in Fig. 6. The circle intersects the line at points A and B.   From points A and B we draw circles with radius AB. Let anguish C be the intersection point of these circles. We got points A and B in the first step, when constructing a circle with an arbitrary radius.

The desired line passes through points C and O.

Fig.6

Known Tasks

1.   Brahmagupta Challenge

Construct an inscribed quadrangle on its four sides. One solution uses the Apollonius circle.   We solve the Apollonius problem using the analogy between a tri-circle and a triangle. How do we find a circle inscribed in a triangle: we build the intersection point of the bisectors, lower the perpendiculars from it to the sides of the triangle, the bases of the perpendiculars (the intersection points of the perpendicular with the side on which it is lowered) and give us three points lying on the desired circle. Draw a circle through these three points - the solution is ready. We will do the same with the Apollonius task.

2. The Apollonius Challenge

Using a compass and a ruler, draw a circle tangent to the three given circles. According to legend, the task was formulated by Apollonius of Perga in about 220 BC. e. in the book “Touch”, which was lost but was restored in 1600 by François Viet, “Gallic Apollonius,” as his contemporaries called it.

If none of the given circles lies inside the other, then this problem has 8 substantially different solutions.

Building regular polygons.

P

equal (or equilateral ) triangle   - this regular polygonwith three sides, the first of the regular polygons. Allside of a regular triangle are equal to each other, and allangles are 60 °. To build an equilateral triangle, you need to divide the circle into 3 equal parts. To do this, it is necessary to draw an arc of radius R of this circle from only one end of the diameter, we obtain the first and second division. The third division is at the opposite end of the diameter. Connecting these points, we get an equilateral triangle.

Regular hexagon   canbuild with a compass and ruler. Belowconstruction method is given   by dividing the circle into 6 parts. We use the equality of the sides of a regular hexagon to the radius of the circumscribed circle. Of the opposite ends of one of the diameters of a circle, we describe arcs of radius R. The intersection points of these arcs with a given circle will divide it into 6 equal parts. Successively connecting the found points, we get a regular hexagon.

Building a regular pentagon.

P
a regular pentagon can bebuilt with a compass and a ruler, or inscribing it into a givencircle, or construction based on a given side. This process is described by Euclid.in his "Beginnings" about 300 BC. e.

Here is one of the methods for constructing a regular pentagon in a given circle:

Construct a circle into which the pentagon will fit and mark its center asO . (This is the green circle in the diagram on the right).

Select a point on the circle.A which will be one of the vertices of the pentagon. Build a line throughO andA .

Draw a line perpendicular to the lineOa passing through a pointO . Designate its intersection with a circle as a pointB .

Build a pointC midway betweenO andB .

C through pointA . Mark its intersection with the lineOB (inside the original circle) as a pointD .

Draw a circle centered inA through point D, the intersection of the given circle with the original (green circle) is denoted as pointsE andF .

Draw a circle centered inE through pointA G .

Draw a circle centered inF through pointA . Mark its other intersection with the original circle as a pointH .

Build a regular pentagonAEGHF .

Insoluble problems

The following three construction tasks were set back in antiquity:

Trisection angle   - split an arbitrary angle into three equal parts.

In other words, it is necessary to construct the trisectrixes of the angle — rays that divide the angle into three equal parts. P.L. Wanzel proved in 1837 that a problem is solvable only when, for example, trisection is feasible for angles α \u003d 360 ° / n, provided that the integer n is not divisible by 3. However, from time to time in the press (incorrect) methods of trisection of the angle by compasses and ruler are published.

Doubling Cube - the classical antique task of constructing with the compass and ruler the edges of a cube, the volume of which is twice the volume of a given cube.

In modern notation, the problem is reduced to solving the equation.   It all comes down to the problem of constructing a length segment. P. Wanzel proved in 1837 that this problem cannot be solved with the help of a compass and a ruler.

Circle squared   - the task is to find the construction with the help of a compass and a ruler of a square, equal in area to this circle.

As you know, using the compass and ruler, you can perform all 4 arithmetic operations and extract the square root; it follows that the quadrature of the circle is possible if and only if, using a finite number of such actions, it is possible to construct a segment of length π. Thus, the unsolvability of this problem follows from the non-algebraic (transcendence) number of π, which was proved in 1882 by Lindeman.

Another well-known insoluble problem with the help of compass and ruler isbuilding a triangle according to three given lengths of bisectors .

Moreover, this task remains unsolvable even in the presence of a trisector.

It was only in the 19th century that it was proved that all three tasks were unsolvable using only the compass and the ruler. The question of the possibility of construction is completely solved by algebraic methods based on Galois theory.

  DO YOU KNOW THAT ...

  (from the history of geometric constructions)

Once upon a time, mystical meaning was put into the construction of regular polygons.

So, the Pythagoreans, followers of the religious and philosophical teachings founded by Pythagoras, and who lived in ancient Greece (VI-I V   centuries BC e.), took as a sign of their union a star-shaped polygon formed by the diagonals of a regular pentagon.

The rules of strict geometric construction of some regular polygons are set out in the book "Beginnings" of the ancient Greek mathematician Euclid, who lived inIII at. BC. To carry out these constructions, Euclid proposed using only a ruler and a compass, which at that time was without an articulated device for connecting the legs (such a limitation in tools was an indispensable requirement of ancient mathematics).

Regular polygons are widely used in ancient astronomy. If Euclid was interested in the construction of these figures from the point of view of mathematics, then for the ancient Greek astronomer Claudius Ptolemy (about 90 - 160 AD), it turned out to be necessary as an auxiliary tool in solving astronomical problems. So, in the 1st book of the Almagest, the entire tenth chapter is devoted to the construction of regular pentagons and decagons.

However, in addition to purely scientific works, the construction of regular polygons was an integral part of books for builders, artisans, and artists. The ability to depict these figures has long been required in architecture, and in jewelry, and in the visual arts.

The Ten Books of Architecture by the Roman architect Vitruvius (who lived around 63-14 BC) says that the city walls should be in the form of a regular polygon, and the towers of the fortress “should be made round or polygonal, because the quadrangle rather destroyed by siege weapons. ”

The layout of the cities was very interesting for Vitruvius, who believed that it was necessary to plan the streets so that the main winds would not blow along them. It was assumed that there were eight such winds and that they were blowing in certain directions.

In the Renaissance, the construction of regular polygons, and in particular the pentagon, was not a simple mathematical game, but was a necessary prerequisite for building fortresses.

The regular hexagon was the subject of a special study of the great German astronomer and mathematician Johannes Kepler (1571-1630), which he talks about in his book "A New Year's Present, or Hexagonal Snowflakes." Discussing the reasons why snowflakes are hexagonal, he notes, in particular, the following: “... a plane can be covered without gaps only with the following figures: equilateral triangles, squares and regular hexagons. Among these figures, the regular hexagon covers the largest area. ”

One of the most famous scientists involved in geometric constructions was the great German artist and mathematician Albrecht Dürer (1471-1528), who devoted a significant part of his book "Guides ..." to them. He proposed rules for constructing regular polygons with 3.4, 5 ... 16 sides. The circle dividing methods proposed by Dürer are not universal, in each case an individual technique is used.

Dürer applied the methods of constructing regular polygons in artistic practice, for example, when creating various kinds of ornaments and patterns for parquet. Sketches of such patterns were made by him during a trip to the Netherlands, where parquet floors were found in many houses.

Dürer made ornaments of regular polygons that are connected into rings (rings of six equilateral triangles, four quadrangles, three or six hexagons, fourteen heptagons, four octagons).

  Conclusion

So,geometric constructions   - This is a way to solve the problem, in which the answer is obtained graphically. The constructions are carried out with drawing tools with the maximum accuracy and accuracy of work, since the correctness of the solution depends on this.

Thanks to this work, I got acquainted with the history of the emergence of the compass, got acquainted in more detail with the rules for performing geometric constructions, gained new knowledge and put them into practice.
  Solving problems with compasses and a ruler is a useful pastime that allows you to take a fresh look at the known properties of geometric shapes and their elements.In this paper, we consider the most pressing problems associated with geometric constructions using a compass and a ruler. The main tasks are considered and their solutions are given. The above problems are of significant practical interest, consolidate the knowledge gained in geometry and can be used for practical work.
  Thus, the goal of the work is achieved, the tasks are completed.

I. Introduction.

II. Main part:

The construction of a segment equal to the product of the other two using a compass and a ruler:

1. the first way to build;

   second construction method;

   third way to build

d) fourth method of construction.

2) The construction of a segment equal to the ratio of the other two using a compass and a ruler:

the first way to build;

second way to build.

Conclusion

Application.

Introduction

Geometrical constructions, or the theory of geometrical constructions, is a branch of geometry where they study questions and methods for constructing geometric figures using certain construction elements. Geometrical constructions are studied both in Euclidean geometry and in other geometries, both on a plane and in space. Classical construction tools are a compass and a ruler (one-sided mathematical), however, there are constructions by other tools: only one compass, only one ruler, if a circle and its center are drawn on the plane, only one ruler with parallel edges, etc.

All construction problems are based on the construction postulates, that is, on the simplest elementary construction problems, and the problem is considered solved if it is reduced to a finite number of these simple postulate problems.

Naturally, each tool has its own constructive power - its own set of postulates. So, it is known that it is impossible to divide a segment using only one ruler into two equal parts, but using a compass is possible.

The art of constructing geometric shapes using compasses and rulers was highly developed in ancient Greece. One of the most difficult construction tasks, which they already knew how to accomplish then, is the construction of a circle tangent to these three circles.

A number of simple constructions are studied at the school with a compass and a ruler (one-sided without divisions): the construction of a line passing through a given point and perpendicular or parallel to this line; halving a given angle, dividing a segment into several equal parts, using the Thales theorem (in fact, dividing a segment by a natural number); the construction of a segment greater than the given by an integer number of times (in fact, multiplying the segment by a natural number). However, we have never encountered a task where it would be necessary to multiply a segment by a segment using a compass and a ruler, that is, to construct a segment equal to the product of two given segments, or to divide a segment into a segment, that is, to construct a segment equal to the ratio of two other segments. This problem seemed very interesting to us, and we decided to research it, try to find a solution and the possibility of applying the found solution method to solving other problems, for example, in mathematics and physics.

When solving construction problems, the traditional methodology recommends us four stages: analysis, construction, proof and research. However, the indicated scheme for solving construction problems is considered to be very academic, and it takes a lot of time to implement it, so often individual steps of the traditional scheme of solving a problem are omitted, for example, the stages of proof and research. Whenever possible, we used all four stages in our work, and even then only where there was a need and expediency in this.

And the last: the method we found for constructing the above segments involves the use, in addition to the compass and ruler, of an arbitrarily selected unit segment. The introduction of a single segment is also dictated by the fact that it is necessary at least in order to confirm the validity of the method we have found for finding a segment using concrete particular examples.

GENERAL PROBLEM I

Using a compass and a ruler, construct a line equal to the product of two other lines.

Note:

supposed to:

The line is one-sided, without divisions.

The unit length is specified.

Study.

1. Consider the lines y \u003d 2x-2 2 and y \u003d 3x-3 2 and try to find the coordinates of the intersection point of these lines by geometric and analytical methods:

and
) geometric method ( Fig. 1) showed that the coordinates of the point A of the intersection of these lines: “5” is the abscissa, “6” is the ordinate, i.e. AE \u003d 5, AD \u003d 6.

b) the analytical method confirms this result, i.e. A (5; 6) is the intersection point of the lines.

Indeed, solving the system of equations

y \u003d 6 A (5; 6) - the point of intersection of lines.

2. Consider the segment: OB \u003d 2, OS \u003d 3, AD \u003d 6, AE \u003d 5.

It can be assumed that HELL \u003d OB × OS, because 6 \u003d 2 × 3; AE \u003d OB + OS, because 5 \u003d 2 + 3, where

2 \u003d OB-angular coefficient of the equation y \u003d 2x-2 2, 3 \u003d OS - the angular coefficient of the equation y \u003d 3x-3 2, AD \u003d y A, OD \u003d x A are the coordinates of point A of the intersection of our lines.

We will verify our assumption on a general example using the analytical method, i.e. on the equations of lines y \u003d mx-m 2 and y \u003d nx-n 2 (where m ≠ n), we verify that the point of intersection of the lines has the coordinates:

y \u003d nx-n 2 nx-n 2 \u003d mx-m 2 x \u003d (m 2 -n 2) ÷ (mn) \u003d m + n and y \u003d mx-m 2 \u003d m (m + n) -m 2 \u003d mn

the coordinates of the point A of the intersection of lines, where m and n are the angular coefficients of these lines,

3. It remains to find a method for constructing the segment. HELL \u003d OB × OS \u003d m ∙ n \u003d y A - the ordinates of the point A of the intersection of the lines Y \u003d mx-m 2 and Y \u003d nx-n 2, where m ≠ n and m \u003d OB, n \u003d OC are the segments laid on the axis Oh. And for this we must find a method for constructing the lines Y \u003d mx-m 2 and Y \u003d nx-n 2. from the reasoning it is clear that these lines must go through points B and C of the segments OB \u003d m and OC \u003d n, which belong to the axis oX.

Remark 1.The above designations of the segments correspond to Fig. 1 “Applications”

First way   constructing the segment AD \u003d mn, where m\u003e 1 unit, n\u003e 1 unit, m ≠ n.

single segment

arbitrary segment, m\u003e 1 unit, n\u003e 1 unit.

n is an arbitrary segment, where m ≠ n.

Building (Fig. 2)

Draw a direct OH

On OX we set OA 1 = m

On OX, we set aside A 1 C 1 \u003d 1 unit

We construct С 1 В 1 \u003d m, where С 1 В 1 ┴ ОХ

Draw a line A 1 B 1, the equation of which is y \u003d mx-m 2 in the coordinate axes of the HOU (the scale on the axes is the same).

Note:

Fig.2

Remark 1.

Indeed, the slope of this straight line tgά 1 \u003d С 1 В 1 / А 1 С 1 \u003d m / 1ед \u003d m, which passes through point А 1 of the segment ОА 1 \u003d m.

We analogously construct a line whose equation is Y \u003d nx-n 2.

6. On the axis OX we postpone OA 2 \u003d n (point A 2 accidentally coincides with point C1).

7.On the axis OX, we postpone A 2 C 2 \u003d 1 unit.

8. Build В 2 С 2 \u003d n, where В 2 С 2 ┴ ОХ.

9. Draw a line B 2 A 2, the equation of which is Y \u003d nx-n 2.

Remark 2.Indeed, the slope of this straight line is tg ά 2 \u003d C 2 B 2 / A 2 C 2 \u003d n / 1ed \u003d n, which passes through t. A 2 of the segment OA 2 \u003d n.

10. Received T.A (m + n; mn) - the intersection point of the lines Y \u003d mx-m 2 and Y \u003d nx-n 2

11. We draw a blood pressure perpendicular to oh, where D belongs to the axis oh.

12. The segment HELL \u003d mn (ordinate t. A), i.e. desired length.

Remark 3.a) indeed, if in our example, n \u003d 4 units, m \u003d 3 units, then there should be BP \u003d mn \u003d 3 units ∙ 4 units \u003d 12 units. And it happened with us: HELL \u003d 12 units .; b) the straight line В 1 В 2 was not used in this construction. In - too.

There are at least three different ways to construct the segment AD \u003d mn.

Second way   plotting the segment HELL \u003dmnwherem\u003e 1 unitn\u003e 1 unitm   andn–– any.

Analysis

The analysis of the previously constructed drawing (Fig. 2), where using the found method of constructing the lines Y \u003d mx-m 2 and Y \u003d nx-n 2, found T.A (m + n; mn) (this is the first method), suggests that T.A (m + n; mn) can be found by constructing any of these lines (Y \u003d mx-m 2 or Y \u003d nx-n 2) and the perpendicular to the AD, where AD is the perpendicular to OX, AD \u003d mn, D belongs to the axis OH. Then the desired point A (m + n; mn) is the intersection point of any of these lines and the perpendicular AD. It is enough to find the slope angles of these lines whose tangents, according to the angular coefficients, are m and n, i.e. tg ά 1 \u003d m and tg ά 2 \u003d n. Considering that tg ά 1 \u003d m / 1ed \u003d m and tg ά 2 \u003d n / 1ed \u003d n, where 1ed is a unit segment, we can easily construct straight lines whose equations are Y \u003d mx-m 2 and Y \u003d nx-n 2.

single segment

n n\u003e 1 unit, m and n are any numbers.

P

construction (Fig. 3)

Fig.3

1. Draw a direct OH.

2. On the axis OX we postpone the segment OA 1 \u003d m.

3. On the axis OX, we postpone the segment A 1 D \u003d n.

4. On the axis OX, we postpone the segment A 1 C 1 \u003d 1 unit.

5. Build С 1 В 1 \u003d m, where С 1 В 1 ┴ ОХ.

6. Draw the straight line A1B1, the equation of which is Y \u003d mx-m2, in the coordinate axes of the HOU (the scale on the axes is the same).

7. Restore the perpendicular to OX at point D.

8. We get the point A (m + n; mn) - the point of intersection of the line Y \u003d mx-m2 and perpendicular AD

9. The segment AD \u003d mn, that is, the desired segment.

Conclusion:This second method is more universal than the first method, since it allows you to find the point A (m + n; mn) even when m \u003d n\u003e 1 unit, then the coordinates of this point are A (2m; m 2) and AD \u003d m 2.

In other words, this method allows you to find the segment equal to the square of the given, whose length is more than 1 unit.

Comment:   Indeed, if in our example m \u003d 3 units, n \u003d 5 units, then there should be AD \u003d mn \u003d 3 units × 5 units \u003d 15 units. And so it happened: AD \u003d 15 units.

Third way   plottingAD= mnwherem\u003e 1 unitn\u003e 1 unit andm≠ n.

Using figure No. 2, we draw a dashed line the straight line В 1 В 2 to the intersection with ОX at the point Е € ОХ, and the straight line В 1 В ┴ В 2 С 2, then

B 1 B \u003d C 1 C 2 \u003d OS 2 -OC 1 \u003d (n + 1 unit) - (m + 1 unit) \u003d nm, and 2 B \u003d B 2 C 2 -B 1 C 1 \u003d mn \u003d\u003e B 1 В \u003d В 2 В \u003d\u003e ΔВ 1 ВВ 2 - isosceles, rectangular\u003e ΔЕС 1 В 1 - isosceles, rectangular \u003d\u003e ά \u003d 45º

Because OS 1 \u003d m + 1 unit, and EU 1 \u003d B 1 C 1 \u003d m, then OE \u003d OS 1 -ES 1 \u003d m + 1 unit-m \u003d 1 unit.

From the arguments it follows that the points B 1 and B 2 can be found differently, because they are the points of intersection of the straight line ЕВ 1 drawn at an angle ά \u003d 45º to the axis ОХ and perpendicular to ОХ: В 1 С 1 and В 2 С 2, and ОЕ \u003d 1ed. Next, using the previous methods, we will have the following construction method.

Single segment.

n n\u003e 1 unit, and m ≠ n.

Building (Fig. 4)

1. Draw a direct OH.

7. Put off OA 2 \u003d n, where A 2 € OX.

8. We postpone A 2 C 2 \u003d 1 unit, where C 2 € OX.

9. Restore the perpendicular C 2 B 2 to the axis OX at point C 2, where B 2 is the point of intersection of the perpendicular with the straight line EB 1.

10. We draw a straight line A 2 B 2, the equation of which is Y \u003d nx-n 2, until it intersects with a straight line A 1 B 1 at point A.

11. Drop the perpendicular to OX from point A and get AD equal to mn, where D € ОX, since in the coordinate planes of the XOU axes the coordinates of point A (m + n; mn).

Fig. 4

Comment:The disadvantage of this method is the same as that of the first construction method, where construction is possible only if m ≠ n.

Fourth way   plottingAD= mnwherem   andn- any large unit length.

Single segment.

n n\u003e 1 unit, m and n are any.

Building (Fig. 5)

Fig.5

1. Draw a direct OH.

2. Set aside OE \u003d 1 unit, where E € OX.

3. Ottzhim EC 1 \u003d m, where C 1 € OH.

4. Restore the perpendicular at the point C 1 to the axis OX.

5. Construct ά \u003d С 1 ЕВ 1 \u003d 45º, where В 1 is the point of intersection of the perpendicular С 1 В 1 with side ά \u003d 45º.

6. Putting aside ОА 1 \u003d m, we draw a straight line А 1 В 1, the equation of which is Y \u003d mx-m 2, А € ОХ.

7. Put A 1 D \u003d n, where D € OX.

8. Restore the perpendicular at point D until it intersects at point A with line A 1 B 1, the equation of which is Y \u003d mx-m 2.

9. The segment of the perpendicular AD \u003d the product of the segments m and n, that is, AD \u003d mn, since A (m + n; mn).

Comment:This method compares favorably with the first and third methods, where m ≠ n, since we are dealing with any segments m and n, a single segment can be less than only one of them that is involved in the beginning of construction (we have m\u003e 1 unit).

General Problem II

Using a compass and a ruler to build a segment equal to the ratio of the other two segments.

Note:

a single segment is less than a divider segment.

The first way to build a segmentn= k/ mwherem\u003e 1 unit

Single segment.

Building (Fig. 6)

2.On the opamp, we set aside OM \u003d k.

3. On OX we postpone OA 1 = m.

4. On OX, we set aside A 1 C 1 \u003d 1 unit.

5. Construct С 1 В 1 \u003d m, where С 1 В 1 ┴ ОХ.

6. Draw a line A 1 B 1, the equation of which is y \u003d mx-m 2 in the coordinate axes of the HOU (the scale on the axes is the same, equal to 1 unit).

7. Restore the perpendicular MA at the point M to the axis OA, where A is the intersection point of MA with the line A 1 B 1 (i.e., A € A 1 B 1).

8. We omit the perpendicular from point A to the axis OX until it intersects with the axis OX at point D. The segment AD \u003d OM \u003d k \u003d mn.

9. Segment A 1 D \u003d n is the desired segment equal to n \u003d k / m.

R is.6

Evidence:

1. The equation of the straight line A 1 B 1 is really Y \u003d mx-m 2, for Y \u003d 0 we have 0 \u003d mx-m 2 \u003d\u003e x \u003d m \u003d OA 1, and the angular coefficient is tg

2. In ∆АDA 1 tg 1 D \u003d AD / A 1 D \u003d B 1 C 1 / A 1 C 1 \u003d\u003e A 1 D \u003d AD × A 1 C 1 / B 1 C 1 \u003d k × 1 unit / m \u003d mn / m \u003d n, i.e. And 1 D \u003d n \u003d k / m is the desired segment.

Comment.Indeed, if in our example m \u003d 3 units, k \u003d 15 units, then there should be A 1 D \u003d n \u003d k / m \u003d 15 units / 3 units \u003d 5 units. And so it happened.

Second way   plottingn= k/ mwherem\u003e 1 unit

Single segment.

Fig. 7

1. Build the coordinate axis of the HOU.

2.On the opamp, we set aside OM \u003d k.

3. We postpone OE \u003d 1 unit, where E € OX.

4. Set aside the EU 1 \u003d m, where C 1 € OX.

5. Restore the perpendicular at point C 1 to the axis OX.

6. Build С 1 ЕВ 1 \u003d 45º, where В 1 is the point of intersection of the perpendicular С 1 В 1 with the side of the angle С 1 ЕВ 1 \u003d 45º.

7. On OX we postpone OA 1 = m.

8. Draw a line A 1 B 1, the equation of which is y \u003d mx-m 2 in the coordinate axes of the HOU (the scale on the axes is the same, equal to 1 unit).

9. Restore the perpendicular MA at the point M to the axis OA, where A is the intersection point of MA with the line A 1 B 1 (i.e., A € A 1 B 1).

10. We omit the perpendicular from point A to the axis OX until it intersects with the axis OX at point D. Line segment AD \u003d OM \u003d k \u003d mn.

11. The segment A 1 D \u003d n is the desired segment equal to n \u003d k / m.

Evidence:

1.∆В 1 С 1 Е - rectangular and isosceles, since С 1 ЕВ 1 \u003d 45º \u003d\u003e В 1 С 1 \u003d EC 1 \u003d m.

2.A 1 C 1 \u003d OS 1 - OA 1 \u003d (OE + EC1) - OA 1 \u003d 1 unit + m-m \u003d 1 unit.

3. The equation of the straight line A 1 B 1 is really Y \u003d mx-m 2, for Y \u003d 0 we have 0 \u003d mx-m 2 \u003d\u003e x \u003d m \u003d OA 1, and the angular coefficient is tg

4.В ∆АDA 1 tg 1 D \u003d AD / A 1 D \u003d B 1 C 1 / A 1 C 1 \u003d\u003e A 1 D \u003d AD × A 1 C 1 / B 1 C 1 \u003d k × 1 unit / m \u003d mn / m \u003d n, i.e. And 1 D \u003d n \u003d k / m is the desired segment.

Conclusion

In our work, we found and investigated various construction methods using a compass and a ruler of a segment equal to the product or the ratio of two other segments, having previously given our definition of these actions with segments, since in no special literature we could find not only the definition of multiplication and division of segments, but even mention of these actions on segments.

Here we used practically all four stages: analysis, construction, proof and research.

In conclusion, we would like to note the possibility of applying the found methods for constructing segments in separate sections of physics and mathematics.

1. If we extend the straight lines y \u003d mx-m 2 and y \u003d nx-n 2 (n\u003e m\u003e 0) to the intersection with the axis of the OS, then we can obtain segments equal to m 2, n 2, n 2 - m 2 (Fig. 8)where OK \u003d m 2, OM \u003d n 2, KM \u003d n 2 - m 2.

R
is.8

Evidence:

If x \u003d 0, then y \u003d 0-m 2 \u003d\u003e OK \u003d m 2.

It is similarly proved that OM \u003d n 2 \u003d\u003e KM \u003d OM-OK \u003d n 2 - m 2.

2. Since the product of two segments is the area of \u200b\u200bthe rectangle with sides equal to these segments, then, finding the segment equal to the product of the other two, we thus represent the area of \u200b\u200bthe rectangle in the form of a segment whose length is numerically equal to this area.

3. In mechanics, thermodynamics there are physical quantities, for example, work (A \u003d FS, A \u003d PV), numerically equal to the areas of rectangles constructed in the corresponding coordinate planes, therefore, in problems where you need, for example, to compare the work on the areas of rectangles, just do this if these areas are represented as segments numerically equal to the areas of the rectangles. And the segments are easy to compare with each other.

4. The considered construction method allows one to construct other segments, for example, using the system of equations y \u003d mx-m 3 and y \u003d nx-n 3, one can construct segments having data m and n such as m 2 + mn + n 2 and mn (m + n), since the point A of the intersection of the lines given by this system of equations has coordinates (m 2 + mn + n 2; mn (m + n), and it is also possible to construct segments n 3, m 3, and the difference n 3 - m 3 obtained at the opamp in the negative region at X \u003d 0.

Artworks. ... help compass   and rulers. Division algorithm segment   AV in half: 1) put the leg compass   to point A; 2) install the solution compass equal   length segment ...

Pythagoras Biography

Biography \u003e\u003e Mathematics

... building   correct geometric shapes with help compass   and rulers. ... help compass   and rulers. Since the occurrence of the task more than two ... is equal to   b / 4 + p, one leg is equal to b / 4, and other   b / 2-p. By the Pythagorean theorem, we have: (b / 4 + p) \u003d (b / 4) + (b / 4-p) or ...

Known since ancient times.

The following operations are possible in building problems:

• Choose any a point   on a plane, a point on one of the constructed lines or the intersection point of two constructed lines.
• Via compass   draw a circle centered at the constructed point with a radius equal to the distance between the two constructed points.
• Via rulers   draw a line passing through two constructed points.

Simple example

Task.   Using a compass and a ruler, break this segment Ab   into two equal parts. One solution is shown in the figure:

• Using a compass, draw a circle centered at A   radius Ab.
• Draw a circle centered at B   radius Ab.
• Find the intersection points P   and Q   two constructed circles.
• Using a ruler, draw a line connecting the points P   and Q.
• Find the intersection point Ab   and Pq. This is the desired midpoint of the segment. Ab.

Regular Polygons

Antique geometers knew ways to build the right ones for n \u003d 2 ^ k \\, \\!, 3 \\ cdot 2 ^ k, 5 \\ cdot 2 ^ k   and 3 \\ cdot5 \\ cdot2 ^ k.

Insoluble problems

The following three construction tasks were set back in antiquity:

•   - split an arbitrary angle into three equal parts.
•   - construct a segment that is an edge of a cube twice as large as a cube with a given edge.
• - build a square equal in area to this circle.

Buildings with one compass and one ruler

By the Mohr – Mascheroni theorem, using one compass, you can build any figure that can be built with a compass and a ruler. Moreover, a straight line is considered to be constructed if two points are given on it.

It is easy to see that with the help of one ruler only projectively invariant constructions can be carried out (see, for example, surface theory ).

In particular, it is impossible to even split a segment into two equal parts. But if there is a previously drawn circle on the plane with a marked center, using the ruler, you can draw the same constructions as the compass and the ruler ( poncelet-Steiner theorem   (Poncelet-Steiner theorem),.

See also

•   - A program that allows you to build using a compass and a ruler.

Literature

• A. Adler. Theory of geometric constructions,   Translated from German by G. M. Fichtenholtz. Third Edition. L., Uchpedgiz, 1940 - 232 p.
• I.I. Alexandrov, Collection of geometric problems for construction,   The eighteenth edition, M., Uchpedgiz, 1950 - 176 p.
• B.I. Argunov, MB Balk. Geometrical constructions on the plane,   A manual for students of pedagogical institutes. Second Edition. M., Uchpedgiz, 1957 - 268 p.
• A.M. Voronets. Compass geometry,   A popular library of mathematics under the general editorship of L.A. Lyusternik, M.-L., ONTI, 1934 - 40 p.
• Yu. I. Manin, On the solvability of construction problems with the help of a compass and a ruler,   Encyclopedia of Elementary Mathematics fourth book (geometry)   , M., Fizmatgiz, 1963 .-- 568s.
• Y. Petersen. Methods and theories for solving geometric construction problems,   Moscow, printing house of E. Lissner and Yu. Roman, 1892 - VIII + 114s.
• V.V. Prasolov Three classic building problems. Doubling a cube, trisection of an angle, squaring a circle,   M .: Nauka, 1992. 80 p. Series of Popular Lectures in Mathematics, Issue 62
• J. Steiner, Geometric constructions performed using a straight line and a fixed circle,   M., Uchpedgiz, 1939 - 80 p.

   Back forward

Attention! The slide preview is used for informational purposes only and may not give an idea of \u200b\u200ball the presentation features. If you are interested in this work, please download the full version.

Textbook:Geometry, 7-9: a textbook for educational institutions / (L.S. Atanasyan, V.F. Butuzov, S.B. Kadomtsev and others) - 16th ed. - M.: Education, 2011.

Lesson Objectives:

1. give an idea of \u200b\u200ba new class of construction problems;
2. consider the simplest construction tasks;
3. to teach students to solve such problems.

Tasks:

Educational aspect:

• give an idea of \u200b\u200ba new class of problems - the construction of geometric ones using a compass and a ruler without scale divisions
• to form practical skills of work;
• expand knowledge of the history of geometry.

Developing aspect:

• development of self-control skills;
• formation of ICT competency;
• the formation of logical thinking.

Educational aspect:

• fostering a responsible attitude to educational work, will and perseverance to achieve final results in the study of the topic;
• fostering interest in the history of mathematics as a science.

Lesson Type:combined.

Forms of organization of educational activities:individual, collective.

Lesson Stages:

• preparation for active learning activities;
• application of knowledge;
• debriefing and reflection;
• homework information.

Equipment:

• Textbook, notebook, pencil, fountain pen, ruler, compass, handout (CMM);
• Computer with minimum technical requirements: Windows 95/98 / ME / NT / 2000 / XP, 7.
• Multimedia projector, screen.

Lesson Resources:

• test items (CMM) annex 1;
• presentation;
• assessment of the degree of assimilation of material application 3.

Lesson plan:

№	Lesson stage	The purpose of the lesson	Time
1.	Organizational moment (slides 1-2)	Lesson topic message; Setting a lesson goal; Message of the lesson stages.	2 minutes.
2.	Reiteration. Checking homework. (Slide 3)	Testing the theoretical knowledge of students on the subject of a circle when performing a test.	5 minutes.
3.	Preparing students for the perception of new material. (Slides 4-8)	Updating reference knowledge	10 minutes.
4.	Learning New Material (Slides 9-19)	Development of skills for solving the simplest tasks for building with compasses and a ruler, discussed in the textbook.	25 minutes
5.	Lesson summary.	Summarizing the lesson.	2 minutes.
6.	Homework. (Slide 20)	Briefing on homework.	1 min.

DURING THE CLASSES

1. Organizational moment:

The topic of today's lesson is “Examples of building tasks” (slide 1).

The purpose of the lesson is to consider the simplest construction problems that can be solved only with the help of a compass and a ruler without divisions; learn to solve them (slide 2).

2. Repetition. Checking homework:

We’ve studied the topic “Circumference” and today we’ll test your knowledge using a test. Perform a test task (each is given KIMs with a test task). For each question, select the correct answer. Evaluate your knowledge yourself by counting the number of correct answers. If the correct answers are 6 - the mark is “5”, if the correct answers are 5 - the mark is “4”, if the correct answers are 4 - the mark is “3”, the smaller the number of correct answers is the mark “2”.

(Correct answers on slide 3 of the presentation).

3. Preparing students for the perception of new material:

Introductory teacher talk:

We already dealt with geometric constructions: we drawn straight lines, laid off segments equal to the data, drew corners, triangles and other figures using various tools. When constructing a segment of a given length, a ruler with millimeter divisions was used, and when constructing the angle of a given degree measure, a protractor was used.

In homework, you had the following task:

Draw a triangle ABC such that AB \u003d 3.6 cm, AC \u003d 2.7 cm, A \u003d 48 °. What in stdid you use the tools to solve this problem?

So, we used a ruler with millimeter divisions and a protractor. But there are tasks in which it is agreed on with what tools you need to build the proposed geometric shape (slide 4-5).

Task 1. Using a compass and a ruler without divisions on a given ray, postpone a segment equal to this from its beginning. Drawing on the screen.

(Students suggest solutions).

Now check your solution (see slide 6)

Thus, many constructions in geometry can be performed using only a compass and a ruler without divisions (slide 7).

In the future, when speaking about construction problems, we will keep in mind precisely such constructions.

Compass and ruler construction tasks are the traditional material studied in the planimetric course. Usually these tasks are solved according to a four-part scheme (see pp. 95–96 of the textbook). First draw (draw) the desired shape and establish the relationship between the task data and the desired elements. This part of the solution is called analysis. It makes it possible to draw up a plan for solving the problem.

Then according to the plan building   compass and ruler.

After that you need to provethat the constructed figure satisfies the conditions of the problem.

Finally necessary explore, for any data, the problem has a solution, and if so, how many solutions.

In cases where the task is quite simple, individual parts, such as analysis or research, can be omitted (slide 8).

In the seventh grade, we will solve the simplest problems of building with compasses and a ruler, in other classes we will solve more complex problems.

4. The study of new material:

And so, our task is to complete the construction tasks using only two tools: a compass and a ruler without large scale divisions.

What can be done with their help? It's clear that the ruler allows you to draw an arbitrary straight line, as well as build a straight line passing through two given points. Using a compass, you can draw a circle of arbitrary radius, as well as a circle centered at a given point and radius equal to this segment(slide 9).

Performing these simple operations, we will be able to solve many interesting construction problems (slide 10):

1. On a given ray from its beginning to postpone a segment equal to this.
2. Set aside a given angle from a given ray.
3. Build a bisector of a given non-expanded angle.
4. Construct a line passing through a given point and perpendicular to the line on which this point lies.
5. Build the middle of this segment.

We have already solved problem number 1.

Now, using a computer, we will consider the solution to problem No. 2. Perform the corresponding constructions in a notebook (slides 11-12).

And now we will consider problems No. 3 - 5 (slide 13-18).

(the corresponding constructions and descriptions of tasks in the notebook are performed)

After completing the work, the teacher draws the attention of students to the fact that such tasks were considered in antiquity(slide 19).

Now let's turn to the history of geometry. Ancient Greek mathematicians achieved extremely great art in geometric constructions using a compass and a ruler. They proved that the angle can be divided into four equal angles. To do this, you need to divide it in half, and then build the bisector of each half. Is it possible to divide the angle into three equal parts using a compass and a ruler? This task, called angle trisection problemsfor many centuries has attracted the attention of mathematicians. However, she did not succumb to their efforts. Only in the last century it was proved that such an construction is impossible for an arbitrary angle.

There are other construction problems, about which it is known that they are insoluble with the help of a compass and a ruler. I suggest that you independently find material containing information to familiarize yourself with these tasks.

5. Summarizing the lesson:

We learned a lot, learned what tasks can be solved only with the help of a compass and a ruler. Each of you has a sheet with questions. Evaluate your work in today's lesson by choosing one of the suggested answer options.

1. Rate the difficulty of the lesson. You were in a lesson:
   • easy;
   • usually;
   • hard
2. Rate the degree of your assimilation of the material:
   • fully learned, I can apply;
   • fully learned, but difficult to apply;
   • partially learned;
   • not learned.

Collect leaflets to assess the degree of assimilation of the material of today's lesson, so that the next lesson will organize the work correctly. Lesson grades are reported, including circle grades.

6. Homework:

• answer questions 17–21 on page 50;
• solve problems No. 153, 154 (slide 20).

In construction problems, we will consider the construction of a geometric figure, which can be done using a ruler and compass.

Using the ruler, you can:

arbitrary straight line;

an arbitrary line passing through a given point;

a line passing through two given points.

Using a compass, you can describe the circumference of a given radius from a given center.

With a compass, you can set aside a line on a given line from a given point.

Consider the main tasks for the construction.

Task 1   Draw a triangle with the given sides a, b, c (Fig. 1).

Decision.   Using a ruler, draw an arbitrary line and take an arbitrary point B on it. With a compass solution equal to a, we describe a circle with center B and radius a. Let C be the point of its intersection with the line. With a solution of the compass equal to c, we describe a circle from the center of B, and a solution of the compass equal to b is a circle from the center of C. Let A be the intersection point of these circles. Triangle ABC has sides equal to a, b, c.

Comment. In order for the three segments of the line to serve as sides of the triangle, it is necessary that the larger of them is less than the sum of the other two (and< b + с).

Task 2

Decision.   This angle with the vertex A and the OM beam are shown in Figure 2.

Draw an arbitrary circle centered at the vertex A of the given angle. Let B and C be the points of intersection of the circle with the sides of the angle (Fig. 3, a). With radius AB we draw a circle centered at point O - the starting point of this ray (Fig. 3, b). The intersection point of this circle with a given ray is denoted by C 1. Let us describe a circle with center C 1 and radius BC. Point B 1 of the intersection of two circles lies on the side of the desired angle. This follows from the equality Δ ABC \u003d Δ OB 1 C 1 (the third sign of the equality of triangles).

Task 3. Construct a bisector of a given angle (Fig. 4).

Decision.   From the vertex A of the given angle, as from the center, draw a circle of arbitrary radius. Let B and C be the points of its intersection with the sides of the angle. From points B and C with the same radius we describe circles. Let D be their intersection point other than A. Ray AD divides the angle A in half. This follows from the equality Δ ABD \u003d Δ ACD (the third sign of the equality of triangles).

Task 4.   Draw the middle perpendicular to this segment (Fig. 5).

Decision.   An arbitrary but identical solution of a compass (large 1/2 AB) describes two arcs with centers at points A and B that intersect at some points C and D. Direct CD will be the desired perpendicular. Indeed, as can be seen from the construction, each of the points C and D is equally distant from A and B; therefore, these points must lie on the middle perpendicular to the segment AB.

Task 5.   Divide this segment in half. It is solved in the same way as task 4 (see Fig. 5).

Task 6.   Through this point, draw a line perpendicular to this line.

Decision.   Two cases are possible:

1) a given point O lies on a given line a (Fig. 6).

From point O we draw a circle with arbitrary radius that intersects the line a at points A and B. From points A and B we draw circles with the same radius. Let O 1 be their intersection point other than O. We obtain OO 1 ⊥ AB. In fact, the points O and O 1 are equidistant from the ends of the segment AB and, therefore, lie on the middle perpendicular to this segment.