Help! They gave it to my granddaughter. Use a compass to construct a regular triangle

In construction problems, a compass and a ruler are considered ideal tools, in particular, a ruler has no divisions and has only one side of infinite length, and a compass can have an arbitrarily large or arbitrarily small opening.

Acceptable constructions. The following operations are allowed in construction tasks:

1. Mark a point:

arbitrary point of the plane;

an arbitrary point on a given line;

an arbitrary point on a given circle;

the point of intersection of two given lines;

points of intersection/tangency of a given line and a given circle;

points of intersection/tangency of two given circles.

2. Using a ruler you can draw a straight line:

an arbitrary straight line on a plane;

an arbitrary straight line passing through this point;

a straight line passing through two given points.

3. Using a compass you can construct a circle:

an arbitrary circle on a plane;

an arbitrary circle with a center at a given point;

an arbitrary circle with a radius equal to the distance between two given points;

a circle with a center at a given point and a radius equal to the distance between two given points.

Solving construction problems. The solution to the construction problem contains three essential parts:

Description of the method for constructing the required object.

Proof that the object constructed in the described way is indeed the desired one.

Analysis of the described construction method for its applicability to different options initial conditions, as well as for the uniqueness or non-uniqueness of the solution obtained by the described method.

Constructing a segment equal to the given one. Let a ray with a beginning at point $O$ and a segment $AB$ be given. To construct a segment $OP = AB$ on a ray, you need to construct a circle with a center at point $O$ of radius $AB$. The point of intersection of the ray with the circle will be the required point $P$.

Constructing an angle equal to a given one. Let a ray with origin at point $O$ and angle $ABC$ be given. With the center at point $B$ we construct a circle with an arbitrary radius $r$. Let us denote the intersection points of the circle with the rays $BA$ and $BC$ as $A"$ and $C"$, respectively.

Let's construct a circle with a center at point $O$ of radius $r$. Let us denote the point of intersection of the circle with the ray as $P$. Let's construct a circle with a center at point $P$ of radius $A"B"$. We denote the intersection point of the circles as $Q$. Let's draw the ray $OQ$.

We get the angle $POQ$, equal to angle$ABC$, since triangles $POQ$ and $ABC$ are equal on three sides.

Constructing the perpendicular bisector to a segment. Let's construct two intersecting circles of arbitrary radius with centers at the ends of the segment. By connecting two points of their intersection, we obtain a perpendicular bisector.

Constructing the bisector of an angle. Let's draw a circle of arbitrary radius with the center at the vertex of the corner. Let's construct two intersecting circles of arbitrary radius with centers at the points of intersection of the first circle with the sides of the angle. By connecting the vertex of an angle with any of the intersection points of these two circles, we obtain the bisector of the angle.

Constructing the sum of two segments. To construct on a given ray a segment equal to the sum of two given segments, you need to apply the method of constructing a segment equal to a given one twice.

Constructing the sum of two angles. In order to subtract an angle from a given ray, equal to the sum two given angles, you need to apply the method of constructing an angle equal to the given one twice.

Finding the midpoint of a segment. In order to mark the middle of a given segment, you need to construct a perpendicular bisector to the segment and mark the point of intersection of the perpendicular with the segment itself.

Constructing a perpendicular line through a given point. Let it be required to construct a line perpendicular to a given point and passing through a given point. We draw a circle of arbitrary radius with a center at a given point (regardless of whether it lies on a line or not), intersecting the line at two points. We construct a perpendicular bisector to a segment with ends at the points of intersection of the circle and the line. This will be the desired perpendicular line.

Constructing a parallel line through a given point. Let it be required to construct a line parallel to a given point and passing through a given point outside the line. We construct a line passing through a given point and perpendicular to a given line. Then we construct a straight line passing through this point, perpendicular to the constructed perpendicular. The resulting straight line will be the required one.

Municipal budget educational institution

average secondary school No. 34 s in-depth study individual items

MAN, physics and mathematics section

“Geometric constructions using compass and ruler”

Completed by: student of grade 7 “A”

Batishcheva Victoria

Head: Koltovskaya V.V.

Voronezh, 2013

3. Constructing an angle equal to the given one.

P Let's draw an arbitrary circle with a center at 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. Draw a circle with radius AB with center at point About the starting point given half-line. Let us denote the point of intersection of this circle with this half-line as C 1 . Let us describe a circle with center C 1 and Fig.3

radius of the aircraft. 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.

We draw a circle with an arbitrary radius r with a center 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 melancholy C be the point of intersection of these circles. We obtained points A and B in the first step, when constructing a circle with an arbitrary radius.

The desired straight line passes through points C and O.

Fig.6

Known Issues

1.Brahmagupta's problem

Construct an inscribed quadrilateral using its four sides. One solution uses the Apollonius circle.Let's solve Apollonius' problem using the analogy between a tricircle and a triangle. How we find a circle inscribed in a triangle: we construct the point of intersection of the bisectors, drop perpendiculars from it to the sides of the triangle, the bases of the perpendiculars (the points of intersection of the perpendicular with the side on which it is dropped) 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 Apollonius' problem.

2. Apollonius' problem

Using a compass and ruler, construct a circle tangent to the three given circles. According to legend, the problem was formulated by Apollonius of Perga around 220 BC. e. in the book "Touch," which was lost, but was restored in 1600 by François Viète, the "Gallic Apollonius," as his contemporaries called him.

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

Construction of regular polygons.

P

correct (or equilateral ) triangle - This regular polygonwith three sides, the first of the regular polygons. All sides of a regular triangle are equal to each other, and all the angles are 60°. To construct 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 get the first and second divisions. The third division is at the opposite end of the diameter. By connecting these points, we get an equilateral triangle.

Regular hexagon Canconstruct using a compass and ruler. Belowthe construction method is giventhrough dividing the circle into 6 parts. We use the equality of the sides of a regular hexagon to the radius of the circumscribed circle. From the opposite ends of one of the diameters of the circle we describe arcs of radius R. The points of intersection of these arcs with a given circle will divide it into 6 equal parts. By sequentially connecting the found points, a regular hexagon is obtained.

Construction of a regular pentagon.

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a regular pentagon can beconstructed using a compass and ruler, or by fitting it into a givencircle, or construction based on a given side. This process is described by Euclidin his Elements about 300 BC. e.

Here is one method for constructing a regular pentagon in a given circle:

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

Select a point on the circleA , which will be one of the vertices of the pentagon. Construct a straight line throughO AndA .

Construct a line perpendicular to the lineO.A. , passing through the pointO . Designate one of its intersections with the circle as a pointB .

Plot a pointC in the middle betweenO AndB .

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

Draw a circle with center atA through point D, mark the intersection of this circle with the original (green circle) as pointsE AndF .

Draw a circle with center atE through the pointA G .

Draw a circle with center atF through the pointA . Label its other intersection with the original circle as a pointH .

Construct a regular pentagonAEGHF .

Unsolvable problems

The following three construction tasks were set in antiquity:

Trisection of an angle - divide an arbitrary angle into three equal parts.

In other words, it is necessary to construct angle trisectors - rays dividing the angle into three equal parts. P. L. Wanzel proved in 1837 that the problem is solvable only when, for example, trisection is feasible for angles α = 360°/n, provided that the integer n is not divisible by 3. However, in the press from time to time (incorrect) methods for trisecting an angle with a compass and ruler are published.

Doubling the cube - classical ancient problem of constructing with a compass and ruler the edge 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 segment of length. P. Wantzel proved in 1837 that this problem could not be solved using a compass and straight edge.

Squaring a circle - a task consisting in finding a construction using a compass and a ruler of a square equal in area to the given circle.

As you know, with the help of a compass and ruler you can perform all 4 arithmetic operations and extract square root; it follows that squaring 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 nature (transcendence) of the number π, which was proved in 1882 by Lindemann.

Another well-known problem that cannot be solved using a compass and ruler isconstructing a triangle using three given bisector lengths .

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

It was only in the 19th century that it was proven that all three problems were unsolvable using only a compass and straightedge. The question of the possibility of construction is completely resolved by algebraic methods based on Galois theory.

DID YOU KNOW THAT...

(from the history of geometric constructions)

Once upon a time, a mystical meaning was invested in the construction of regular polygons.

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

The rules for the strict geometric construction of some regular polygons are set out in the book “Elements” by the ancient Greek mathematician Euclid, who lived inIIIV. BC To carry out these constructions, Euclid proposed using only a ruler and a compass, which at that time did not have a hinged device for connecting the legs (such a limitation in instruments was an immutable requirement of ancient mathematics).

Regular polygons were 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, craftsmen, and artists. The ability to depict these figures has long been required in architecture, jewelry, and fine arts.

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

The layout of cities was of great interest to Vitruvius, who believed that it was necessary to plan the streets so that the main winds did not blow along them. It was assumed that there were eight such winds and that they blew in certain directions.

During the Renaissance, constructing regular polygons, and in particular the pentagon, was not an easy task. math game, but was a necessary prerequisite for the construction of fortresses.

The regular hexagon was the subject of a special study by the great German astronomer and mathematician Johannes Kepler (1571-1630), which he talks about in his book “New Year's Gift, or Hexagonal Snowflakes.” Discussing the reasons why snowflakes have a hexagonal shape, 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 Durer (1471 -1528), who dedicated a significant part of his book “Manuals...” to them. He proposed rules for constructing regular polygons with 3, 4, 5... 16 sides. The methods for dividing a circle proposed by Dürer are not universal; in each specific case an individual technique is used.

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

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

Conclusion

So,geometric constructions is a method of solving a problem in which the answer is obtained graphically. Constructions are carried out using drawing tools with maximum precision and accuracy of work, since the correctness of the solution depends on this.

Thanks to this work, I became acquainted with the history of the origin of the compass, became more familiar with the rules for performing geometric constructions, gained new knowledge and applied it in practice.
Solving construction problems with compasses and a ruler is a useful pastime that allows you to take a fresh look at known properties geometric shapes and their elements.This paper discusses the most pressing problems associated with geometric constructions using compasses and rulers. The main problems are considered and their solutions are given. The given problems are of significant practical interest, consolidate the acquired knowledge in geometry and can be used for practical work.
Thus, the goal of the work has been achieved, the assigned tasks have been completed.

How to construct a triangle using a compass A compass is not only a tool for constructing a circle. It also allows you to set aside equal segments of a given length. This will help us use it to construct a triangle.

You will need: a sheet of paper, a compass, a ruler. Instructions. 1. Take any piece of paper. Place a dot in the center of the sheet. This will be the first vertex A of the triangle being created. A

Instructions 2. Open the compass to a distance corresponding to the required side of the triangle being created. Firmly fix the legs of the compass in this position.

Instructions 3. Place the compass needle at the marked point. Using a leg with a stylus, draw an arc of a circle with a measured radius.

Instructions 4. Place a dot anywhere along the circumference of the drawn arc. This will be the second vertex B of the triangle being created.

Instructions 5. In a similar way, place the leg on the second top. Draw another circle so that it intersects the first.

Instruction 6. At the intersection point of both drawn arcs is the third vertex C of the created triangle. Mark it on the picture.

How to draw a triangle?

Construction of various triangles is a mandatory element of the school geometry course. For many, this task causes fear. But in fact, everything is quite simple. The following article describes how to draw any type of triangle using a compass and ruler.

There are triangles

• versatile;
• isosceles;
• equilateral;
• rectangular;
• obtuse-angled;
• acute-angled;
• inscribed in a circle;
• described around a circle.

Construction of an equilateral triangle

An equilateral triangle is one in which all sides are equal. Of all the types of triangles, equilateral triangles are the easiest to draw.

1. Using a ruler, draw one of the sides at a given length.
2. Measure its length using a compass.
3. Place the point of the compass at one end of the segment and draw a circle.
4. Move the point to the other end of the segment and draw a circle.
5. We got 2 points of intersection of the circles. By connecting any of them to the edges of the segment, we get an equilateral triangle.

Construction of an isosceles triangle

This type of triangles can be constructed using the base and sides.

An isosceles triangle is one in which two sides are equal. In order to draw an isosceles triangle using these parameters, you must perform the following steps:

1. Using a ruler, mark off a segment equal in length to the base. We denote it with the letters AC.
2. Using a compass, measure the required side length.
3. We draw from point A, and then from point C, circles whose radius equal to length side.
4. We get two intersection points. By connecting one of them with points A and C, we obtain the required triangle.

Constructing a right triangle

A triangle with one right angle is called a right triangle. If we are given a leg and a hypotenuse, drawing a right triangle is not difficult. It can be constructed using a leg and a hypotenuse.

Constructing an obtuse triangle using an angle and two adjacent sides

If one of the angles of a triangle is obtuse (more than 90 degrees), it is called obtuse. To draw an obtuse triangle using the specified parameters, you must do the following:

1. Using a ruler, mark off a segment equal in length to one of the sides of the triangle. Let's denote it by the letters A and D.
2. If an angle has already been drawn in the assignment, and you need to draw the same one, then on its image put two segments, both ends of which lie at the vertex of the angle, and the length is equal to the indicated sides. Connect the resulting dots. We have the desired triangle.
3. To transfer it to your drawing, you need to measure the length of the third side.

Construction of an acute triangle

An acute triangle (all angles less than 90 degrees) is constructed using the same principle.

1. Draw two circles. The center of one of them lies at point D, and the radius is equal to the length of the third side, and the center of the second is at point A, and the radius is equal to the length of the side indicated in the task.
2. Connect one of the intersection points of the circle with points A and D. The required triangle is constructed.

Inscribed triangle

In order to draw a triangle in a circle, you need to remember the theorem, which states that the center of the circumscribed circle lies at the intersection of the perpendicular bisectors:

For an obtuse triangle, the center of the circumscribed circle lies outside the triangle, while for a right triangle it lies at the middle of the hypotenuse.

Draw a circumscribed triangle

A circumscribed triangle is a triangle in the center of which a circle is drawn touching all its sides. The center of the incircle lies at the intersection of the bisectors. To build them you need:

How to construct an isosceles triangle? This is easy to do with a ruler, pencil and notebook cells.

Construction isosceles triangle we start from the foundation. To make the pattern even, the number of cells at the base must be an even number.

Divide the segment - the base of the triangle - in half.

The vertex of the triangle can be chosen at any height from the base, but always exactly above the middle.

How to construct an acute isosceles triangle?

The angles at the base of an isosceles triangle can only be acute. In order for an isosceles triangle to be acute, the angle at the vertex must also be acute.

To do this, select the vertex of the triangle higher, away from the base.

The higher the apex, the smaller the apex angle. The angles at the base increase accordingly.

How to construct an obtuse isosceles triangle?

As the vertex of an isosceles triangle approaches the base, the degree measure of the angle at the vertex increases.

This means that in order to construct an isosceles obtuse triangle, we select a lower vertex.

How to construct an isosceles right triangle?

To construct an isosceles right triangle, you need to select a vertex at a distance equal to half the base (this is due to the properties of an isosceles right triangle).

For example, if the length of the base is 6 cells, then we place the vertex of the triangle at a height of 3 cells above the middle of the base. Please note: in this case, each cell at the corners at the base is divided diagonally.

The construction of an isosceles right triangle can be started from the vertex.

We select a vertex, and from it at right angles we lay equal segments up and to the right. This - sides triangle.

Let's connect them and get an isosceles right triangle.

We will consider the construction of an isosceles triangle using a compass and a ruler without divisions in another topic.