Lesson "construction with compasses and ruler". Constructing, using a compass and a ruler, a segment equal to the product or ratio of two others - creative work Constructing figures using a compass
Instructions
Place the compass needle at the marked point. Using a leg with a stylus, draw an arc of a circle with a measured radius.
Place a dot anywhere along the circumference of the drawn arc. This will be the second vertex B of the triangle being created.
Place the leg on the second peak in a similar way. Draw another circle so that it intersects the first.
The third vertex C of the created triangle is located at the intersection point of both drawn arcs. Mark it on the picture.
Having received all three vertices, connect them with straight lines using any flat surface (preferably a ruler). Triangle ABC is constructed.
If a circle touches all three sides of a given triangle and its center is inside the triangle, then it is called inscribed in the triangle.
You will need
- ruler, compass
Instructions
From the vertices of the triangle (the side opposite the angle being divided), circular arcs of arbitrary radius are drawn with a compass until they intersect with each other;
The point of intersection of the arcs along the ruler is connected to the vertex of the divisible angle;
The same is done with any other angle;
The radius of a circle inscribed in a triangle will be the ratio of the area of the triangle and its semi-perimeter: r=S/p, where S is the area of the triangle, and p=(a+b+c)/2 is the semi-perimeter of the triangle.
The radius of a circle inscribed in a triangle is equidistant from all sides of the triangle.
Sources:
- http://www.alleng.ru/d/math/math42.htm
Let us consider the problem of constructing a triangle, provided that its three sides or one side and two angles are known.
You will need
- - compass
- - ruler
- - protractor
Instructions
Let's say there are three sides: a, b and c. Using it is not difficult with such sides. First, let's select the longest of these sides, let it be side c, and draw it. Then we set the opening of the compass to the value of the other side, side a, and draw a circle with a compass of radius a with the center at one of the ends of side c. Now set the opening of the compass to the size of side b and draw a circle with the center at the other end of side c. The radius of this circle is b. Let's connect the point of intersection of the circles with the centers and get a triangle with the required sides.
To draw a triangle with a given side and two adjacent angles, use a protractor. Draw a side of the specified length. At its edges, mark the corners with a protractor. At the intersection of the sides of the angles, get the third vertex of the triangle.
Video on the topic
note
For the sides of a triangle, the following statement is true: the sum of the lengths of any two sides must be greater than the third. If this is not met, then it is impossible to construct such a triangle.
The circles in step 1 intersect at two points. You can choose any one, the triangles will be equal.
A regular triangle is one in which all sides are the same length. Based on this definition, constructing this type of triangle is not a difficult task.
You will need
- Ruler, sheet of lined paper, pencil
Instructions
Using a ruler, connect the points marked on the piece of paper sequentially, one after another, as shown in Figure 2.
note
In a regular (equilateral) triangle, all angles are equal to 60 degrees.
Helpful advice
An equilateral triangle is also an isosceles triangle. If a triangle is isosceles, this means that 2 of its 3 sides are equal, and the third side is considered the base. Any regular triangle is isosceles, while the converse is not true.
Any equilateral triangle has the same not only sides, but also angles, each of which is equal to 60 degrees. However, a drawing of such a triangle, constructed using a protractor, will not be highly accurate. Therefore, to construct this figure, it is better to use a compass.
You will need
- Pencil, ruler, compass
Instructions
Then take a compass, place it at the ends (the future vertex of the triangle) and draw a circle with a radius equal to the length of this segment. You don’t have to draw the entire circle, but only draw a quarter of it, from the opposite edge of the segment.
Now move the compass to the other end of the segment and again draw a circle of the same radius. Here it will be enough to construct a circle passing from the far end of the segment to the intersection with the already constructed arc. The resulting point will be the third vertex of your triangle.
To complete the construction, take the ruler and pencil again and connect the intersection point of the two circles with both ends of the segment. You will get a triangle with all three sides absolutely equal - this can be easily checked with a ruler.
Video on the topic
A triangle is a polygon that has three sides. An equilateral or regular triangle is a triangle in which all sides and angles are equal. Let's look at how to draw a regular triangle.
You will need
- Ruler, compass.
Instructions
Using a compass, draw another circle, the center of which will be at point B, and the radius will be equal to the segment BA.
The circles will intersect at two points. Choose any of them. Call it C. This will be the third vertex of the triangle.
Connect the vertices together. The resulting triangle will be correct. Make sure of this by measuring its sides with a ruler.
Let's consider a way to construct a regular triangle using two rulers. Draw a segment OK, it will be one of the sides of the triangle, and points O and K will be its vertices.
Without moving the ruler after constructing the segment OK, attach another ruler perpendicular to it. Draw a straight line m intersecting the segment OK in the middle.
Using a ruler, measure a segment OE equal to a segment OK so that one end coincides with point O and the other is on straight line m. Point E will be the third vertex of the triangle.
Complete the construction of the triangle by connecting points E and K. Check the correctness of the construction using a ruler.
note
You can make sure that the triangle is regular using a protractor by measuring the angles.
Helpful advice
An equilateral triangle can also be drawn on a checkered sheet of paper using one ruler. Instead of using another ruler, use perpendicular lines.
Sources:
- Classification of triangles. Equilateral triangles
- What is a triangle
- constructing a regular triangle
An inscribed triangle is one whose vertices are all on the circle. You can build it if you know at least one side and angle. The circumcircle is called the circumcircle, and it will be the only one for this triangle.
You will need
- - circle;
- - side and angle of a triangle;
- - paper;
- - compass;
- - ruler;
- - protractor;
- - calculator.
Instructions
From point A, use a protractor to plot the given angle. Continue the side of the angle until it intersects with the circle and place point C. Connect points B and C. You have a triangle ABC. It can be of any type. The center of the circle for an acute triangle is outside, for an obtuse triangle it is outside, and for a rectangular triangle it is on the hypotenuse. If you are given not an angle, but, for example, three sides of a triangle, calculate one of the angles from the radius and the known side.
Much more often you have to deal with the reverse construction, when you are given a triangle and you need to describe a circle around it. Calculate its radius. This can be done using several formulas, depending on what is given to you. The radius can be found, for example, by the side and sine of the opposite angle. In this case, it is equal to the length of the side divided by twice the sine of the opposite angle. That is, R=a/2sinCAB. It can also be expressed through the product of the sides, in this case R=abc/√(a+b+c)(a+b-c)(a+c-b)(b+c-a).
Determine the center of the circle. Divide all sides in half and draw perpendiculars to the midpoints. The point of their intersection will be the center of the circle. Draw it so that it intersects all the vertices of the corners.
The two short sides of a right triangle, which are usually called legs, by definition must be perpendicular to each other. This property of the figure greatly facilitates its construction. However, it is not always possible to accurately determine perpendicularity. In such cases, you can calculate the lengths of all sides - they will allow you to construct a triangle in the only possible, and therefore correct, way.
You will need
- Paper, pencil, ruler, protractor, compass, square.
Encyclopedic YouTube
1 / 5
✪ 7th grade, lesson 22, Constructions with compasses and ruler
✪ Geometry 7 Circle Constructions with compass and ruler
✪ Constructing a triangle using two sides and the angle between them
✪ Geometry 7 Examples of construction problems
✪ 7th grade, lesson 23, Examples of construction problems
Subtitles
Examples
Bisection problem. Use a compass and ruler to divide this segment AB into two equal parts. One of the solutions is shown in the figure:
- Using a compass we draw circles with centers at points A And B radius AB.
- Finding intersection points P And Q two constructed circles (arcs).
- Using a ruler, draw a segment or line passing through the points P And Q.
- Finding the desired midpoint of the segment AB- point of intersection AB And PQ.
Formal definition
In construction problems, many of the following objects are considered: all points of the plane, all straight lines of the plane, and all circles of the plane. In the conditions of the problem, a certain set of objects is initially specified (considered constructed). It is allowed to add (build) to the set of constructed objects:
- arbitrary point;
- 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;
- an arbitrary straight line passing through a given point;
- a straight line passing through two given points;
- 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.
It is required, using a finite number of these operations, to construct another set of objects that is in a given relationship with the original set.
The solution to the construction problem contains three essential parts:
- Description of the method for constructing a given set.
- Proof that the set constructed in the described way is indeed in a given relationship with the original set. Usually the proof of the construction is carried out as a regular proof of the theorem, based on axioms and other proven theorems.
- Analysis of the described construction method for its applicability to different versions of the initial conditions, as well as for the uniqueness or non-uniqueness of the solution obtained by the described method.
Known Issues
Another well-known and insoluble problem using a compass and ruler is constructing a triangle using three given lengths of bisectors. This problem remains insoluble even with a tool that performs trisection of an angle, such as a tomahawk.
Acceptable segments for construction using a compass and ruler
Using these tools, it is possible to construct a segment whose length is:
To construct a segment with a length numerically equal to the product, quotient and square root of the lengths of given segments, it is necessary to specify a unit segment on the construction plane (that is, a segment of length 1). Extracting roots from segments with other natural powers that are not powers of 2 is impossible using a compass and ruler. So, for example, it is impossible to construct a segment of length from a unit segment using a compass and a ruler. From this fact, in particular, it follows that the problem of doubling a cube is unsolvable.
Possible and impossible constructions
From a formal point of view, the solution to any construction problem is reduced to a graphical solution of some algebraic equation, and the coefficients of this equation are related to the lengths of given segments. Therefore, we can say that the construction task comes down to finding the real roots of some algebraic equation.
Therefore, it is convenient to talk about constructing a number - a graphical solution to an equation of a certain type.
Based on the possible constructions of segments, the following constructions are possible:
- Construction of solutions to linear equations.
- Construction of solutions to equations that reduce to solutions of quadratic equations.
In other words, it is possible to construct only segments equal to arithmetic expressions using the square root of the original numbers (given lengths of the segments).
It is important to note that it is essential that the decision must be expressed using square roots, not radicals of arbitrary degrees. Even if an algebraic equation has a solution in radicals, then it does not follow that it is possible to construct a segment equal to its solution with a compass and ruler. The simplest equation is: x 3 − 2 = 0 , (\displaystyle x^(3)-2=0,) associated with the famous problem of doubling the cube, which reduces to this cubic equation. As mentioned above, the solution to this equation ( 2 3 (\displaystyle (\sqrt[(3)](2)))) cannot be constructed with a compass and ruler.
The ability to construct a regular 17-gon follows from the expression for the cosine of the central angle of its side:
cos (2 π 17) = − 1 16 + 1 16 17 + 1 16 34 − 2 17 + (\displaystyle \cos (\left((\frac (2\pi )(17))\right))=- (\frac (1)(16))\;+\;(\frac (1)(16))(\sqrt (17))\;+\;(\frac (1)(16))(\sqrt (34-2(\sqrt (17))))\;+\;) + 1 8 17 + 3 17 − 34 − 2 17 − 2 34 + 2 17 , (\displaystyle +(\frac (1)(8))(\sqrt (17+3(\sqrt (17))-(\ sqrt (34-2(\sqrt (17))))-2(\sqrt (34+2(\sqrt (17)))))),) which, in turn, follows from the possibility of reducing an equation of the form x F n − 1 = 0 , (\displaystyle x^(F_(n))-1=0,) Where F n (\displaystyle F_(n))- any prime number Fermat, using a change of variable to a quadratic equation.Variations and generalizations
- Constructions using one compass. According to the Mohr-Mascheroni theorem, with the help of one compass you can construct any figure that can be constructed with a compass and a ruler. In this case, a straight line is considered constructed if two points are specified on it.
- Constructions using one ruler. Obviously, with the help of a single ruler, only projective-invariant constructions can be carried out. In particular,
- it is impossible even to divide a segment into two equal parts,
- It is also impossible to find the center of a given circle.
- If there is a pre-drawn circle on the plane with a marked center and one ruler, you can carry out the same constructions as with a compass and a ruler (
If it is quite natural that with the allowance of a greater variety of tools it becomes possible to solve a larger set of construction problems, then one could foresee that, on the contrary, with the restrictions imposed on the tools, the class of solvable problems will be narrowed. All the more remarkable is the discovery made by the Italian Mascheroni (1750-1800): all geometric constructions that can be done with a compass and a ruler can be done with just a compass. It should, of course, be noted that it is impossible to actually draw a straight line through two given points without a ruler, so this basic construction is not covered by Mascheroni’s theory. Instead, we have to assume that a line is given if two of its points are given. But with the help of just a compass it is possible to find the point of intersection of two lines defined in this way, or the point of intersection of a line with a circle.
Probably the simplest example of Mascheroni's construction is doubling a given segment. The solution was already given on page 185. Further, on page 186 we learned how to divide a given segment in half. Let's now see how to bisect an arc of a circle with center O. Here is a description of this construction. With a radius we draw two arcs with centers. From point O we plot two such arcs on these arcs and that Then we find the point of intersection of the arc with the center P and the radius and the arc with the center and radius. Finally, taking as the radius a segment we describe the arc with the center P or to the intersection with the intersection point of the arc is the desired midpoint of the arc. We leave the proof to the reader as an exercise.
Rice. 48. Intersection of a circle and a line not passing through the center
It would be impossible to prove Mascheroni's main statement by showing, for every construction that can be done with a compass and straightedge, how it can be done with just a compass: after all, the possible constructions are innumerable. But we will achieve the same goal if we establish that each of the following basic constructions is feasible using a single compass:
1. Draw a circle if the center and radius are given.
2. Find the intersection points of two circles.
3. Find the intersection points of the line and the circle.
4. Find the point of intersection of two lines.
Any geometric construction (in the usual sense, with the assumption of a compass and straightedge) is composed of a finite sequence of these elementary constructions. That the first two of them can be done using a single compass is immediately clear. More difficult constructions 3 and 4 are performed using the properties of inversion discussed in the previous paragraph.
Let us turn to construction 3: we will find the points of intersection of a given circle C with a line passing through these points. We will draw arcs with centers and radii that are respectively equal and, except for point O, they will intersect at point P. Then we will construct a point inverse to point P relative to circle C (see. construction described on page 186). Finally, let's draw a circle with a center and a radius (it will certainly intersect with C): its points of intersection with circle C will be the desired ones. To prove it, it is enough to establish that each of the points is at the same distances from (as for the points, their similar property immediately follows from the construction). Indeed, it is enough to refer to the fact that the reciprocal point of the point is separated from the points at a distance equal to the radius of the circle C (see page 184). It is worth noting that the circle passing through the points is the inverse straight line in inversion relative to the circle C, since this circle and the straight line intersect
Rice. 49. The intersection of a circle and a line passing through the center
with C at the same points. (When inverted, the points on the main circle remain stationary.)
The indicated construction is not feasible only if the straight line passes through the center C. But then the points of intersection can be found by means of the construction described on page 188, as obtained when we draw an arbitrary circle with center B intersecting C at points Method of drawing the inverse circle of a straight line connecting two given points immediately gives a construction that solves problem 4. Let the straight lines be given by points (Fig. 50).
Rice. 50. Intersection of two lines
Let's draw an arbitrary circle C and, using the above method, construct circles inverse to straight lines and. These circles intersect at point O and at another point. Point X, the inverse of the point, is the desired intersection point: how to construct it has already been explained above. That X is the desired point is clear from the fact that there is a unique point inverse to a point that simultaneously belongs to both lines and, therefore, the point X, the inverse, must lie simultaneously on both
These two constructions complete the proof of the equivalence between Mascheroni's constructions, in which it is allowed to use only a compass, and ordinary geometric constructions with a compass and a ruler.
We did not care about the elegance of solving the individual problems we considered here, since our goal was to clarify the inner meaning of Mascheroni’s constructions. But as an example we will also indicate the construction of a regular pentagon; more precisely, we are talking about finding some five points on the circle that can serve as the vertices of a regular inscribed pentagon.
Let A be an arbitrary point on the circle K. Since the side of a regular inscribed hexagon is equal to the radius of the circle, it will not be difficult to plot points on K such that
Known since ancient times.
The following operations are possible in construction tasks:
- Mark any point on a plane, a point on one of the constructed lines, or the intersection point of two constructed lines.
- By using compass draw a circle with a center at the constructed point and a radius equal to the distance between the two already constructed points.
- By using rulers draw a straight line passing through the two constructed points.
In this case, a compass and a ruler are considered ideal tools, in particular:
1. Simple example
Dividing a segment in half
Task. Use a compass and ruler to divide this segment AB into two equal parts. One of the solutions is shown in the figure:
- Using a compass we construct a circle with the center at a point A radius AB.
- Constructing a circle with a center at a point B radius AB.
- Finding intersection points P And Q two constructed circles.
- Use a ruler to draw a line connecting the points P And Q.
- Finding the intersection point AB And P.Q. This is the desired midpoint of the segment AB.
2. Regular polygons
Ancient geometers knew methods for constructing correct n-gons for and .
4. Possible and impossible constructions
All constructions are nothing more than a solution to some equation, and the coefficients of this equation are related to the lengths of given segments. Therefore, it is convenient to talk about constructing a number - a graphical solution to an equation of a certain type.
Within the framework of the building requirements, the following buildings are possible:
In other words, you can only construct numbers equal to arithmetic expressions using the square root of the original numbers (the lengths of the segments). For example,
5. Variations and generalizations
6. Fun facts
- GeoGebra, Kig, KSEG - programs that allow you to perform constructions using compasses and rulers.
Literature
- A. Adler. Theory of geometric constructions, Translation from German by G. M. Fikhtengolts. Third edition. L., Navchpedvid, 1940-232 p.
- I. Alexandrov, Collection of geometric construction problems, Eighteenth edition, M., Navchpedvid, 1950-176 p.
- B. I. Argunov, M B Balk.
The material in this paragraph can be used in elective classes. It can be presented to students, both in the form of a lecture and in the form of student reports.
Problems that have been known since ancient times as “famous problems of antiquity” have attracted much attention for many centuries. Three famous problems usually appeared under this name:
1) squaring the circle,
2) trisection of the angle,
3) doubling the cube.
All these tasks arose in ancient times from the practical needs of people. At the first stage of their existence, they acted as computational problems: using some “recipes”, approximate values of the desired quantities (area of a circle, circumference, etc.) were calculated. At the second stage of the history of these problems, significant changes in their nature occur: they become geometric (constructive) problems.
In Ancient Greece during this period they were given classical formulations:
1) build a square equal in size to the given circle;
2) divide this angle into three equal parts;
3) construct an edge of a new cube, the volume of which would be twice that of the given cube.
All these geometric constructions were proposed to be carried out using a compass and ruler.
The simplicity of the formulation of these problems and the “insurmountable difficulties” encountered on the way to solving them contributed to the growth of their popularity. In an effort to provide rigorous solutions to these problems, ancient Greek scientists “along the way” obtained many important results for mathematics, which contributed to the transformation of disparate mathematical knowledge into an independent deductive science (the Pythagoreans, Hippocrates of Chios and Archimedes left a particularly noticeable mark at that time).
The problem of doubling the cube.
The problem of doubling a cube is as follows: knowing the edge of a given cube, construct an edge of a cube whose volume would be twice the volume of the given cube.
Let a be the length of the edge of a given cube, x be the length of the edge of the desired cube. Let be the volume of a given cube, and be the volume of the desired cube, then, according to the formula for calculating the volume of a cube, we have that: =, and since, according to the conditions of the problem, we arrive at the equation.
It is known from algebra that the rational roots of the given equation with integer coefficients can only be integer and be contained among the divisors of the free term of the equation. But the only divisors of the number 2 are the numbers +1, - 1, +2, - 2, and none of them satisfies the original equation. Therefore, the equation has no rational roots, which means that the problem of doubling a cube cannot be solved using a compass and ruler.
The problem of doubling a cube using a compass and ruler can only be solved approximately. Here is one of the simplest ways to approximately solve this problem.
Let AB=BC=a, and ABC. We construct AD=AC, then CD with an accuracy of 1%. Indeed, CD 1.2586…. At the same time =1.2599….
The problem of squaring the circle.
Justification of the unsolvability of the problem using a compass and ruler.
The problem of squaring a circle is as follows: construct a square equal in size to the circle.
Let be the radius of the given circle, and let be the length of the side of the desired square. Then, from here.
Consequently, the problem of squaring the circle will be solved if we construct a segment of length. If the radius of a given circle is taken as a unit segment (=1), then the matter will be reduced to constructing a segment of length from a unit segment.
As is known, knowing a unit segment, we can use a compass and ruler to construct only those segments whose lengths are expressed in terms of rational numbers using a finite set of rational operations and extracting square roots and, therefore, are algebraic numbers. In this case, not all algebraic numbers will be used. For example, you cannot construct a segment of length, etc.
In 1882, Lindemann proved that it is transcendental. It follows that it is impossible to construct a segment of length with a compass and a ruler and, therefore, with these means the problem of squaring a circle is unsolvable.
Approximate solution of the problem using a compass and ruler.
Let's consider one of the techniques for approximate construction of length segments. This technique is as follows. A quarter of the circle AB with a center at point O and a radius equal to one is divided in half by point C. On the continuation of the diameter CD, we lay off a segment DE equal to the radius. From point E we draw rays EA and EB until they intersect with the tangent at point C. The cut segment AB is approximately equal to the length of the arc AB, and the doubled segment is equal to the semicircle.
The relative error of this approximation does not exceed 0.227%.
Angle trisection problem.
Justification of the unsolvability of the problem using a compass and ruler.
The angle trisection problem is as follows: Divide this angle into three equal parts.
Let us limit ourselves to solving the problem for angles not exceeding 90. If is an obtuse angle, then =180-, where<90, так что, и поэтому задача о трисекции тупого угла сводится к задаче о трисекции острого угла.
Note that (in the presence of a unit segment) the problem of constructing the angle (90) is equivalent to the problem of constructing the segment x=cos. In fact, if the angle is constructed, then the construction of the segment x = cos is reduced to the construction of a right triangle using the hypotenuse and an acute angle.
Back. If a segment x is constructed, then the construction of an angle such that x = cos is reduced to the construction of a right triangle using the hypotenuse and leg.
Let be the given angle and be the desired angle, so =. Then cos=cos 3. It is known that cos 3= 4cos-3cos. Therefore, assuming cos = and cos =, we arrive at the equation:
cos =4cos-3cos,
A segment, and therefore an angle, can be constructed only if this equation has at least one rational root. But this does not happen for everyone, and therefore the problem of trisection of an angle, generally speaking, cannot be solved using a compass and a ruler. For example. For =60 we get =1 and the found equation takes the form: . It is easy to verify that this equation does not have any rational root, which means that it is impossible to divide an angle of 60 into three equal parts using a compass and ruler. Thus, the problem of trisection of an angle cannot be solved with a compass and a ruler in general form.
Approximate solution of the problem using a compass and ruler.
Let's consider one of the methods for approximate solution of the problem using a compass and ruler, proposed by Albert Durer (1471-1528).
Let angle ASB be given. From vertex S we describe a circle with an arbitrary radius and connect the points of intersection of the sides of the angle with the circle by chord AB. We divide this chord into three equal parts at points R and R (A R = R R = RB). From points A and B, as from centers, with radii A R = RB we describe arcs intersecting the circle at points T and T. Let us carry out RSAB. With radii A S= BS we draw arcs intersecting AB at points U and U. Arcs AT, SS and TB are equal to each other, since they are subtended by equal chords.
To find the trisection points of the angle X and X, Dürer divides the segments RU and RU into three equal parts by the points PV and PV. Then we draw arcs with radii AV and BV that intersect the circle at points X and X. By connecting these points with S, we obtain the division of this angle into three equal parts with a good approximation to the true values.