Definition

A section is a flat figure that is formed when a three-dimensional figure is intersected by a plane and whose boundary lies on the surface of the three-dimensional figure.

Comment

To construct sections of various spatial figures, it is necessary to remember the basic definitions and theorems on parallelism and perpendicularity of lines and planes, as well as the properties of spatial figures. Let us recall the main facts.
For a more detailed study, it is recommended to read the topics “Introduction to solid geometry. Parallelism” and “Perpendicularity. Angles and distances in space”.

Important definitions

1. Two lines in space are parallel if they lie in the same plane and do not intersect.

2. Two straight lines intersect in space if a plane cannot be drawn through them.

4. Two planes are parallel if they do not have common points.

5. Two straight lines in space are called perpendicular if the angle between them is \(90^\circ\) .

6. A line is called perpendicular to a plane if it is perpendicular to any line lying in this plane.

7. Two planes are called perpendicular if the angle between them is \(90^\circ\) .

Important Axioms

1. Through three points that do not lie on one straight line, a plane passes, and moreover, only one.

2. A plane passes through a line and a point not lying on it, and moreover, only one.

3. A plane passes through two intersecting lines, and moreover, only one.

Important theorems

1. If a line \(a\) , not lying in the plane \(\pi\) , is parallel to some line \(p\) , lying in the plane \(\pi\) , then it is parallel to the given plane.

2. Let the line \(p\) be parallel to the plane \(\mu\) . If the plane \(\pi\) passes through the line \(p\) and intersects the plane \(\mu\) , then the line of intersection of the planes \(\pi\) and \(\mu\) is the line \(m\) - parallel to the line \(p\) .

3. If two intersecting lines from one plane are parallel to two intersecting lines from another plane, then such planes will be parallel.

4. If two parallel planes \(\alpha\) and \(\beta\) are intersected by a third plane \(\gamma\) , then the intersection lines of the planes are also parallel:

\[\alpha\parallel \beta, \ \alpha\cap \gamma=a, \ \beta\cap\gamma=b \Longrightarrow a\parallel b\]

5. Let the line \(l\) lie in the plane \(\lambda\) . If the line \(s\) intersects the plane \(\lambda\) at a point \(S\) not lying on the line \(l\) , then the lines \(l\) and \(s\) intersect.

6. If a line is perpendicular to two intersecting lines lying in a given plane, then it is perpendicular to this plane.

7. Three perpendiculars theorem.

Let \(AH\) be the perpendicular to the plane \(\beta\) . Let \(AB, BH\) be an oblique and its projection onto the plane \(\beta\) . Then the line \(x\) in the plane \(\beta\) will be perpendicular to the oblique line if and only if it is perpendicular to the projection.

8. If a plane passes through a straight line perpendicular to another plane, then it is perpendicular to this plane.

Comment

Another important fact, often used to build sections:

in order to find the point of intersection of a line and a plane, it is enough to find the point of intersection of a given line and its projection onto this plane.

To do this, from two arbitrary points \(A\) and \(B\) of the line \(a\) we draw perpendiculars to the plane \(\mu\) – \(AA"\) and \(BB"\) (points \ (A", B"\) are called the projections of the points \(A, B\) onto the plane). Then the line \(A"B"\) is the projection of the line \(a\) onto the plane \(\mu\) . The point \(M=a\cap A"B"\) is the intersection point of the line \(a\) and the plane \(\mu\) .

Note that all points \(A, B, A", B", M\) lie in the same plane.

Example 1

Given a cube \(ABCDA"B"C"D"\) . \(A"P=\dfrac 14AA", \ KC=\dfrac15 CC"\). Find the point of intersection of the line \(PK\) and the plane \(ABC\) .

Solution

1) Because cube edges \(AA", CC"\) are perpendicular to \((ABC)\) , then the points \(A\) and \(C\) are projections of the points \(P\) and \(K\) . Then the line \(AC\) is the projection of the line \(PK\) onto the plane \(ABC\) . We extend the segments \(PK\) and \(AC\) beyond the points \(K\) and \(C\), respectively, and obtain the point of intersection of the lines - the point \(E\) .

2) Find the relation \(AC:EC\) . \(\triangle PAE\sim \triangle KCE\) two corners ( \(\angle A=\angle C=90^\circ, \angle E\)- general), that is, \[\dfrac(PA)(KC)=\dfrac(EA)(EC)\]

If we denote the edge of the cube by \(a\) , then \(PA=\dfrac34a, \ KC=\dfrac15a, \ AC=a\sqrt2\). Then:

\[\dfrac(\frac34a)(\frac15a)=\dfrac(a\sqrt2+EC)(EC) \Rightarrow EC=\dfrac(4\sqrt2)(11)a \Rightarrow AC:EC=4:11\ ]

Example 2

Given a regular triangular pyramid \(DABC\) with base \(ABC\) , whose height is equal to the side of the base. Let the point \(M\) divide the side edge of the pyramid in the ratio \(1:4\) , counting from the top of the pyramid, and \(N\) the height of the pyramid in the ratio \(1:2\) , counting from the top of the pyramid. Find the point of intersection of the line \(MN\) with the plane \(ABC\) .

Solution

1) Let \(DM:MA=1:4, \DN:NO=1:2\) (see figure). Because the pyramid is regular, then the height falls to the point \(O\) of the intersection of the medians of the base. Find the projection of the line \(MN\) onto the plane \(ABC\) . Because \(DO\perp (ABC)\) , then so is \(NO\perp (ABC)\) . Hence, \(O\) is a point belonging to this projection. Let's find the second point. Let us drop the perpendicular \(MQ\) from the point \(M\) to the plane \(ABC\) . The point \(Q\) will lie on the median \(AK\) .
Indeed, since \(MQ\) and \(NO\) are perpendicular to \((ABC)\) , then they are parallel (that is, they lie in the same plane). Therefore, since points \(M, N, O\) lie in the same plane \(ADK\) , then the point \(Q\) will also lie in this plane. But also (by construction) the point \(Q\) must lie in the plane \(ABC\) , therefore, it lies on the line of intersection of these planes, and this is \(AK\) .

Hence, the line \(AK\) is the projection of the line \(MN\) onto the plane \(ABC\) . \(L\) is the point of intersection of these lines.

2) Note that in order to draw the drawing correctly, it is necessary to find the exact position of the point \(L\) (for example, in our drawing the point \(L\) lies outside the segment \(OK\) , although it could lie and inside it; but how is it right?).

Because by condition, the side of the base is equal to the height of the pyramid, then we denote \(AB=DO=a\) . Then the median is \(AK=\dfrac(\sqrt3)2a\) . Means, \(OK=\dfrac13AK=\dfrac 1(2\sqrt3)a\). Let's find the length of the segment \(OL\) (then we will be able to understand whether the point \(L\) is inside or outside the segment \(OK\) : if \(OL>OK\) - then outside, otherwise - inside).

a) \(\triangle AMQ\sim \triangle ADO\) two corners ( \(\angle Q=\angle O=90^\circ, \ \angle A\)- general). Means,

\[\dfrac(MQ)(DO)=\dfrac(AQ)(AO)=\dfrac(MA)(DA)=\dfrac 45 \Rightarrow MQ=\dfrac 45a, \ AQ=\dfrac 45\cdot \dfrac 1(\sqrt3)a\]

Means, \(QK=\dfrac(\sqrt3)2a-\dfrac 45\cdot \dfrac 1(\sqrt3)a=\dfrac7(10\sqrt3)a\).

b) Denote \(KL=x\) .
\(\triangle LMQ\sim \triangle LNO\) two corners ( \(\angle Q=\angle O=90^\circ, \ \angle L\)- general). Means,

\[\dfrac(MQ)(NO)=\dfrac(QL)(OL) \Rightarrow \dfrac(\frac45 a)(\frac 23a) =\dfrac(\frac(7)(10\sqrt3)a+x )(\frac1(2\sqrt3)a+x) \Rightarrow x=\dfrac a(2\sqrt3) \Rightarrow OL=\dfrac a(\sqrt3)\]

Therefore, \(OL>OK\) , which means that the point \(L\) really lies outside the segment \(AK\) .

Comment

Do not be afraid if, when solving a similar problem, you get that the length of the segment is negative. If in the conditions of the previous problem we got that \(x\) is negative, this would just mean that we incorrectly chose the position of the point \(L\) (that is, that it is inside the segment \(AK\) ) .

Example 3

Given a regular quadrangular pyramid \(SABCD\) . Find the section of the pyramid by the plane \(\alpha\) passing through the point \(C\) and the midpoint of the edge \(SA\) and the parallel line \(BD\) .

Solution

1) Denote the midpoint of the edge \(SA\) by \(M\) . Because the pyramid is regular, then the height \(SH\) of the pyramid falls to the intersection point of the diagonals of the base. Consider the plane \(SAC\) . The segments \(CM\) and \(SH\) lie in this plane, let them intersect at the point \(O\) .

For the plane \(\alpha\) to be parallel to the line \(BD\) , it must contain some line parallel to \(BD\) . The point \(O\) is located together with the line \(BD\) in the same plane - in the plane \(BSD\) . Draw a line \(KP\parallel BD\) (\(K\in SB, P\in SD\) ) in this plane through the point \(O\) . Then, by connecting the points \(C, P, M, K\) , we obtain a section of the pyramid by the plane \(\alpha\) .

2) Find the relation in which the points \(K\) and \(P\) divide the edges \(SB\) and \(SD\) . Thus, we completely define the constructed section.

Note that since \(KP\parallel BD\) , then by the Thales theorem \(\dfrac(SB)(SK)=\dfrac(SD)(SP)\). But \(SB=SD\) , so also \(SK=SP\) . So only \(SP:PD\) can be found.

Consider \(\triangle ASC\) . \(CM, SH\) are the medians in this triangle, therefore, the intersection point is divided in relation \(2:1\) , counting from the top, i.e. \(SO:OH=2:1\) .

Now by the Thales theorem from \(\triangle BSD\) : \(\dfrac(SP)(PD)=\dfrac(SO)(OH)=\dfrac21\).

3) Note that, by the three perpendiculars theorem, \(CO\perp BD\) as an oblique (\(OH\) ​​is a perpendicular to the plane \(ABC\) , \(CH\perp BD\) is a projection). So \(CO\perp KP\) . Thus, a section is a quadrilateral \(CPMK\) whose diagonals are mutually perpendicular.

Example 4

Given a rectangular pyramid \(DABC\) with an edge \(DB\) perpendicular to the plane \(ABC\) . At the base lies right triangle with \(\angle B=90^\circ\) , with \(AB=DB=CB\) . Draw a plane through the line \(AB\) perpendicular to the face \(DAC\) , and find the section of the pyramid by this plane.

Solution

1) The plane \(\alpha\) will be perpendicular to the face \(DAC\) if it contains a line perpendicular to \(DAC\) . Draw a perpendicular from the point \(B\) to the plane \(DAC\) - \(BH\) , \(H\in DAC\) .

Draw auxiliary \(BK\) - median in \(\triangle ABC\) and \(DK\) - median in \(\triangle DAC\) .
Because \(AB=BC\) , then \(\triangle ABC\) is isosceles, so \(BK\) is the height, i.e. \(BK\perp AC\) .
Because \(AB=DB=CB\) and \(\angle ABD=\angle CBD=90^\circ\), then \(\triangle ABD=\triangle CBD\), hence \(AD=CD\) , hence \(\triangle DAC\) is also isosceles and \(DK\perp AC\) .

Let's apply the theorem on three perpendiculars: \(BH\) is a perpendicular to \(DAC\) ; oblique \(BK\perp AC\) , hence the projection \(HK\perp AC\) . But we have already determined that \(DK\perp AC\) . Thus, the point \(H\) lies on the segment \(DK\) .

Connecting the points \(A\) and \(H\) , we get the segment \(AN\) , along which the plane \(\alpha\) intersects with the face \(DAC\) . Then \(\triangle ABN\) is the desired section of the pyramid by the plane \(\alpha\) .

2) Determine the exact position of the point \(N\) on the edge \(DC\) .

Denote \(AB=CB=DB=x\) . Then \(BK\) , as the median dropped from the top right angle in \(\triangle ABC\) is \(\frac12 AC\) , hence \(BK=\frac12 \cdot \sqrt2 x\) .

Consider \(\triangle BKD\) . Find the relation \(DH:HK\) .

Note that since \(BH\perp (DAC)\) , then \(BH\) is perpendicular to any line from this plane, so \(BH\) is the height in \(\triangle DBK\) . Then \(\triangle DBH\sim \triangle DBK\), hence

\[\dfrac(DH)(DB)=\dfrac(DB)(DK) \Rightarrow DH=\dfrac(\sqrt6)3x \Rightarrow HK=\dfrac(\sqrt6)6x \Rightarrow DH:HK=2:1 \]

Consider now \(\triangle ADC\) . The medians of an exact intersection triangle are divided by \(2:1\) , counting from the vertex. Hence, \(H\) is the intersection point of the medians in \(\triangle ADC\) (because \(DK\) is a median). That is, \(AN\) is also a median, so \(DN=NC\) .

Let's figure out how to build a section of a pyramid, on concrete examples. Since there are no parallel planes in the pyramid, the construction of the line of intersection (trace) of the secant plane with the plane of the face most often involves drawing a straight line through two points lying in the plane of this face.

In the simplest tasks, it is required to construct a section of the pyramid by a plane passing through the given points already lying in one face.

Example.

Construct Plane Section (MNP)

Triangle MNP - Pyramid Section

Points M and N lie in the same plane ABS, so we can draw a line through them. The trace of this line is the segment MN. It is visible, so we connect M and N with a solid line.

Points M and P lie in the same ACS plane, so we draw a straight line through them. The trace is the segment MP. We do not see it, so we draw the segment MP with a stroke. We construct the trace PN in a similar way.

Triangle MNP is the required section.

If the point through which it is required to draw a section lies not on an edge, but on a face, then it will not be the end of the trace-segment.

Example. Construct a section of the pyramid by a plane passing through the points B, M and N, where the points M and N belong, respectively, to the faces ABS and BCS.

Here points B and M lie on the same face of ABS, so we can draw a line through them.

Similarly, we draw a straight line through the points B and P. We have obtained, respectively, the traces of BK and BL.

Points K and L lie on the same face of ACS, so we can draw a line through them. Its trace is the segment KL.

Triangle BKL is the required section.

However, it is not always possible to draw a straight line through the data in the point condition. In this case, you need to find a point lying on the line of intersection of the planes containing the faces.

Example. Construct a section of the pyramid by a plane passing through the points M, N, P.

Points M and N lie in the same plane ABS, so a straight line can be drawn through them. We get the trace MN. Similarly - NP. Both traces are visible, so we connect them with a solid line.

Points M and P lie in different planes. Therefore, we cannot connect them directly.

We continue the line NP.

It lies in the plane of the BCS face. NP intersects only with lines lying in the same plane. We have three such lines: BS, CS and BC. There are already intersection points with lines BS and CS - these are just N and P. So, we are looking for the intersection of NP with line BC.

The intersection point (let's call it H) is obtained by continuing the lines NP and BC until the intersection.

This point H belongs both to the plane (BCS), since it lies on the line NP, and to the plane (ABC), since it lies on the line BC.

Thus, we have received one more point of the secant plane lying in the plane (ABC).

Through H and a point M lying in the same plane, we can draw a straight line.

We get the trace MT.

T is the point of intersection of lines MH and AC.

Since T belongs to the line AC, we can draw a line through it and the point P, since they both lie in the same plane (ACS).

The quad MNPT is the required section of the pyramid by the plane passing through the given points M,N,P.

We have worked with the line NP, extending it to find the point of intersection of the cutting plane with the plane (ABC). If we work with the straight line MN, we arrive at the same result.

We argue as follows: the line MN lies in the plane (ABS), so it can intersect only with lines lying in the same plane. We have three such lines: AB, BS and AS. But with lines AB and BS there are already points of intersection: M and N.

Hence, extending MN, we are looking for the point of its intersection with the straight line AS. Let's call this point R.

The point R lies on the line AS, so it also lies in the plane (ACS) to which the line AS belongs.

Since the point P lies in the plane (ACS), we can draw a line through R and P. We get the trace of PT.

The point T lies in the plane (ABC), so we can draw a line through it and the point M.

Thus, we got the same MNPT cross section.

Let's consider another example of this kind.

Construct a section of the pyramid by a plane passing through the points M, N, P.

Draw a straight line through points M and N lying in the same plane (BCS). We get the trace MN (visible).

Draw a straight line through points N and P lying in the same plane (ACS). We get the trace PN (invisible).

We cannot draw a straight line through points M and P.

1) The line MN lies in the plane (BCS), where there are three more lines: BC, SC and SB. There are already points of intersection with lines SB and SC: M and N. Therefore, we are looking for the point of intersection of MN with BC. Continuing these lines, we get the point L.

The point L belongs to the line BC, which means that it lies in the plane (ABC). Therefore, through L and P, which also lies in the plane (ABC), we can draw a straight line. Her footprint is PF.

F lies on the line AB, and hence in the plane (ABS). Therefore, through F and the point M, which also lies in the plane (ABS), we draw a straight line. Her track is FM. The quadrilateral MNPF is the required section.

2) Another way is to continue straight PN. It lies in the plane (ACS) and intersects lines AC and CS lying in this plane at points P and N.

So, we are looking for the intersection point of PN with the third straight line of this plane - with AS. We continue AS and PN, at the intersection we get the point E. Since the point E lies on the line AS, which belongs to the plane (ABS), then through E and the point M, which also lies in (ABS), we can draw a line. Her track is FM. Points P and F lie on the water plane (ABC), we draw a straight line through them and get the trace PF (invisible).

Problems on the construction of sections of polyhedra occupy a significant place both in the school geometry course for senior classes and in exams. different levels. The solution of this type of problems contributes to the assimilation of the axioms of stereometry, the systematization of knowledge and skills, the development of spatial representation and constructive skills. Difficulties that arise in solving problems on the construction of sections are well known.

From early childhood, we are faced with sections. We cut bread, sausage and other products, cut a stick or pencil with a knife. The secant plane in all these cases is the plane of the knife. Sections (sections of pieces) are different.

The section of a convex polyhedron is a convex polygon, the vertices of which, in the general case, are the points of intersection of the cutting plane with the edges of the polygon, and the sides are the lines of intersection of the cutting plane with the faces.

To construct a line of intersection of two planes, it is enough to find two common points these planes and draw a straight line through them. This is based on the following statements:

1. if two points of a straight line belong to a plane, then the whole line belongs to this plane;

2.if two various planes have a common point, then they intersect along a line passing through this point.

As I already said, the construction of sections of polyhedra can be carried out on the basis of the axioms of stereometry and theorems on the parallelism of lines and planes. At the same time, there are certain methods for constructing plane sections of polyhedra. The following three methods are the most effective:

trace method

Internal design method

Combined method.

In the study of geometry and, in particular, those sections of it where images of geometric figures are considered, images of geometric figures help to use computer presentations. With the help of a computer, many geometry lessons become more visual and dynamic. Axioms, theorems, proofs, problems for construction, problems for constructing sections can be accompanied by successive constructions on the monitor screen. Computer-generated drawings can be saved and pasted into other documents.

I want to show a few slides on the topic: "Construction of sections in geometric bodies"

To construct the point of intersection of a line and a plane, find a line in the plane that intersects the given line. Then the desired point is the point of intersection of the found line with the given one. Let's see it on the next slides.

Task 1.

Two points M and N are marked on the edges of the tetrahedron DABC; M GAD, N b DC. Pick the point of intersection of the line MN with the plane of the base.

Solution: in order to find the point of intersection of the line MN with the plane

base we will continue AC and segment MN. We mark the point of intersection of these lines through X. The point X belongs to the line MN and the face AC, and AC lies in the base plane, which means that the point X also lies in the base plane. Therefore, the point X is the point of intersection of the line MN with the plane of the base.

Let's consider the second problem. Let's complicate it a little.

Task 2.

Given a tetrahedron DABC of points M and N, where M € DA, N C (DBC). Find the point of intersection of the line MN with the plane ABC .

Solution: The point of intersection of the line MN with the plane ABC must lie in the plane that contains the line MN and in the plane of the base. We continue the segment DN to the point of intersection with the edge DC. We mark the point of intersection through E. We continue the line AE and MN to the point of their intersection. Note X. The point X belongs to MN, so it lies on the plane that contains the line MN and X belongs to AE, and AE lies on the plane ABC. So X also lies in the plane ABC. Hence X is the point of intersection of the line MN and the plane ABC.

Let's complicate the task. Consider a section of geometric figures by planes passing through three given points.

Task 3

Points M, N and P are marked on the edges AC, AD and DB of the tetrahedron DABC. Construct a section of the tetrahedron by the plane MNP.

Solution: construct a straight line along which the plane MNP. Intersects face plane ABC. Point M is a common point of these planes. To build another common point, we continue the segment AB and NP. We mark the intersection point through X, which will be the second common point of the plane MNP and ABC. So these planes intersect along the straight line MX. MX intersects the edge BC at some point E. Since E lies on MX and MX is a line belonging to the plane MNP, it follows that PE belongs to MNP. The quadrilateral MNPE is the required section.

Task 4

We construct a section of a straight prism ABCA1B1C1 by a plane passing through the points P , Q,R, where R belongs to ( AA 1C 1C), R belongs V 1C1,

Q belongs to AB

Solution: All three points P,Q,R lie in different faces, so we cannot yet construct a line of intersection of the secant plane with any face of the prism. Let's find the intersection point of PR with ABC. Let us find the projections of the points P and R onto the base plane PP1 perpendicular to BC and RR1 perpendicular to AC. Line P1R1 intersects line PR at point X. X is the point of intersection of line PR with plane ABC. It lies in the desired plane K and in the plane of the base, like the point Q. XQ is a straight line intersecting K with the plane of the base. XQ intersects AC at point K. Therefore, KQ is the segment of the intersection of the plane X with the face ABC. K and R lie in the X plane and in the plane of the AA1C1C face. Draw a line KR and mark the point of intersection with A1Q E. KE is the line of intersection of the plane X with this face. Find the line of intersection of the X plane with the plane of the faces BB1A1A. KE intersects with A1A at point Y. The line QY is the line of intersection of the secant plane with the plane AA1B1B. FPEKQ - desired section.

The entire history of geometry and some other branches of mathematics is closely connected with the development of the theory of geometric constructions. The most important axioms of geometry, formulated by Euclid around 300 BC, clearly show the role played by geometric constructions in the formation of geometry.

There are special topics in school geometry that you look forward to, anticipating a meeting with an incredibly beautiful material. Such topics include "Polyhedra and the construction of their sections." Here, not only opens wonderful world geometric bodies with unique properties, but also interesting scientific hypotheses. And then the geometry lesson becomes a kind of study of unexpected aspects of the usual school subject.

At the geometry lessons this year we went through the topic “Construction of sections of polyhedra”. As part of the program, we studied one method for constructing sections, but I became interested in what methods still exist.

The purpose of my work: Learn all methods of constructing sections of polyhedra.

None of the geometric bodies possess such perfection and beauty as polyhedra. "There are defiantly few polyhedrons," L. Carroll once wrote, "but this detachment, which is very modest in number, managed to get into the very depths of various sciences."

At present, the theory of geometric constructions is a vast and deeply developed area of ​​mathematics associated with the solution of various fundamental questions that go into other branches of mathematics.

1. History of descriptive geometry

Even in ancient times, a person drew and painted images of things, trees, animals and people on rocks, stones, walls and household items. He did this to satisfy his needs, including aesthetic ones. At the same time, the main requirement for such images was that the image evoked the correct visual representation of the shape of the depicted object.

With the growth of practical and technical applications images (in the construction of buildings and other civil and military structures, etc.) they began to impose such requirements that the image could be used to judge the geometric properties, sizes and relative positions individual elements a certain subject. Such requirements can be judged by many ancient monuments that have survived to this day. However, strict geometric grounded rules and methods for depicting spatial figures (with respect to perspective) began to be systematically developed by artists, architects and sculptors only in the Renaissance: Leonardo da Vinci, Dürer, Raphael, Michelangelo, Titian, etc.

Descriptive geometry as a science was created at the end of the 18th century by the great French geometer and engineer Gaspard Monge (1746-1818). In 1637, the French geometer and philosopher Rene Descartes (1596 - 1650) created the coordinate method and laid the foundations of analytical geometry, and his compatriot, engineer and mathematician Girard Desag (1593 - 1662), used this coordinate method to build perspective projections and substantiated the theory axonometric projections.

In the 17th century, technical drawings were successfully developed in Russia, made in the form of plans and profiles to scale. Here, first of all, we should name the drawings of the outstanding Russian mechanic and inventor I.P. Kulibin (1735 - 1818). In his project for a wooden arch bridge, orthogonal projections were first used (1773). (Orthogonal projection of a plane onto a line lying in it, or of space onto a plane, is a special case of parallel projection, in which the projection direction is perpendicular to the line or plane being projected onto.)

A great contribution to the development of orthogonal projections was made by the French engineer A. Frezier (1682–1773), who was the first to consider the projection of an object onto two planes - horizontal and frontal.

The greatest merit of G. Monge was the generalization of all the scientific works of his predecessors, the whole theory of methods for depicting spatial figures and the creation of a unified mathematical science of orthogonal projection - descriptive geometry.

The birth of this new science almost coincided with the foundation in St. Petersburg of Russia's first higher transport educational institution- Institute of the Corps of Railway Engineers (December 2, 1809)

Graduates of this institute, its professors and scientists have made a significant contribution to the development of geometric methods of representation, to the theory and practice of descriptive geometry.

1. Definitions of polyhedra

In stereometry, figures in space are studied, called bodies . Visually, a (geometric) body must be imagined as a part of space occupied by physical body and limited surface.

Polyhedron - this is a body whose surface consists of several flat polygons. The polyhedron is called convex , if it is located on one side of the plane of every planar polygon on its surface. The common part of such a plane and the surface of a convex polyhedron is called edge . The faces of a convex polyhedron are flat convex polygons. The sides of the faces are callededges of the polyhedron, and the vertices vertices of the polyhedron.

cross section polyhedron is called a plane geometric figure, which is the set of all points in space that simultaneously belong to the given polyhedron and plane; the plane is called the secant plane.

The surface of a polyhedron consists of edges, segments and faces of flat polygons. Since a line and a plane intersect at a point, and two planes intersect along a straight line, the section of a polyhedron by a plane isflat polygon; the vertices of this polygon are the points of intersection of the cutting plane with the edges of the polyhedron, and the sides are the segments along which the cutting plane intersects its faces. This means that to construct the desired section of a given polyhedron by the plane α, it suffices to construct the points of its intersection with the edges of the polyhedron. Then sequentially connect these points with segments, while highlighting with solid lines the visible and dashed invisible sides of the resulting polygon of the section.

III. Methods for constructing sections of polyhedra

The method of sections of polyhedra in stereometry is used in construction problems. It is based on the ability to build a section of a polyhedron and determine the type of section.

This material is characterized by the following features:

• The section method is used only for polyhedra, since various complex (inclined) types of sections of bodies of revolution are not included in the secondary school curriculum.
• The tasks mainly use the simplest polyhedra.
• The tasks are presented mostly without numerical data in order to create the possibility of their multiple use.

To solve the problem of constructing a section of a polyhedron, the student must know:

• What does it mean to construct a section of a polyhedron by a plane;
• How can a polyhedron and a plane be located relative to each other;
• How the plane is set;
• When the problem of constructing a section of a polyhedron by a plane is considered solved.

Since the plane is defined:

• three points;
• Straight and dot;
• two parallel lines;
• two intersecting lines,

The construction of the section plane takes place depending on the assignment of this plane. Therefore, all methods for constructing sections of polyhedra can be divided into methods.

3.1 Construction of sections of polyhedra based on the system of stereometry axioms

Task 1 . Construct a section of the pyramid RABC by the plane α = (MKH), where M, K and H are the internal points of the ribs PC, RV and AB, respectively (Fig. 1, a).

Solution .

1st step . The points M and K lie in each of the two planes α and PBC. Therefore, according to the axiom of the intersection of two planes, the plane α intersects the plane RVS along the straight line MK. Consequently, the segment MK is one of the sides of the desired section (Fig. 1, b).

2nd step . Similarly, the KN segment is the other side of the desired section (Fig. 1, c).

3rd step . Points M and H do not simultaneously lie in any of the faces of the pyramid RABC, therefore the segment MH is not a side of the section of this pyramid. Straight lines KH and RA lie in the plane of the ABP face and intersect. Let us construct the point T= KN ∩AR (Fig. 1d).

Since the straight line KN lies in the plane α, the point T also lies in the plane α. Now we see that the planes α and APC have common points M and T. Therefore, according to the axiom of the intersection of two planes, the plane α and the plane APC intersect along the straight line MT, which, in turn, intersects the edge AC at the point R (Fig. 1, e).

4th step . Now, just as in step 1, we establish that the plane α intersects the faces ACP and ABC along the segments MR and HR, respectively. Therefore, the desired section is the quadrilateral MKHR (Fig. 1, f).

Rice. 2

Task 2. Construct a section of the pyramid MABCD by the plane α = (PRC), where K, H and P are the internal points of the edges MA, MB and MD, respectively (Fig. 2, a).

Solution. The first two steps are the same as steps 1 and 2 of the previous problem. As a result, we obtain the sides KR and KH (Fig. 2, b) of the desired section. Let's build the remaining vertices and sides of the polygon - sections.

3rd step . Let us continue the segment KR until it intersects with the line AD at point F (Fig. 2, c). Since the line KP lies in the secant plane α, the point F= KP ∩ AD = KP ∩ (ABC) is common to the planes α and ABC.

4th step . Let us continue the segment KH until it intersects with the straight line AB at the point L (Fig. 2, d). Since the straight line KN lies in the secant plane α, then the point L = KN ∩ AB = KN ∩ (ABC) is common for the planes α and ABC.

In this way , the points F and L are common for the planes α and ABC. This means that the plane α intersects the plane ABC of the base of the pyramid along the straight line FL.

5th step . Let's draw a straight line FL. This line intersects the edges BC and DC, respectively, at the points R and T (Fig. 2e), which serve as the vertices of the required section. This means that the plane α intersects the face of the base ABCD along the segment RT - the side of the desired section.

6th step . Now we draw the segments RH and PT (Fig. 2, f), along which the plane α intersects the faces of the BMC and MCD of this pyramid. We get the pentagon PKHRT - the desired section of the pyramid MABCD (Fig. 2, f).

Let's consider a more complex problem.

Task 3 . Construct a section of the pentagonal pyramid PABCDE by the plane α = (KQR), where K, Q are the internal points of the edges PA and PC, respectively, and the point R lies inside the face DPE (Fig. 3, a).

Solution . The lines (QK and AC lie in the same plane ASR (according to the axiom of the line and the plane) and intersect at some point T1, (Fig. 3b), while T1 є α, since QК є α.

The straight line PR intersects DE at some point F (Fig. 3, c), which is the intersection point of the plane AR and the side DE of the base of the pyramid. Then the lines KR and AF lie in the same plane AR and intersect at some point T2 (Fig. 3, d), while T2 є α, as a point of the line KR є α (according to the axiom of the line and the plane).

Received: line T1 T2 lies in the secant plane α and in the plane of the base of the pyramid (according to the axiom of the line and the plane), while the line intersects the sides DE and AE of the base ABCDE of the pyramid, respectively, at points M and N (Fig. 3, e), which are the points of intersection planes α with edges DE and AE of the pyramid and serve as the vertices of the desired section.

Further , the line MR lies in the plane of the face DPE and in the secant plane α (according to the axiom of the line and the plane), while intersecting the edge PD at some point H - another vertex of the desired section (Fig. 3, f).

Further, let us construct a point Т3 - Т1Т2 ∩ AB (Fig. 3, g), which, like a point of the straight line Т1Т2 є α, lies in the plane a (according to the axiom of the line and the plane). Now the plane of the face RAB contains two points T3 and K of the secant plane α, which means that the line T3K is the line of intersection of these planes. The straight line Т3К intersects the edge РВ at the point L (Fig. 3, h), which serves as the next vertex of the required section.

Rice. 3

Thus, the "chain" of the sequence of constructing the desired section is as follows:

one . Т1 = QK ∩AC;

2. F = PR ∩ DE;

3. Т2 = KR ∩ AF;

4 . M = T1T2 ∩ DE;

5 . N = T1T2 ∩ AE;

6. H = MR ∩ PD;

7. T3 = T1T2 ∩ AB;

eight . L = T3K ∩ PB.

Hexagon MNKLQH - desired section.

The section of the pyramid in fig. 1 and the section of the cube in fig. 2 are built on the basis of only the axioms of stereometry.

At the same time, a section of a polyhedron with parallel faces (prism, parallelepiped, cube) can be built using the properties of parallel planes.

3.2 The trace method in constructing plane sections of polyhedra

The line along which the cutting plane α intersects the plane of the base of the polyhedron is called the trace of the plane α in the plane of this base.

From the definition of the trace, we obtain: at each of its points, lines intersect, one of which lies in the secant plane, the other in the plane of the base. It is this property of the trace that is used in the construction of plane sections of polyhedra by the trace method. Moreover, in the cutting plane, it is convenient to use such straight lines that intersect the edges of the polyhedron.

First, we define the cutting plane by its trace in the plane of the base of the prism (pyramid) and a point belonging to the surface of the prism (pyramid).

Task 1 . Construct a section of the prism ABCBEA1B1C1D1E1 by the plane α, which is given by the trace l in the plane ABC of the base of the prism and the point M belonging to the edge DD1.

Solution. Analysis . Let us assume that the pentagon MNPQR is the desired section (Fig. 4). To construct this flat pentagon, it is enough to construct its vertices N, P, Q, R (point M is given) - the points of intersection of the secant plane α with the edges CC1, BB1, AA1, EE1 of the given prism, respectively.

E1 D1

To construct the point N =α ∩ CC1, it suffices to construct the line of intersection of the secant plane α with the plane of the face CDD1C1. To do this, in turn, it suffices to construct in the plane of this face one more point belonging to the secant plane α. How to build such a point?

Since the line l lies in the plane of the base of the prism, it can intersect the plane of the CDD1C1 face only at a point that belongs to the line CD = (CDD1) ∩ (ABC), i.e. the point X = l ∩ CD = l ∩ (CDD1) belongs to the secant plane α. Thus, to construct the point N = α ∩ CC1, it suffices to construct the point X = l ∩ CD.

Similarly, to construct the points P= α ∩ BB1, Q = α ∩ AA1, and R = α ∩ EE1, it suffices to construct the points Y = l ∩ BC, Z = 1 ∩ AB, and T = 1 ∩ AE, respectively.

Building . We build (Fig. 5):

1. X = l ∩ CD (Fig. 5b);

2. N = МХ ∩ СС1 (Fig. 5, c);

3. Y = l ∩ BC (Fig. 5d);

4. P = NY ∩ BB1 (Fig. 5e);

5. Z = 1 ∩ AB (Fig. 5, f);

6. Q= PZ ∩ AA1 (Fig. 5, g);

7. T= l ∩ AE (Fig. 5, h);

8. R= QT ∩ EE1 (Fig. 5i).

The pentagon MNPQR is the desired section (Fig. 5, j).

Proof. Since the line l is the trace of the secant plane α, then the points X = l ∩ CD, Y = l ∩ BC, Z = 1 ∩ AB and T= l ∩ AE belong to this plane.

Therefore we have:

М Є α, X Є α => МХ є α, then МХ ∩ СС1 = N є α, hence, N = α ∩ СС1;

N Є α, Y Є α => NY Є α, then NY ∩ BB1= P Є α, hence P = α ∩ BB1;

Р Є α, Z Є α => РZ Є α, then PZ ∩ AA1 = Q Є α, hence Q = α ∩ AA1;

Q Є α, T Є α => QТ Є α, then QТ ∩ EE1 =R Є α, hence R = α ∩ EE1.

Therefore, MNPQR is the required section.

Study. The trace l of the secant plane α does not intersect the base of the prism, and the point M of the secant plane belongs to the side edge DD1 of the prism. Therefore, the cutting plane α is not parallel to the side edges. Therefore, points N, P, Q and R of intersection of this plane with the side edges of the prism (or extensions of these edges) always exist. And since, in addition, the point M does not belong to the trace l, the plane α defined by them is unique. This means that the problem has (always) a unique solution.

3.3 Internal design method in constructing plane sections of polyhedra

In some textbooks, the method of constructing sections of polyhedra, which we will now consider, is called the method of internal design or the method of correspondences, or the method of diagonal sections.

Task 1 . Construct a section of the pyramid PABCDE by the plane α = (MFR) if the points M, F and R are internal points of the edges PA, PC and PE, respectively. (Fig. 6)

Solution . We denote the plane of the base of the pyramid by β. To construct the desired section, we construct the points of intersection of the secant plane α with the edges of the pyramid.

Let us construct the point of intersection of the secant plane with the edge РD of the given pyramid.

The planes APD and CPE intersect the plane β along the lines AD and CE, respectively, which intersect at some point K. The line PK = (APD) ∩ (CPE) intersects the line FR є α at some point K1: K1 = PK ∩ FR, with this K1 є α. Then: M є α, K1 є α => straight line MK є a. Therefore, the point Q = MK1 ∩ PD is the point of intersection of the edge PD and the secant plane: Q = α ∩ PD. Point Q is the vertex of the desired section. Similarly, we construct the intersection point of the plane α and the edge РВ. The planes BPE and APD intersect the plane β along the lines BE and AD, respectively, which intersect at the point H. The line PH = (BPE) ∩ (APD) intersects the line MQ at the point H1. Then the line RN1 intersects the edge PB at the point N = α ∩ PB - the top of the section.

In this way , the sequence of steps for constructing the desired section is as follows:

one . K = AD ∩ EC; 2. K1 = RK ∩ RF;

3 . Q = MK1 ∩ PD; 4. H = BE ∩ AD;

5 . H1 = PH ∩ MQ; 6. N = RН1 ∩ РВ.

The pentagon MNFQR is the required section.

3.4 Combined method in constructing plane sections of polyhedra

The essence of the combined method for constructing sections of polyhedra is as follows. At some stages of constructing a section, either the method of traces or the method of internal design is used, and at other stages of constructing the same section, the studied theorems on parallelism, perpendicularity of lines and planes are used.

To illustrate the application of this method, consider the following problem.

Task1.

Construct a section of the parallelepiped ABCDА1В1С1D1 by the plane α given by the points P, Q and R, if the point P lies on the diagonal A1C1, the point Q on the edge BB1 ​​and the point R on the edge DD1. (Fig. 7)

Solution

We will solve this problem using the trace method and parallelism theorems for lines and planes.

First of all, we construct the trace of the secant plane α = (PQR) on the plane ABC. To do this, we construct the points T1 = PQ ∩ P1B (where PP1 ║AA1,P1є AC) and T2 = RQ ∩ BD. Having built the trace T1T2, we notice that the point P lies in the plane A1B1C1, which is parallel to the plane ABC. This means that the plane α intersects the plane A1B1C1 along a straight line passing through the point P and parallel to the straight line T1T2. Draw this line and denote by M and E the points of its intersection with the edges A1B1 and A1D1, respectively. We get: M = α ∩ A1B1, E = α ∩ A1D1. Then the segments ER and QM are the sides of the required section.

Further, since the plane BCC1 is parallel to the plane of the face ADD1A1, then the plane α intersects the face BCC1B1 along the line QF (F= α ∩ CC1) parallel to the line ER. Thus, the pentagon ERFQM is the required section. (Point F can be obtained by doing RF║ MQ)

Let's solve this problem using the method of internal design and theorems on the parallelism of lines and planes.(Fig. 8)

Rice. eight

Let H=AC ∩ BD. Drawing the line HH1 parallel to the edge BB1 ​​(H1 є RQ), we construct the point F: F=РН1 ∩ CC1. The point F is the point of intersection of the plane α with the edge CC1, since РН1 є α. Then the segments RF and QF, along which the plane α intersects, respectively, the faces CC1D1D and BCC1B1 of this parallelepiped, are the sides of its required section.

Since the plane ABB1 is parallel to the plane CDD1, the intersection of the plane α and the face ABB1A1 is the segment QM (M Є A1B1), parallel to the segment FR; segment QM - section side. Further, the point E = MP ∩ A1D1 is the point of intersection of the plane α and the edge A1D1, since MP є α. Therefore, point E is another vertex of the desired section. Thus, the pentagon ERFQM is the required section. (Point E can be constructed by drawing a line RE ║ FQ. Then M = PE ∩ A1B1).

IV. Conclusion

Thanks to this work, I summarized and systematized the knowledge gained in this year's geometry course, got acquainted with the rules for creative work acquired new knowledge and put it into practice.

I would like to use my newly acquired knowledge in practice more often.

Unfortunately, I have not considered all methods for constructing sections of polyhedra. There are many more special cases:

• construction of a section of a polyhedron by a plane passing through given point parallel to a given plane;
• construction of a section passing through a given line parallel to another given line;
• construction of a section passing through a given point parallel to two given skew lines;
• construction of a section of a polyhedron by a plane passing through a given line perpendicular to a given plane;
• construction of a section of a polyhedron by a plane passing through a given point perpendicular to a given line, etc.

In the future, I plan to expand my research and supplement my work with an analysis of the above particular cases.

I believe that my work is relevant as it can be used by middle and high school students to self-study to the Unified State Examination in mathematics, for in-depth study of the material on electives and for self-education of young teachers. High school graduates must not only master the material school programs, but also be able to creatively apply it, find a solution to any problem.

V. Literature

1. Potoskuev E.V., Zvavich L.I. Geometry. Grade 10: Textbook for general educational institutions with in-depth and profile study of mathematics. - M.: Bustard, 2008.
2. Potoskuev E.V., Zvavich L.I. Geometry. Grade 10: Task book for educational institutions with in-depth and profile study of mathematics. - M.: Bustard, 2008.
3. Potoskuev E.V. The image of spatial figures on the plane. Construction of sections of polyhedra. Tutorial for students of the Faculty of Physics and Mathematics of the Pedagogical University. - Tolyatti: TSU, 2004.
4. Scientific and practical journal for high school students "Mathematics for schoolchildren", 2009, No. 2 / No. 3,1-64.
5. Geometry in tables - Textbook for high school students - Nelin E.P.
6. Geometry, grades 7-11, Reference materials, Bezrukova G.K., Litvinenko V.N., 2008.
7. Mathematics, Help Guide, For high school students and those entering universities, Ryvkin A.A., Ryvkin A.Z., 2003.
8. Algebra and geometry in tables and diagrams, Roganin A.N., Dergachev V.A., 2006.