Tetrahedron. Problems for constructing sections in a tetrahedron

Definition

The section is flat figure, which is formed when the spatial figure intersects with a plane and whose boundary lies on the surface of the spatial 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 that is 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 \(ABC\) .

Decision

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 \(ABC\) .

Decision

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 can 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 \(BD\) .

Decision

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

Dana rectangular pyramid\(DABC\) with an edge \(DB\) perpendicular to the plane \(ABC\) . The base is a 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.

Decision

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 \(DN=NC\) .

The task itself usually goes like this: "build natural look section figures". Of course, we decided not to leave this question aside and try, if possible, to explain how the oblique section is constructed.

In order to explain how an oblique section is built, I will give a few examples. Of course, I will start with the elementary, gradually increasing the complexity of the examples. I hope that after analyzing these examples of section drawings, you will understand how this is done and will be able to complete your learning task yourself.

Consider a "brick" with dimensions of 40x60x80 mm by an arbitrary inclined plane. The cutting plane cuts it along the points 1-2-3-4. I think everything is clear here.

Let's move on to the construction of a natural form of the sectional figure.
1. First of all, let's draw the axis of the section. The axis should be drawn parallel to the section plane - parallel to the line into which the plane is projected on the main view - usually it is on the main view that the task is set for construction of an oblique section(I will always refer to main view, bearing in mind that this is almost always the case in training drawings).
2. On the axis, we set aside the length of the section. In my drawing, it is designated as L. The size L is determined in the main view and is equal to the distance from the point where the section enters the part to the point where it exits.
3. From the resulting two points on the axis perpendicular to it, we set aside the section widths at these points. The width of the section at the point of entry into the part and at the point of exit from the part can be determined in the top view. AT this case both segments 1-4 and 2-3 are equal to 60 mm. As you can see from the picture above, the edges of the section are straight, so we simply connect our two resulting segments, getting a rectangle 1-2-3-4. This is - the natural view of the figure of the section of our brick with an inclined plane.

Now let's complicate our detail. Let's put a brick on the base 120x80x20 mm and add stiffeners to the figure. Let's draw a cutting plane so that it passes through all four elements of the figure (through the base, brick and two stiffeners). In the picture below you can see three views and a realistic image of this part.

Let's try to build a natural view of this slanted section. Let's start again with the section axis: draw it parallel to the section plane indicated on the main view. On it we set aside the length of the section equal to A-E. Point A is the entry point of the section into the part, and in a particular case, the entry point of the section into the base. The exit point from the base is point B. Let's mark point B on the axis of the section. The same way we also note the entry-exit points to the edge, to the "brick" and to the second edge. From points A and B perpendicular to the axis, we set aside segments equal to the width of the base (on each side of the axis, 40, only 80 mm). Let's connect the extreme points - we get a rectangle, which is a natural view of the section of the base of the part.

Now it's time to build a piece of section, which is a section of the edge of the part. From points B and C, we set aside perpendiculars of 5 mm in each direction - we will get segments of 10 mm. Connect the extreme points and get the cross section of the rib.

From points C and D we set aside perpendicular segments equal to the width of the "brick" - completely similar to the first example of this lesson.

Having set aside the perpendiculars from points D and E equal to the width of the second edge and connecting the extreme points, we obtain a natural view of its section.

It remains to erase the jumpers between the individual elements of the resulting section and apply hatching. You should get something like this:

If, according to a given section, we divide the figure, then we will see the following view:

I hope you are not intimidated by the tedious paragraphs of the description of the algorithm. If you have read all of the above and still do not fully understand, how to draw a cross section, I strongly advise you to take a piece of paper and a pencil in your hands and try to repeat all the steps after me - this will almost 100% help you learn the material.

Once I promised the continuation of this article. Finally I'm ready to present to you step by step construction oblique section of the part, closer to the level of homework. Moreover, the oblique section is defined in the third view (the oblique section is defined in the left view)

or write down our phone number and tell your friends about us - someone is probably looking for a way to make drawings

or create a note about our lessons on your page or blog - and someone else will be able to master the drawing.

Yes, everything is fine, but I would like to see how the same thing is done on a more complex part, with chamfers and a cone-shaped hole, for example.

Thank you. But aren't the stiffeners hatched on the cuts?
Exactly. It is they who do not hatch. Because they are general rules making cuts. However, they are usually hatched when making cuts in axonometric projections - isometry, dimetry, etc. When performing inclined sections, the area related to the stiffener is also shaded.

Thank you, very accessible. Can you tell me if the oblique section can be done in the top view, or in the left view? If so, I would like to see the simplest example. Please.

It is possible to make such cuts. But unfortunately I don't have an example at hand right now. And there's another one interesting point: on the one hand, there is nothing new there, but on the other hand, in practice, it is really more difficult to draw such sections. For some reason, everything starts to get confused in the head and most students have difficulties. But don't give up!

Yes, everything is fine, but I would like to see how the same thing is done, but with holes (through and non-through), otherwise they never turn into an ellipse in my head

help me with a complex problem

It's a pity that you wrote here. We would write in the mail - maybe we could have time to discuss everything.

You explain well. What if one of the sides of the part is semicircular? Also, there are holes in the part.

Ilya, use the lesson from the section on descriptive geometry "Section of a cylinder by an inclined plane". With it, you can figure out what to do with holes (they are also cylinders in fact) and with a semicircular side.

I thank the author for the article! Brief and understandable. About 20 years ago I myself gnawed at the granite of science, now I help my son. I forgot a lot, but your article returned a fundamental understanding of the topic. I’ll go with the inclined section of the cylinder to deal with)

Add your comment.

1. The concept of a positional problem. Recall that the plane is called cutting plane polyhedron if there are points of the polyhedron on both sides of this plane. Cross section of a polyhedron A plane is a polygon whose sides are segments along which the cutting plane intersects the faces of the polyhedron.

On fig. 30 pictured triangular prism. (In this projection drawing, the dot images are denoted by the same letters as the corresponding original dots). Imagine that we need to mark the points: a) M lying on the edge ; b) N lying in the face ; c) lying inside the prism.

If we depict these points as it is done in figure a), then only about the point M we can conditionally say that it lies on the edge . Point position N and K cannot be determined from this figure. Figure b) already allows us to conclude that the point N lies in the face , and the point is

inside the prism. On what basis can these conclusions be drawn? The fact is that in the second figure we set the projections of the points N and K on the base plane parallel to the side edges of the prism. Strictly speaking, in order to be sure that the point M lies on the edge, visual perceptions alone are also not enough. (In the design with which the image of the prism was performed, the point M serves as a projection of any point on a straight line parallel to the design direction and passing through it.)

If, on the other hand, we indicate that when projecting parallel to the side edges of the prism, the point M projected on a base to a point BUT, then such confidence appears.

A similar situation is shown in Fig. 31. Here it is necessary to mark the points: a) M on the side edge SA; b) N- on the verge SAB;
in) To- inside the pyramid. The difference lies in the fact that the right figure uses the central projection of marked points onto the plane of the base of the pyramid from its top. S.

In order to make the image visual, in the considered examples it is necessary to use not one projection, but two. The first projection, with which the image of a polyhedron is made, is called external. The second design is of an auxiliary nature. It is associated with the figure itself - as a rule, it is a projection onto a plane containing one of the faces of the polyhedron. We will deal only with prisms and pyramids, and most often choose the plane of their base as such a plane. Assisted design is called internal. It can be seen from the considered examples that it is convenient to use internal parallel projection for a prism, and central one for a pyramid.

Let be F 0 - some figure in space, which is projected parallel to the plane p(external design). In order for the image of the figure to be visual, we choose some plane in space, different from the plane p, and consider a new projection, parallel or central, of the points of the figure F 0 to this plane (internal design).

Consider a point in space M 0 and its projection onto the plane p 0 ¢ in interior design. Both these points are projected onto the plane p. At the same time, the projection M points M 0 is called basic(or just a projection), and the projection М¢ points - secondary.

If for a point M 0 figures F 0 its projection and secondary projection are known, then from the image we can judge the position of this point on the original. In this case, the point is said to be M 0 , belonging to the shape F 0 is given on the projection drawing. figure image F 0 , on which each point of the figure is given, is called complete.

In projection drawings, you often have to solve problems about finding the intersection of various shapes. Such tasks are called positional. If some image is complete, then any positional problem is solvable on this image.

In conclusion, we note the following. If a M 0 ¢ , N 0 ¢, K 0 ¢, ... – images of points M 0 , N 0 , K 0 , ... with internal design, then with external design (parallel) images MM¢, NN¢, KK¢, ... parallel lines M 0 M 0 ¢, N 0 N 0 ¢, K 0 K 0 ¢, ... on surface p will also be parallel. If M 0 ¢, N 0 ¢, K 0 ¢, ... – images of points M 0 , N 0 , K 0 , ... for internal central design with center S 0 , then images MM¢, NN¢, KK¢, ... direct M 0 M 0 ¢, N 0 N 0 ¢, K 0 K 0 ¢, ... intersect on the plane during external design p at one point S. This point will be the image of the point S 0 .

Among positional problems, we will be interested only in problems related to the construction of sections of polygons. Let us consider the main methods for constructing such sections. Usually, when solving stereometric problems, the images of the points of the figure in the projection drawing are denoted by the same letters as the corresponding points on the original figure. We will also adhere to this rule in the future.

2. Construction of sections based on the properties of parallel lines and planes. This method is especially often used when constructing sections of parallelepipeds. This is because the opposite faces of the parallelepiped are parallel. According to the theorem on the intersection of parallel planes by the third plane, the lines of intersection of parallel faces are parallel segments.

Task 1. The base of a quadrangular pyramid SABCD is a parallelogram. Construct a section of the pyramid by a plane passing through a point lying on the side edge AS, parallel to the diagonal BD grounds.

How many such planes can be built? What figures can be obtained in section?

Decision. Draw an arbitrary straight line in the plane of the base of the pyramid a, parallel to the diagonal BD. A plane passes through this line and point. a, and the only one. On the basis of parallelism of a straight line and a plane and, therefore, a plane a is desired.

In the plane of the base there are infinitely many lines parallel to the line BD, therefore, there are infinitely many planes that satisfy the condition of the problem.

The type of polygon obtained in the section depends on the number of faces that intersect the plane a. Since a quadrangular pyramid has five faces, triangles, quadrangles and pentagons can be obtained in section.

On fig. 32 shows various cases of the location of the line a relative to the parallelogram ABCD. Obviously, depending on this location, the type of polygon-section will be determined.

On the left in fig. 33 the case is considered when the straight line a 1 crosses sides AD,AB at points M, N respectively, and lies with the point in the same half-space with the boundary BSD. Here the cross section is a triangle MKN.

The figure on the right shows the case when the line a 3 lies with a dot on different sides from the plane BSD and crosses the sides DC, BC bases in points M, N respectively. Denote by X point of intersection of lines AD and a 3 . Since the straight line AD lies in the plane of the face ASD, then this face also contains a point X. On the other hand, the point X belongs to the line a 3 lying in the cutting plane. Therefore, the straight line will be the line of intersection of the secant plane and the plane of the face A.S.D. This allows you to find the point R=SDÇ KX. Similarly, a point allows you to build a vertex TÎ BS desired section. In the considered case, the cutting plane intersects all the faces of the pyramid and the section is a pentagon.

Other cases relative position straight a and consider the bases of the pyramid yourself.

Consider special methods for constructing sections.

4. Method of traces. If the cutting plane is not parallel to the face of the polyhedron, then it intersects the plane of this face in a straight line. The line along which the secant plane intersects the plane of the face of the polyhedron is called following the cutting plane on the plane of this face. One of the methods for constructing sections of polyhedra is based on using the trace of a cutting plane on the plane of one of its faces. Most often, when constructing sections of a prism and a truncated pyramid, the plane of the lower base is chosen as such a plane, and in the case of a pyramid, the plane of its base.

Consider the construction of sections by the trace method using examples.

Task 2. An image of a quadrangular prism is given ABCDA 1 B 1 C 1 D one . Specify three points belonging to its different side faces, and construct a section passing through these three points.

Decision. Recall that to set a point in a projection drawing, you must set its primary and secondary projections. In the case of a prism, we agreed to use internal parallel projection to define secondary projections. Therefore, to set the point M lying on the edge ABB 1 BUT 1 , specify its projection M 1 on the base plane parallel to the side edges of the prism. The points are set in the same way. N and K lying in the faces AD 1 DA 1 , CDD 1 C 1 respectively (Fig. 34). Let's build a trace of the secant plane on the plane of the lower base of the prism. Parallel lines MM 1 , lie in the same plane and, therefore, in the general case, the lines intersect at some point X. Since the straight line lies in the secant plane, and the straight line lies in the plane of the lower base, then the point X belongs to the trace of the cutting plane on the plane of the lower base of the prism. Likewise, dots K, N and their secondary projections K 1 , N 1 allow you to find the second point Y belonging to the desired trace.

Straight AB lying on the edge ABB 1 BUT 1 , intersects the trace XY at the point Z, so the straight line MZ lies as in the plane of the face ABB 1 BUT 1 and in the cutting plane. Line segment TR, where T=MZÇ AA 1 , P=MZÇ BB 1 , will be the side of the polygon-section. Next, we sequentially build its sides TR and RQ passing through these points N and K respectively. Finally, building a side PQ.

Task 3 . Given an image of a pentagonal pyramid SABCDE. Set points N and K, belonging to the lateral edges SC, SD respectively and the point M lying on the edge A.S.E. Construct a section through given points.

Decision. To set points K,N and M Let's use the internal central projection with the center at the top of the pyramid. In this case, the projections of points K and N there will be points D and C, and the projection of the point M- dot (Fig. 35).

The lines and lying in the plane generally intersect at a point X lying in the cutting plane. On the other hand, the point X lies in the plane of the base, and hence it belongs to the trace of the secant plane on the plane of the base. The second point of the desired trace will be the point . Straight AE lying on the edge ASE pyramids, crosses the trail XY at the point Z. Drawing a straight line ZM, find the side LP polygon-section. In order to find the vertex of the section, we build a point and then a straight line.

5. Internal design method. The essence of this method lies in the fact that here, with the help of internal projection, the points of the section are searched for by their known secondary projections. The method of internal design is especially convenient to apply in cases where the trace of the cutting plane is far removed from the given figure. This method is also indispensable when some of the lines containing the sides of the base of the polyhedron intersect the trace outside the drawing. Consider the application of the method with examples.

Problem 4. Given an image of a hexagonal prism and three points lying in three side faces, no two of which are adjacent. Construct a section of a prism by a plane passing through the given points.

Decision. Let the given points M,L,K lie on the faces , , , and M¢,L¢,K¢– their secondary projections
(Fig. 36).

Find the point where the cutting plane intersects the side edge. To do this, using the internal projection for the point, we find the main projection X lying in the cutting plane. Search point X is the point of intersection of the line passing through the point X¢ parallel to the side edges of the prism, and straight ML lying in the cutting plane. Dot X lets build the top and then the side QR sections. Similarly, using the point , we construct the point Y, direct KY and find the top R sections. Next, the sides are built PQ and PO sections.

The remaining constructions are performed in the following sequence:

1) build a point Z¢=AK¢Ç BD;

2) find a point Z (ZÎ PK);

3) draw a straight line oz and find the top S (SÎ DD 1) sections;

4) sequentially build sides SR,ST and TO sections.

Task 5 . An image of a quadrangular pyramid and three points lying on its lateral edges is given. Construct a section passing through given points.

Decision. Let be SABCD- this pyramid, M,N, K- these points (Fig. 37). Secondary point projections M, N, K in interior central design from top S on the base plane are the points A, C and D respectively. Note that in this problem the sides and KN sections are built immediately. It remains to find only the vertex of the section L lying on the side edge SB. To do this, we will construct a point and “raise” it into the cutting plane using internal projection. preimage of a point X¢ with this central design there will be a point X=X¢SÇ MN. Vertex L, belonging to the edge SB, lies on the line KX.

6. Combined method. The essence of this method is to combine the method of traces or the method of internal design with constructions based on the properties of parallel lines and planes.

Consider the following example.

Problem 6. Point M is the midpoint of the edge AD Cuba ABCDA 1 B 1 C 1 D one . Construct a section of a cube by a plane passing through a point M parallel to the diagonal BD bases and diagonals AB 1 side face AA 1 AT 1 AT.

Decision. cutting plane a parallel to the diagonal BD bases and passes through the point M, also lying at the base, so it intersects the base in a straight line
(Fig. 38).

Straight l will be the trace of the plane a on the plane of the bottom base of the cube. Let's denote . Track m plane a on the face plane ABB 1 BUT 1 is constructed in a similar way. This trace passes through the point N, in parallel AB one . Let's denote .

You can continue building the section without resorting to special methods. However, we will use the trace method. Let the line sun crosses the track l at the point X. points X and desired plane a lie in the plane of the face VSS 1 AT 1 . Denote by L point of intersection of a line and an edge AT 1 With one . Further, it is convenient to use the theorem on the intersection of two parallel planes by a third plane. By virtue of this theorem, . Here RÎ DD 1 ,PÎ C 1 D 1 .

Prove that the hexagon obtained in the section is regular.

Image of a circle

1. Ellipse and its properties. When depicting a cylinder, a cone and a ball (sphere), we will have to draw ellipses. Ellipse can be defined different ways. We give the definition by means of contraction of the plane to a straight line.

Ellipse a line is called, which is the image of a circle when the plane is compressed to a straight line passing through the center of the circle (Fig. 39).

Given a circle, a straight line passing through its center, and a contraction factor, it is easy to construct an image of any point of a given circle using the above definition. Having completed the construction of several image points and connecting them with a smooth line, you can draw an ellipse, which is the image of a circle.

Oxy so that its axis Ox matched with direct compression l, and the beginning O was the center of the circle w radius a(Fig. 40). In this coordinate system, the circle w is determined by the equation: or

This means that any point whose coordinates satisfy equation (1) belongs to the circle w, and the point whose coordinates do not satisfy (1) does not belong.

Let be is the compression ratio, is an arbitrary point of the plane, and M 0 - its projection on a straight line l. When compressed to a point M goes to a point such that . Since the straight line MM 1 parallel to the axis Oy, then , and the projection M 0 of these points on the compression line Ox determined by the coordinates.

Hence , . Therefore, the compression formulas look like

Conversely, formulas (2) determine the compression of the plane to the axis Ox with compression ratio , where the point goes to the point .

From these formulas, . Substituting x and y into equation (1), we get: . So the coordinates of the point M 1 , which is the image of a point on a circle, satisfy the equation

where . This is the equation in the system Oxy defines an ellipse g, which is obtained by squeezing the circle w to the axis Ox. Recall that equation (3) is called the canonical equation of the ellipse.

Using the canonical equation of an ellipse, one can study its geometric properties. Let's recall some concepts related to the ellipse and its properties.

Let the ellipse g is given in a rectangular coordinate system by the canonical equation (3). As x and y are included in this equation to the second degree, we can draw the following conclusions.

If , then О g(Fig. 41). It follows that the origin of coordinates O is the center of symmetry of the ellipse. The center of symmetry of an ellipse is called its center.

If , then , . It follows from this that the lines Ox and Oy are the axes of symmetry of the ellipse. The axes of symmetry of an ellipse are called its axes. Each of the axes intersects the ellipse at two points. Axis Ox has the equation , therefore, from equation (3) for the abscissas of the points BUT 1 , A 2 we have intersections. From here BUT 1 (a;0), BUT 2 (–a;0). Similarly, we find that the axis Oy intersects the ellipse at points AT 1 (0;b) and AT 2 (0;–b). The points of intersection of an ellipse with its axes are called peaks ellipse. Segments BUT 1 BUT 2 and AT 1 AT 2 also called ellipse axes. Ellipse Center O is the common midpoint of each of these segments.

A line segment whose endpoints are in an ellipse is called chord this ellipse. The chord of an ellipse passing through its center is called ellipse diameter. Means, the axes of the ellipse are its mutually perpendicular diameters.

Note that for , we have . In this case A 1 A 2 >B 1 B 2 and segments A 1 A 2 , B 1 B 2 are named respectively major and minor axes ellipse. In this case, the numbers are called respectively major and minor semiaxes ellipse. For , on the contrary, . Here the names of the axes change accordingly.

Let us consider parametric equations of an ellipse and a method based on them for constructing points of an ellipse.

Let the segments BUT 1 BUT 2 and AT 1 AT 2 are the axes of the ellipse. We build on them, as on diameters, concentric circles w 1 and w 2 respectively (Fig. 42). Consider a beam h starting at point O. This ray traverses circles w 1 and w 2 in points M 1 and M 2. Through the dot M 1 draw a line parallel to the minor axis AT 1 AT 2 , and through the dot M 2 - straight line parallel to the major axis BUT 1 BUT 2. Let us show that the point M the intersection of these lines belongs to an ellipse with given axes.

We choose a rectangular coordinate system Oxy starting at point O. Let in this system the point M has coordinates ( x;y). Next, let the beam h forms with beam OA 1 corner t. If , then , . Because the dots M and M 1 have equal abscissas, and points M and M 2 - equal ordinates,

From equalities (4) , , therefore, due to the main trigonometric identity we have , i.e. the constructed point belongs to an ellipse with semiaxes a and b.

For any value tÎ}