Mathematics teacher of the Shchelkovsky branch of the State Budgetary Educational Institution of the Moscow Region "Krasnogorsk College" Artemiev Vasily Ilyich.
The study of the topic "Solving problems for the construction of sections" begins in the 10th grade or in the first year of NGO institutions. If the mathematics classroom is equipped with multimedia tools, then the solution to the problem of studying is facilitated with the help of various programs. One of these programs is software dynamic mathematics GeoGebra 4.0.12. It is suitable for studying and learning at any stage of education, facilitates the creation of mathematical constructions and models by students, which allow interactive research when moving objects and changing parameters.
Consider the application of this software product on a specific example.
A task. Construct a section of the pyramid by the plane PQR, if the point P lies on the line SA, the point Q lies on the line SB, the point R lies on the line SC.
Solution. Let's consider two cases. Case 1. Let the point P belong to the edge SA.
1. Using the Point tool, mark arbitrary points A, B, C, D. Right-click on point D, select "Rename". Rename D to S and set the position of this point as shown in Figure 1.
2. Using the tool "Segment by two points" we will construct the segments SA, SB, SC, AB, AC, BC.
3. Right-click on segment AB and select "Properties" - "Style". Set up a dotted line.
4. Mark the points P, Q, R on the segments SA, SB, CS.
5. Use the tool "Line by two points" to construct a line PQ.
6. Consider the line PQ and the point R. Question for students: How many planes pass through the line PQ and the point R? Justify the answer. (Answer. A plane passes through a line and a point not lying on it, and moreover, only one).
7. We build direct PR and QR.
8. Select the Polygon tool and click on the PQRP points one by one.
9. Use the Move tool to change the position of the points and observe the changes in the section.
Picture 1.
10. Right-click on the polygon and select "Properties" - "Color". Fill the polygon with some gentle color.
11. On the object panel, click on the markers and hide the lines.
12. As an additional task, you can measure the cross-sectional area.
To do this, select the "Area" tool and left-click on the polygon.
Case 2. The point P lies on the line SA. To consider the solution of the problem for this case, you can use the drawing of the previous problem. Let's hide only the polygon and the point P.
1. Use the tool "Line by two points" to construct a straight line SA.
2. Mark a point P1 on line SA as shown in figure 2.
3. Draw a line P1Q.
4. Select the tool "Intersection of two objects", and left-click on the straight lines AB and P1Q. Let's find the point of their intersection K.
5. Let's draw a line P1R. Find the point of intersection M of this line with the line AC.
Question for students: how many planes can be drawn through lines P1Q and P1R? Justify the answer. (Answer. A plane passes through two intersecting lines, and moreover, only one).
6. Let's draw direct KM and QR. Question for students. Which planes simultaneously belong to the points K, M? Intersection of which planes is the straight line KM?
7. Construct the polygon QRKMQ. Fill with a gentle color and hide the auxiliary lines.
Figure 2.
Using the "Move" tool, we move the point along the straight line AS. We consider various positions of the section plane.
Tasks for constructing sections:
1. Construct a section defined by parallel lines AA1 and CC1. How many planes pass through the parallel lines? 
2. Construct a section passing through intersecting lines. How many planes pass through the intersecting lines? 
3. Construction of sections using the properties of parallel planes:
a) Construct a section of a parallelepiped by a plane passing through the point M and the line AC. 
b) Construct a section of the prism by a plane passing through the edge AB and the middle of the edge B1C1.
c) Construct a section of the pyramid by a plane passing through the point K and parallel to the plane of the bases of the pyramid.
4. Construction of sections by the trace method:
a) Given a pyramid SABCD. Construct a section of the pyramid by a plane passing through the points P, Q and R.
5) Draw the line QF and find the point H of intersection with the edge SB.
6) Let's draw direct HR and PG.
7) Select the resulting section with the "Polygon" tool and change the fill color.
b) Construct a section of the parallelepiped ABCDA1B1C1D1 on your own by a plane passing through points P, K and M. List of sources.
1. Electronic resource http://www.geogebra.com/indexcf.php
2. Electronic resource http://geogebra.ru/www/index.php (Site of the Siberian GeoGebra Institute)
3. Electronic resource http://cdn.scipeople.com/materials/16093/projective_geometry_geogebra.PDF
4. Electronic resource. http://nesmel.jimdo.com/geogebra-rus/
5. Electronic resource http://forum.sosna24k.ru/viewforum.php?f=35&sid=(GeoGebra forum for teachers and schoolchildren).
6. Electronic resource www.geogebratube.org (Interactive materials for working with the program)
Problems on the construction of sections of polyhedra occupy a significant place both in the school geometry course for senior classes and in exams at various 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 of these planes and draw a 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 different planes have a common point, then they intersect along a straight line passing through this point.
As I have 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, tasks for construction, tasks 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 AT 1C1,
Q belongs to AB
Solution: All three points P, Q, R lie on 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.
There are 2 main methods for constructing sections of polyhedra:
Axiomatic method for constructing sections
1. Method of traces
Example 1
On the edges AA" and B"C" of the prism ABCA"B"C" we set the points P and Q, respectively. We construct a section of the prism by a plane (PQR), the point R of which we set in one of the following faces:
a) BCCB "C";
b) A "B" C";
c) ABC
Solution.
a) 1) Since the points Q and R lie in the plane (BCC"), then the line QR lies in this plane. Draw it. This is the trace of the plane (PQR) onto the plane (BCC"). (fig.1)
2) Find points B"" and C", at which the line QR intersects the lines BB" and CC, respectively. Points B" and C" are traces of the plane (PQR), respectively, on the lines BB" and CC".
3) Since the points B "" and P lie in the plane (ABB"), then the line B "" P lies in this plane. Let's draw it. The segment B ** P is the trace of the plane (PQR) on the face ABB "A".
4) Since the points P and C lie in the plane (ACC"), then the line PC"" lies in this plane. Draw it. This is the trace of the plane (PQR) on the plane (ACC").
5) Find the point V at which the straight line PC"" intersects the edge A"C". This is the trace of the plane (PQR) on the edge A "C".
6) The wheelbarrow as the points Q and V lie in the plane (A "B" C "), then the line QV lies in this plane. Let's draw the line QV. The segment QV is the trace of the plane (PQR) on the face ABC. So, we got the polygon QB ""PV - the required section.
b) 1) Since the points Q and R lie in the plane (A "B" C "), then the line QR lies in this plane. Let's draw it. This is the trace of the plane (PQR) on the plane (A" B "C"). (Fig. .2)
2) Find the points D" and E", at which the line QR intersects the lines A"B" and B"C", respectively. Since the point D" lies on the edge A"B", the segment QD" is the trace of the plane (PQR) on the face A"B"C".
3) Since the points D "and P lie in the plane (ABB"), then the line D "P lies in this plane. Let's draw it. This is the trace of the plane (PQR) on the plane (ABB"), and the segment D "P is the trace plane (PQR) on face ABB"A".
4) Since the points P and E" lie in the plane (ACC"), then the line PE lies in this plane. Let's draw it. This is the trace of the plane (PQR) on the plane (ACC").
5) Find the point C""=PE""CC". Since the point C"" lies on the edge CC", then the segment PC"" is the trace of the plane (PQR) on the face ACC"A".
6) Since the points Q and C "" lie in the plane (BCC"), then the line QC "" lies in this plane. Draw it. This is the trace of the plane (PQR) on the plane (BCC"), and the segment QC "" - plane trace (PQR) on face BCC"B". So, we got the polygon QD"PC"" - this is the desired section.
in) 1) Of the three given points P, Q and R, no two lie in any one of the planes of the faces of the prism, so we find the main trace of the plane (PQR) (i.e., the line of intersection of the plane (PQR) with the plane (ABC), chosen as the main). To do this, we first find the projections of the points P, Q and R onto the plane (ABC) in the direction parallel to the side edge of the prism. Since the point P lies on the edge AA", then the point P" coincides with the point A. Since the point Q lies in the plane (BCC"), then in this plane through the point Q we draw a line parallel to the line BB", and find the point Q ", in which the drawn line intersects the line BC. Since the point R, by condition, lies in the plane chosen as the main one, the point R" coincides with the point R. (Fig. 3)
2) Parallel lines PP" and QQ" define the plane. Let us draw the lines PQ and P"Q" in this plane and find the point S=PQ intersects P"Q". Since the point S" lies on the line PQ, it lies in the plane (PQR), and since the point S" lies on the line P"Q", it lies in the plane (ABC). Thus, the point S "is a common point of the planes (PQR) and (ABC). This means that the planes (PQR) and (ABC) intersect along a straight line passing through the point S".
3) Since the point R coincides with the point R", then the point R is another common point of the planes (PQR) and (ABC). Thus, the line S "R is the main trace of the plane (PQR). Let's draw this line. As you can see from the figure, the straight line S "R intersects the edges AB and BC of the base of the prism, respectively, at the points S" "and S" "".
4) Since the points S""" and Q lie in the plane (BCC"), then the line S""" Q lies in this plane. Draw it. This is the trace of the plane (PQR) on the plane (BCC"). And the segment S""" Q, is the trace of the plane (PQR) on the face of BCC"B".
5) Similarly, we find the segment S "" P - the trace of the plane (PQR) on the face ABB "A".
7) We find the point F=PC"" intersects A"C" and then we get the segment PF - the trace of the plane (PQR) on the face of ACC"A".
8) The points Q and F lie in the plane A"B"C", so the line QF lies in the plane (A"B"C"). Let's draw a straight line QF, we get a segment QF - a trace of a plane (PQR) on the face A "B" C. So, we got a polygon QS "" "S" "PF - the desired section.
3 note. Let us show another way of finding the point C"", in which we do not find the point of intersection of the line S""" Q with the line C"C"". We will argue as follows. If the trace of the plane (PQR) on the line CC" is some point V, then its projection onto the plane (ABC) coincides with the point C. Then the point S""""= V"P"intersects VP lies on the main trace S"R of the plane (PQR). We build this point S"""" as the intersection point of the lines V"P" (this is the line CA) and S"R. And then we draw the line S""""P. It intersects the line CC" at the point V.
Example 2
On the edge MB of the pyramid MABCD, we set the point P, on its face MCD we set the point Q. We construct a section of the pyramid by the plane (PQR), the point R of which we set:
a) on the MC edge;
b) on the verge of MAD;
c) in the plane (MAS), outside the pyramid.
Solution.
a) The trace of the plane (PQR) on the face of the MBC is the segment PR, and its trace on the face of the MCD is the segment RD", where the point D" is the point of intersection of the line RQ with the edge MD. It is clear that the plane (PQR) has traces on the faces MAD and MAB (since the plane (PQR) has common points with these faces). Find the plane trace (PQR) on the line MA. Let's do it like this:
1) Let's build the points P", Q" and R" - the projections of the points P, Q and R from the center M onto the plane (ABC), thus taken as the main plane. (Fig. 4)
3) If the plane (PQR) intersects the line MA at some point V, then the point V" coincides with the point A and the point S"""= VQ intersects V"Q" lies on the line S"S"". In other words, at the point S""" three lines intersect: VQ, V"Q"" and S" S"". The last two lines of these three are already on the drawing. Therefore, we construct the point S""" as the point of intersection of the lines V"Q" and SS"".
4) Draw the line QS""" (it coincides with the line VQ, since the line VQ must pass through the point S""", i.e. the points V, Q and S""" lie on the same line).
5) Find the point V at which the line QS"" "intersects the line MA, Point V is the trace of the plane (PQR) on the edge MA. Further, it is clear that the segments PV and VD" are the traces of the plane (PQR), respectively, on the faces of the MAB and M.A.D. Thus, the polygon PRD "V is the required section.
b) 1) We take the plane (ABC) as the main plane and build the points P", Q" and R" - the projections of the points P, Q and R, respectively, onto the plane (ABC). The center of this internal projection is the point M. (Fig. 5.)
2) We build a straight line S"S"" - the main trace of the plane (PQR).
3) If the plane (PQR) intersects the line MA at the point V, then the point V "- the projection of the point V onto the plane (ABC) from the center M- coincides with the point A, and the lines S"S"", V"R" and the line VR, the point V of which we have not yet built, intersect at the point S""". Find this point S"""=V"R" intersects S"S"" ."", and find the point V=RS""" intersects MA. Further construction is clear. The required section is the polygon PVD"T.
in)
(Fig.6.) Let the point R be located in the plane (MAS) as shown in Figure 6.
1) We take the plane (ABC) as the main plane and build the points P", Q" and R" - the projections of the points P, Q and R, respectively, onto the plane (ABC). (The center of the projection is the point M.)
2) We build a straight line S"S"", - the main trace of the plane (PQR).
3) Find the point V - the trace of the plane (PQR) on the line MA. Point V" - the projection of the point V onto the plane (ABC) from the center M - coincides in this case with point A.
4) Find the point S"""= P"V" intersects S"S"", and then the point V =PS""" intersects MA.
5) We get the trace PV of the plane (PQR) on the plane (MAB).
6) Find the point T - the trace of the plane (PQR) on the line MO. It is clear that the point T" in this case coincides with the point D. To construct the point T, we construct the point S""""=Q"T" intersects S"S"", and then the point T = QS""" "intersects MT" .
7) The set of traces PV, VT, TC", and C "P, i.e., the polygon PVTC" - the required section.
Combined sectioning method
The essence of the combined method for constructing sections of polyhedra is the application of theorems on the parallelism of lines and planes in space in combination with the axiomatic method.
Example number 1.
On the edges AB and AD of the pyramid MABCD, we set the points P and Q, respectively, the midpoints of these edges, and on the edge MC, we set the point R. Let's construct a section of the pyramid by a plane passing through the points P, Q, and R.
Solution
(picture 14):
one). It is clear that the main trace of the plane PQR is the line PQ.
2). Find the point K at which the MAC plane intersects the line PQ. Points K and R belong to both the PQR plane and the MAC plane. Therefore, by drawing the straight line KR, we get the line of intersection of these planes.
3). Let's find the point N=AC BD, draw the line MN and find the point F=KR MN.
four). The point F is a common point of the planes PQR and MDB, that is, these planes intersect along a straight line passing through the point F. At the same time, since PQ is the midline of the triangle ABD, then PQ is parallel to BD, that is, the line PQ is also parallel to the plane MDB. Then the plane PQR passing through the line PQ intersects the plane MDB along the line parallel to the line PQ, that is, parallel to the line BD. Therefore, in the plane MDB through the point F we draw a line parallel to the line BD.
5). Further constructions are clear from the figure. As a result, we get the polygon PQD"RB" - the required section.
1. Construction of a section passing through a given line parallel to another given line.
Let, for example, it is required to construct a section of a polyhedron by a plane @ passing through a given line p parallel to the second given line q. In the general case, the solution of this problem requires some preliminary constructions, which can be carried out according to the following plan:
one). Through the second line q and some point W of the first line p we draw the betta plane (Fig.
2). Draw a line q" parallel to q in the betta plane through the point W.
3). Intersecting lines p and q". The plane @ is defined. This completes the preliminary constructions and we can proceed to the construction of a direct section of the polyhedron by the plane @. In some cases, the features of a particular problem allow us to implement a more short solution plan. Let's consider examples.
Example number 2.
On the edges BC and MA of the pyramid MABC, we define the points P and Q, respectively. We construct a section of the pyramid by the plane @ passing through the line PQ parallel to the line AR, the point R, which we define as follows: a). On edge MB; b). It coincides with point B; in). On the verge of MAB.
Solution:
a)
.(Figure The plane passing through the second line, that is, the line AR, and the point Q taken on the first line, is already on the image. This is the MAB plane.
2). In the plane MAB through the point Q we draw a line QF parallel to AR.
3). The intersecting lines PQ and QF determine the plane @ (this plane PQF) - the plane of the desired section. Let us construct this section by the trace method.
four). Point B coincides with point F" - the projection of point F onto the plane ABC (from the center M), and point A coincides with point Q" - the projection of point Q onto this plane. Then the point S"=FQ F"Q" lies on the main trace of the secant plane @. Since the point P lies on the main trace of the secant plane, the line S"P is the main trace of the plane @, and the segment S""P is the trace of the plane @ on the edge of ABC. Further, it is clear that the point P should be connected to the point F. As a result, we get the quadrilateral PFQS"" - the required section.
b)
(Figure The plane passing through the line AB and the point P of the line PQ has already been built on the image. This is the plane ABC. Let's continue the construction according to the above plan.
2). In the plane ABC through the point P we draw a line PD parallel to the line AB.
3). The intersecting lines PQ and PD define the alpha plane (this is the PQD plane) - the plane of the desired section. Let's build this section.
four). It is clear that the trace of the alpha plane on the MAC face is the segment DQ.
5). We carry out further constructions, taking into account the following considerations. Since line PD is parallel to line AB, line PD is parallel to plane MAB. Then the plane alpha, passing through the line PD, intersects the plane MAB along the line parallel to the line PD, that is, the line AB. So, in the plane MAB through the point Q we draw a straight line QE parallel to AB. Segment QE is a trace of the alpha plane on the face of MAB.
6). Let's connect the point P with the point E. The segment PE is the trace of the alpha plane on the face of the MBC. Thus, the quadrilateral PEQD is the required section. coincides with point A, and point L" coincides with R"=MR BC. Then the point S "=LQ L"Q" lies on the main trace of the secant plane alpha. This main trace is the straight line S"P, and the trace of the alpha plane on the face of ABC is the segment S""P. Further, the straight line PL is the trace of the alpha plane on the MBC plane, and the segment PN is the trace of the alpha plane on the MBC face. So, the quadrilateral PS""QN is the desired section.
Example 3
On the diagonals AC and C"E" of the bases of the prism ABCDEA"B"C"D"E" we set the points P and Q, respectively. Let's construct a section of the prism by the plane alpha passing through the line PQ parallel to one of the following lines: a). AB; b) .ac"; in). BC Solution:
a)
(Figure Plane passing through the line AB - the second given line and the point P, taken on the first line, has already been built. This is the plane ABC.
2). In the plane ABC through the point P we draw a line parallel to the line AB, and find the points K and L at which this line intersects the lines BC and AE, respectively. B"C" are also parallel to each other. Taking into account that KL is parallel to AB and A"B" is parallel to AB, we draw a line in the plane A"B"C" through the point Q parallel to the line A"B", and find the points F and T at which this line intersects, respectively straight lines C"D" and A"E". Next, we get the segment TL - the trace of the alpha plane on the face AEE"A", the point S"=KL CD, the straight line S"F - the trace of the alpha plane on the plane CDD", the segment FC"" - the trace of the alpha plane on the face CDD"C" and, finally, the segment C""K" - the trace of the alpha plane on the face BCC"B". As a result, we get the polygon KLTFC"" - the required section. 
b)
(Figure Let's draw a plane through the line AC "- the second given line, and the point P taken on the first line. This is the plane ACC".
2). In the plane ACC" through the point P we draw a line parallel to the line AC" and find the point C"" at which this line intersects the line CC".
3). The intersecting lines PQ and PC"" determine the alpha plane (plane C""PQ) - the plane of the desired section. Let us construct this section, for example, by the trace method. One point belonging to the trace of the alpha plane on the ABC plane, which we take as the main one, is already on the drawing. This is point P. Let's find one more point of this trace.
four). The projection of the point C "" on the plane ABC is the point C, and the projection of the point Q is the point Q "- the point of intersection of the line CE with the line passing in the plane CEE" through the point Q parallel to the line EE ". Point S" \u003d C ""Q CQ " is the second point of the main trace of the alpha plane. So, the main trace of the alpha plane is the line S "P. It intersects the sides BC and AE of the base of the prism, respectively, at the points S"" and S"""... Then the segment S""S""" is the trace of the secant plane alpha on the face ABCDE. And the segment S""C"" is a trace of the alpha plane on the face of BCC"B". It is easy to see that the lines C"" Q and EE" lie in the same plane. Find the point E"" =C""Q EE". Then it is clear to obtain further traces of the alpha plane: S"""S"", S"""T, TF and FC"". As a result, we get the polygon S""S"""TFC"" - the required section.
in)
(drawing Through the second given line - the line BC "- and, for example, through the point P, which lies on the first given line, let's move the plane. Let's do this by the trace method. It is easy to establish that the main trace of this plane BC" P is the line BP. Then we find the point S"=BP CD and trace S"C" of plane BC"P and plane CDD".
2). In the plane BC "P through the point P we draw a line parallel to the line BC". The point of intersection of the drawn line with the line S"C" is denoted by V.
3). The intersecting lines PQ and PV define the plane alpha (plane PQV) - the plane of the desired section. Let's build this section.
four). We find the points Q "and V" - the projections of the points Q and V, respectively, onto the plane ABC, which we take as the main plane. Then we find the point S""=QV Q"V". This is one of the points of the main trace of the alpha plane. And there is already one more point of this trace. This is a given point P. So, the line S "" P is the main trace of the alpha plane, and the resulting segment S "" "S" """ is the trace of the alpha plane on the face of ABCDE. The further course of construction is clear: S "" "" "=S""P CD, S"""""V, points C""=S"""""V CC" and F=S"""""V C"D", then FQ and point T= FQ A"E" and finally TS"""". As a result, we get the polygon S"""C""FTS"""" - the required section.
Note: Let us briefly outline the course of solving example 3, c, in which the point Q was taken on the first given line, and not the point P (Figure 22).
one). We build the plane BC"Q (this is the plane BC"E").
2). The plane BC"Q intersects the plane ABC along the straight line BN parallel to C"E" (for construction, you can use the fact that BN is parallel to CE).
3). In the plane BC"Q through the point Q we draw a line QM parallel to BC" (M=QM BN).
four). We build a section of a prism by a plane defined by intersecting lines PQ and QM. This can be done in the following order: MP, S"=MP AE and S""=MP BC, S""""=MP CE, C""=S""""Q CC", S"""C" ", F=S"""C"" C"D", FQ, T=FQ A"E", TS. Polygon S""C""FTS"- desired section.
2. Construction of a section passing through a given point parallel to two given skew lines.
Let it be required to construct a section of a polyhedron by a plane passing through a given point K parallel to two given skew lines l and m. At background:#FFCCCC; border:outset #CC33FF 1.5pt">
1. Let's choose some point W. (This point may lie on one of the given skew lines, it may coincide with point K.)
2. Draw lines l" and m" through the point W. (Naturally, if the point W lies on one of the lines, for example, on the line l, then the line l" coincides with the line l.)
3. The intersecting lines l "and m" define the betta plane - the plane of the auxiliary section of the polyhedron. We build a section of the polyhedron by the betta plane.
4. Construct sections of the polyhedron by the alpha plane passing through the point K and parallel to the beta plane.
Consider examples of the application of the outlined plan.
EXAMPLE 4
On the edges AD and C"D" of the prism ABCDA"B"C"D", we set the points P and Q, respectively, and on the edge DD" we set the point K. Let's construct a section of the prism with a plane alpha passing through the point K parallel to the line PQ and one of the following straight lines: a) AB; b) A "B; c) BR, whose point R is given on the edge A"D".
Solution. a)
(Fig. 2 Let point W coincide with point P.
2) In the plane ABC through the point P draw a line parallel to the line AB. Find the point E at which the line drawn intersects the line BC.
3) The intersecting lines PQ and PE define the betta plane - the plane of the auxiliary section. Let us construct a section of the prism by the betta plane. Direct PE and points C"" and D"" are traces of the betta plane, respectively, on the straight lines CC" and DD". Then we build a straight line D "" P and get a point F on the edge A "D". Thus, the section of the prism by the betta plane is - I polygon PEC "" QF.
4) We now build a section of the prism by the alpha plane passing through the point K parallel to the beta plane. As a result, we get the triangle KLN - the required section.
b)
(Fig. Let the point W coincide with the point Q. To draw a line parallel to the line A "B" through the point Q, first draw the gamma plane through the line A "B and the point Q. Let's do it this way. Find the point Q" - the projection of the point Q onto plane ABC and draw the line AQ". It is clear that AQ" is parallel to A"Q. Now through the point B in the plane ABC we draw the line l" parallel to AQ". The intersecting lines A"B and l" define the gamma plane. In the gamma plane through the point Q draw a line l"" parallel to A"B.
3) Intersecting straight lines PQ and l "", the betta plane is determined - the plane of the auxiliary section of the prism. Let's build this section. To do this, we find the point S"=l" intersects l"", and then the line PS" - the main trace of the betta plane. Further we find the point s""=PS" intersects CD and draw the line S""Q - the trace of the betta plane on the CDD plane ". We get the point D"" - the trace of the betta plane on the line DD". The point D"" and the point P lie in the plane ADD". Therefore, the line PD"" is the trace of the betta plane on the plane ADD", and the segment PF is the trace of the betta plane on the face ADD"A". Thus, the section of the prism by the betta plane is the quadrilateral PS "" QF. (Please note: QF is parallel to PS "". And this, of course, is so. After all, the bases of the prism lie in parallel planes. This circumstance could be used when constructing a section of the prism by the betta plane.)
4) Now we build a section of the prism by the alpha plane passing through the point K parallel to the beta plane. This build is easy to do. As a result, we get the triangle KLN - the required section.
in)
(Fig. Let's choose point Q as point W.
2) Draw the gamma plane through the line BR and the point Q. The gamma plane intersects the plane ABC along the straight line l "parallel to QR. To construct the straight line l" we build the points R "and Q" - the projections of the points R and Q, respectively, onto the plane ABC - and draw the straight line Q "R", and then in the plane ABC through the point In we draw the line l" parallel to Q"R". In the gamma plane through the point Q we draw the line l"" parallel to BR. We get the point S"=l" intersects l"".
3) The intersecting lines PQ and l "" define the betta plane - the plane of the auxiliary section of the prism. Let's build this section. It is clear that the line PS" is the main trace of the betta plane. Next, we find the points S""= PS" intersects CD, S"""= PS" intersects BC and C"" = QS"" intersects CC". We obtain segments PS"" ", S""C"" and C""Q- traces of the betta plane, respectively, on the faces ABCD, BCC"B and CDD"C". Next, either draw a line in the plane A "B" C "parallel to the trace PS", and get the point F, or find the point D "" \u003d S "" Q intersects DD "and draw the line D "" P. This line intersects the line A "D" at the point F. Thus, we get two more traces of the betta plane: QF n FP. So, the polygon PS"""C""QF is the section of the prism by the betta plane.
4) Now let's build a section of the prism by the alpha plane passing through the point K parallel to the beta plane. As a result, we get the triangle KLN - the required section.
EXAMPLE 5.
On the edges MB and MA of the pyramid MABCD, we set the points P and K, respectively, and on the segment AC, we set the point Q. We construct a section of the pyramid by the plane alpha passing through the point K parallel to the line PQ and one of the following lines: a) CD; b) MS; c) RV, the points R and V of which we set respectively on the edges AB and MC of the pyramid.
Solution.
a)
(Fig. 2In the plane ABC through the point Q we draw a line parallel to the line CD, and find the points S". S"" and S""", at which this line intersects the lines BC, AD and AB, respectively.
2) The intersecting lines PQ and S"S"" define the betta plane - the plane of the auxiliary section of the pyramid. Let's construct this section. The main trace of the betta plane is the line S"S"". The segment PS" is the trace of the betta plane on the MBC face, the straight line PS""" is its trace on the MAB plane, the segment PA" is on the MAB face, the segment A"S"" is on the MAD face.
b)
(Fig. 27.) Let's build the given section in the following order:
1) In the MAC plane through
point Q we draw the line QA parallel to MC
2) We construct an auxiliary section of the pyramid by a plane, which is determined by. To this end, we find the point S"=PA" intersects AB, draw the line S"Q, which is the main trace of the plane PQA", get the points S""=S"Q intersects AD and S"""=S"Q intersects BC and connect point A" with point S"", and point P with point S""". Quadrangle PA"S""S""" is an auxiliary section of the pyramid. The plane of this section is parallel to lines PQ and MC, but does not pass through point K .
3) Now let's construct a section of the pyramid by a plane passing through the point K parallel to the plane PQA ". As a result, we get a quadrilateral B" KFE - the desired section.
a)
(Fig. 28.) Let's construct a given section of the pyramid, first constructing an auxiliary section by its plane passing through the line PQ parallel to the line RV. Let's do it in the following order:
1) Construct a point S "=PV intersects BC and draw a line S" R.
2) The intersecting lines S "V and S" R determine the plane. In this plane, draw a line PS"" through the point P parallel to RV.
3) The intersecting lines PQ and PS"" define the plane of the auxiliary section of the pyramid. Let's build this section. We find successively the straight line S "" Q - the main trace of the plane of the auxiliary section, then the points T "=S" "Q intersects BC, T" "= S" "Q intersects AB and T" "" \u003d S" "Q intersects CD, Let's draw then the line T"P and find the point E \u003d T"P intersects "MC. We connect the point P with the point T"", and the point E with T""". The quadrangle PT ""T" "" E is an auxiliary section of the pyramid. The plane of this section is parallel to the lines PQ and RV, but does not pass through the point K. Now we will construct a section of the pyramid with a plane passing through the point K parallel to the plane of the auxiliary section. As a result, we obtain a quadrilateral KV "C" D" - the desired section.
Finding the cross-sectional area in polyhedra.
Task number 1.
Task #2
Task number 3.
Task number 4.
Task number 5.
Task number 6.
Task #7
Task number 8.
Using properties of similar triangles.
Therefore, below are a few simple problems in which similar triangles play a major role, especially since they also need to be built (and seen!!!) using the standard stereometric technique: intersect one plane with another plane and construct their line of intersection along two points common to planes.
Task number 1.
Task #2
Task #3
Task #4
Task #5
There are four main ways to find the distance between skew lines:
1) Finding the length of the common perpendicular of two intersecting lines, that is, a segment with ends on these lines and perpendicular to both.
2) Finding the distance from one of the intersecting lines to a plane parallel to it passing through the other line.
3) Finding the distance between two parallel planes passing through given skew lines.
4) Finding the distance from a point - which is the projection of one of the intersecting lines onto a plane perpendicular to it - to the projection of another line onto the same plane.
Task #18
Task #19
Present 4 options for solving this problem and choose the most rational of them. Justify your choice.
Task #20
Task #21
Task #22
Finding the distance and angle between skew lines in a polyhedron.
Task number 1.
Task number 2.
Task number 3.
passing through the side edge and the median of the base intersecting with it, and a plane passing through the same median and the middle of any other side edge.
Sections.
Task number 1.
Task number 2.
Task number 3.
Two opposite edges of a tetrahedron are perpendicular, and their lengths are equal to a and b, the distance between them is equal to c. A cube is inscribed in the tetrahedron, four edges of which are perpendicular to these two edges of the tetrahedron, and exactly two vertices of the cube lie on each face of the tetrahedron. Find the edge of the cube.
Task number 4.
Task number 5.
Task number 6.
Task number 7.
Task number 8.
Task number 9.
The ratio of the volumes of the parts of a polyhedron.
Task number 1.
Task number 2.
Task number 3.
Task number 4.
Projections and sections of regular polyhedra.
Task number 1.
Show that the projections of the dodecahedron and icosahedron on planes parallel to their faces are regular polygons.
Task number 2.
Show that the projection of a dodecahedron onto a plane perpendicular to a straight line passing through its center and the midpoint of an edge is a hexagon (not a decagon).
Task number 3.
a) show that the projection of the icosahedron onto the plane. perpendicular to the line passing through its center and vertex is a regular decagon. b). Prove that the projection of a dodecahedron onto a plane perpendicular to a straight line passing through its center and vertex is an irregular dodecahedron.
Task number 4.
Does there exist a section of a cube that is a regular t hexagon?
Task number 5.
Is there a section of the octahedron that is a regular hexagon?
Task number 6.
Is there a section of the dodecahedron that is a regular hexagon?
Task number 7.
All faces ABC and ABD of an icosahedron have a common edge AB. A plane parallel to the plane ABC is drawn through the vertex D. Is it true that the section of the icosahedron by this plane is a regular hexagon?
Answers to tasks by topic:
4. Angle between planes.
5. Sections
6. The ratio of the volumes of the parts of the polyhedron.
7. Projections and sections of regular polyhedra.
1. Finding the cross-sectional area in polyhedra.
The solution of the problem
№1 №2 №3 №4 №5 №6 №7 №8
Task number 1.
https://pandia.ru/text/78/375/images/image040_59.gif" width="597" height="292 src=">
Task number 2.
https://pandia.ru/text/78/375/images/image042_56.gif" width="577" height="277 src=">
Task number 3.
https://pandia.ru/text/78/375/images/image044_53.gif" width="630" height="275 src=">
Task number 4.
https://pandia.ru/text/78/375/images/image046_49.gif" width="641" height="332 src=">
Task number 5.
https://pandia.ru/text/78/375/images/image048_46.gif" width="642" height="245 src=">
Task number 6.
https://pandia.ru/text/78/375/images/image050_46.gif" width="680" height="340 src=">
Task number 7.
https://pandia.ru/text/78/375/images/image052_47.gif" width="659" height="340 src=">left" style="margin-left: 6.75pt; margin-right: 6.75 pt">
2. Using the properties of similar triangles.
The solution of the problem
№1 №2 №3 №4 №5
Task number 1.
https://pandia.ru/text/78/375/images/image055_46.gif" width="605" height="254">
2nd case 
Task number 2.
https://pandia.ru/text/78/375/images/image058_41.gif" width="683" height="260 src=">
Task number 3.
https://pandia.ru/text/78/375/images/image061_42.gif" width="536" height="203">
https://pandia.ru/text/78/375/images/image063_41.gif" width="341" height="107 src=">MsoNormalTable">
The point C belongs to the plane CB"A"D (since CD" is perpendicular to C"D as the diagonal of the square, and since B"C" is perpendicular to the plane CC"D"D, which implies that B"C" is perpendicular to CE), we obtain CE is perpendicular to B"C" and CE is perpendicular to C"D). Then we draw EF perpendicular to B"D and then we get B"D perpendicular to CF (by the three perpendiculars theorem: CF is inclined with respect to the plane AB"C"D, CE - the perpendicular and EF - the projection of the oblique CF; then it is also perpendicular to the oblique CF itself.) Since EF and CF belong respectively to both planes, the angle phi (angle CFE) is the required one.
This justification is followed by a simple computational part.
"B" EF and D ""C" EF), as a result of which the perpendiculars A "" M and D "" M, drawn in both figures to their intersection line, will fall into one point M, moreover, inside and not outside the prism , since angles B"A""D and C"D""A are obtuse (B"D and more BD=AC=A""C"" and C"A more than AC=BD=B""D"" ). Further, having found the diagonals and sides of the rhombuses, one can find the segments A "" M and D "" M using, for example, two formulas for the area of the rhombus 
Note: Of course, in this and similar problems, no dimensions of the polyhedron (for example, "a") are needed, therefore, when choosing the numerical values of the parameter "k" for various variants of the problem, the content of its condition in the appropriate place should be formulated, for example, as follows: "... in a prism whose height is so many times greater than the side of the base ... ", etc.
3. Finding the distance and angle between skew lines in a polyhedron.
The solution of the problem
№1 №2 №3 №4 №5
Task number 1.
MsoNormalTable">
№1
Solving the problem in the first way
suggests:
- a difficult justification that the required perpendicular (h skr.) with ends on two given intersecting lines is located inside the cube (and not outside it);
- approximate determination of the location of this perpendicular;
- guess that to find the length of the segment h skr. it is necessary, using the theorem on three perpendiculars, to project it onto adjacent faces of the cube, to which the intersecting lines (diagonals) belong, and only then approach a simple solution:
| 2.
Solving the problem in the second way involves
the following actions: |
The construction of another secant plane passing through the diagonal B "D" and intersecting the second of the skew lines A "C"; this plane is convenient to choose the diagonal section BB "D" D to this sign of the perpendicularity of two planes of the plane BB "D" D is perpendicular to the plane ACD ", since the plane BB"D"D passes through a straight line (B"D) perpendicular to another plane (ACD"). Next, a line of intersection of both planes is built along their 2 common points (D "O) and is fixed by the intersection of this line with a diagonal B" D (point N);
-and finally, according to the theorem that if a plane is perpendicular to one of the parallel lines, then it is also perpendicular to the other, from the point O "belongs to A"C" we draw in the section plane BB"D"D to the intersection with D"O the segment O "M is parallel to B"D; in this case, O "M will be perpendicular to the plane ACD" and therefore O "M \u003d h crt.;
- then in the computational part of the solution, having considered the section BB "D'D and in it - the right-angled triangle OO'D', we find: As you can see, both of the first methods are of little use for tasks that represent at least some complexity
| 3.
Solving the problem in the third way involves
: |
And, finally, in the computational part of the solution, you can use the trick from the previous solution method or resort to the similarity of triangles:
| 4.
Solving the problem in the fourth way involves: |
Task number 3.
In this problem, for choosing a solution method, the determining factor is the perpendicularity of the line AC to the diagonal plane ВB'D'D (because AC is perpendicular to ВD and AC is perpendicular to BB'), to which another line B'F belongs, i.e., the secant plane BB' D'D is convenient for choosing it as the projection plane. And then the simple calculation part follows:
one). From the similarity of triangle DFT and triangle D'FB' we find DT = kd;
2). From the similarity of triangle NOT and triangle BB'T we find ON: 
Task number 4.
This problem is presented here to demonstrate the application of the second method (constructing a perpendicular from the first line to a parallel plane containing the second line) to the simplest situations of locating skew lines in such a complex polyhedron as a regular hexagonal prism.
https://pandia.ru/text/78/375/images/image077_33.gif" width="186" height="87 src=">
Task number 5.
https://pandia.ru/text/78/375/images/image079_29.gif" width="347" height="326 src=">
5. Sections.
The solution of the problem
№1 №2 №3 №4 №5 №6
Task number 1.
In any case, the points A, B and C lie in the same plane, and therefore we can consider the section by the plane containing these points. Since the section plane passes through the point of contact of the spheres (the spheres of the plane), and the section turns out to be tangent to the circle (circle and line). Let O' and 0'' be the centers of the first and second circles. Since O'A || 0''B and points O', C and 0'' lie on one straight line, angle AO'C = angle BO''C. Therefore, the angle ACO' = the angle BCO'', i.e. points A, B and C lie on the same straight line.
Task number 2.
The axial section of this truncated cone is the circumscribed trapezoid ABCD with bases AD = 2R and BC = 2r. Let P be the tangent point of the inscribed circle with the side AB, O be the center of the inscribed circle. In triangle ABO, the sum of the angles at vertices A and B is 90 °, so it is right-angled. Therefore, AP: RO - RO: BP, i.e., PO'2 = AP * BP. It is also clear that AP = R and BP = r. Therefore, the radius RO of a sphere inscribed in a cone is equal to the square root of the product of R and r, and hence S = 4п(R2 + Rr+ r2). Expressing the volume of this truncated cone by formulas, we obtain that the area of its total surface is equal to 2n(R2 + Rr + r2) = S/2 (it should be taken into account that the height of the truncated cone is equal to twice the radius of the sphere around which it is described).
Task number 3.
The common perpendicular to these edges is divided by the planes of the faces of the cube parallel to them into segments of length y, x and r (x is the length of the edge of the cube; the segment of length y is adjacent to the edge a). The planes of the faces of the cube, parallel to these edges, intersect the tetrahedron in two rectangles. The smaller sides of these rectangles are equal to the edge of the cube x. Since the sides of these rectangles are easy to calculate, we get x = bu/c and x = az/c. Therefore, c=x+y+r=x+cx/b + ex/a, i.e. x=abc/(ab + bc + ca).
Task number 4.
Each side of the resulting polygon belongs to one of the faces of the cube, so the number of its sides does not exceed 6. In addition, the sides belonging to the opposite faces of the cube are parallel, since the lines of intersection of the plane with two parallel planes are parallel. Therefore, the section of a cube cannot be a regular pentagon, since it does not have parallel sides. It is easy to check that a regular triangle, a square, and a regular hexagon can be sections of a cube.
Task number 5.
Consider a circle that is a section of a given body and draw a line l through its center perpendicular to its plane. This line intersects the given body along some segment AB. All sections passing through the line l are circles with diameter AB.
Task number 6.
Consider an arbitrary section passing through the vertex A. This section is a triangle ABC, and its sides AB and AC are generators of the cone, i.e., have constant length. Therefore, the cross-sectional area is proportional to the sine of the angle BAC. The BAC angle changes from 0° to φ,
MsoNormalTable">
Task number 2.
Consider a cube whose vertices are located at the vertices of the dodecahedron. In our problem, we are talking about a projection onto a plane parallel to the face of this cube. Now it is easy to verify that the projection of the dodecahedron is indeed a hexagon (Fig. 70).
Task number 3.
a) The considered projection of the icosahedron passes into itself upon rotation by 36° (in this case, the projections of the upper faces pass into the projections of the lower faces). Therefore, it is a regular 10-coal (Fig. 71, a).
b) The considered projection of the dodecahedron is a 12-gon turning into itself when rotated by 60° (Fig. 71. b). Half of its sides are projections of edges parallel to the projection plane, and the other half of the sides are projections of edges not parallel to the projection plane. Therefore, this 12-gon is irregular.
MsoNormalTable">
Task number 4.
Exists. The middle of the indicated in Fig. The 72 edges of a cube are the vertices of a regular hexagon. This follows from the fact that the sides of this hexagon are parallel to the sides of a regular triangle PQR, and their lengths are half the lengths of the sides of this triangle.
Task number 6. Exists. We take three pentagonal faces with a common vertex A and consider a section by a plane intersecting these faces and parallel to the plane in which three pairwise common vertices of the faces under consideration lie (Fig. 74). This section is a hexagon with pairwise parallel opposite sides. When rotated by 120 ° about the axis passing through the vertex A and perpendicular to the cutting plane, the dodecahedron and the cutting plane pass into themselves. Therefore, the section is a convex hexagon with angles of 120°, the lengths of which, alternating, take two values. In order for this hexagon to be regular, it is enough that these two values \u200b\u200bare equal. As the cutting plane moves from one of its extreme positions to the other, moving away from the vertex A, the first of these values increases from 0 to d, and the second decreases from d to a, where a is the length of the edge of the dodecahedron. (d is the length of the diagonal of the face (d is greater than a). Therefore, at some point, these values \u200b\u200bare equal, that is, the section is a regular hexagon. |
|
Task number 7.
No, not true. Consider the projection of the icosahedron onto the plane ABC. It is a regular hexagon (see Fig. 69). Therefore, the section under consideration would be a regular hexagon only if all 6 vertices connected by edges to points A, B and C (and different from A, B and C) lay in the same plane. But, as you can easily see, this is not true (otherwise it would turn out that all the vertices of the icosahedron are located on three parallel planes).
TASKS
|
2. Using the properties of similar triangles.
The solution of the problem
№1 №2 №3 №4 №5
Task number 1.
https://pandia.ru/text/78/375/images/image055_46.gif" width="605" height="254">
2nd case
Task number 2.
https://pandia.ru/text/78/375/images/image058_41.gif" width="683" height="260 src=">
Task number 3.
https://pandia.ru/text/78/375/images/image060_43.gif" width="570" height="264 src=">
Task number 4.
https://pandia.ru/text/78/375/images/image063_41.gif" width="341" height="107 src=">right">
During the lesson, everyone will be able to get an idea about the topic "Problems for constructing sections in a parallelepiped. First, we will repeat the four main supporting properties of the box. Then, using them, we will solve some typical problems for constructing sections in a parallelepiped and for determining the cross-sectional area of a parallelepiped.
Topic: Parallelism of lines and planes
Lesson: Tasks for constructing sections in a parallelepiped
During the lesson, everyone will be able to get an idea about the topic. "Problems for constructing sections in a parallelepiped".
Consider the parallelepiped ABCDА 1 B 1 C 1 D 1 (Fig. 1). Let's remember its properties.
Rice. 1. Properties of a parallelepiped
1) Opposite faces (equal parallelograms) lie in parallel planes.
For example, the parallelograms ABCD and A 1 B 1 C 1 D 1 are equal (that is, they can be superimposed) and lie in parallel planes.
2) The lengths of parallel edges are equal.
For example, AD = BC = A 1 D 1 = B 1 C 1 (Fig. 2).
Rice. 2. The lengths of the opposite edges of the parallelepiped are equal
3) The diagonals of the parallelepiped intersect at one point and bisect this point.
For example, the diagonals of the parallelepiped BD 1 and B 1 D intersect at one point and bisect this point (Fig. 3).
4) In the section of a parallelepiped there can be a triangle, a quadrangle, a pentagon, a hexagon.
Problem on section of a parallelepiped
For example, consider solving the following problem. Given a parallelepiped ABCDА 1 B 1 C 1 D 1 and points M, N, K on the edges AA 1 , A 1 D 1 , A 1 B 1, respectively (Fig. 4). Construct sections of the parallelepiped by the MNK plane. The points M and N simultaneously lie in the plane AA 1 D 1 and in the cutting plane. Hence, MN is the line of intersection of the two indicated planes. Similarly, we obtain MK and KN. That is, the section will be the triangle MKN.
1. Geometry. Grade 10-11: a textbook for students of educational institutions (basic and profile levels) / I. M. Smirnova, V. A. Smirnov. - 5th edition, corrected and supplemented - M.: Mnemozina, 2008. - 288 p.: ill.
Tasks 13, 14, 15 p. 50
2. Given a parallelepiped ABCDА 1 B 1 C 1 D 1 . M and N are the midpoints of the edges DC and A 1 B 1 .
a) Construct the points of intersection of the lines AM and AN by the plane of the face BB 1 C 1 C.
b) Construct the line of intersection of the planes AMN and BB 1 C 1
3. Construct sections of the parallelepiped ABCDА 1 B 1 C 1 D 1 by a plane passing through BC 1 and the midpoint M of the edge DD 1 .
In this method, the first step (after finding the secondary projections of these points) is to build a trace of the cutting plane on the plane of the upper or lower base of the prism or truncated pyramid or on the base of the pyramid
ass 2. Given an image of a triangular prism ABCA 1 B 1 C 1 and three pointsM, N, P, which lie respectively on the edge CC 1 and faces ABB 1 A 1 , BCC 1 B 1 . Construct a section of a prism by a plane, passing through M, N, P.
Solution. We already have one point on the upper base of the prism, so we will build the trace on the upper base. We build secondary projections of points N and P to the top base. Then: 1 .NPN 3 P 3 =X; 2 .MX=p-track; 3 .pB 1 C 1 =D.
Further steps have already been shown above in the drawing.
ass 3. Dec. We will build a cutting plane trace on the lower base of the prism.
Building:1. MNED=X, MPEP 3 =Y;
2. p=XY- trace; 3. pBC=G, pDC=H.
We need to find a point on the edge BB 1 or on edge AA 1 .
AT 
facets ABB 1 A 1 we already have one point P. Therefore, the lower edge of this face, i.e. AB, we continue until the intersection with the trace.
4. ABp=Z.
5. PZAA 1 =F; PZBB 1 =K.Further actions are already shown above.
If it turns out that the line AB does not intersect with the trace, then the desired FK will also be parallel. ass 4. Dec. 1. PNP o N o= X;
2. MNCN o= Y;3. p=XY- trace;
3. CBp=Z;4. ZMSB=E;
5. ENSA=G 6. GEMF- section claim.
17. Construction of a section of a cylinder.
If the cutting plane is given by three points, then we can always find its trace on the plane of the base of the cylinder or cone and the point ( P, O) on its axis. Therefore, we consider that the cutting plane is given by these elements.
FROM 
the beginning of the race is the case when the plane intersects only the lateral surface of the cylinder. Then the section of the cylinder will be an ellipse (;¯ and its image is also an ellipse. We know how to construct an ellipse if two of its conjugate diameters are known. We will now show how to find the image of the main diameters of an ellipse (;¯.
Let and 1 be ellipses representing the lower and upper bases of the cylinder, O and O 1 - their centers. Let's draw a diameter A 3 B 3 lower base, parallel to the trace and its conjugate diameter C 3 D 3 . For building C 3 D 3 we use a chord K 3 L 3 , one end of which belongs to the contour generatrix. Recall that A 3 B 3 and C 3 D 3 depict perpendicular diameters. Let's continue C 3 D 3 to the intersection with the trace. Let's get the point X. Straight PX call it the axis of the section.
Let's raise the points C 3 and D 3 to the axis of the section. Get C and D. Line segment CD is an image of a large-diameter section. Let's raise the segment A 3 B 3 to the height OP. We get a segment AB, which is an image of a small section diameter. Negative AB and CD – mating dia. ellipse .
H 
Find more points at which the ellipse passes from the visible side of the cylinder to the invisible one, which means that the solid line turns into a dotted line. These are the points of intersection of the secant plane with the contour generators. Let Y 3 =K 3 L 3 C 3 D 3 . Let's raise Y 3 to the axis of the section. Let's get a point Y. Let's raise the chord K 3 L 3 to the height YY 3 . We get a segment KL. We have found the required point K, and along the way, one more additional point L. Dot M, depicting the intersection of the secant plane and with the second contour generatrix is symmetrical to the point K relative to the point P.Additionally, we will construct a point N, symmetrical L
relation-points P
Let's show a way how you can find any number of points on a section without using these diameters.
choose any. point V 3 on an ellipse . We carry diameter V 3 T 3 and continue it until the intersection with the trace. We get a point U. We raise the points V 3 and T 3 to straight U.P.. We get two points V and T on the section. Choosing instead V 3 another point, we get another 2 points per section. If you select a point K 3 lying on the contour generatrix, we will find points K and M, in which the solid line on the section should turn into a dashed one.
