In this lesson, we will look at the tetrahedron and its elements (tetrahedron edge, surface, faces, vertices). And we will solve several problems for constructing sections in a tetrahedron using the general method for constructing sections.
Topic: Parallelism of lines and planes
Lesson: Tetrahedron. Problems for constructing sections in a tetrahedron
How to build a tetrahedron? Take an arbitrary triangle ABC. Arbitrary point D not lying in the plane of this triangle. We get 4 triangles. The surface formed by these 4 triangles is called a tetrahedron (Fig. 1.). The internal points bounded by this surface are also part of the tetrahedron.
Rice. 1. Tetrahedron ABCD
Elements of a tetrahedron
A,B,
C,
D - vertices of a tetrahedron.
AB,
AC,
AD,
BC,
BD,
CD - edges of a tetrahedron.
ABC,
ABD,
bdc,
ADC - faces of a tetrahedron.
Comment: you can take the plane ABC behind tetrahedron base, and then the point D is top of a tetrahedron. Each edge of the tetrahedron is the intersection of two planes. For example, rib AB is the intersection of planes ABD And ABC. Each vertex of the tetrahedron is the intersection of three planes. Vertex A lies in the planes ABC, ABD, ADWITH. Dot A is the intersection of the three marked planes. This fact is written as follows: A= ABC ∩ ABD ∩ AUD.
Tetrahedron definition
So, tetrahedron is a surface formed by four triangles.
Edge of a tetrahedron- the line of intersection of two planes of the tetrahedron.
Make 4 equal triangles from 6 matches. It is not possible to solve the problem on a plane. And in space it is easy to do. Let's take a tetrahedron. 6 matches are its edges, four faces of a tetrahedron and will be four equal triangles. Problem solved.
Dan tetrahedron ABCD. Dot M belongs to the edge of the tetrahedron AB, dot N belongs to the edge of the tetrahedron IND and dot R belongs to the edge DWITH(Fig. 2.). Construct a section of a tetrahedron by a plane MNP.
Rice. 2. Drawing for task 2 - Construct a section of a tetrahedron by a plane
Solution:
Consider the face of a tetrahedron DSun. In this edge of the point N And P faces belong DSun, and hence the tetrahedron. But by the condition of the point N, P belong to the cutting plane. Means, NP is the line of intersection of two planes: the face planes DSun and cutting plane. Let's assume that the lines NP And Sun are not parallel. They lie in the same plane DSun. Find the point of intersection of the lines NP And Sun. Let's denote it E(Fig. 3.).
Rice. 3. Drawing for task 2. Finding point E
Dot E belongs to the section plane MNP, since it lies on the line NP, and the straight line NP lies entirely in the plane of the section MNP.
Also dot E lies in the plane ABC because it lies on a line Sun out of plane ABC.
We get that EAT- line of intersection of planes ABC And MNP, because the points E And M lie simultaneously in two planes - ABC And MNP. Connect the dots M And E, and continue the line EAT to the intersection with the line AU. point of intersection of lines EAT And AU denote Q.
So in this case NPQM- desired section.
Rice. 4. Drawing for problem 2. Solution of problem 2
Consider now the case when NP parallel BC. If straight NP parallel to some line, for example, a line Sun out of plane ABC, then the straight line NP parallel to the whole plane ABC.
The desired section plane passes through a straight line NP, parallel to the plane ABC, and intersects the plane in a straight line MQ. So the line of intersection MQ parallel to a straight line NP. We get NPQM- desired section.
Dot M lies on the side ADIN tetrahedron ABCD. Construct a section of a tetrahedron by a plane that passes through a point M parallel to base ABC.
Rice. 5. Drawing for task 3 Construct a section of a tetrahedron by a plane
Solution:
cutting plane φ
parallel to the plane ABC by condition, then this plane φ
parallel to straight lines AB, AU, Sun.
In plane ABD through a point M let's draw a straight line PQ parallel AB(Fig. 5). Straight PQ lies in the plane ABD. Similarly in plane AUD through a point R let's draw a straight line PR parallel AU. got a point R. Two intersecting lines PQ And PR plane PQR are respectively parallel to two intersecting lines AB And AU plane ABC, hence the planes ABC And PQR are parallel. PQR- desired section. Problem solved.
Dan tetrahedron ABCD. Dot M- internal point, point of a tetrahedron face ABD. N- internal point of the segment DWITH(Fig. 6.). Construct a point of intersection of a line NM and plane ABC.
Rice. 6. Drawing for task 4
Solution:
To solve, we construct an auxiliary plane DMN. Let the line DM intersects the line AB at a point TO(Fig. 7.). Then, SCD is a section of the plane DMN and a tetrahedron. In plane DMN lies and straight NM, and the resulting line SC. So if NM not parallel SC, then they intersect at some point R. Dot R and will be the desired point of intersection of the line NM and plane ABC.
Rice. 7. Drawing for problem 4. Solution of problem 4
Dan tetrahedron ABCD. M- internal point of the face ABD. R- internal point of the face ABC. N- internal point of the edge DWITH(Fig. 8.). Construct a section of a tetrahedron by a plane passing through the points M, N And R.
Rice. 8. Drawing for task 5 Construct a section of a tetrahedron by a plane
Solution:
Consider the first case, when the line MN not parallel to the plane ABC. In the previous problem, we found the point of intersection of the line MN and plane ABC. This is the point TO, it is obtained using the auxiliary plane DMN, i.e. we do DM and get a point F. We spend CF and at the intersection MN get a point TO.
Rice. 9. Drawing for task 5. Finding point K
Let's draw a straight line KR. Straight KR lies both in the plane of the section and in the plane ABC. Getting points R 1 And R 2. Connecting R 1 And M and on continuation we get a point M 1. Connecting the dot R 2 And N. As a result, we obtain the desired cross section R 1 R 2 NM 1. The problem in the first case is solved.
Consider the second case, when the line MN parallel to the plane ABC. Plane MNP goes through a straight line MN parallel to the plane ABC and crosses the plane ABC along some line R 1 R 2, then the straight line R 1 R 2 parallel to this line MN(Fig. 10.).
Rice. 10. Drawing for problem 5. Desired section
Now let's draw a line R 1 M and get a point M 1.R 1 R 2 NM 1- desired section.
So, we have considered the tetrahedron, solved some typical tasks on the tetrahedron. In the next lesson, we will look at the box.
1. I. M. Smirnova, V. A. Smirnov. - 5th edition, corrected and supplemented - M .: Mnemosyne, 2008. - 288 p. : ill. Geometry. Grade 10-11: textbook for students of general educational institutions (basic and profile levels)
2. Sharygin I. F. - M.: Bustard, 1999. - 208 p.: ill. Geometry. Grade 10-11: Textbook for general educational institutions
3. E. V. Potoskuev, L. I. Zvalich. - 6th edition, stereotype. - M. : Bustard, 008. - 233 p. :ill. Geometry. Grade 10: Textbook for general educational institutions with in-depth and profile study of mathematics
Additional web resources
2. How to construct a section of a tetrahedron. Mathematics ().
3. Festival of pedagogical ideas ().
Do homework tasks on the topic "Tetrahedron", how to find the edge of the tetrahedron, the faces of the tetrahedron, the vertices and the surface of the tetrahedron
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 18, 19, 20 p. 50
2. Point E mid-rib MA tetrahedron IAWS. Construct a section of a tetrahedron by a plane passing through the points B, C And E.
3. In the MAVS tetrahedron, the point M belongs to the AMB face, the P point to the BMC face, and the K point to the AC edge. Construct a section of a tetrahedron by a plane passing through the points M, R, K.
4. What figures can be obtained as a result of the intersection of a tetrahedron by a plane?
The tetrahedron, or triangular pyramid, is the simplest of the polyhedra, just as the triangle is the simplest of the polygons in the plane. The word "tetrahedron" is formed from two Greek words: tetra - "four" and hedra - "base", "face". A tetrahedron is given by its four vertices - points that do not lie in the same plane; faces of a tetrahedron - four triangles; The tetrahedron has six edges. In contrast to an arbitrary -angular pyramid (at ), any of its faces can be chosen as the base of the tetrahedron.
Many properties of tetrahedra are similar to those of triangles. In particular, 6 planes drawn through the midpoints of the edges of the tetrahedron perpendicular to them intersect at one point. At the same point, 4 straight lines intersect, drawn through the centers of the circles circumscribed near the faces perpendicular to the planes of the faces, and is the center of the sphere circumscribed near the tetrahedron (Fig. 1). Similarly, 6 bisector half-planes of the tetrahedron, i.e., half-planes that divide the dihedral angles at the edges of the tetrahedron in half, also intersect at one point - at the center of the sphere inscribed in the tetrahedron - a sphere that touches all four faces of the tetrahedron. Any triangle has, in addition to the inscribed, 3 more excircles (see Triangle), but the tetrahedron can have any number - from 4 to 7 - excircles, i.e. spheres touching the planes of all four faces of the tetrahedron. There are always 4 spheres inscribed in truncated trihedral angles, one of which is shown in Fig. 2, right. Another 3 spheres can be inscribed (not always!) in the truncated dihedral angles at the edges of the tetrahedron - one of them is shown in fig. 2, left.
For a tetrahedron, there is another possibility of its mutual arrangement with a sphere - contact with a certain sphere with all its edges (Fig. 3). Such a sphere - sometimes called "semi-inscribed" - exists only when the sums of the lengths of the opposite edges of the tetrahedron are equal: (Fig. 3).
For any tetrahedron, an analogue of the theorem on the intersection of medians of a triangle at one point is valid. Namely, 6 planes drawn through the edges of the tetrahedron and the midpoints of the opposite edges intersect at one point - at the centroid of the tetrahedron (Fig. 4). 3 “middle lines” also pass through the centroid - segments connecting the midpoints of three pairs of opposite edges, and they are divided by a point in half. Finally, 4 “medians” of the tetrahedron also pass through - segments connecting the vertices with the centroids of opposite faces, and they are divided at a point in a ratio of 3:1, counting from the vertices.
The most important property of a triangle - equality (or) - has no reasonable "tetrahedral" analogue: the sum of all 6 dihedral angles of a tetrahedron can take any value between and. (Of course, the sum of all 12 plane angles of a tetrahedron - 3 at each vertex - is independent of the tetrahedron and equals .)
Triangles are usually classified according to their degree of symmetry: regular or equilateral triangles have three axes of symmetry, isosceles - one. The classification of tetrahedra according to the degree of symmetry is richer. The most symmetrical tetrahedron is regular, bounded by four regular triangles. It has 6 planes of symmetry - they pass through each edge perpendicular to the opposite edge - and 3 axes of symmetry passing through the midpoints of the opposite edges (Fig. 5). Less symmetrical are regular triangular pyramids (3 symmetry planes, Fig. 6) and isohedral tetrahedra (i.e., tetrahedra with equal faces - 3 symmetry axes, Fig. 7).
2) 
,
where is the dihedral angle at the edge. There are other formulas for calculating the volume of a tetrahedron.
Tetrahedron in Greek means "tetrahedron". This geometric figure has four faces, four vertices and six edges. The edges are triangles. In fact, the tetrahedron is the First mention of polyhedra appeared long before the existence of Plato.
Today we will talk about the elements and properties of the tetrahedron, and also learn the formulas for finding area, volume and other parameters for these elements.
Elements of a tetrahedron
A segment released from any vertex of the tetrahedron and lowered to the intersection point of the medians of the opposite face is called the median.
The height of the polygon is a normal segment dropped from the opposite vertex.
A bimedian is a segment connecting the centers of crossing edges.
Properties of a tetrahedron
1) Parallel planes that pass through two skew edges form a circumscribed parallelepiped.
2) A distinctive property of the tetrahedron is that the medians and bimedians of the figure meet at one point. It is important that the latter divides the medians in a ratio of 3:1, and bimedians - in half.
3) The plane divides the tetrahedron into two parts equal in volume if it passes through the middle of two crossing edges.
Types of tetrahedron
The species diversity of the figure is quite wide. A tetrahedron can be:
- correct, that is, at the base is an equilateral triangle;
- isohedral, in which all faces are the same in length;
- orthocentric, when the heights have a common intersection point;
- rectangular, if the flat angles at the top are normal;
- proportionate, all bi heights are equal;
- wireframe, if there is a sphere that touches the edges;
- incentric, that is, the segments lowered from the vertex to the center of the inscribed circle of the opposite face have a common intersection point; this point is called the center of gravity of the tetrahedron.
Let us dwell in detail on the regular tetrahedron, the properties of which practically do not differ.
Based on the name, you can understand that it is called so because the faces are regular triangles. All the edges of this figure are congruent in length, and the faces are congruent in area. A regular tetrahedron is one of five similar polyhedra.
Tetrahedron formulas
The height of a tetrahedron is equal to the product of the root of 2/3 and the length of the edge.
The volume of a tetrahedron is found in the same way as the volume of a pyramid: the square root of 2 divided by 12 and multiplied by the length of the cubed edge.
The remaining formulas for calculating the area and radii of circles are presented above.
Plan for the preparation and conduct of the lesson:
I. Preparatory stage:
- Repetition of the known properties of the triangular pyramid.
- Putting forward hypotheses about possible, not previously considered, features of the tetrahedron.
- Formation of groups for conducting research on these hypotheses.
- Distribution of tasks for each group (taking into account desire).
- Distribution of responsibilities for the task.
II. Main stage:
- Hypothesis solution.
- Consultations with a teacher.
- Work form.
III. The final stage:
- Presentation and defense of the hypothesis.
Lesson objectives:
- generalize and systematize the knowledge and skills of students; study additional theoretical material on the specified topic; to teach how to apply knowledge in solving non-standard problems, to see simple components in them;
- to form the skill of students working with additional literature, to improve the ability to analyze, generalize, find the main thing in what they read, prove new things; develop students' communication skills;
- cultivate a graphic culture.
Preparatory stage (1 lesson):
- Student's message "Secrets of the Great Pyramids".
- Introductory speech of the teacher about the diversity of types of pyramids.
- Discussion questions:
- On what grounds can irregular triangular pyramids be combined
- What do we mean by the orthocenter of a triangle, and what can be called the orthocenter of a tetrahedron
- Does a rectangular tetrahedron have an orthocenter?
- Which tetrahedron is called isohedral What properties can it have
- As a result of considering various tetrahedra, discussing their properties, the concepts are clarified and a certain structure appears:
- Consider the properties of a regular tetrahedron. (Appendix)
Properties 1-4 are proven verbally using Slide 1.
Property 1: All edges are equal.
Property 2: All planar angles are 60°.
Property 3: The sums of plane angles at any three vertices of a tetrahedron are 180°.
Property 4: If the tetrahedron is regular, then any of its vertices is projected into the orthocenter of the opposite face.
Given:
ABCD is a regular tetrahedron
AH - height
Prove:
H - orthocenter
Proof:
1) the point H can coincide with any of the points A, B, C. Let H ?B, H ?C
2) AH + (ABC) => AH + BH, AH + CH, AH + DH,
3) Consider ABH, BCH, ADH
AD - general => ABH, BCH, ADH => BH =CH = DH
AB \u003d AC \u003d AD t. H - is the orthocenter of ABC
Q.E.D.
- In the first lesson Properties 5-9 are formulated as hypotheses that require proof.
Each group gets their own homework:
Prove one of the properties.
Prepare a rationale with a presentation.
II. Main stage (within a week):
- Hypothesis solution.
- Consultations with a teacher.
- Work form.
III. Final stage (1-2 lessons):
Representation and defense of the hypothesis using presentations.
When preparing the material for the final lesson, students come to the conclusion about the features of the point of intersection of heights, we agree to call it an “amazing” point.
Property 5: The centers of the circumscribed and inscribed spheres coincide.
Given:
DABC is a regular tetrahedron
About 1 - the center of the described sphere
O - the center of the inscribed sphere
N is the point of contact of the inscribed sphere with the face ABC
Prove: O 1 = O
Proof:
Let OA = OB =OD = OC be the radii of the circumscribed circle
Drop ON + (ABC)
AON = CON - rectangular, along the leg and hypotenuse => AN = CN
Omit OM + (BCD)
COM DOM - rectangular, along the leg and hypotenuse => CM = DM
From paragraph 1 CON COM => ON = OM
ОN + (ABC) => ON,OM - radii of the inscribed circle.
The theorem has been proven.
For a regular tetrahedron, there is the possibility of its mutual arrangement with a sphere - contact with a certain sphere with all its edges. Such a sphere is sometimes called a “semi-inscribed” sphere.
Property 6: The segments connecting the midpoints of opposite edges and perpendicular to these edges are the radii of a semi-inscribed sphere.
Given:
ABCD is a regular tetrahedron;
AL=BL, AK=CK, AS=DS,
BP=CP, BM=DM, CN=DN.
Prove:
LO=OK=OS=OM=ON=OP
Proof.
Tetrahedron ABCD - regular => AO= BO = CO = DO
Consider triangles AOB, AOC, COD, BOD, BOC, AOD.
AO=BO=>?AOB – isosceles =>
OL - median, height, bisector
AO=CO=>?AOC– isosceles =>
OK - median, height, bisector
CO=DO=>?COD– isosceles =>
ON– median, height, bisector AOB=> AOC= COD=
BO=DO=>?BOD–isosceles => BOD=BOC=AOD
OM– median, height, bisector
AO=DO=>?AOD– isosceles =>
OS - median, height, bisector
BO=CO=>?BOC– isosceles =>
OP– median, height, bisector
AO=BO=CO=DO
AB=AC=AD=BC=BD=CD
3) OL, OK, ON, OM, OS, OP - heights in equal OL,OK,ON,OM,OS, OP radii
isosceles triangles of the sphere
Consequence:
A regular tetrahedron contains a semi-inscribed sphere.
Property 7: if the tetrahedron is regular, then every two opposite edges of the tetrahedron are mutually perpendicular.
Given:
DABC is a regular tetrahedron;
H - orthocenter
Prove:
Proof:
DABC - regular tetrahedron =>? ADB - equilateral
(ADB) (EDC) = ED
ED - ADB height => ED +AB,
AB + CE ,=> AB+ (EDC) => AB + CD.
The perpendicularity of other edges is proved similarly.
Property 8: Six planes of symmetry intersect at one point. Four straight lines intersect at the point O, drawn through the centers of the circles circumscribed near the faces perpendicular to the planes of the faces, and the point O is the center of the circumscribed sphere.
Given:
ABCD is a regular tetrahedron
Prove:
O is the center of the described sphere;
6 planes of symmetry intersect at point O;
Proof.
CG + BD BCD - equilateral => GO + BD (by the theorem of three GO + BD perpendiculars)
BG = GD, because AG - ABD median
ABD (ABD)=> ? BOD - isosceles => BO=DO
ED + AB, as ABD - equilateral => OE + AD (by the three perpendiculars theorem)
BE = AE, because DE - median?ABD
ABD (ABD) =>?AOB - isosceles =>BO=AO
(AOB) (ABD) = AB
ON + (ABC) OF + AC (by the three
BF + AC, because ABC - equilateral perpendiculars)
AF = FC, because BF - median? ABC
ABC (ABC) => AOC - isosceles => AO = CO
(AOC) ?(ABC) = AC
BO = AO =>AO = BO = CO = DO are sphere radii,
AO = CO circumscribed about tetrahedron ABCD
(ABR) (ACG) = AO
(BCT) (ABR) = BO
(ACG) (BCT) = CO
(ADH) (CED) = DO
AB + (ABR)(ABR)(BCT)(ACG)(ADH)(CED) (BDF)
Hence:
Point O is the center of the circumscribed sphere,
6 planes of symmetry intersect at point O.
Property 9: The obtuse angle between the perpendiculars passing through the vertices of the tetrahedron to the orthocenters is 109°28"
Given:
ABCD is a regular tetrahedron;
O is the center of the described sphere;
Prove:
Proof:
1)AS - height
ASB = 90 o OSB rectangular
2) (according to the property of a regular tetrahedron)
3)AO=BO - radii of the circumscribed sphere
4) 
70°32"
6) AO=BO=CO=DO =>?AOD=?AOC=?AOD=?COD=?BOD=?BOC
(according to the property of a regular tetrahedron)
=>AOD=AOC=AOD=COD=BOD=BOC=109°28"
This is what needed to be proven.
An interesting fact is that some organic substances have such an angle: silicates and hydrocarbons.
As a result of working on the properties of a regular tetrahedron, the students came up with the idea to call the work “An amazing point in a tetrahedron”. There were proposals to consider the properties of rectangular and isohedral tetrahedra. Thus, the work went beyond the lesson.
Conclusions:
The “surprising” point in a regular tetrahedron has the following features:
- is the point of intersection of the three axes of symmetry
- is the point of intersection of the six planes of symmetry
- is the intersection point of the heights of a regular tetrahedron
- is the center of the inscribed sphere
- is the center of the semi-inscribed sphere
- is the center of the circumscribed sphere
- is the centroid of the tetrahedron
- is the vertex of four equal regular triangular pyramids with bases - faces of a tetrahedron.
Conclusion.
(The teacher and students sum up the lesson. One of the students speaks with a brief report on tetrahedra, as a structural unit of chemical elements.)
The properties of a regular tetrahedron and its “surprising” point are studied.
It was found that the shape of only such a tetrahedron, which has all of the above properties, as well as an “ideal” point, can be occupied by molecules of silicates and hydrocarbons. Or molecules can consist of several regular tetrahedra. At present, the tetrahedron is known not only as a representative of ancient civilization, mathematics, but also as the basis of the structure of substances.
Silicates are salt-like substances containing silicon compounds with oxygen. Their name comes from the Latin word "silex" - "flint". The basis of silicate molecules is atomic radicals, having the form of tetrahedra.
Silicates are sand, and clay, and brick, and glass, and cement, and enamel, and talc, and asbestos, and emerald, and topaz.
Silicates make up more than 75% of the earth's crust (and together with quartz about 87%) and more than 95% of igneous rocks.
An important feature of silicates is the ability for mutual combination (polymerization) of two or more silicon-oxygen tetrahedra through a common oxygen atom.
The same form of molecules have saturated hydrocarbons, but they consist, unlike silicates, of carbon and hydrogen. General formula of molecules
Hydrocarbons include natural gas.
It is necessary to consider the properties of rectangular and isohedral tetrahedra.
Literature.
- Potapov V.M., Tatarinchik S.N. "Organic Chemistry", Moscow 1976.
- Babarin V.P. “Secrets of the Great Pyramids”, St. Petersburg, 2000
- Sharygin I. F. “Problems in geometry”, Moscow, 1984
- Big encyclopedic dictionary.
- “School Directory”, Moscow, 2001.
