The term "pyramid" is borrowed from the Greek "pyramis" or "pyramidos". The Greeks, in turn, borrowed this word, it is believed, from the Egyptian language. In the Ahmes papyrus the word “pyramus” appears in the sense of the edge of a regular pyramid. Others believe that the term originates from the shapes of breads in Ancient Greece(“piros” - rye). Due to the fact that the shape of the flame sometimes resembles the image of a pyramid, some medieval scholars believed that the term comes from the Greek word "pir" - fire. That is why in the geometry textbooks of the 16th century. the pyramid is called the “fireform body”.
IN Ancient Egypt The tombs of the pharaohs were shaped like pyramids. In the 3rd millennium BC. the Egyptians built step pyramids made of stone blocks; Later Egyptian pyramids acquired a geometrically correct shape - for example, the Cheops pyramid, whose height reaches almost 147 m, and others. Euclid defines a pyramid as a solid figure bounded by planes that from one plane (the base) converge at one point (the apex). This definition was criticized already in antiquity. For example, Heron, who proposed the following definition of a pyramid: it is a figure bounded by triangles converging at one point, and the base of which is a polygon. The most important drawback of this definition is the use of a vague concept of foundation. Taylor defined a pyramid as a polyhedron in which all but one of its faces meet at one point. Legendre, in his Elements of Geometry, defines a pyramid as follows: “A solid figure formed by triangles meeting at one point and ending on different sides of a flat base.” After this formulation, the concept of foundation is explained. Legendre's definition is clearly redundant, that is, it contains features that can be derived from others. And here is another definition that appeared in textbooks of the 19th century: a pyramid is a solid angle intersected by a plane.
The first direct calculation of the volume of the pyramid that has come down to us is found in Heron of Alexandria. It is interesting to note that in ancient documents there are rules for determining the volume of a truncated pyramid, but there are no rules for calculating the volume full pyramid. In the Moscow Papyrus there is a problem entitled “Actions with a truncated pyramid,” which sets out the correct calculation of the volume of one truncated pyramid. The Babylonian cuneiform tablets also do not contain calculations for the volume of a pyramid, but they do contain many examples of calculating the volume of a truncated pyramid. The first formula for the volume of a pyramid was apparently discovered by the ancient Egyptians. After all, they needed to be able to at least approximately calculate how much stone would be needed to build a particular pyramid.
According to Archimedes, back in the 5th century. BC. Democritus of Abdera established that the volume of a pyramid is equal to one third of the volume of a prism with the same base and the same height. A complete proof of this theorem was given by Eudoxus of Cnidus in the 4th century. BC.
Calculating the volume of a pyramid whose base is a certain polygon comes down to calculating the volumes of tetrahedrons (triangular pyramids). Therefore, the main attention below will be focused on calculating the volumes of these simplest polyhedra.
Typically the formula for calculating the volume of a tetrahedron is ABCD motivated by the following considerations. Firstly, it is believed that the formula for calculating the volume of a prism (in particular, a triangular one) is already known. Secondly, one way or another, the validity of the statement is substantiated that two different tetrahedrons with one common face and equal corresponding heights have equal volumes. Most in a simple way to substantiate the second statement is the use of the so-called Cavalieri principle.
In Figure 1, tetrahedrons ABCD And ABCD" with a common edge ABC have equal heights, planes π 1, π 2 are parallel (plane π 1 contains a triangle ABC, and the vertices D And D" these tetrahedra belong to the π plane 2). It is easy to show that triangles A"B"C" And A""B""C"", resulting from the intersection of these tetrahedra with an arbitrary plane π 3 parallel to the first two planes, are equal (Prove this!) and thus have equal areas. Hence the conclusion is drawn that the volumes of such tetrahedra are equal (“two stacks of equal pancakes fill one volume”).
With such agreements basic formula to calculate the volume of a tetrahedron:
Where S is the area of one of the faces of the tetrahedron, and N- the length of the height lowered onto this face is obtained by building up this tetrahedron to a prism as shown in Figure 2, in which the planes ABC And A"B"C" parallel. Then this prism, the volume of which is equal to the product of the area of its base and its height, is composed of three equal-sized (according to the Cavalieri principle) tetrahedra ABCA", A"B"C.B. And A"B"CC"(the last two are equal in size, since they have a common edge A"B"C and equal heights lowered to this face).
Another possibility for calculating the volume of a tetrahedron is provided by another important construction related to the tetrahedron, the so-called described parallelepiped. It is obtained by drawing three pairs of parallel planes, each of which is drawn through the opposite edges of the tetrahedron (Fig. 3)
The opposite edges of the tetrahedron are the diagonals of the opposite faces of the described parallelepiped. In addition to the original tetrahedron, the parallelepiped contains four more “small” equal-sized tetrahedra, the volumes of which are equal to one-sixth of the volume of the described parallelepiped (the equal size of these tetrahedra follows from the basic formula). It follows that the volume of the original tetrahedron is equal to a third of the volume of the described parallelepiped. Thus, to calculate the volume of a tetrahedron we obtain the following formula:
Where A = BD, b = A"C", d And φ - distance and angle, respectively, between intersecting lines BD And A"C". Indeed, the value is equal to the area of the face ABCD, A d is the length of the height of this parallelepiped.
From here, in particular, it follows that if two opposite edges of a tetrahedron move along two given crossing lines ( d And φ - are given) without changing their lengths, then the volume of the tetrahedron does not change
Note 1. The fact that in any tetrahedron the product of the area of a face and the height drawn to it does not depend on the choice of base and height can be proven directly. To do this, we choose two arbitrary faces of the tetrahedron with areas S 1 and S 2, corresponding heights H 1 and H 2 and a common edge of length a(Fig. 4).
We need to prove that the relation is denoted by h 1 and h 2 heights drawn to a common edge in the “first” and “second” faces. From the likeness right triangles(first with leg H 1 and hypotenuse h 2, and the second - with a leg H 2 and hypotenuse h 1 ; the similarity follows from the fact that acute angles are linear angles of the dihedral angle at the edge a) we have that
Besides, 
From these two equalities we obtain the desired statement.
Note 2. In his famous speech at the International Mathematical Congress in Paris in August 1900, D. Hilbert, among the 23 problems he posed, under number three, mentioned a problem closely related to the issues of teaching the theory of volumes of polyhedra. It draws attention to the fact that when deriving a formula for calculating the volume of a tetrahedron, one has to use a rather complex passage to the limit (the so-called “Devil's staircase”) or use the Cavalieri principle, as was done above. Hilbert drew attention to this circumstance and put forward a hypothesis about the possibility of rigorously proving that without the operation of passing to the limit, the theory of volumes of polyhedra cannot be constructed in the same way as is done in planimetry in the theory of areas of polygons. In the same year, 1900, the German mathematician Max Dehn, a student of Hilbert, confirmed the hypothesis expressed by his teacher, proving that there are polyhedra of equal volume that are not equally composed. In other words, one of them cannot be divided into polyhedra that could be used to form another polyhedron (it is on the idea of equicomposition that the theory of polygon areas is built). One of the pairs of such equal-sized polyhedra is a cube and a regular tetrahedron. For more details see.
In connection with the motivation of the basic formula, a prism consisting of three equal tetrahedra was used to calculate the volume of a tetrahedron.
And what triangular prisms can be divided into three equal tetrahedron? From which three equal tetrahedrons can a prism be constructed?
To answer the first question, consider (see Fig. 2) a triangular prism ABCA"B"C" and suppose that the tetrahedrons ABCA", A"B"C"C And A"B"B.C. are equal to each other (the last of them lies between the other two).
Then tetrahedrons ABCA" And A"B"B.C. have a common edge A"B.C., and since they are equal, then the edges coming from the vertices are also equal A And B". But AA" = BB" How lateral ribs prisms, AB = A"B" as the corresponding sides of the bases of the prism, therefore A.C. = B"C.
Let us consider tetrahedrons in a similar way A"B"C"C And A"B"B.C. like tetrahedrons with a common face A"B"C and fourth peaks B And C" respectively. Compare the side ribs again: BB" = C"C like the side ribs of a prism; C"B" = B.C. as the corresponding sides of the bases of the prism. Then from the equality of these tetrahedra it follows that A"B = A"C".
So, if a prism is composed of three equal tetrahedra as shown in Figure 2, then
A.C. = A"C" = A"B = B"C. (*)
This prism can be divided into three tetrahedrons in other ways. (There are 6 such methods in total. Prove it yourself!) In each such partition, similar to what was just considered, we make sure that the relation (*) is satisfied. Thus, if a prism can be divided into three equal tetrahedrons, then non-intersecting diagonals of the two lateral faces of the prism must be equal to each other and equal to the side of the base, lying in the third lateral face.
Before giving a detailed answer to the second question, let us dwell on one method of constructing a prism from three equal tetrahedra.
Let us denote by a the plane passing through the point A. Consider an equilateral triangle AMN in this plane. Let us restore perpendiculars to a at its vertices. On these lines we select three points A", B, C such that A.A." = 3l,M.B. = l, NC = 2l, Where l- an arbitrary positive number (see Fig. 5). Let us now complete the resulting tetrahedron ABCA" to a prism with a base ABC and side ribs BB", CC". The prism thus obtained satisfies condition (*) and, therefore, consists of three equal tetrahedra: ABCA", A"B"B.C., A"B"C"C. Indeed, denoting for a triangle side length AMN and using the Pythagorean theorem, it is easy to calculate that
The full answer to the question posed is contained in task 2.
Tasks and exercises
1. Is it possible to cut a cube: a) into 5 triangular pyramids; b) into 4 triangular pyramids?
2. Which three equal tetrahedra can be used to make a prism? Find the relationship that its edges must satisfy.
Note. Let a tetrahedron be given ABCA" and the lengths of all its edges are known. Build it up to a prism ABCA"B"C" and express BC" through known quantities.
3. Prove that it is impossible to construct a prism from three regular tetrahedra.
4. Let all the faces of the tetrahedron be equal triangles whose sides are equal a, b And c. Calculate the volume of the tetrahedron.
5. Given three parallel lines p, q And r, not lying in the same plane. The edge of the tetrahedron moves freely in a straight line p(without changing its length), and the remaining two vertices - in straight lines q And r. Prove that the volume of the tetrahedron remains constant.
6. Parallel lines p, q, r And l, not lying in the same plane, intersect the first plane at points A, B, C And D, and the second - at points A", B", C" And D" respectively. Prove that tetrahedrons AB"C"D" And A"BCD have the same volumes.
7 * . (All-Russian Olympiad, 1988 .) a) Is it possible to place two non-intersecting regular tetrahedrons with edge 1 inside a cube with edge 1?
b) Which greatest number regular tetrahedrons with edge 1 are placed inside a cube with edge 1 if the tetrahedra can only touch their faces?
2. Up to a permutation of vertices, the equality must be satisfied AB = C.A.", as well as one of the following two relations:
(AB) 2 = (A.C.) 2 + (B.A.") 2 – (B.C.) 2 – (A.A.") 2 ,
3(AB) 2 = (A.C.) 2 + (B.A.") 2 + (B.C.) 2 + (A.A.") 2 .
7. a) It is possible; b) 3 tetrahedrons.
Literature
1. Boltyansky V. G. Hilbert's third problem. - M.: Nauka, 1977.
2. Gilbert D., Vossen K. Visual geometry. - M.: Nauka, 1981.
3. Ponarin Y.P. Elementary geometry: In 2 volumes. - T. 1. Planimetry, plane transformations. - M.: MTsNMO, 2004.
4. Ponarin Y.P. Elementary geometry: In 2 volumes. - T. 1. Stereometry, space transformations. - M.: MTsNMO, 2006.
5. Hadamard J. Elementary geometry. - Part 2: Stereometry. - M.: Education, 1957.
In geometry tetrahedron is a regular polyhedron that has four faces that are equilateral triangles. It follows from this that all edges tetrahedron have the same length, and all its faces have the same area. This geometric body and its basic properties are studied in school geometry lessons, but in life it “ V pure form "Doesn't happen very often. Or rather, tetrahedron often simply not as noticeable and obvious as, for example, a sphere or a parallelepiped.
Nevertheless, in technology this geometric body is found quite often. For example, the form tetrahedrons have optical elements that are the basis for the design of reflectors. Due to the peculiarities of the arrangement of the edges tetrahedrons reflect light to the same point from which it comes, and therefore they appear to glow themselves. Reflectors have found very wide application as road safety devices.
Finding the volume of a tetrahedron
| V |
a- edge of a tetrahedron
V- volume of the tetrahedron
Because the tetrahedron is by its nature an exclusively rigid static form, this property is quite widely used in technology. For example, the rods of many load-bearing metal structures are arranged precisely in the form of tetrahedrons, and thanks to this, engineers are able to create lightweight and exceptionally strong trusses for bridges and floors of various structures.
The crystal lattices of many durable natural minerals also have the shape tetrahedron. One of them is diamond, in which the atoms are located precisely at the vertices of this geometric body. Interestingly, graphite is also composed of carbon atoms, meaning its chemical composition is similar chemical composition diamond, however, in terms of strength characteristics it is very significantly inferior to the latter precisely because the shape of its crystal lattice is different. Therefore, the production of artificial diamonds from graphite consists precisely in ordering carbon atoms in such a way that they form tetrahedrons.
The arrangement of the fruits of some plants in clusters also has the shape of this geometric body. For example, walnuts are often positioned such that their centers are at the vertices of a tetrahedron.
Now in Russia and some foreign countries milk packaging is produced, also shaped tetrahedron. The basis for its manufacture is a pipe made of a special material, reminiscent of that used in the manufacture of so-called “ tetrapacks" As it is filled with milk or cream, special devices seal it in such a way that adjacent seams are perpendicular to each other, and as a result the finished bags have the shape of a tetrahedron.
Classic tetrahedron is also a puzzle known as " Rubik's pyramid», « Japanese tetrahedron" And " Moldavian pyramid" The famous Hungarian architect and inventor, however, has nothing to do with it, although the principle on which it is based is practically the same as that used in his famous cube. In fact, this toy was developed by the German Uwe Meffert in 1972, then, independently of him, invented by the Moldovan engineer A.A. Ordynets, and since 1981 produced by the company Tomy Toys, whose headquarters are located in Japan.