“Sphere of politics” - Relations of social actors regarding state power. Scientific and theoretical. The process of interaction between politics and economics. Together with the state. Regulation of social relations is conditioned by social interests. The process of interaction between politics and morality. The power of the state, persuasion, stimulation.

“Prism geometry” - Given a right quadrangular prism ABCDA1B1C1D1. Euclid probably considered it a matter of practical guides in geometry. A straight prism is a prism whose side edge is perpendicular to the base. Prism in geometry. According to the property of 2 volumes, V=V1+V2, that is, V=SABD h+SBDC h=(SABD+SBDC) h. So triangles A1B1C1 and ABC are equal on three sides.

“Volume of a prism” - How to find the volume of a straight prism? The volume of the original prism is equal to the product S · h. Basic steps in proving the direct prism theorem? Area S of the base of the original prism. Holding altitude triangle ABC. Task. Straight prism. Lesson objectives. The concept of a prism. Volume of a straight prism. The solution of the problem. The prism can be divided into straight triangular prisms with height h.

“Sphere surface” - Mars. Is the ball a ball? Ball and sphere. Earth. Encyclopedia. We support our school baseball team. Venus. Uranus. Is there a ball in the picture? A little bit of history. Atmosphere. I decided to spend a little research……. Saturn. Are you ready to answer the questions?

The topic “Different problems on polyhedra, cylinder, cone and ball” is one of the most difficult in the 11th grade geometry course. Before solving geometric problems, they usually study the relevant sections of the theory that are referred to when solving problems. In the textbook by S. Atanasyan and others on this topic (p. 138) one can only find definitions of a polyhedron described around a sphere, a polyhedron inscribed in a sphere, a sphere inscribed in a polyhedron, and a sphere described around a polyhedron. IN methodological recommendations this textbook (see the book “Studying geometry in grades 10–11” by S.M. Saakyan and V.F. Butuzov, p. 159) says what combinations of bodies are considered when solving problems No. 629–646, and addresses attention to the fact that “when solving a particular problem, first of all, it is necessary to ensure that students have a good understanding of the relative positions of the bodies indicated in the condition.” The following is the solution to problems No. 638(a) and No. 640.

Considering all of the above, and the fact that the most difficult problems for students are the combination of a ball with other bodies, it is necessary to systematize the relevant theoretical principles and communicate them to students.

Definitions.

1. A ball is called inscribed in a polyhedron, and a polyhedron described around a ball if the surface of the ball touches all faces of the polyhedron.

2. A ball is said to be circumscribed about a polyhedron, and a polyhedron is said to be inscribed in a ball, if the surface of the ball passes through all the vertices of the polyhedron.

3. A ball is said to be inscribed in a cylinder, truncated cone (cone), and a cylinder, truncated cone (cone) is said to be circumscribed about the ball if the surface of the ball touches the bases (base) and all the generatrices of the cylinder, truncated cone (cone).

(From this definition it follows that the great circle of a ball can be inscribed into any axial section of these bodies).

4. A ball is said to be circumscribed about a cylinder, a truncated cone (cone), if the circles of the bases (base circle and apex) belong to the surface of the ball.

(From this definition it follows that around any axial section of these bodies the circle of a larger circle of the ball can be described).

General notes on the position of the center of the ball.

1. The center of a ball inscribed in a polyhedron lies at the point of intersection of the bisector planes of all dihedral angles of the polyhedron. It is located only inside the polyhedron.

2. The center of a ball circumscribed about a polyhedron lies at the intersection point of planes perpendicular to all edges of the polyhedron and passing through their midpoints. It can be located inside, on the surface or outside the polyhedron.

Combination of a sphere and a prism.

1. A ball inscribed in a straight prism.

Theorem 1. A sphere can be inscribed into a straight prism if and only if a circle can be inscribed at the base of the prism, and the height of the prism is equal to the diameter of this circle.

Corollary 1. The center of a sphere inscribed in a right prism lies at the midpoint of the altitude of the prism passing through the center of the circle inscribed in the base.

Corollary 2. A ball, in particular, can be inscribed in straight lines: triangular, regular, quadrangular (in which the sums of the opposite sides of the base are equal to each other) under the condition H = 2r, where H is the height of the prism, r is the radius of the circle inscribed in the base.

2. A sphere circumscribed about a prism.

Theorem 2. A sphere can be described around a prism if and only if the prism is straight and a circle can be described around its base.

Corollary 1. The center of a sphere circumscribed about a straight prism lies at the midpoint of the height of the prism drawn through the center of a circle circumscribed about the base.

Corollary 2. A ball, in particular, can be described: near a right triangular prism, near a regular prism, near a rectangular parallelepiped, near a right quadrangular prism, in which the sum of the opposite angles of the base is equal to 180 degrees.

From the textbook by L.S. Atanasyan, problems No. 632, 633, 634, 637(a), 639(a,b) can be suggested for the combination of a ball and a prism.

Combination of a ball with a pyramid.

1. A ball described near a pyramid.

Theorem 3. A ball can be described around a pyramid if and only if a circle can be described around its base.

Corollary 1. The center of a sphere circumscribed about a pyramid lies at the point of intersection of a straight line perpendicular to the base of the pyramid passing through the center of a circle circumscribed about this base and a plane perpendicular to any lateral edge drawn through the middle of this edge.

Corollary 2. If lateral ribs pyramids are equal to each other (or equally inclined to the plane of the base), then a ball can be described around such a pyramid. The center of this ball in this case lies at the point of intersection of the height of the pyramid (or its extension) with the symmetry axis of the side edge, which lies in the plane of the side edge and height.

Corollary 3. A ball, in particular, can be described: near a triangular pyramid, near a regular pyramid, near a quadrangular pyramid in which the sum of opposite angles is 180 degrees.

2. A ball inscribed in a pyramid.

Theorem 4. If the side faces of the pyramid are equally inclined to the base, then a ball can be inscribed into such a pyramid.

Corollary 1. The center of a ball inscribed in a pyramid whose side faces are equally inclined to the base lies at the point of intersection of the height of the pyramid with the bisector of the linear angle of any dihedral angle at the base of the pyramid, the side of which is the height of the side face drawn from the top of the pyramid.

Corollary 2. You can fit a ball into a regular pyramid.

From the textbook by L.S. Atanasyan, problems No. 635, 637(b), 638, 639(c), 640, 641 can be suggested for the combination of a ball with a pyramid.

Combination of a ball with a truncated pyramid.

1. A ball circumscribed about a regular truncated pyramid.

Theorem 5. A sphere can be described around any regular truncated pyramid. (This condition is sufficient, but not necessary)

2. A ball inscribed in a regular truncated pyramid.

Theorem 6. A ball can be inscribed into a regular truncated pyramid if and only if the apothem of the pyramid is equal to the sum of the apothems of the bases.

There is only one problem for the combination of a ball with a truncated pyramid in L.S. Atanasyan’s textbook (No. 636).

Combination of ball with round bodies.

Theorem 7. A ball can be described around a cylinder, a truncated cone (straight circular), or a cone.

Theorem 8. A ball can be inscribed into a (straight circular) cylinder if and only if the cylinder is equilateral.

Theorem 9. You can fit a ball into any cone (straight circular).

Theorem 10. A ball can be inscribed into a truncated cone (straight circular) if and only if its generatrix is ​​equal to the sum of the radii of the bases.

From the textbook by L.S. Atanasyan, problems No. 642, 643, 644, 645, 646 can be suggested for the combination of a ball with round bodies.

To more successfully study the material on this topic, it is necessary to include oral tasks in the lessons:

1. The edge of the cube is equal to a. Find the radii of the balls: inscribed in the cube and circumscribed around it. (r = a/2, R = a3).

2. Is it possible to describe a sphere (ball) around: a) a cube; b) rectangular parallelepiped; c) an inclined parallelepiped with a rectangle at its base; d) straight parallelepiped; e) an inclined parallelepiped? (a) yes; b) yes; c) no; d) no; d) no)

3. Is it true that a sphere can be described around any triangular pyramid? (Yes)

4. Is it possible to describe a sphere around any quadrangular pyramid? (No, not near any quadrangular pyramid)

5. What properties must a pyramid have in order to describe a sphere around it? (At its base there should be a polygon around which a circle can be described)

6. A pyramid is inscribed in a sphere, the side edge of which is perpendicular to the base. How to find the center of a sphere? (The center of the sphere is the intersection point of two geometric loci of points in space. The first is a perpendicular drawn to the plane of the base of the pyramid, through the center of a circle circumscribed around it. The second is a plane perpendicular to a given side edge and drawn through its middle)

7. Under what conditions can you describe a sphere around a prism, at the base of which is a trapezoid? (Firstly, the prism must be straight, and secondly, the trapezoid must be isosceles so that a circle can be described around it)

8. What conditions must a prism satisfy in order for a sphere to be described around it? (The prism must be straight, and its base must be a polygon around which a circle can be described)

9. A sphere is described around a triangular prism, the center of which lies outside the prism. Which triangle is the base of the prism? (Obtuse triangle)

10. Is it possible to describe a sphere around an inclined prism? (No you can not)

11. Under what condition will the center of a sphere circumscribed about a right triangular prism be located on one of the lateral faces of the prism? (The base is a right triangle)

12. The base of the pyramid is an isosceles trapezoid. The orthogonal projection of the top of the pyramid onto the plane of the base is a point located outside the trapezoid. Is it possible to describe a sphere around such a trapezoid? (Yes, you can. The fact that the orthogonal projection of the top of the pyramid is located outside its base does not matter. What is important is that at the base of the pyramid lies isosceles trapezoid- a polygon around which a circle can be described)

13. About regular pyramid the sphere is described. How is its center located relative to the elements of the pyramid? (The center of the sphere is on a perpendicular drawn to the plane of the base through its center)

14. Under what condition does the center of a sphere described around a right triangular prism lie: a) inside the prism; b) outside the prism? (At the base of the prism: a) an acute triangle; b) obtuse triangle)

15. A sphere is described around a rectangular parallelepiped whose edges are 1 dm, 2 dm and 2 dm. Calculate the radius of the sphere. (1.5 dm)

16. What truncated cone can a sphere fit into? (In a truncated cone, into the axial section of which a circle can be inscribed. The axial section of the cone is an isosceles trapezoid, the sum of its bases must be equal to the sum of its lateral sides. In other words, the sum of the radii of the bases of the cone must be equal to the generator)

17. A sphere is inscribed in a truncated cone. At what angle is the generatrix of the cone visible from the center of the sphere? (90 degrees)

18. What property must a straight prism have in order for a sphere to be inscribed into it? (Firstly, at the base of a straight prism there must be a polygon into which a circle can be inscribed, and, secondly, the height of the prism must be equal to the diameter of the circle inscribed in the base)

19. Give an example of a pyramid that cannot fit a sphere? (For example, a quadrangular pyramid with a rectangle or parallelogram at its base)

20. At the base of a straight prism is a rhombus. Is it possible to fit a sphere into this prism? (No, it’s impossible, since in general it’s impossible to describe a circle around a rhombus)

21. Under what condition can a sphere be inscribed into a right triangular prism? (If the height of the prism is twice the radius of the circle inscribed in the base)

22. Under what condition can a sphere be inscribed into a regular quadrangular truncated pyramid? (If the cross-section of a given pyramid is a plane passing through the middle of the side of the base perpendicular to it, it is an isosceles trapezoid into which a circle can be inscribed)

23. A sphere is inscribed in a triangular truncated pyramid. Which point of the pyramid is the center of the sphere? (The center of the sphere inscribed in this pyramid is at the intersection of three bisectral planes of angles formed by the lateral faces of the pyramid with the base)

24. Is it possible to describe a sphere around a cylinder (right circular)? (Yes, you can)

25. Is it possible to describe a sphere around a cone, a truncated cone (straight circular)? (Yes, you can, in both cases)

26. Is it possible to fit a sphere into any cylinder? What properties must a cylinder have in order to fit a sphere into it? (No, not every time: the axial section of the cylinder must be square)

27. Can a sphere be inscribed into any cone? How to determine the position of the center of a sphere inscribed in a cone? (Yes, absolutely. The center of the inscribed sphere is at the intersection of the altitude of the cone and the bisector of the angle of inclination of the generatrix to the plane of the base)

The author believes that out of the three planning lessons on the topic “Different problems on polyhedra, cylinder, cone and ball”, it is advisable to devote two lessons to solving problems on combining a ball with other bodies. It is not recommended to prove the theorems given above due to insufficient time in class. You can invite students who have sufficient skills for this to prove them by indicating (at the teacher’s discretion) the course or plan of the proof.

Polyhedra circumscribed about a sphere A polyhedron is said to be circumscribed about a sphere if the planes of all its faces touch the sphere. The sphere itself is said to be inscribed in the polyhedron. Theorem. A sphere can be inscribed into a prism if and only if a circle can be inscribed at its base, and the height of the prism is equal to the diameter of this circle. Theorem. You can fit a sphere into any triangular pyramid, and only one.






Exercise 1 Erase the square and draw two parallelograms representing the top and bottom faces of the cube. Connect their vertices with segments. Obtain an image of a sphere inscribed in a cube. Draw a sphere inscribed in a cube, as on the previous slide. To do this, draw an ellipse inscribed in a parallelogram obtained by compressing a circle and a square by 4 times. Mark the poles of the sphere and the tangent points of the ellipse and parallelogram.
























Exercise 1 A sphere is inscribed in a line quadrangular prism, at the base of which is a rhombus with side 1 and acute angle 60 o. Find the radius of the sphere and the height of the prism. Solution. The radius of the sphere is equal to half the height of the DG base, i.e. The height of the prism is equal to the diameter of the sphere, i.e.






Exercise 4 A sphere is inscribed in a right quadrangular prism, at the base of which is a quadrilateral, perimeter 4 and area 2. Find the radius r of the inscribed sphere. Solution. Note that the radius of the sphere is equal to the radius of the circle inscribed at the base of the prism. Let's take advantage of the fact that the radius of a circle inscribed in a polygon is equal to the area of ​​this polygon divided by its semi-perimeter. We get,














Exercise 3 Find the radius of a sphere inscribed in a regular triangular pyramid, the side of the base is 2, and the dihedral angles at the base are 60°. Solution. Let us take advantage of the fact that the center of the inscribed sphere is the intersection point of the bisector planes of the dihedral angles at the base of the pyramid. For the radius of the sphere OE the following equality holds: Therefore,


Exercise 4 Find the radius of a sphere inscribed in a regular triangular pyramid, the side edges of which are equal to 1, and the plane angles at the apex are equal to 90°. Answer: Solution. In the tetrahedron SABC we have: SD = DE = SE = From the similarity of triangles SOF and SDE we obtain an equation by solving which we find




Exercise 1 Find the radius of a sphere inscribed in a regular quadrangular pyramid, all edges of which are equal to 1. Let us use the fact that for the radius r of a circle inscribed in a triangle, the formula holds: r = S/p, where S is the area, p is the semi-perimeter of the triangle . In our case, S = p = Solution. The radius of the sphere is equal to the radius of the circle inscribed in the triangle SEF, in which SE = SF = EF=1, SG = Therefore,


Exercise 2 Find the radius of a sphere inscribed in a regular quadrangular pyramid, the base side of which is 1, and the side edge is 2. Let us use the fact that for the radius r of a circle inscribed in a triangle, the formula holds: r = S/p, where S – area, p – semi-perimeter of the triangle. In our case, S = p = Solution. The radius of the sphere is equal to the radius of the circle inscribed in the triangle SEF, in which SE = SF = EF=1, SG = Therefore,


Exercise 3 Find the radius of a sphere inscribed in a regular quadrangular pyramid, the side of the base is 2, and the dihedral angles at the base are 60°. Solution. Let us take advantage of the fact that the center of the inscribed sphere is the intersection point of the bisector planes of the dihedral angles at the base of the pyramid. For the radius of the sphere OG, the following equality holds:


Exercise 4 The unit sphere is inscribed in a regular quadrangular pyramid, the side of the base is 4. Find the height of the pyramid. Let us use the fact that for the radius r of a circle inscribed in a triangle, the formula holds: r = S/p, where S is the area, p is the semi-perimeter of the triangle. In our case, S = 2h, p = Solution. Let us denote the height SG of the pyramid as h. The radius of the sphere is equal to the radius of the circle inscribed in the triangle SEF, in which SE = SF = EF=4. Consequently, we have an equality from which we find




Exercise 1 Find the radius of a sphere inscribed in a regular hexagonal pyramid, the edges of which are equal to 1, and the side edges are equal to 2. Let us use the fact that for the radius r of a circle inscribed in a triangle, the formula holds: r = S/p, where S – area, p – semi-perimeter of the triangle. In our case, S = p = Therefore, Solution. The radius of the sphere is equal to the radius of the circle inscribed in the triangle SPQ, in which SP = SQ = PQ= SH =


Exercise 2 Find the radius of a sphere inscribed in a regular hexagonal pyramid whose base edges are equal to 1 and the dihedral angles at the base are equal to 60°. Solution. Let us take advantage of the fact that the center of the inscribed sphere is the intersection point of the bisector planes of the dihedral angles at the base of the pyramid. For the radius of the sphere OH, the following equality holds:
Exercise Find the radius of a sphere inscribed in a unit octahedron. Answer: Solution. The radius of the sphere is equal to the radius of the circle inscribed in the rhombus SESF, in which SE = SF = EF=1, SO = Then the height of the rhombus, lowered from the vertex E, will be equal to The required radius is equal to half the height, and is equal to O




Exercise Find the radius of a sphere inscribed in a unit icosahedron. Solution. Let us take advantage of the fact that the radius OA of the circumscribed sphere is equal to and the radius AQ of the circumscribed circle about an equilateral triangle with side 1 is equal to According to the Pythagorean theorem applied to right triangle OAQ, we get Exercise Find the radius of the sphere inscribed in the unit dodecahedron. Solution. Let us take advantage of the fact that the radius OF of the circumscribed sphere is equal to and the radius FQ of the circle circumscribed about an equilateral pentagon with side 1 is equal to. By the Pythagorean theorem applied to the right triangle OFQ, we obtain


Exercise 1 Is it possible to fit a sphere into a truncated tetrahedron? Solution. Note that the center O of a sphere inscribed in a truncated tetrahedron must coincide with the center of a sphere inscribed in a tetrahedron, which coincides with the center of a sphere half-inscribed in a truncated tetrahedron. Distances d 1, d 2 from point O to hexagonal and triangular faces are calculated using the Pythagorean theorem: where R is the radius of a half-inscribed sphere, r 1, r 2 are the radii of circles inscribed in a hexagon and triangle, respectively. Since r 1 > r 2, then d 1 r 2, then d 1