A sphere inscribed in a straight prism. Polyhedra circumscribed around a sphere are called circumscribed polyhedra. Combination of a ball with a truncated pyramid

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 4 Is it possible to inscribe a sphere into a rectangular parallelepiped other than a cube? Answer: No.

Exercise 5 Is it possible to inscribe a sphere into an inclined parallelepiped, all of whose faces are rhombuses? Answer: No.

Exercise 1 Is it possible to inscribe a sphere into an inclined triangular prism with a regular triangle at its base? Answer: No.

Exercise 2 Find the height of a regular triangular prism and the radius of the inscribed sphere if the edge of the base of the prism is 1. 3 3 , . 3 6 h r Answer:

Exercise 3 A sphere of radius 1 is inscribed into a regular triangular prism. Find the side of the base and the height of the prism. 2 3, 2. a h Answer:

Exercise 4 A sphere is inscribed into a prism, at the base of which is a right triangle with legs equal to 1. Find the radius of the sphere and the height of the prism. 2 2 , 2 2. 2 r h The area of triangle ABC is, perimeter Let's use the formula r = S / p. We get 2 2. 1 ,

Exercise 5 A sphere is inscribed into a prism, at the base of which is an isosceles triangle with sides 2, 3, 3. Find the radius of the sphere and the height of the prism. 2 , 2. 2 r h The area of triangle ABC is equal The perimeter is 8. Let's use the formula r = S / p. We get 2 2.

Exercise 1 A sphere is inscribed in a right quadrangular prism, at the base of which is a rhombus with side 1 and an acute angle of 60 degrees. 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. 3. 4 r 3. 2 h

Exercise 2 A unit sphere is inscribed in a right quadrangular prism, at the base of which is a rhombus with an acute angle of 60 degrees. Find the side of the base a and the height of the prism h. Answer: 4 3 , 2. 3 a h

Exercise 3 A sphere is inscribed in a right quadrangular prism, at the base of which is a trapezoid. The height of the trapezoid is 2. Find the height of the prism h and the radius r of the inscribed sphere. Answer: 1, 2. r h

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. 1. r 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 1 Find the height of a regular hexagonal prism and the radius of the inscribed sphere if the side of the base of the prism is 1. 3 3, . 2 h r Answer:

Exercise 2 A sphere of radius 1 is inscribed into a regular hexagonal prism. Find the side of the base and the height of the prism. 2 3 , 2. 3 a h Answer:

Exercise 1 Find the radius of a sphere inscribed in a unit tetrahedron. 6. 12 r 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 3 , 2 3 , 6 6. 3 6 3 3: : , 3 6 2 r r 6. 12 r

Exercise 2 A unit sphere is inscribed in a regular tetrahedron. Find the edge of this tetrahedron. 2 6. a Answer:

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°. 3 1 30. 3 3 r tg 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, . OE DE tg O

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°. 3 3. 6 r 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 2 , 2 6 , 6 3. 3 3 6 2: : , 3 6 2 r r 3 3. 6 r

Exercise 1 Find the radius of a sphere inscribed in a regular quadrangular pyramid, all edges of which are equal to 1. 6 2. 4 r 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 – semi-perimeter of the triangle. In our case, S = p = 3, 2 2. 2 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 = 2, 4 Therefore, 1 3.

Exercise 2 Find the radius of a sphere inscribed in a regular quadrangular pyramid, the side of the base is 1, and the side edge is 2. 14 (15 1). 28 r Let us take advantage of 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 = 15, 214. 2 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 = 14, 4 Therefore, 1 15.

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°. 3 30. 3 r tg 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: Therefore, . OG FG tg OFG

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 = 2 h, p = 2 4 2. h. 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. 2 4, h 8. 3 h Therefore, we have an equality from which we find 2 4 2 2, h h

Exercise 1 Find the radius of a sphere inscribed in a regular hexagonal pyramid, whose base edges are equal to 1, and the side edges are equal to 2. 15 3. 4 r 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 = 3, 2 Therefore, 15 3. 2 15, 2 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 = 3.

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°. 3 1 30. 2 2 r tg 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: Therefore, . OH HQ tg OQH

Exercise Find the radius of a sphere inscribed in a unit octahedron. 6. 6 r Answer: Solution. The radius of the sphere is equal to the radius of the circle inscribed in the rhombus SES'F, 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 6. 66. 3 2 .2 3 , 2 O

Exercise Find the radius of a sphere inscribed in a unit icosahedron. 1 7 3 5. 2 6 r 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 around an equilateral triangle with side 1 is equal to. By the Pythagorean theorem applied to the right triangle OAQ, we obtain 10 2 5, 4 3.

Exercise Find the radius of a sphere inscribed in a unit dodecahedron. 1 25 11 5. 2 10 r Solution. Let us use 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 18 6 5, 4 5 5.

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< d 2 и, следовательно, сферы, вписанной в усеченный тетраэдр, не существует. 2 2 1 1 2 2 , d R r

Exercise 2 Is it possible to fit a sphere into a truncated cube? Answer: No. The proof is similar to the previous one.

Exercise 3 Is it possible to fit a sphere into a truncated octahedron? Answer: No. The proof is similar to the previous one.

Exercise 4 Is it possible to fit a sphere into a cuboctahedron? Answer: No. The proof is similar to the previous one.

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 mutual arrangement bodies specified 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 called circumscribed about a polyhedron, and a polyhedron 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 inscribed around 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 the side edges of the pyramid 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 axis of symmetry of the side edge lying in the plane lateral 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 in 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 sphere 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 generator 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.

For more successful study material on this topic, it is necessary to include oral tasks in the course of 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. It is important that at the base of the pyramid lies an isosceles trapezoid - a polygon around which a circle can be described)

13. A sphere is described near a regular pyramid. 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, quadrangular pyramid, the base of which is a rectangle or parallelogram)

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. Can a sphere be inscribed 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.

Ball and sphere

The body obtained by rotating a semicircle around a diameter is called a ball. The surface formed in this case is called a sphere.A ball is a body that consists of all points in space located at a distance not greater than a given one from a given point. This point is called the center of the ball, and this distance is called the radius of the ball.The boundary of a ball is called a spherical surfaceor a sphere. Any segment connecting the center of a ball with a point on the spherical surface is called a radius.A segment connecting two points on a spherical surface and passing through the center of the ball is called a diameter.The ends of any diameter are called diametrically opposite points of the ball. Any section of the balla plane is a circle. The center of this circle is the base of the perpendicular dropped from the center onto the secant plane. The plane passing through the center of the ball is called the diametral plane. The section of a ball by the diametrical plane is called a great circle, and the cross section of the sphere is a great circle.Any diametrical plane of a ball is its plane of symmetry. The center of the ball is its center of symmetry.The plane passing through a point on the spherical surface and perpendicular to the radius drawn to this point is called the tangent plane. This point called the point of contact.The tangent plane has only one common point with the ball - the point of contact. A straight line passing through given point spherical surface perpendicular to the radius drawn to this point is called the tangent.An infinite number of tangents pass through any point on the spherical surface, and all of them lie in the tangent plane of the ball. A spherical segmentthe part of the ball cut off from it by a plane is called the spherical layercalled the part of the ball located between two parallel planes intersecting the ball. Spherical sectoris obtained from a spherical segment and a cone. If the spherical segment is smaller than a hemisphere, then the spherical segment is supplemented with a cone, the vertex of which is in the center of the ball, and the base is the base of the segment. If the segment is larger than the hemisphere, then the specified cone is removed from it. Basic formulasBall (R = OB - radius): S b = 4πR 2 ; V = 4πR 3 / 3. Ball segment (R = OB - radius of the ball, h = SK - height of the segment, r = KV - radius of the base of the segment): V segm = πh 2 (R - h/3)or V segm = πh(h 2 + 3r 2 ) / 6;S segm = 2πRh. Ball sector (R = OB - radius of the ball, h = SC - segment height): V = V segm ±V con , “+” - if the segment is smaller, “-” - if the segment is larger than the hemisphere.or V = V segm +V con = πh 2 (R - h/3) + πr 2 (R - h) / 3. Spherical layer (R 1 and R 2 - radii of the bases of the spherical layer; h = SC - height of the spherical layer or distance between bases):V w/sl = πh 3 / 6 + πh(R 1 2 + R 2 2 ) / 2;S w/sl = 2πRh. Example 1. The volume of the ball is 288π cm 3 . Find the diameter of the ball. SolutionV = πd 3 / 6288π = πd 3 / 6πd 3 = 1728πd 3 = 1728d = 12 cm. Answer: 12. Example 2. Three equal spheres of radius r touch each other and some plane. Determine the radius of the fourth sphere tangent to the three data and the given plane. SolutionLet O 1 , ABOUT 2 , ABOUT 3 - the centers of these spheres and O - the center of the fourth sphere touching the three data and the given plane. Let A, B, C, T be the points of contact of the spheres with a given plane. The points of contact of two spheres lie on the line of the centers of these spheres, therefore O 1 ABOUT 2 = O 2 ABOUT 3 = O 3 ABOUT 1 = 2r. The points are equidistant from the plane ABC, so ABO 2 ABOUT 1 , AVO 2 ABOUT 3 , AVO 3 ABOUT 1 - equal rectangles, therefore, ∆ABC is equilateral with side 2r. Let x be the desired radius of the fourth sphere. Then OT = x. Hence, Likewise This means that T is the center of an equilateral triangle. That's why From hereAnswer: r / 3. A sphere inscribed in a pyramid A sphere can be inscribed in every regular pyramid. The center of the sphere lies at the height of the pyramid at the point of its intersection with the bisector of the linear angle at the edge of the base of the pyramid. Note. If a sphere can be inscribed in a pyramid, not necessarily regular, then the radius r of this sphere can be calculated using the formula r = 3V / S pp , where V is the volume of the pyramid, S pp - its total surface area. Example 3. A conical funnel, the radius of the base R and the height H, is filled with water. A heavy ball is lowered into the funnel. What should be the radius of the ball so that the volume of water displaced from the funnel by the immersed part of the ball is maximum? Solution Let's draw a section through the center of the cone. This section forms an isosceles triangle.If there is a ball in the funnel, then the maximum size of its radius will be equal to the radius of the circle inscribed in the resulting isosceles triangle. The radius of the circle inscribed in the triangle is equal to: r = S / p, where S is the area of the triangle, p is its semi-perimeter. The area of the isosceles triangle is equal to half height (H = SO) multiplied by the base. But since the base is twice the radius of the cone, then S = RH. The semi-perimeter is equal to p = 1/2 (2R + 2m) = R + m.m is the length of each of the equal sides of the isosceles triangle; R is the radius of the circle constituting the base of the cone. Find m by Pythagorean theorem: , whereBriefly it looks like this:Answer:Example 4. In a regular triangular pyramid with a dihedral angle at the base equal to α, there are two balls. The first ball touches all the faces of the pyramid, and the second ball touches all the side faces of the pyramid and the first ball. Find the ratio of the radius of the first ball to the radius of the second ball if tgα = 24/7. Solution
Let RABC be a regular pyramid and point H be the center of its base ABC. Let M be the midpoint of edge BC. Then - linear dihedral angle , which by condition is equal to α, and α< 90°. Центр первого шара, касающегося всех граней пирамиды, лежит на отрезке РН в точке его пересечения с биссектрисой .Let NN 1 - the diameter of the first ball and the plane passing through point H 1 perpendicular to the straight line RN, intersects the side edges RA, PB, RS, respectively, at points A 1 , IN 1 , WITH 1 . Then N 1 will be the center of the correct ∆A 1 IN 1 WITH 1 , and the pyramid RA 1 IN 1 WITH 1 will be similar to the RABC pyramid with similarity coefficient k = RN 1 / RN. Note that the second ball, with center at point O 1 , is inscribed in the RA pyramid 1 IN 1 WITH 1 and therefore the ratio of the radii of the inscribed balls is equal to the similarity coefficient: OH / OH 1 = RN / RN 1 . From the equality tgα = 24/7 we find:Let AB = x. Then Hence the desired OH/O ratio 1 N 1 = 16/9. Answer: 16/9. Sphere inscribed in a prism The diameter D of a sphere inscribed in a prism is equal to the height H of the prism: D = 2R = H. The radius R of a sphere inscribed in a prism is equal to the radius of a circle inscribed in a perpendicular section prism. If a sphere is inscribed in a straight prism, then a circle can be inscribed at the base of this prism. The radius R of a sphere inscribed in a straight prism is equal to the radius of the circle inscribed in the base of the prism. Theorem 1 Let a circle be inscribed at the base of a straight prism, and the height H of the prism equal to the diameter D of this circle. Then a sphere with diameter D can be inscribed into this prism. The center of this inscribed sphere coincides with the middle of the segment connecting the centers of the circles inscribed at the bases of the prism. ProofLet ABC...A 1 IN 1 WITH 1 ... is a straight prism and O is the center of a circle inscribed at its base ABC. Then point O is equidistant from all sides of the base ABC. Let O 1 - orthogonal projection of point O onto base A 1 IN 1 WITH 1 . Then Oh 1 equidistant from all sides of base A 1 IN 1 WITH 1 , and OO 1 || AA 1 . It follows that direct OO 1 parallel to each plane of the lateral face of the prism, and the length of the segment OO 1 equal to the height of the prism and, by convention, the diameter of the circle inscribed at the base of the prism. This means that the points of the segment OO 1 are equidistant from the lateral faces of the prism, and the middle F of the segment OO 1 , equidistant from the planes of the bases of the prism, will be equidistant from all faces of the prism. That is, F is the center of a sphere inscribed in a prism, and the diameter of this sphere is equal to the diameter of a circle inscribed in the base of the prism. The theorem is proven. Theorem 2 Let a circle be inscribed in the perpendicular section of an inclined prism, and the height of the prism is equal to the diameter of this circle. Then a sphere can be inscribed into this inclined prism. The center of this sphere divides the height passing through the center of a circle inscribed in a perpendicular section in half. Proof
Let ABC...A 1 IN 1 WITH 1 ... is an inclined prism and F is the center of a circle with radius FK inscribed in its perpendicular section. Since the perpendicular section of a prism is perpendicular to each plane of its lateral face, the radii of the circle inscribed in the perpendicular section drawn to the sides of this section are perpendicular to the lateral faces of the prism. Consequently, point F is equidistant from all lateral faces. Let us draw a straight line OO through point F 1 , perpendicular to the plane of the bases of the prism, intersecting these bases at points O and O 1 . Then OO 1 - prism height. Since according to the OO condition 1 = 2FK, then F is the middle of the segment OO 1 :FK = OO 1 / 2 = FO = FO 1 , i.e. point F is equidistant from the planes of all faces of the prism without exception. This means that a sphere can be inscribed into a given prism, the center of which coincides with point F - the center of a circle inscribed in that perpendicular section of the prism that divides the height of the prism passing through point F in half. The theorem is proven. Example 5. A ball of radius 1 is inscribed in a rectangular parallelepiped. Find the volume of the parallelepiped. SolutionDraw the top view. Or from the side. Or from the front. You will see the same thing - a circle inscribed in a rectangle. Obviously, this rectangle will be a square, and the parallelepiped will be a cube. The length, width and height of this cube are twice the radius of the sphere. AB = 2, and therefore the volume of the cube is 8. Answer: 8. Example 6. In a regular triangular prism with a side of the base equal to , there are two balls. The first ball is inscribed in the prism, and the second ball touches one base of the prism, its two side faces and the first ball. Find the radius of the second ball. Solution
Let ABCA 1 IN 1 WITH 1 - correct prism and points P and P 1 - the centers of its bases. Then the center of the ball O inscribed in this prism is the midpoint of the segment PP 1 . Consider the plane RVV 1 . Since the prism is regular, PB lies on the segment BN, which is the bisector and height ΔABC. Therefore, the plane and is the bisector plane of the dihedral angle at the lateral edge of the explosive 1 . Therefore, any point of this plane is equidistant from the side faces of AA 1 BB 1 and SS 1 IN 1 B. In particular, the perpendicular OK, lowered from point O to the face ACC 1 A 1 , lies in the RVV plane 1 and is equal to the segment OP. Note that KNPO is a square, the side of which is equal to the radius of the ball inscribed in a given prism. Let O 1 - the center of the ball touching the inscribed ball with center O and the lateral faces AA 1 BB 1 and SS 1 IN 1 Into prisms. Then point O 1 lies on the RVV plane 1 , and its projection P 2 on the plane ABC lies on the segment PB. According to the condition, the side of the base is equal , therefore, PN = 2 and therefore the radius of the ball OR inscribed in the prism is also equal to 2. Since the balls with centers at points O and O 1 touch each other, then the segment OO 1 = OR + O 1 R 2 . Let us denote OP = r, O 1 R 2 = x. Consider ΔOO 1 T, where In this triangle OO 1 = r + x, OT = r - x. That's why Since the figure is O 1 R 2 RT is a rectangle, then Further, by the property of medians of a triangle РВ = 2r, and Р 2 B = 2x, because in right triangle and P 2 L = x. Since PB = PP 2 + R 2 B, then we get the equation , from which, taking into account the inequality x< r, находим Substituting the value r = 2, we finally find Answer:Sphere circumscribed about a polyhedron
The sphere is said to be circumscribed about the polyhedron, if all its vertices lie on this sphere. In this case, the polyhedron is said to be inscribed in the sphere.From the definition it follows that if a polyhedron has a circumscribed sphere, then all its faces are inscribed polygons and, therefore, not every polyhedron has a circumscribed sphere around it. For example, an inclined parallelepiped does not have a circumscribed sphere, because It is impossible to describe a circle around a parallelogram. The center of a sphere circumscribed about a right prism is the middle of the segment connecting the centers of circles described about the bases of a right prism. Example 7. Find the radius of a sphere circumscribed about a cube if the volume of the cube is 27. Write the answer in the form Solution Volume of cube edge of the cube a = 3. According to the Pythagorean theorem, the diagonal of the cube Then we find the radius as half the diagonal of the cube: Let's write the answer in the form Answer: 1.5. Example 8. One of the bases of a regular triangular prism belongs to the great circle of a ball of radius R, and the vertices of the other base belong to the surface of this ball. Determine the height of the prism at which its volume will be greatest. Solution
Perpendicular to plane A 1 IN 1 WITH 1 drawn from the center of the circle circumscribed about this triangle passes through the center of the ball. Let us denote OB 1 = R, OB = R 1 , BB 1 = h = x. Then Let's find the derivative and equate it to zero. We get:Answer:

XV CITY OPEN CONFERENCE OF STUDENTS

"INTELLECTUALS OF THE XXI CENTURY"

Section: MATHEMATICS

The described area at the Olympiads and the Unified State Examination

Kiyaeva Anna Anatolevna

Orenburg – 2008

1.2 Scope described

1.2.1 Basic properties and definitions

1.2.2 Pyramid combination

1.2.3 Combination with prism

1.2.4 Combination with cylinder

1.2.5 Combination with cone

2 Examples of Olympiad tasks

2.1 Examples of Olympiad tasks with a pyramid

2.2 Examples of Olympiad tasks with a prism

2.3 Examples of Olympiad tasks with a cylinder

2.4 Examples of Olympiad tasks with a cone

3.3 Examples of Unified State Exam tasks with a cylinder

3.4 Examples of Unified State Exam tasks with a cone

Introduction

This work is being carried out as part of a project to create a mathematical page for schoolchildren on the website of the boarding lyceum and will be posted in the “Mathematical Methods” section.

Target work - creating a reference book dedicated to the solution method geometric problems with the described sphere at the Olympiads and the Unified State Examination.

To achieve this goal, we needed to solve the following tasks :

1) become familiar with the concept of the described sphere;

2) study the features of combinations of the described sphere with a pyramid, prism, cylinder and cone;

3) among geometric problems, select those that contain the condition for the presence of a described sphere;

4) analyze, systematize and classify the collected material;

5) make a selection of problems for independent solution;

6) present the research result in the form of an abstract.

During the research, we found out that problems with the described area are quite often offered to schoolchildren on the Unified State Exam, so the ability to solve problems of this type plays a very important role in successful completion exams. Also, problems with the described area are often found at mathematics olympiads at various levels. Relevant examples are given in our work. This topic is relevant, since tasks of this type usually cause difficulties for schoolchildren.

Practical significance– the materials we have prepared can be used in preparing schoolchildren for Olympiads, the Unified State Exam and subsequent studies at a university.

1 Sphere and ball

1.1 Sphere and ball: basic concepts and definitions

Sphere is a surface consisting of all points in space located at a given distance from a given point.

This point is called center of the sphere(dot ABOUT in Fig. 1), and this distance radius of the sphere. Any segment connecting the center and any point of the sphere is also called the radius of the sphere. A line segment connecting two points on a sphere and passing through its center is called sphere diameter(line segment DC in Fig. 1). Note that a sphere can be obtained by rotating a semicircle around its diameter.

Ball is called a body bounded by a sphere. The center, radius and diameter of a sphere are also called center , radius And ball diameter. Obviously, a ball of radius R centered at ABOUT contains all points in space that are located from the point ABOUT at a distance not exceeding R(including point ABOUT), and does not contain other points. Ball also called the figure of rotation of a semicircle around its diameter. Ball segment- part of the ball cut off from it by some plane. Every section of a ball by a plane is a circle. The center of this circle is the base of the perpendicular drawn from the center of the ball onto the cutting plane. The plane passing through the center of the ball is called diametrical plane. The section of a ball by the diametral plane is called big circle, and the section of the sphere is large circle. Ball sector – a geometric body that is obtained by rotating a circular sector with an angle less than 90° around a straight line containing one of the radii limiting the circular sector. The spherical sector consists of a spherical segment and a cone with a common base.

Surface area of a sphere:

S = 4π R 2 ,

Where R– radius of the ball, S- area of the sphere.

Sphere volume

Where V– volume of the ball

Ball sector volume

V – volume of the spherical segment.

Segmental surface area

- segment height, segmental surface area

Segment base radius

, - segment base radius, - segment height, 0<H < 2R .

Spherical surface area of a ball segment

- area of the spherical surface of the spherical segment.

In space for a ball and a plane, three cases are possible:

1) If the distance from the center of the ball to the plane is greater than the radius of the ball, then the ball and plane do not have common points.

2) If the distance from the center of the ball to the plane is equal to the radius of the ball, then the plane has only one common point with the ball and the sphere bounding it.

3) If the distance from the center of the ball to the plane is less than the radius of the ball, then the intersection of the ball with the plane is a circle. The center of this circle is the projection of the center of the ball onto a given plane. The intersection of the plane with the sphere is the circumference of the specified circle.

1.2 Described sphere

1.2.1 Definitions and properties

The sphere is called described around the polyhedron(and the polyhedron is included in the sphere), if all the vertices of the polyhedron lie on the sphere.

Two facts follow from the definition of the described sphere:

1) all vertices of a polyhedron inscribed in a sphere are equidistant from a certain point (from the center of the circumscribed sphere);

2) each face of a polyhedron inscribed in a sphere is a polygon inscribed in a certain circle, precisely in the circle that is obtained in the section of the sphere by the plane of the face; in this case, the base of the perpendiculars lowered from the center of the circumscribed sphere on the plane of the faces are the centers of circles circumscribed about the faces.

Theorem 1 . A sphere can be described around a polyhedron if and only if any of the following conditions are met:

a) a circle can be described around any face of a polyhedron, and the axes of the circles described around the faces of the polyhedron intersect at one point;

b) planes perpendicular to the edges of the polyhedron and passing through their midpoints intersect at one point;

c) there is a single point equidistant from all vertices of the polyhedron.

Proof.

Necessity. Let a sphere be described around the polyhedron. Let us prove that condition a) is satisfied. Indeed, since the plane of a given face of a polyhedron intersects a sphere along a circle, then the vertices of the face belonging to the sphere and the plane of the face belong to the line of their intersection - the circle. Since the center of the sphere is equidistant from all vertices of a given face, it lies on a perpendicular to this face drawn through the center of the circle circumscribed about the face.

Adequacy. Let condition a) be satisfied. Let us prove that a sphere can be described around a polyhedron. In fact, since the common point of perpendiculars to the faces drawn through the centers of the circles circumscribed about the faces is equidistant from all the vertices of the polyhedron, a sphere with the center at this point is described around the polyhedron.

Condition a) in this case is equivalent to conditions b) and c).

If a sphere is circumscribed about a polyhedron, then: a) the base of a perpendicular dropped from the center of the sphere to any face is the center of a circle circumscribed about this face (like the base of the height of a pyramid with equal lateral edges - the radii of the sphere drawn from its center to the vertices of a given face ); b) the center of a sphere circumscribed about a polyhedron can be located inside the polyhedron, on its surface (at the center of a circle circumscribed about a face, in particular, in the middle of some edge), outside the polyhedron.

1.2.2 Circumscribed sphere and pyramid

Theorem 2 . A sphere can be described around a pyramid if and only if a circle can be described around its base.

Proof. Let a circle be described around the base of the pyramid. Then this circle and a point outside the plane of this circle - the top of the pyramid - define a single sphere, which will be circumscribed around the pyramid. And back. If a sphere is circumscribed about a pyramid, then the section of the sphere by the plane of the base of the pyramid is a circle circumscribed about the base.

Corollary 1. A sphere can be described around any tetrahedron.

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. The methodological recommendations for this textbook (see the book “Studying Geometry in Grades 10–11” by S.M. Sahakyan and V.F. Butuzov, p. 159) say what combinations of bodies are considered when solving problems No. 629–646 , and attention is drawn 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.

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 called circumscribed about a polyhedron, and a polyhedron inscribed in a ball, if the surface of the ball passes through all the vertices of the polyhedron.

(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.

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.

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.

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 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 in such a 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 sphere 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 generator 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: