Polyhedra inscribed in a sphere. Sphere inscribed in a polyhedron Setting homework
Polyhedra inscribed in a sphere A polyhedron is said to be inscribed in a sphere if all its vertices belong to this sphere. The sphere itself is said to be circumscribed about the polyhedron. Theorem. A sphere can be described around a pyramid if and only if a circle can be described around the base of this pyramid.
Polyhedra inscribed in a sphere Theorem. A sphere can be described near a straight prism if and only if a circle can be described near the base of this prism. Its center will be point O, which is the midpoint of the segment connecting the centers of the circles described around the bases of the prism. The radius of the sphere R is calculated by the formula where h is the height of the prism, r is the radius of the circle circumscribed around the base of the prism.
Exercise 3 The base of the pyramid is a regular triangle, the side of which is equal to 3. One of the side edges is equal to 2 and is perpendicular to the plane of the base. Find the radius of the circumscribed sphere. Solution. Let O be the center of the circumscribed sphere, Q the center of the circumscribed circle around the base, E the midpoint of SC. Quadrilateral CEOQ is a rectangle in which CE = 1, CQ = Therefore, R=OC=2. Answer: R = 2.
Exercise 4 The figure shows the pyramid SABC, for which the edge SC is equal to 2 and is perpendicular to the plane of the base ABC, the angle ACB is equal to 90 o, AC = BC = 1. Construct the center of the sphere circumscribed about this pyramid and find its radius. Solution. Through the middle D of edge AB we draw a line parallel to SC. Through the middle E of edge SC we draw a straight line parallel to CD. Their intersection point O will be the desired center of the circumscribed sphere. In the right triangle OCD we have: OD = CD = By the Pythagorean theorem, we find
Exercise 5 Find the radius of a sphere circumscribed about a regular triangular pyramid, the side edges of which are equal to 1, and the plane angles at the apex are equal to 90 degrees. Solution. In the tetrahedron SABC we have: AB = AE = SE = In the right triangle OAE we have: Solving this equation for R, we find
Exercise 4 Find the radius of a sphere circumscribed about a right triangular prism, at the base of which is a right triangle with legs equal to 1, and the height of the prism equal to 2. Answer: Solution. The radius of the sphere is equal to half the diagonal A 1 C of the rectangle ACC 1 A 1. We have: AA 1 = 2, AC = Therefore, R =
Exercise Find the radius of a sphere circumscribed about a regular 6-gonal pyramid, the edges of which are equal to 1, and the side edges are equal to 2. Solution. Triangle SAD is equilateral with side 2. The radius R of the circumscribed sphere is equal to the radius of the circle circumscribed about triangle SAD. Hence,
Exercise Find the radius of the sphere circumscribed about the unit icosahedron. Solution. In rectangle ABCD, AB = CD = 1, BC and AD are the diagonals of regular pentagons with sides 1. Therefore, BC = AD = By the Pythagorean theorem, AC = The required radius is equal to half of this diagonal, i.e.
Exercise Find the radius of a sphere circumscribed about a unit dodecahedron. Solution. ABCDE is a regular pentagon with side In the rectangle ACGF AF = CG = 1, AC and FG are the diagonals of the pentagon ABCDE and, therefore, AC = FG = By the Pythagorean theorem FC = The required radius is equal to half of this diagonal, i.e.
Exercise The figure shows a truncated tetrahedron obtained by cutting off the corners of a regular tetrahedron of triangular pyramids, the faces of which are regular hexagons and triangles. Find the radius of a sphere circumscribed about a truncated tetrahedron whose edges are equal to 1.
Exercise The figure shows a truncated octahedron obtained by cutting off triangular pyramids from the corners of the octahedron, the faces of which are regular hexagons and triangles. Find the radius of the sphere circumscribed about a truncated octahedron whose edges are equal to 1. Exercise The figure shows a truncated icosahedron obtained by cutting off the corners of the icosahedron of pentagonal pyramids, the faces of which are regular hexagons and pentagons. Find the radius of a sphere circumscribed about a truncated icosahedron whose edges are equal to 1.
Exercise The figure shows a truncated dodecahedron obtained by cutting off triangular pyramids from the corners of the dodecahedron, the faces of which are regular decagons and triangles. Find the radius of a sphere circumscribed about a truncated dodecahedron whose edges are equal to 1.
Exercise Find the radius of a sphere circumscribed about a unit cuboctahedron. Solution. Recall that a cuboctahedron is obtained from a cube by cutting off regular triangular pyramids with vertices at the vertices of the cube and side edges equal to half the edge of the cube. If the edge of the octahedron is equal to 1, then the edge of the corresponding cube is equal to The radius of the circumscribed sphere is equal to the distance from the center of the cube to the middle of its edge, i.e. is equal to 1. Answer: R = 1.
Open lesson on the topic “Inscribed and circumscribed polyhedra”
Lesson topic: A sphere inscribed in a pyramid. A sphere described near a pyramid.
Lesson type: Lesson on introducing new material. Lesson objectives:- Introduce the concept of a sphere inscribed in a polyhedron; sphere circumscribed about the polyhedron. Compare the circumcircle and the circumscribed sphere, the inscribed circle and the inscribed sphere. Analyze the conditions for the existence of an inscribed sphere and a circumscribed sphere. Develop problem solving skills on the topic. Developing student skills independent work.
Development of logical thinking, algorithmic culture, spatial imagination, development of mathematical thinking and intuition, creativity at the level necessary for continuing education and for independent activity in the field of mathematics and its applications in future professional activities;
- Interactive whiteboard
Presentation “Inscribed and described sphere”
Conditions of the problems in the drawings on the board. Handouts (supporting notes).
- Planimetry. Inscribed and circumscribed circle. Stereometry. Inscribed sphere Stereometry. Described sphere
- Setting lesson goals (2 minutes). Preparation for learning new material by repetition (frontal survey) (6 minutes). Explanation of new material (15 minutes) Understanding the topic by independently compiling notes on the topic “Stereometry. Described area” and application of the topic in solving problems (15 minutes). Summing up the lesson by checking knowledge and understanding of the topic studied (frontal survey). Evaluating student responses (5 minutes). Staging homework(2 minutes). Reserve jobs.
- Introduce the concept of a sphere inscribed in a polyhedron; sphere circumscribed about the polyhedron. Compare the circumcircle and the circumscribed sphere, the inscribed circle and the inscribed sphere. Analyze the conditions for the existence of an inscribed sphere and a circumscribed sphere. Develop problem solving skills on the topic.
- What circle is called inscribed in a polygon? What is the name of the polygon in which a circle is inscribed? Which point is the center of a circle inscribed in a polygon? What property does the center of a circle inscribed in a polygon have? Where is the center of a circle inscribed in a polygon? Which polygon can be described around a circle, under what conditions?
- What circle is called the circumscribed circle of a polygon? What is the name of the polygon around which the circle is circumscribed? What point is the center of the circle circumscribed about the polygon? What property does the center of a circle circumscribed about a polygon have? Where can the center of a circle circumscribed about a polygon be located? Which polygon can be inscribed in a circle and under what conditions?
- Formulate the definition of a sphere inscribed in a polyhedron. What is the name of a polyhedron into which a sphere can be inscribed? What property does the center of a sphere inscribed in a polyhedron have? What represents the set of points in space equidistant from the faces of a dihedral angle? (trihedral angle?) Which point is the center of a sphere inscribed in a polyhedron? In which polyhedron can a sphere be inscribed, under what conditions?
In a regular quadrangular pyramid, the side of the base is equal to A, the height is h. Find the radius of the sphere inscribed in the pyramid.
D. Students solve the problem.
Task. In a regular triangular pyramid, the side of the base is 4, the side faces are inclined to the base at an angle of 60 0. Find the radius of the sphere inscribed in this pyramid.
4. Understanding the topic when independently compiling notes on “Sphere circumscribed about a polyhedron"and application in problem solving.
A. U Students independently fill out notes on the topic “A sphere described around a polyhedron.” Answer the following questions:
Formulate the definition of a sphere circumscribed about a polyhedron.
What is the name of the polyhedron around which a sphere can be described?
What property does the center of a sphere circumscribed about a polyhedron have?
What is the set of points in space that are equidistant from two points?
Which point is the center of the sphere circumscribed about the polyhedron?
Where can the center of the sphere described around the pyramid be located? (polyhedron?)
Around which polyhedron can a sphere be described?
IN. Students solve the problem independently.
Task. In a regular triangular pyramid, the side of the base is equal to 3, and the side ribs are inclined to the base at an angle of 60 0. Find the radius of the sphere circumscribed around the pyramid.
WITH. Checking the compiled outline and analyzing the solution to the problem.
5. Summing up the lesson by checking knowledge and understanding of the topic studied (frontal survey). Evaluating student responses.
A. Students independently summarize the lesson.
IN. Answer additional questions.
Is it possible to describe a sphere around a quadrangular pyramid, at the base of which lies a rhombus that is not a square?
Is it possible to describe a sphere around a rectangular parallelepiped? If so, where is its center?
Where the theory learned in class is applied in real life (architecture, cellular telephony, geostationary satellites, GPS detection system).
6. Setting homework.
A. Make a note on the topic “A sphere described around a prism. A sphere inscribed in a prism." (Look at problems in the textbook: No. 632,637,638)
B. Solve problem No. 640 from the textbook.
S. From the manual of B.G. Ziv “Didactic materials on geometry grade 10” solve problems: Option No. 3 C12 (1), Option No. 4 C12 (1).
D. Additional task: Option No. 5 C12 (1).
7. Reserve tasks.
From the manual of B.G. Ziv “Didactic materials on geometry grade 10” solve problems: Option No. 3 C12 (1), Option No. 4 C12 (1).
Educational and methodological kit
Geometry, 10-11: Textbook for educational institutions. Basic and profile levels/ L.S. Atanasyan, V.F. Butuzov, S.B. Kadomtsev et al., M.: Education, 2010.
B.G. Ziv “Didactic materials on geometry grade 10”, M.: Education.
Math teacher
GBOU lyceum-boarding school "DPC"
Nizhny Novgorod
Polyhedra inscribed in a sphere A convex polyhedron is called inscribed if all its vertices lie on some sphere. This sphere is called described for a given polyhedron. The center of this sphere is a point equidistant from the vertices of the polyhedron. It is the point of intersection of planes, each of which passes through the middle of the edge of the polyhedron perpendicular to it.
Formula for finding the radius of a circumscribed sphere Let SABC be a pyramid with equal lateral edges, h is its height, R is the radius of the circle circumscribed around the base. Let's find the radius of the circumscribed sphere. Note the similarity of right triangles SKO1 and SAO. Then SO 1 /SA = KS/SO; R 1 = KS · SA/SO But KS = SA/2. Then R 1 = SA 2 /(2SO); R 1 = (h 2 + R 2)/(2h); R 1 = b 2 /(2h), where b is a side edge.
A parallelepiped inscribed in a sphere Theorem: A sphere can be described around a parallelepiped if and only if the parallelepiped is rectangular, since in in this case it is straight and around its base - a parallelogram - a circle can be described (since the base is a rectangle).
Problem 1 Find the radius of a sphere circumscribed about a regular tetrahedron with edge a. Solution: SO 1 = SA 2 /(2SO); SO = = = a SO 1 = a 2 /(2 a) = a /4. Answer: SO 1 = a /4. Let us first construct an image of the center of a circumscribed ball using the image of a regular tetrahedron SABC. Let's draw the apothems SD and AD (SD = AD). In the isosceles triangle ASD, each point of the median DN is equidistant from the ends of the segment AS. Therefore, point O 1 is the intersection of the height SO and the segment DN. Using the formula from R 1 = b 2 /(2h), we get:
Problem 2 Solution: Using the formula R 1 =b 2 /(2h) to find the radius of the circumscribed ball, we find SC and SO. SC = a/(2sin(α /2)); SO 2 = (a/(2sin(α /2)) 2 – (a /2)2 = = a 2 /(4sin 2 (α /2)) – 2a 2 /4 = = a 2 /(4sin 2 ( α /2)) · (1 – 2sin 2 (α /2)) = = a 2 /(4sin 2 (α /2)) · cos α In a regular quadrangular pyramid, the side of the base is equal to a, and the plane angle at the apex is equal to α . Find the radius of the circumscribed ball. R 1 = a 2 /(4sin 2 (α /2)) · 1/(2a/(2sin(α /2))) =a/(4sin(α /2) ·). : R 1 = a/(4sin(α /2) ·).
Polyhedra circumscribed about a sphere A convex polyhedron is called circumscribed if all its faces touch some sphere. This sphere is called inscribed for a given polyhedron. The center of an inscribed sphere is a point equidistant from all faces of the polyhedron.
Position of the center of an inscribed sphere Concept of a bisector plane of a dihedral angle. A bisector plane is a plane that divides a dihedral angle into two equal dihedral angles. Each point of this plane is equidistant from the faces of the dihedral angle. In the general case, the center of a sphere inscribed in a polyhedron is the intersection point of the bisector planes of all dihedral angles of the polyhedron. It always lies inside the polyhedron.
A pyramid circumscribed around a ball A ball is said to be inscribed in an (arbitrary) pyramid if it touches all faces of the pyramid (both lateral and base). Theorem: If the side faces are equally inclined to the base, then a ball can be inscribed in such a pyramid. Since the dihedral angles at the base are equal, their halves are also equal and the bisectors intersect at one point at the height of the pyramid. This point belongs to all bisector planes at the base of the pyramid and is equidistant from all faces of the pyramid - the center of the inscribed ball.
Formula for finding the radius of an inscribed sphere Let SABC be a pyramid with equal lateral edges, h is its height, r is the radius of the inscribed circle. Let's find the radius of the circumscribed sphere. Let SO = h, OH = r, O 1 O = r 1. Then, by the property of the bisector of the internal angle of a triangle, O 1 O/OH = O 1 S/SH; r 1 /r = (h – r 1)/; r 1 = rh – rr 1 ; r 1 · (+ r) = rh; r 1 = rh/(+ r). Answer: r 1 = rh/(+ r).
A parallelepiped and a cube described around a sphere Theorem: A sphere can be inscribed into a parallelepiped if and only if the parallelepiped is straight and its base is a rhombus, and the height of this rhombus is the diameter of the inscribed sphere, which, in turn, is equal to the height of the parallelepiped. (Of all the parallelograms, only a circle can be inscribed in a rhombus) Theorem: A sphere can always be inscribed in a cube. The center of this sphere is the point of intersection of the diagonals of the cube, and the radius is equal to half the length of the edge of the cube.
Combinations of figures Inscribed and circumscribed prisms A prism circumscribed about a cylinder is a prism whose base planes are the planes of the bases of the cylinder, and the side faces touch the cylinder. A prism inscribed in a cylinder is a prism whose base planes are the planes of the bases of the cylinder, and the side edges are the generators of the cylinder. A tangent plane to a cylinder is a plane passing through the generatrix of the cylinder and perpendicular to the plane of the axial section containing this generatrix.
Inscribed and circumscribed pyramids A pyramid inscribed in a cone is a pyramid whose base is a polygon inscribed in the circle of the base of the cone, and the apex is the vertex of the cone. The lateral edges of a pyramid inscribed in a cone form the cone. A pyramid circumscribed around a cone is a pyramid whose base is a polygon circumscribed around the base of the cone, and the apex coincides with the apex of the cone. The planes of the side faces of the described pyramid are tangent to the plane of the cone. A tangent plane to a cone is a plane passing through the generatrix and perpendicular to the plane of the axial section containing this generatrix.
Other types of configurations A cylinder is inscribed in a pyramid if the circle of one of its bases touches all the lateral faces of the pyramid, and its other base lies on the base of the pyramid. A cone is inscribed in a prism if its vertex lies on the upper base of the prism, and its base is a circle inscribed in a polygon - the lower base of the prism. A prism is inscribed in a cone if all the vertices of the upper base of the prism lie on the lateral surface of the cone, and the lower base of the prism lies on the base of the cone.
Problem 1 In a regular quadrangular pyramid, the side of the base is equal to a, and the plane angle at the apex is equal to α. Find the radius of the ball inscribed in the pyramid. Solution: Let's express the sides of SOK in terms of a and α. OK = a/2. SK = KC cot(α /2); SK = (a · ctg(α /2))/2. SO = = (a/2) Using the formula r 1 = rh/(+ r), we find the radius of the inscribed ball: r 1 = OK · SO/(SK + OK); r 1 = (a/2) · (a/2) /((a/2) · ctg(α /2) + (a/2)) = = (a/2) /(ctg(α /2) + 1) = (a/2)= = (a/2) Answer: r 1 = (a/2)
Conclusion The topic “Polyhedra” is studied by students in grades 10 and 11, but in curriculum there is very little material on the topic “Inscribed and circumscribed polyhedra,” although it is of great interest to students, since the study of the properties of polyhedra contributes to the development of abstract and logical thinking, which will later be useful to us in study, work, and life. While working on this essay, we studied all the theoretical material on the topic “Inscribed and circumscribed polyhedra,” examined possible combinations of figures and learned to apply all the studied material in practice. Problems involving the combination of bodies are the most difficult question in the 11th grade stereometry course. But now we can say with confidence that we will not have problems solving such problems, since during our research work we established and proved the properties of inscribed and circumscribed polyhedra. Very often, students have difficulties when constructing a drawing for a problem in this topic. But, having learned that to solve problems involving the combination of a ball with a polyhedron, the image of the ball is sometimes unnecessary and it is enough to indicate its center and radius, we can be sure that we will not have these difficulties. Thanks to this essay, we were able to understand this difficult but very fascinating topic. We hope that now we will not have any difficulties in applying the studied material in practice.
Definition. The sphere is called inscribed in a polyhedron, if the planes of all faces of the polyhedron touch the sphere in wheelbarrows located inside these faces. In this case, the polyhedron is said to be circumscribed about a sphere.
Theorem 1.A sphere (ball) can be inscribed into an arbitrary tetrahedron.
The set of points equidistant from the lateral faces of the tetrahedron is the straight line of intersection of two bisector planes of dihedral angles at two lateral edges. This line will be intersected by the bisector plane of the dihedral angle at the base. The resulting point is equidistant from all faces of the tetrahedron.
In the tetrahedron ABCD, the planes CDN and ADM are bisector planes of the dihedral angles at the lateral edges CD and AD. They intersect along straight line OD. Plane AKC is the bisector plane of the dihedral angle at the base (edge AC). This plane will intersect the line OD at point S (P is the point of intersection of lines DM and KC, belonging to the planes AKC and ADM at the same time, therefore point S is the point of intersection of AP and OD), which will be a point equidistant from all faces of the tetrahedron and, therefore, will be the center of a sphere inscribed in the tetrahedron ABCD.
Example 1. Find the radius of a sphere inscribed in a regular tetrahedron.
Let's consider similar triangles DPS and DOK (by two angles: angle D is common, angles DPS and DOK are right angles).
Then PS:KO=DS:DK,
taking into account that PS=r=SO and DS=DO-SO=DO-r,
, , That 
.
Answer: the radius of a sphere inscribed in a regular tetrahedron is equal to
Theorem 2. IN correct pyramid you can enter a sphere.
Theorem 3. A sphere can be inscribed into a regular truncated pyramid if and only if its apothem is equal to the sum of the radii of the circles inscribed at its bases.
Theorem 4. A sphere can be inscribed into any prism if a circle can be inscribed into its perpendicular section, the radius of which is equal to half the height of the prism.
Theorem 5. A sphere can be inscribed into a regular prism if and only if the height of the prism is equal to the diameter of the circle inscribed at its base.
Spheres described around a cylinder, cone and
Truncated cone.
Definition. The sphere is called described about the cylinder or truncated cone, if all points of the base circles belong to the sphere; The sphere is called described around the cone, if all points of the base circle, as well as the vertex of the cone, belong to the sphere.
In these cases, the cylinder, truncated cone, or cone is said to be inscribed in a sphere.
Theorem 1.A sphere can be described around an arbitrary cylinder.
O 1 and O 2 are the centers of the lower and upper bases, respectively. Straight line O 1 O 2 is perpendicular to the planes of the base. Let us draw a plane passing through the middle of the generatrix of the cylinder perpendicular to this generatrix. This plane will be parallel to the planes of the base and intersect the straight line O 1 O 2 at point O, which will be the center of the sphere circumscribed about the cylinder. The distance from point O to all points of the base will be equal, since O 1 O 2 is the GMT, equidistant from the circle (a straight line passing through the center of the circle and perpendicular to the plane of the circle). This means that point O is the center of a sphere with radius OA, circumscribed around a cylinder.
Theorem 2. A sphere can be described around a truncated cone.
O 1 and O 2 are the centers of the lower and upper bases, respectively. Straight line O 1 O 2 is perpendicular to the planes of the base. Let us consider the generatrix of a truncated cone AB. Let's find the GMT, equidistant from wheelbarrows A and B. They will be a plane passing through point P - the middle of AB and perpendicular to this straight line. This plane will intersect O 1 O 2 at point O, which will be equidistant from points A and B. It is also obvious that point O will be equidistant from all points of the bases of the truncated cone. Consequently, this point O will be the center of a sphere with radius OA, described around a truncated cone.
Theorem 3. A sphere can be described around a cone.
Similar to the previous theorem, OA is the height of the cone, which is the HMT, equidistant from the circle. Let's consider the generatrix AB and find the GMTs equidistant from A and B. The resulting plane (according to the previous problem) will intersect OA at point O 1, which will be equidistant from points A and B, as well as from any points of the base of the cone. Thus, we obtained that point O 1 is the center of a sphere with radius O 1 A, described around a cone.
GEOMETRY
Section II. STEREOMETRY
§23. COMBINATIONS OF GEOMETRIC BODIES.
5. Polyhedron inscribed in a sphere.
A polyhedron is said to be inscribed in a ball if all its vertices lie on the surface of the ball.
In this case, the ball is called circumscribed around the polyhedron.
The main properties of a prism inscribed in a ball are as follows (Fig. 511):
1) A sphere can be described around a straight prism if its base is a polygon around which a circle can be described.
2) The center of the ball is the midpoint of the height of the prism connecting the centers of the circles circumscribed around the polygons of the prism bases.
3) The bases of the prism are inscribed at the level of parallel sections of the ball.
Example 1. A sphere is described around a regular triangular prism, the side of which is 5 cm. The radius of the sphere is 13 cm. Find the height of the prism.
Solutions. 1) Let a sphere be described around a regular triangular prism ABCA I B 1 C 1 (Fig. 511).
2) QB = R ABC - radius of the circle described around∆ ABC. 
Where a = 5 cm - side of the base of regular triangle ABC.
Then 
3) In ∆ OQB: OB = R = 13 cm - radius of the ball, OQB = 90°.
We have
4) Since point O is the mid-height of the prism QQ 1 then QQ 1 = 2 ∙ 12 = 24 (cm).
The main properties of a pyramid inscribed in a ball are as follows (Fig. 512).
1) A ball can be described around a pyramid if its base is a polygon around which a circle can be described. The center of the sphere circumscribed around the pyramid lies on a perpendicular to the plane of the base drawn through the center of the circle circumscribed around the base.
2) The center of a ball circumscribed around a regular pyramid lies on a line containing the height of the pyramid.
3) The center of a ball circumscribed around a regular pyramid coincides with the center of a circle circumscribed around an isosceles triangle, the side of which is the lateral edge of the pyramid, and the height is the height of the pyramid. The radius of the ball is equal to the radius of this circle.
Note that the center of the circumscribed ball may belong to the height of the pyramid, or lie on its extension (that is, it is located either inside the pyramid or outside it). When solving problems using the method proposed below, there is no need to consider two cases. With the chosen untying method, the location of the center of the ball (inside or outside the pyramid) is not taken into account.
Example 2. Prove that the radius of the ball R , described around the correctpyramids can be found using the formula
where H is the height of the pyramid, r - the radius of the circle described around the base of the pyramid.
Solutions. 1) Let point O be the center of a sphere circumscribed correctly: a pyramid with a height Q K (Fig. 512). By condition Q K = I, KA = r - the radius of the circle described around the base.
2) Continue Q to the second intersection with the bullet at the point Q 1. Then QQ 1 = 2 R - diameter of the circle, and therefore Q A Q 1 = 90° and QQ 1 - hypotenuse of a right triangle Q A Q 1.
4) By the property of the leg of a right triangle in∆ Q A Q 1 we obtain A Q 2 = QQ 1 ∙ Q K, i.e. A Q 2 = 2 R ∙ N.
5) So, A Q 2 = H 2 + r 2 and A Q 2 = 2 R H. Hence H 2 + r 2 = 2 R H; R = (r 2 + H 2 )/2 H , which was what needed to be proven.
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