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By Frederick A. Valentine

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During this quantity, that is devoted to H. Seifert, are papers in response to talks given on the Isle of Thorns convention on low dimensional topology held in 1982.

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The volumes of polyhedrons with equal functions S(t) are equal (by Cavalieri’s principle). It is easy to verify that any of the new simple polyhedrons can be split into tetrahedrons whose vertices lie in given planes. 4). 15. 4). 16. The projection to the plane perpendicular to given lines sends a, b and c into points A, B and C, respectively. Let s be the area of triangle ABC; KS the edge of the tetrahedron moving along line a. 5 the volume of the considered tetrahedron is equal to 31 sKS. 17.

Hence, the length of arc ⌣ CK is equal to 12 (πR ± l). 19. Let O be the center of the sphere. Take point E so that {CE} = {AB}. Since ∠OCE = 60◦ and CE = 1 = OC, it follows that OE = 1. , with the midpoint of segment BC. Segment OO1 is a midline of triangle CBD, therefore, BD = 2OO1 = 2 OC 2 − BC 2 =2 4 1− AB 2 + AC 2 = 1. 20. Let A and B be two points of the given circle, A1 and B1 be the other intersection points of lines P A and P B with the sphere; l the tangent to the circle circumscribed about triangle P AB at point P .

6 1 1 √ − 3 2 Therefore, . 36. Let us consider a cube with edge 2 2. A sphere of radius 2 whose center coincides with that of the cube √ is tangent to all its edges and its intersections with the faces are circles of radius 2. The surface of the sphere is divided by the surface of the cube into 6 spherical segments and 8 curvilinear triangles. Let x be the area of a spherical segment and y the area of a curvilinear triangle. Then the areas in question are equal to y and 16π − y − 3x, respectively, where 16π is the surface area of√the sphere of radius 2.

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