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In the mathematical field of topology, a section (or cross section) [1] of a fiber bundle is a continuous right inverse of the projection function. In other words, if E {\displaystyle E} is a fiber bundle over a base space , B {\displaystyle B} :
For instance, while all the cross-sections of a ball are disks, [2] the cross-sections of a cube depend on how the cutting plane is related to the cube. If the cutting plane is perpendicular to a line joining the centers of two opposite faces of the cube, the cross-section will be a square, however, if the cutting plane is perpendicular to a ...
The number of points (n), chords (c) and regions (r G) for first 6 terms of Moser's circle problem. In geometry, the problem of dividing a circle into areas by means of an inscribed polygon with n sides in such a way as to maximise the number of areas created by the edges and diagonals, sometimes called Moser's circle problem (named after Leo Moser), has a solution by an inductive method.
In algebraic geometry, a lemniscate (/ l ɛ m ˈ n ɪ s k ɪ t / or / ˈ l ɛ m n ɪ s ˌ k eɪ t,-k ɪ t /) [1] is any of several figure-eight or ∞-shaped curves. [ 2 ] [ 3 ] The word comes from the Latin lēmniscātus , meaning "decorated with ribbons", [ 4 ] from the Greek λημνίσκος ( lēmnískos ), meaning "ribbon", [ 3 ] [ 5 ...
In the tables of knots and links in Dale Rolfsen's 1976 book Knots and Links, extending earlier listings in the 1920s by Alexander and Briggs, the Borromean rings were given the Alexander–Briggs notation "6 3 2", meaning that this is the second of three 6-crossing 3-component links to be listed.
The inverse problem for earth sections is: given two points, and on the surface of the reference ellipsoid, find the length, , of the short arc of a spheroid section from to and also find the departure and arrival azimuths (angle from true north) of that curve, and . The figure to the right illustrates the notation used here.
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1. A cone and a cylinder have radius r and height h. 2. The volume ratio is maintained when the height is scaled to h' = r √ π. 3. Decompose it into thin slices. 4. Using Cavalieri's principle, reshape each slice into a square of the same area. 5. The pyramid is replicated twice. 6. Combining them into a cube shows that the volume ratio is 1:3.