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Packing circles in simple bounded shapes is a common type of problem in recreational mathematics. The influence of the container walls is important, and hexagonal packing is generally not optimal for small numbers of circles. Specific problems of this type that have been studied include: Circle packing in a circle; Circle packing in a square
Circle packing in a circle is a two-dimensional packing problem with the objective of packing unit circles into the smallest possible larger circle. Table of solutions, 1 ≤ n ≤ 20 [ edit ]
Square packing in a circle is a related problem of packing n unit squares into a circle with radius as small as possible. For this problem, good solutions are known for n up to 35. Here are the minimum known solutions for up to n =12: [ 11 ] (Only the cases n=1 and n=2 are known to be optimal)
A college student just solved a seemingly paradoxical math problem—and the answer came from an incredibly unlikely place. Skip to main content. 24/7 Help. For premium support please call: 800 ...
The related circle packing problem deals with packing circles, possibly of different sizes, on a surface, for instance the plane or a sphere. The counterparts of a circle in other dimensions can never be packed with complete efficiency in dimensions larger than one (in a one-dimensional universe, the circle analogue is just two points). That is ...
Pages in category "Unsolved problems in geometry" The following 48 pages are in this category, out of 48 total. This list may not reflect recent changes. A.
Many mathematical problems have been stated but not yet solved. These problems come from many areas of mathematics, such as theoretical physics, computer science, algebra, analysis, combinatorics, algebraic, differential, discrete and Euclidean geometries, graph theory, group theory, model theory, number theory, set theory, Ramsey theory, dynamical systems, and partial differential equations.
In general, the same inversion transforms the given line L and given circle C into two new circles, c 1 and c 2. Thus, the problem becomes that of finding a solution line tangent to the two inverted circles, which was solved above. There are four such lines, and re-inversion transforms them into the four solution circles of the Apollonius problem.