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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.
A circle (C 3) centered at B' with radius |B'B| meets the circle (C 2) at A'. A circle (C 4) centered at A' with radius |A'A| meets the circle (C 1) at E and E'. Two circles (C 5) centered at E and (C 6) centered at E' with radius |EA| meet at A and O. O is the sought center of |AD|. The design principle can also be applied to a line segment AD.
The same inversion transforms the third circle into another circle. The solution of the inverted problem must either be (1) a straight line parallel to the two given parallel lines and tangent to the transformed third given circle; or (2) a circle of constant radius that is tangent to the two given parallel lines and the transformed given circle.
The hexagonal packing of circles on a 2-dimensional Euclidean plane. These problems are mathematically distinct from the ideas in the circle packing theorem.The related circle packing problem deals with packing circles, possibly of different sizes, on a surface, for instance the plane or a sphere.
The outer coin makes two rotations rolling once around the inner coin. The path of a single point on the edge of the moving coin is a cardioid.. The coin rotation paradox is the counter-intuitive math problem that, when one coin is rolled around the rim of another coin of equal size, the moving coin completes not one but two full rotations after going all the way around the stationary coin ...
The most famous of these problems, squaring the circle, otherwise known as the quadrature of the circle, involves constructing a square with the same area as a given circle using only straightedge and compass. Squaring the circle has been proved impossible, as it involves generating a transcendental number, that is, √ π.
In general, the same inversion transforms the given circle C 1 and C 2 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 original Apollonius ...
Malfatti's assumption that the two problems are equivalent is incorrect. Lob and Richmond (), who went back to the original Italian text, observed that for some triangles a larger area can be achieved by a greedy algorithm that inscribes a single circle of maximal radius within the triangle, inscribes a second circle within one of the three remaining corners of the triangle, the one with the ...