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Circle packing in a circle – Two-dimensional packing problem; Circle packing in an equilateral triangle – Two-dimensional packing problem; Circle packing in an isosceles right triangle – Two-dimensional packing problem; Circle packing theorem – Describes the possible tangency relations between circles with disjoint interiors
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 ...
In this right triangle: sin A = a/h; cos A = b/h; tan A = a/b. Trigonometric ratios are the ratios between edges of a right triangle. These ratios depend only on one acute angle of the right triangle, since any two right triangles with the same acute angle are similar. [31]
The lune of Hippocrates is the upper left shaded area. It has the same area as the lower right shaded triangle. In geometry, the lune of Hippocrates, named after Hippocrates of Chios, is a lune bounded by arcs of two circles, the smaller of which has as its diameter a chord spanning a right angle on the larger circle.
Every triangle has an infinitude of inscribed ellipses. One of them is a circle, and one of them is the Steiner inellipse which is tangent to the triangle at the midpoints of the sides. Every acute triangle has three inscribed squares. In a right triangle two of them are merged and coincide with each other, so there are only two distinct ...
Malfatti's problem is to carve three cylinders from a triangular block of marble, using as much of the marble as possible. In 1803, Gian Francesco Malfatti conjectured that the solution would be obtained by inscribing three mutually tangent circles into the triangle (a problem that had previously been considered by Japanese mathematician Ajima Naonobu); these circles are now known as the ...
Given any triangle ABC, and any point M on BC, construct the incircle and circumcircle of the triangle. Then construct two additional circles, each tangent to AM, BC, and to the circumcircle. Then their centers and the center of the incircle are collinear. [3] [4] Until 2003, academia thought this third problem of Thébault the most difficult ...
This list of triangle topics includes things related to the geometric shape, either abstractly, as in idealizations studied by geometers, or in triangular arrays such as Pascal's triangle or triangular matrices, or concretely in physical space.