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In an equilateral triangle the area of the Malfatti circles (left) is approximately 1% smaller than the three area-maximizing circles (right). Gian Francesco Malfatti posed the problem of cutting three cylindrical columns out of a triangular prism of marble, maximizing the total volume of the columns. He assumed that the solution to this ...
Napoleon's theorem: If the triangles centered on L, M, N are equilateral, then so is the green triangle.. In geometry, Napoleon's theorem states that if equilateral triangles are constructed on the sides of any triangle, either all outward or all inward, the lines connecting the centres of those equilateral triangles themselves form an equilateral triangle.
Circle packing in an equilateral triangle is a packing problem in discrete mathematics where the objective is to pack n unit circles into the smallest possible equilateral triangle. Optimal solutions are known for n < 13 and for any triangular number of circles, and conjectures are available for n < 28. [1] [2] [3]
In geometry, Lemoine's problem is a straightedge and compass construction problem posed by French mathematician Émile Lemoine in 1868: [1] [2]. Given one vertex of each of the equilateral triangles placed on the sides of a triangle, construct the original triangle.
An equilateral triangle is a triangle in which all three sides have the same length, and all three angles are equal. Because of these properties, the equilateral triangle is a regular polygon, occasionally known as the regular triangle. It is the special case of an isosceles triangle by modern definition, creating more special properties.
For any interior point P, the sum of the lengths of the perpendiculars s + t + u equals the height of the equilateral triangle.. Viviani's theorem, named after Vincenzo Viviani, states that the sum of the shortest distances from any interior point to the sides of an equilateral triangle equals the length of the triangle's altitude. [1]
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 ...
Given an equilateral triangle ABC in the plane, and a point P in the plane of the triangle ABC, the lengths PA, PB, and PC form the sides of a (maybe, degenerate) triangle. [1] [2] Proof of Pompeiu's theorem with Pompeiu triangle ′ The proof is quick. Consider a rotation of 60° about the point B.