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The converse of the triangle inequality theorem is also true: if three real numbers are such that each is less than the sum of the others, then there exists a triangle with these numbers as its side lengths and with positive area; and if one number equals the sum of the other two, there exists a degenerate triangle (that is, with zero area ...
The parameters most commonly appearing in triangle inequalities are: the side lengths a, b, and c;; the semiperimeter s = (a + b + c) / 2 (half the perimeter p);; the angle measures A, B, and C of the angles of the vertices opposite the respective sides a, b, and c (with the vertices denoted with the same symbols as their angle measures);
The reverse inequality follows from the same argument as the standard Minkowski, but uses that Holder's inequality is also reversed in this range. Using the Reverse Minkowski, we may prove that power means with p ≤ 1 , {\textstyle p\leq 1,} such as the harmonic mean and the geometric mean are concave.
Rewriting the inequality above allows for a more concrete geometric interpretation, which in turn provides an immediate proof. [1]+ +. Now the summands on the left side are the areas of equilateral triangles erected over the sides of the original triangle and hence the inequation states that the sum of areas of the equilateral triangles is always greater than or equal to threefold the area of ...
Finding all right triangles with integer side-lengths is equivalent to solving the Diophantine equation + =.. In mathematics, a Diophantine equation is an equation, typically a polynomial equation in two or more unknowns with integer coefficients, for which only integer solutions are of interest.
3.2 Functions of several variables. 4 Operations that preserve convexity. ... Every norm is a convex function, by the triangle inequality and positive homogeneity.
The Alzheimer's Association released a list of gift recommendations for individuals with the disease at every stage of dementia. Experts offer insights on how to navigate the holiday amid dementia.
The minimum-weight perfect matching can have no larger weight, so w(M) ≤ w(C)/2. Adding the weights of T and M gives the weight of the Euler tour, at most 3w(C)/2. Thanks to the triangle inequality, even though the Euler tour might revisit vertices, shortcutting does not increase the weight, so the weight of the output is also at most 3w(C)/2 ...