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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 ...
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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.
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 right side is the area of triangle ABC, but on the left side, r + z is at least the height of the triangle; consequently, the left side cannot be smaller than the right side. Now reflect P on the angle bisector at C. We find that cr ≥ ay + bx for P's reflection. Similarly, bq ≥ az + cx and ap ≥ bz + cy. We solve these inequalities for ...
According to the triangle inequality, for every three vertices u, v, and x, it should be the case that w(uv) + w(vx) ≥ w(ux). Then the algorithm can be described in pseudocode as follows. [1] Create a minimum spanning tree T of G. Let O be the set of vertices with odd degree in T. By the handshaking lemma, O has an even number of vertices.
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 ...
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.