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The reverse triangle inequality is an equivalent alternative formulation of the triangle inequality that gives lower bounds instead of upper bounds. For plane geometry, the statement is: [ 19 ] Any side of a triangle is greater than or equal to the difference between the other two sides .
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);
Barrow's proof of this inequality was published in 1937, as his solution to a problem posed in the American Mathematical Monthly of proving the Erdős–Mordell inequality. [1] This result was named "Barrow's inequality" as early as 1961. [4] A simpler proof was later given by Louis J. Mordell. [5]
Pages in category "Triangle inequalities" The following 8 pages are in this category, out of 8 total. This list may not reflect recent changes. *
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 ...
Erdős–Mordell inequality. Let be an arbitrary point P inside a given triangle , and let , , and be the perpendiculars from to the sides of the triangles. (If the triangle is obtuse, one of these perpendiculars may cross through a different side of the triangle and end on the line supporting one of the sides.)
The closely related Dido's problem asks for a region of the maximal area bounded by a straight line and a curvilinear arc whose endpoints belong to that line. It is named after Dido, the legendary founder and first queen of Carthage. The solution to the isoperimetric problem is given by a circle and was known already in Ancient Greece. However ...
Let G = (V,w) be an instance of the travelling salesman problem. That is, G is a complete graph on the set V of vertices, and the function w assigns a nonnegative real weight to every edge of G. 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).
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