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Three examples of the triangle inequality for triangles with sides of lengths x, y, z.The top example shows a case where z is much less than the sum x + y of the other two sides, and the bottom example shows a case where the side z is only slightly less than x + y.
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);
Triangle inequality: If a, b, and c are the lengths of the sides of a triangle then the triangle inequality states that +, with equality only in the degenerate case of a triangle with zero area. In Euclidean geometry and some other geometries, the triangle inequality is a theorem about vectors and vector lengths :
This inequality fails for general triangles (to which Ono's original conjecture applied), as shown by the counterexample =, =, =, = / The inequality holds with equality in the case of an equilateral triangle , in which up to similarity we have sides 1 , 1 , 1 {\displaystyle 1,1,1} and area 3 / 4. {\displaystyle {\sqrt {3}}/4.}
Fuss' theorem for the relation among the same three variables in bicentric quadrilaterals; Poncelet's closure theorem, showing that there is an infinity of triangles with the same two circles (and therefore the same R, r, and d) Egan conjecture, generalization to higher dimensions; List of triangle inequalities
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 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 ...
For four points in order around a circle, Ptolemy's inequality becomes an equality, known as Ptolemy's theorem: ¯ ¯ + ¯ ¯ = ¯ ¯. In the inversion-based proof of Ptolemy's inequality, transforming four co-circular points by an inversion centered at one of them causes the other three to become collinear, so the triangle equality for these three points (from which Ptolemy's inequality may ...