Semiperimeter

Abstract: Heron of Alexandria showed that the areaK of a triangle with sidesa,b, andc is given by $$K = \sqrt {s(s - a)(s - b)(s - c)} ,$$ wheres is the semiperimeter (a+b+c)/2. Brahmagupta gave a generalization to quadrilaterals inscribed in a circle. In this paper we derive formulas giving the areas of a pentagon or hexagon inscribed in a circle in terms of their side lengths. While the pentagon and hexagon formulas are complicated, we show that each can be written in a surprisingly compact form related to the formula for the discriminant of a cubic polynomial in one variable.

Abstract: Heron's remarkable formula, K = ^s(s ? a)(s ? b)(s ? c), for the area K of a triangle with side lengths a, b, and c, and semiperimeter s = (a -\-b + c)/2, can be proved by a number of methods. Algebraic, geometric, trigonometric and functiontheoretic proofs can be found in [1], [2], [4], [5], [6], [7], [9], [10], [12], and [14]. The purpose of this Capsule is to use "proofs without words" to establish two lemmas (which are of interest in their own right) that reduce the proof of Heron's formula to elementary algebra. It is based on the proof found in [2] (which reappears in [4] and [10]). Let AABC be a triangle with sides a, b, c, as in Figure 1(a), and bisect each angle to locate the center of the incircle (as did Heron). Extending an inradius (length r) to each side now partitions the triangle into six smaller right triangles, with side lengths as indicated in Figure 1(b).

Abstract: We consider the problem of computing a minimum perimeter triangle enclosing a convex polygon This problem defied a linear-time solution due to the absence of a property called the interspersing property This property was crucial in the linear-time solution for the minimum area triangle enclosing a convex polygon We have discovered a non-trivial interspersing property for the minimum perimeter problem This resulted in an optimal solution to the minimum perimeter triangle problem