Show that a triangle's perimeter is more than 10 times the radius of its incircle.

Let A, p, and r represent the area, perimeter, and radius of the triangle's incircle, respectively. Also, let a, b, and c represent the lengths of the triangle's sides.

Since r = 2A/p, p > 10r is equivalent to p^2 > 20A.

Since the arithmetic mean is greater or equal to the geometric mean, ( (p-2a)+(p-2b)+(p-2c) )/3 >= the cube root of ( (p-2a)(p-2b)(p-2c) ) --> p^3/27 >= (p-2a)(p-2b)(p-2c). 

From Heron's formula, it follows that

 16A^2 = p(p-2a)(p-2b)(p-2c)

So p^4/27 >= 16A^2 --> p^2 >= 12sqrt(3)A > 20A.

Posted by Dennis on 2008-02-26 14:33:10