Two medians of a triangle have lengths of 12 and 15. How long is the third median when the area of the triangle is a maximum? (Try to solve this without calculus.)

Let the vertices of triangle ABC have coordinates A(0,0), B(2b,2c), and C(2a,0). Also let median AD=12 and median BE=15.

So (a+b)^2 + c^2 =144 --> b = -a + sqrt(144-c^2)

Also, (2b-a)^2 + 4c^2 = 225 -->

c =(3/(4a))sqrt(178a^2 - a^4 -1521) 

But area = 2ac = 1.5sqrt(-x^2 + 178x - 1521) where x=a^2.

The quadratic is maximized when x=89 --> a=sqrt(89), b=7/sqrt(89), and c=60/sqrt(89).

Since the third median has length

 sqrt(4a^2 + b^2 + c^2 - 4ab), it has length

sqrt(369) = 3sqrt(41).

Posted by Dennis on 2008-02-03 13:53:18