16 Can't see a neat geometric solution to this, but here's a co-ordinate geometry one.
Set the triangle up so that B is at the origin, BA lies along the x-axis and BC along the y-axis.
Then, AX^2 + BX^2 + CX^2 = {(x-3)^2 + (y-0)^2} + {(x-0)^2 + (y-0)^2} + {(x-0)^2 + (y-4)^2}
= 3x^2 + 3y^2 - 6x - 8y + 25
and, (completing the square on both x and y),
= 3(x - 1)^2 + 3(y - 4/3)^2 + 50/3,
so the minimum will be 50/3 when x = 1 and y = 4/3.
Tiggsy
