ABC and XYZ are similar triangles and the circumcircle of the triangle XYZ is the incircle of the triangle ABC. If k = Area of ABC/Area of XYZ, then find the minimum value of k.

let the side lengths of XYZ be x,y,z
then its area is
(1/4)*sqrt[(x+y+z)(x-y+z)(y-x+z)(z-x+y)]

let the side lengths of ABC be a,b,c
since ABC and XYZ are similar then we have
a=tx b=ty c=tz
and the area of ABC is
(1/4)*sqrt[(a+b+c)(a-b+c)(b-a+c)(c-a+b)]
(t^2/4)*sqrt[(x+y+z)(a-b+c)(b-a+c)(c-a+b)]
thus k=t^2 so we wish to minimize t to minimize k

now from the equations for the radii of both the incircle and circumcircle we have
r=sqrt[(x+y+z)(x-y+z)(y-x+z)(z-x+y)]/(2x+2y+2z)
r=t(xyz)/sqrt[(x+y+z)(x-y+z)(y-x+z)(z-x+y)]
thus
(x-y+z)(y-x+z)(z-x+y)=2t(xyz)
and
t=(x-y+z)(y-x+z)(z-x+y)/(2xyz)
now here I simply used Mathematica's minimize function
with the constraints that x>0 y>=x z>=y x+y>z x+z>y and y+z>x to restrict it to values of x,y,z that give a valid triangle. And it gave a minimum for t of 1/2 when x=y=z=1
and this gives a minimum of k of 1/sqrt(2)

Posted by Daniel on 2009-05-26 12:39:20