The center of a circle having radius 1 is denoted by O. The triangle ABC is inscribed within the circle such that the respective areas of the circular segments described by the sides AB, BC and AC are in the ratio 3:4:5.

A circle with its center located at P is inscribed within the triangle.

Determine the distance OP.

I'm going to take my life in my hands and guess 0.2:

For the ratios of the circular segments to be so nearly the same, the triangle must very nearly be equilateral. If we take the middle side to be approximately 1.7, the required condition is satisfied if the other sides are approximately 1.6 and 1.8. Now, the product of the incircle radius and circumcircle radius of a triangle with sides a,b,c is (a*b*c)/(2(a+b+c)) = .48, while the distance, d, between the circumcentre and the incentre is: d^2 = R(R-2r) ; here d^2 = 1(1-.96), so d = 0.2

I won't be surprised to be proved wrong!

 

Posted by broll on 2010-09-19 12:25:56