I (radius=r) is the incircle of triangle ABC (right angled at B) and O (radius=R) is another circle which touches the extended side BC and AC. Find the ratio (R/r) if the point of intersection of circles I and O is midpoint of AC.
In order for the incircle to touch the hypotenuse, AC, at its midpoint, the triangle must be an isosceles right triangle.
Sides BC and BA can be extended far enough to form sides of a square with the center of circle O as the fourth corner. The diagonal of this square, starting at the center of circle O, consists of three parts: a radius of circle O extending to the point of tangency of circle O with hypotenuse AC and with circle I, of length R; the radius of circle I to that point, of length r; and the diagonal of a smaller square, with side of length r, making the diagonal of length r*sqrt(2). Thus the entire diagonal has length R + r*(1 + sqrt(2)).
The side of the square has length R, so its diagonal must be of length R*sqrt(2).
R + r*(1+sqrt(2)) = R*sqrt(2)
1 + (r/R)*(1+sqrt(2)) = sqrt(2)
(r/R)*(1+sqrt(2)) = sqrt(2) - 1
r/R = (sqrt(2) - 1) / (sqrt(2) + 1)
R/r = (sqrt(2) + 1) / (sqrt(2) - 1) ~= 5.828427124746188
Posted by Charlie
on 2007-08-20 11:48:05