 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Find the Ratio (Posted on 2007-08-20) 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.

 See The Solution Submitted by Praneeth Rating: 3.2500 (4 votes) Comments: ( Back to comment list | You must be logged in to post comments.) solution | Comment 2 of 3 | 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 Please log in:

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