Posted by dan on Wednesday, January 12, 2011 at 4:39am.
First, the first circle that "just fits" inside the rectangle is tangent to only three sides, the two long sides and one short side. Call its radius R, and the centre A.
The second circle will be tangent to the opposite short side, one of the long sides and the first circle. Call the radius of the second circle r, and the centre B.
Join the centres of the two circles, AB, and drop perpendiculars parallel to the sides of the rectangle until they meet. Call this point C.
Thus we have a right triangle ABC, right-angled at C.
We have the following information:
R=(short side)/2=1
AB=R+r
AC=R-r
BC=sqrt(AB²-AC²)
By equating the length of the long side with the sum of R,BC and r, you can solve for r:
R+BC+r = long side = 3
Post if you need a diagram.
Thanks for the reply.
I'm still stumbling over where point C is. "Drop perpendiculats parallel to the sides of the rectangle until they meet".
A diagram would most definately be helpful.
Once again, thank you for your help.
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