Circle reasoning
A big circle is painted in the corner of a room. It touches the two walls. If you were to paint a small circle that touches the big circle and the two walls, how big would it be?
Problem source: dansmath.
A big circle is painted in the corner of a room. It touches the two walls. If you were to paint a small circle that touches the big circle and the two walls, how big would it be?
Problem source: dansmath.
March 3rd, 2010 at 12:35 am
If the big circle has radius R, and the small circle has radius r, then the small circle has its center at the point (r,r) from the corner, and a distance of r+R along the line from the corner to the center of the large circle. By inspection, we have
R sqrt(2) = R + r(1+sqrt(2))
and r = R[ (sqrt(2)-1)/(sqrt(2)+1) ]
Of course, we could repeat the process with smaller circles still, recursively. Will the centers ever make it to the corner, or do they stop short? If so, where?
March 3rd, 2010 at 1:48 am
Last question: The centers will never make it to the corner and they will never stop aproaching. There are an infinite number of smaller circles.
March 3rd, 2010 at 3:39 am
Nothing to add to Mike Anderson’s reply re the geometry, and only this as to the algebra:
r/R = (sqrt(2)-1)/(sqrt(2)+1)
=(sqrt(2)-1)²/[(sqrt(2)+1)(sqrt(2)-1)]
=(2-2 sqrt(2)+1)/(2-1)
=3-2 sqrt(2)
March 3rd, 2010 at 9:35 pm
I am a visual learner and approached this in what appears to be a slightly different way. Same result though. Here’s my approach (link):
http://s3.amazonaws.com/ember/yQrYOSjYuyfF0ojrdCAC42MZBu1zyxK0_o.png
In response to Mike’s follow up question (do they ever stop short and if so, where?), lim(x>?) [3-2sqrt(2)]^x = 0. Obviously as has already been stated, it never truly gets there, but it does approach 0 very quickly.
If I can get past this cold that’s been driving me crazy for a couple of days, I’d like to find the area of the square (0,0) (0,R) (R,R) (R,0) that falls outside all circles to infinity.
March 3rd, 2010 at 9:38 pm
The question marks in my comment are “infinity”. I guess infinity isn’t an allowable character.
March 4th, 2010 at 5:34 am
Richard:
Thanks for the reduction; some of us statistics instructors have a bad? weird? habit of not reducing fractions.
MM:
The infinite sequence of circles have radii in the ratio of a=3-2 sqrt(2), so they have areas in the ratio of a^2. This means that the sum of their areas is just a geometric series, giving 1/(1-a^2). However, you want the uncovered area of the upper-left quadrant, which only contains 1/4 of the original circle. So adjust for that, and the fact that our geometric series starts at 1 (rather than 0) to get
A = R^2 – pi R^2 / 4 – R^2 a^2/(1-a^2) = R(1-pi/4 – a^2/(1-a^2))
which is about 18.4% of the RxR square.