 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Circles Abound (2) (Posted on 2004-10-17) I've a black circle of radius 5.

I wish to create 5 identical white circles of some lesser radius, which I will place on top of the black circle and completely obscure (cover) the black circle.

What is the smallest radius which I can make the smaller circles and still meet my requirement?

 No Solution Yet Submitted by SilverKnight Rating: 3.3333 (3 votes) Comments: ( Back to comment list | You must be logged in to post comments.) I think this is right | Comment 4 of 11 | Well, I�m not sure if this is the best arrangement, but I am going to place my circles such that each one is tangent to the center of the black circle. Also, they will be evenly spaced around the circle � in other words, where one circle intersects the black circle, the neighboring circle intersects the black circle at that same exactly point.

This means that there are 5 intersection points around the black circle. These are the vertices of a regular pentagon that is inscribed inside the black circle. So what I want to find is the radius of a circle that passes through three points: the center of the black circle, and two adjacent vertices of the inscribed pentagon.

Let�s call the center of the black circle O, and two adjacent vertices of the inscribed pentagon A and B. Now I�m going to focus my attention on Triangle OAB. Notice that OA and OB are both length R, so Triangle OAB is isosceles. Also, since the inscribed pentagon was regular, this means that Angle AOB is simply 360/5 = 72.

Well, we are trying to find a circle that passes through these 3 points. That means the circle will circumscribe Triangle OAB. Let�s call the midpoint of AB point M. By symmetry, the center of the circle is somewhere on the line OM. Notice that since Triangle OAB is isosceles, OM bisects Angle AOB. So Angle AOM = MOB = (AOB)/2 = 72/2 = 36.

I�m going to just pick a point on OM as the center of the white circle, and label it C. If this is the center, then CO = CA = CB = r. This means that Triangle COA is isosceles. The base of Triangle COA is R, and the two legs are r. Since Angle AOM = AOC = 36, and Triangle AOC is isosceles, this means that Angle OAC = 36 as well, which means Angle ACO = 180-2*36 = 108.

Ok, well, by law of sines, R/sin(108) = r/sin(36). So:
r = R* sin(36)/sin(108) = R* 0.618 = 5* 0.618
r = 3.09017

Like I said, though, I don�t know if my arrangement is optimal.

 Posted by nikki on 2004-10-18 13:08:24 Please log in:
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