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Circles Abound (2) (Posted on 2004-10-17) Difficulty: 4 of 5
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)

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If the five white circles are to meet at the center of the black circle, with their centers arranged at equal angles around the black circle's center and be just large enough so that their pairwise intersection points lie just on the circumference of the black circle (at vertices of a regular pentagon), then one such white circle's intersections form a triangle, with the two sides that meet at the center being of length 5 with an angle of 72 degrees between them, making the third side (the chord opposite the 72-degree angle) 2*5*sin(36 deg) = 5.8778525... . 

Per MathWorld, the circumradius of a triangle is given by abc/sqrt((a+b+c)*(b+c-a)*(c+a-b)*(a+b-c)).  In this instance that works out to 3.090169943749.... Perhaps there's a form in terms of radicals of integers.

However, it might be possible to move one of the white circles closer, covering an area around the center, allowing the opposite two circles to move out a bit, as well as the two adjacent circles to move out, possibly a different amount, to allow all four of the circles that have moved outward to be able to cover more of the outer circumference of the black circle even as the one that moved in covers less.  If the increased coverage of the other four more than offsets the decreased coverage of the one that moved inward, then a smaller radius would suffice.

This sounds like a complicated geometric problem to then be optimized by calculus.


  Posted by Charlie on 2004-10-17 15:26:47
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