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PROBLEM of APOLLONIUS in the January 2010 Issue of American Scientist D. Mackenzie Discusses the Apollonian Gasket Which Involve

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PROBLEM of APOLLONIUS in the January 2010 Issue of American Scientist D. Mackenzie Discusses the Apollonian Gasket Which Involve PROBLEM OF APOLLONIUS In the January 2010 issue of American Scientist D. Mackenzie discusses the Apollonian Gasket which involves finding the radius of the largest circle which just fits into the space between three tangent circles of different radii. The problem was first solved in detail by Rene Descartes in 1643. Descartes found and the article states that – 2 2 2 2 2 1 1 1 1 1 1 1 1 1 + + + = + + + a b c d 2 a b c d where a, b, c are the radii of the three outer circles and d is the radius of the circle just fitting into the gap formed by them when brought into contact. A derivation is not given since the article is intended for the typical Sigma Xi reader and not aimed directly at mathematicians. We remedy this situation here by deriving this result and also determine the area of the typical triangular space into which the fourth circle just fits. Also some additional variations on the problem are considered and worked out. Our starting point is the four circle diagram and the triangle formed by connecting the center of the three outer circles by straight lines – The semi-perimeter of the triangle is s=a+b+c so that the Heron formula (known to Descartes) yields- AreaTriangle = s(s − b − c)(s − a − b)(s − a − c) = abc(a + b + c) This area in turn must match the area of the three triangles bordered by the blue lines in the figure. These areas again can be obtained via the Heron formula and read- AreaABO = (a + b + d)(dab) , AreaBCO = (b + c + d)(dbc) , AreaACO = (a + c + d)(dac) Equating the areas we obtain the result- (a + b + c) (a + b + d) (a + c + d) (b + c + d) = + + d c b a Although this result looks different from that given by Descartes, it gives the same answer. For example, if a=b=c=1 then we find d=[2/sqrt(3)]-1= 0.1547005…If a=-1,b=1/2, and c=1/2, as in the example given in the article, the value of d is found to be d=1/3. Next we look at the related problem of the area of the triangle with curved sides formed by the three outer circles. This area equals the area of triangle ABC minus the sectors of the circles whose outer arc form the sides of the inner triangle. One has- 2 2 2 AreaInner = (abc)(a + b + c) − (a θ A + b θB + c θC )/ 2 where the angles are obtained from the law of cosines and read- (a + b)2 + (a + c)2 − (b + c)2 θ A = arccos 2(a + c)(a + b) (a + b)2 + (c + b)2 − (a + c)2 θB = arccos 2(a + b)(b + c) (a + c)2 + (b + c)2 − (a + b)2 θC = arccos 2(a + c)(b + c) For the simplest case one takes a=b=c=1. This produces the angles θA=θB=θC =arcos(0.5)= π/3 so that the area becomes AreaInner= sqrt(3)-π/2=0.161254…We have used this result in some earlier investigations on oscillatory heat transfer where one was interested in determining the fraction of fluid being transported through the intersticial spaces in a bundle of open ended capillaries. One may generalize this problem by asking for the diameter of the largest circle which just fits into the star-shaped area formed by the surrounding of n circles and also determine the area of this star shaped area. By demanding a n/2π rotational symmetry of the star-shaped area, one has the simplification that each of the surrounding circles has the same radius R=a. A schematic of the problem is as shown- One has a guiding regular polygon with sides of length 2a at whose vertexes we place circles of radius R=a. Lines drawn inward from the middle of each of the polygon sides intersect at point D. The length of these lines is h=a tan[π(1/2-1/n)]. Next lines are drawn from the center of each circle to the same point D. This results in n isosceles triangles with each area equal to – 1 2 1 1 AreaIsosceles = 2ah = a tan π ( − ) 2 2 n If one now subtracts the area of the two sectors at the bottom of one of the isosceles triangles and multiplied the result by n, the area of the star region formed inside the surrounding circles will be- 2 π 2 π 2 AreaStar = na tan[ (1− )] − [1− ] 2 n 2 n and the radius of the largest circle centered at D will be- π 2 R = a[sec( [1− ]) −1] 2 n The circumference of the star will be 2naθ. Working these numbers out for n=6 , where the guiding polygon is a hexagon, we find- 2 π AreaStar = 6a 3 − , Radius = a , Circumference = 4πa 3 We show you here a constructed figure of this n=6 case- We point out that the edges of the star do not correspond the classical asteroid figure which is generated by a point on the periphery of a small cylinder rolling within a larger cylinder. The difference between an astroid and one of our n=4 stars is that the scallops are somewhat deeper for the star periphery as shown- A problem associated with these discussions deals with the space left when a cylinder of radius R=a is placed into a right angled corner. This space can be looked as one fourth of the star area formed by four surrounding cylinders of radius R=a each. One finds- 1 2 π π 2 π 2 AreaCorner = []4a (tan( ) − ) = a 1− = 0.2146...a 4 4 4 4 an obvious result. January 1, 2010 .
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