254 MISCELLANEOUS PROBLEMS [CH. XII 



time, five smaller equal circular tin discs i3 familiar to frequenters 

 of English fairs. Its effectiveness depends on making the tin 

 discs as small as possible, and therefore leads to the interesting 

 geometrical question of finding the size of the smallest tin discs 

 which can be used for the purpose. 



The problem is soluble if the radius of each tin disc is just 

 greater than three-fifths of the radius of the red circle. Of course 

 in a show, the visitor is not allowed to move the discs when once 

 he has put them down, and it is only rarely, very rarely, that he 

 succeeds empirically in placing them correctly. 



The rule works out thus. If is the centre of the red circle, a 

 its radius, and A OB a diameter of it, we take on OA a point P 

 such that approximately OP = a/40. We then place the first tin 

 disc with its centre on OB, and so that P is on its edge: suppose 

 that its edge meets the edge of the red circle in C and C, points 

 on opposite sides of AB. Place the next two discs so that in 

 each case AP is a chord of the disc. Suppose that these discs 

 meet the edge of the red circle in D and D' respectively, and 

 D being on the same side of AB. Then if we place the two re- 

 maining discs so that CD and CD' are chords of them, the 

 problem will be solved. The minimum discs are not obtained, 

 as one might at first guess, by the overlapping of five discs placed 

 at the vertices of an inscribed pentagon. 



In practice and for simplicity it is desirable to make the tin 

 discs a trifle larger than the theory requires, and to treat P as 

 coincident with 0. 



The mathematical discussion on this problem is too technical 

 and long to insert here. Probably a bare statement of the results 

 such as is given above, is all that most readers will want. 

 Should closer approximations be desired, here they are*. If the 

 radius of the red circle be taken as one foot, the problem is im- 

 possible if the radius of each disc is less than '609180 foot, and is 

 soluble if it exceeds -609185 foot. Also OP = -028545 foot, hence 

 lies very near but not quite on the circumference of the first disc 



* See Solutions of Numerical Functional Equations, by E. H. Neville, Pro- 

 ceedings of the London Mathematical Society, 1915, second series, vol. xiv, 

 pp. 308—326. 



