Prof. Tait on Listing's Topologie. 39 



If it lie in any other region, the required circle has D for one 

 extremity of a diameter, and the most distant of A, B, for 

 the other. 



(/3) When there is an obtuse angle, at C say (fig. 3) . Make 

 the same construction as before, but, in addition, describe the 

 circle whose diameter is A B. All is as before, except that 

 A B is the circle required, if D lie within it ; and that if D lie 

 within the middle portion of the larger of the two lunes formed 

 the required circle is A B D. 



[In figs. 2, 3, 4, which refer to these two cases in order, and 

 to the intermediate case in which the triangle is right-angled at 

 0, each region is denoted by three or by two letters. When 

 there are three, the meaning is that the required circle passes 

 through the corresponding points; when there are but two, 

 these are the ends of a diameter. The separate regions are, 

 throughout, bounded by full lines; the dotted lines merely 

 indicate constructions.] 



(11) A very celebrated question, directly connected with 

 our subject, is to make a Knight (at chess) move to each square 

 on the board once only till it returns to its original position. 

 From the time of Euler onwards numerous solutions have been 

 given. To these I need not refer further. 



A much simpler question is the motion of a Rook, and to this 

 the lately celebrated American " 15-puzzle " is easily reduced. 

 For any closed path of a rook contains an even number of 

 squares, since it must pass from white to black alternately. 

 [This furnishes a good instance of the extreme simplicity which 

 often characterizes the solutions of questions in our subject 

 which, at first sight, appear formidable.] And in the American 

 puzzle every piece necessarily moves like a rook. Hence if an 

 even number of interchanges of pieces will give the required 

 result, the puzzle can be solved; if not, the arrangement is 

 irreducible. 



(12) A few weeks ago, in a railway-train, I saw the follow- 

 ing problem proposed: — Place four sovereigns and four shil- 

 lings in close alternate order in a line. Required, in four 

 moves, each of two contiguous pieces (without altering the rela- 

 tive position of the two), to form a continuous line of four 

 sovereigns followed by four shillings. Let sovereigns be re- 

 presented by the letter B, shillings by A. 



One solution is as follows: — 



Before starting:— . .ABABABAB 



1st move BAABABA. .B 



2nd „ BAAB. .AABB 



3rd „ B. .BAAAABB 



4th „ BBBBAAAA . . 



