:I0S MR ROBERT BRODTE ON 



The maximum of 2 will be seen in Figs. 11, 12. The symmetrical lines may be 

 drawn to cut portions off 2, as shown by the dotted lines. 



Referring to Fig. 9, it is plain that 4 might be placed on the opposite side of 3 as 

 in Fig. 13. 



As in Figs. 9, 10, it is plain that an indefinite number of solutions can be obtained by 

 laying off BN = BX, and drawing lines symmetrical to XN from X and N (Figs. 13, 14). 



The maximum of 4 is seen in Figs. 13, 14. Joining XN or ZP will give two other 

 solutions. Another solution will be seen in Figs. 15, 16. 



This last gives the same shape to 4 as could be got by cutting a portion off 3 

 in Figs. 9, 10. 



Returning to Figs. 9, 10, the portion 4 may be placed alongside of 2 as in Figs. 17, 

 18. 4 is against 2 both in gnomon and square. If FS be cut off=EY, then any lines 

 drawn from S and E, straight or curved and symmetrical with SE, will give solutions. 

 The semicircle gives one solution. 



The maximum of 2 is seen in Figs. 19, 20, and is Professor Kelland's Case VIII. 

 Considering this fig., it is manifest that 4 might be cut off the bottom of 3, as in Figs. 

 21, 22, which is Professor Kelland's Case IX., and here the original 4 has vanished 

 altogether. 



Considering this last figure, 3 and 4 may be joined into one, and a piece cut off 2 as 

 in Figs. 23, 24, which is Professor Kelland's Case XL In this example 3 has taken quite 

 a new position in the square. 



Returning to Figs. 17, 18, and drawing symmetrical lines so as to give the maximum 

 of 4, Figs. 25, 26 are obtained, and 2 is an equilateral triangle. 



As ACE is in contact with 1, if AD = CB, an indefinite number of solutions are 

 obtained by drawing lines from C and D symmetrical with CD. 



The maximum of 2 is seen in Figs. 27, 28. The lines marked x are all the same 

 length. 



The minimum of 2 is shown by dotted line in Figs. 25, 26. 



Considering Fig. 25, it appears that if EAbe produced a triangle will be cut off equal 

 to 2, and if 2 be added to 4 and cut off from 1 the solution of Figs. 29, 30 is obtained. 



If SF be laid off equal to OL, an indefinite number of solutions are obtained by 

 symmetrical lines from F and S. The dotted line shows the maximum of 4. 



In Figs. 31, 32, 2 is annexed to 1, which is the same as Professor Kelland's Case 

 VII. There are four solutions according to the corner of 3 cut off. 



In this arrangement the original piece 2 is absorbed altogether. 



If 2 be annexed to 3 and cut off from 1, Figs. 33, 34 are obtained, which are Pro- 

 fessor Kelland's Case X. 



The following are varieties :— Figs. 35, 36 ; 37, 38 ; 39, 40 ; 41, 42. 



Professor Kelland's Case VI. gives the maximum of 4. 



The solution in Professor Kelland's Case XII. is obtained on a different system, 

 and does not appear to admit of any distinct variety. 



