PROFESSOR KELLAND ON SUPERPOSITION. 273 



have never met with it, and I do not suppose, at any rate, that half the solutions 

 have been dreamt of before. About these solutions I have not many words to 

 say. It may suffice if I describe the first, and indicate the others by a letter or 

 two placed in the figures. (See Plate V.) 



Problem. From a given square one quarter is cut off, to divide the remaining 

 gnomon into four such parts that they shall be capable of forming a square. 



Let ABE be the gnomon. Let AX be a side of the square which is equal to 

 this gnomon (Euc, ii, 14) : call it x, and call AB a. In Figure I. draw HK paral- 

 lel to AB and equal to x— a. 



Then (1), (2), and (4) will unite as in Figure II. Also (3) will fit in between 

 (1), (2), and (4). Now, if the four do not make up a square, either (3) will reach 

 below P, or (2) will not reach so low as P, or vice versa, seeing that the area is 

 equal to the square of MP. But neither of these circumstances can happen, be- 

 cause then HK + GE would be unequal to MR or x, which it is not. It follows 

 that the four parts exactly make up a square. 



The second method of making the sections differs from the first only in the 

 different method of producing the portions (2) and (4). 



The third differs from the first in the way of producing (2) and (3). This me- 

 thod may be modified ad libitum. It is only necessary that the portion which (3) 

 takes from (2) shall be symmetrical with respect to C and K. 



The fourth, like the second, is a slight modification of (2) and (4) in the first. 

 An isosceles triangle is taken out of (4) into (2). 



The fifth is a modification of the second. A figure equal to (4) is cut out of 

 (2) and remains in (3), whilst (4) itself replaces that figure in (2). 



The sixth is a new form : (3) and (4) instead of being, as in the first method, 

 cut from the right side, are cut from the left and reversed. 



The seventh is again a new form. 



The eighth has two pieces in common with the first method. In other respects 

 it is new. 



The ninth is a slight modification of the eighth, the piece (4) being cut from 

 the top instead of from the bottom. 



The tenth is a modification of the seventh. 



The eleventh is another modification of the eighth ; the piece is now cut out of 

 the top of (3). 



The twelfth is new ; and the demonstration is effected by showing that the 

 figures form a rectangular parallelogram, of which one side is x. 



I must not omit to add, that the major part of these solutions are due to the 

 ingenuity of my students. 



