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XII. — Professor Kelland's Problem on Superposition. By Eobert Beodie. 



(With Two Plates, L, II.) 



(Read February 16, 1891.) 



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. 



[See Trans. Roy. Soc. JEdin., xxi. 271.] 



Preliminary problem, viz. : To cut a rectangle into three parts capable of forming a 

 square (Figs. 1, 2). 



Lay off AX and CY = side of square, join BX, draw YZ II AB, and 1, 2 and 3 will 

 be the required pieces, as is evident. 



The problem will be impossible when the length of the rectangle is greater than four 

 times its breadth; four or more pieces may then be required. The problem can be solved 

 in four different ways by drawing the sloping line BX from each of the four angles. 



The above is a particular case of the more general problem, viz., to cut a rectangle 

 or a parallelogram into three pieces so as to form another with a given side, as AB 

 (Figs. 3, 4). 



In the gnomon (Figs. 5, 6), suppose the square EC placed below in the position BK, 

 then the rectangle AK is formed, and can be divided into three pieces by FX and YZ, 

 so as to form a square as before. 



YZ + AF = side of square = KG + G Y ; .-. YZ = G Y and GZ bisects L G 



If the piece HXZYK be turned round on ZG it will occupy the position DEZSC, and 

 will cut a portion off piece 2 exactly the same as BXZS. This is Professor Kelland's 

 Case V. 



If piece 2 be left untouched, then BXZYG must be so cut that when EDCGZ is 

 turned upon ZG it will coincide with the lower square. 



It is manifest that any lines, straight or curved, drawn from S and Y symmetrical 

 with ZG will give solutions and an indefinite number (Figs. 7, 8). 



If YZ and SZ be drawn giving maximum of 4, this is Case II. ; YS joined is Case IV. ; 

 YG and SG or maximum of 3 is Case I., &c. 



Case I. considered. 



In Case I. (Fig. 9) EY a portion of 4, coincides with 2, and it will also coincide in 

 the square if 4 be turned over (Fig. 10). There will here also be an indefinite number 

 of solutions by any number of lines drawn symmetrically from E and Y. 



VOL. XXXVI. PART II. (NO. 12). 3 A 



