CH. XII] 



MISCELLANEOUS PROBLEMS 



257 



a right-angled triangle. In fact, however, if we are given any 

 two squares we can always by three cuts divide them into five 

 pieces which can be put together to make a square. There are 

 various solutions. Here are two which answer the purpose*. 



The first of these is as follows: Place the larger and smaller 

 squares AG and CE side by side as in figure i below, and take 

 AB equal to CD. Then a cut BH and a cut BE (i.e. a cut BJ on 

 the larger square and another cut JE on the smaller square) 

 will divide the squares into five pieces which can be put together 

 to make one square of which BH and BE are sides. 



A more symmetrical, though less simple, five-part solution 

 (made up of the smaller square together with a four-part division 



C D 



Figure i. 



Figure ii. 



of the larger square) can be effected as follows. Place the larger 



square (denoted by the letters A,B,C, D) next the smaller square 



(denoted by E), as in figure ii, with their bases in the same line. 



Bisect this line in one point a and on their common side take 



another point b whose distance above their bases is half the sum 



of their sides. Through these points draw lines perpendicular 



to each other, crossing in the centre of the larger square, and 



terminating in its sides. This divides the larger square into 



four equal parts A,B,C, D. Produce the two lines drawn through 



a and b for half their length beyond the common base and side 



of the squares, and through their extremities draw two other 



lines perpendicular to them. These four lines will form another 



* H. Perigal, Messenger of Mathematics, 1873, vol. ii, N.S. pp. 103 — 106; 

 H. E. Dudeney, Amusements, London, 1917, p. 32. 



17 



B. K. 



