PEOFESSOR KELLAND ON SUrERPOSITION. 473 



0, H, and J draw perpendiculars to GL, produce those through and J, their own 

 length, to X and Y ; and complete the figure. 



The three perpendiculars are each equal to ^ai, and the portions (4) and (iv.). 

 (3) and (iii.), (2) and (ii.), are respectively equal ; hence the conclusion. 



The division may, in this instance, be effected by two cuts. 



XXIII. MakeBY = -2;, CH = AY; draw KHO parallel to BE; make EF = KL, 

 and complete the rectangle JCOP ; YL, KH, and LJ, JF are the cutting lines. 



Because PO = CJ = a; = BY ; triangle PKO = YLB, also because KO = BE. 

 KH + OQ=«; andBJ = iBF = i(2a— EF) = i (2a— KL) = tf— KH = QC; therefore 

 triangle EQO = JBF; hence the conclusion. 



XXIV. Draw AK, making an angle of 30° with AB ; make AF = EK, and from 

 F, E and H (the fourth corner of the square) draw perpendiculars to AK AK, 

 FG, and JL are the cutting lines. 



For HL = ^, and because AB = 2PH, AK = 2HJ, therefore BK = 2PJ = H J. and 

 FC = iAF = iEK = J (2a-BK) = «-PJ = GJ. Again FB = 2a- AF = 2a-FK = 

 BK = HJ; hence GJHO is equal to (4), and DO = PJ ; DE = AP, consequently 

 DETO is equal to (2) ; hence the conclusion. 



The division may, in this instance, be effected by two cuts, as in No. XXII. 



It will be observed that the angle of section is in all cases either 30'", or are 

 such that the tangent of the angle is a function of VS, and of no other surd. 



2j3 3 



For instance, in No. XVIII. we shall find tan. UAS^^jg^o 



I have only to add, that a large number of the solutions are due to various 

 friends, including students, to whom my best thanks are tendered. It would not 

 be easy to fix the authorship of each solution with certainty, on which account I 

 shall not attempt it. 



