580 



EEPORT — 1888. 



Complete the parallelograms BALC, B'A'LC. Then, since BO : B'C' = p : p' 

 -=BM : MB', CC passes through M; and CM : M.O'=p : p' = IjG : L'C; there- 

 fore LM is equally inclined to LC, LC. 



A similar demonstration holds if L„ M, he the points dividing AA', BB' 

 externally in the ratio p : p'. The lines LM, LjMj intersecting at right angles at 

 S (see fig. 3), and {A'LALj} being harmonic, we see that SL bisects the angle 

 ASA' and similarly BSB'. And S A : S A' = AL : LA' =p : p'. Thus AB may 

 be transformed into A'B' by a half-turn round LM with a stretch from S. 



Again, AB may be otherwise transformed into A'B', thus: let the circum- 

 circles EAA', EBB' (E being the intersection of AB, A'B') meet in I; then lAB, 

 lA'B' are directly similar and a twist round and stretch from 1 suffices. From 

 this property I, as well as S, lies on the circles described on LLj, MMj as 

 diameters. 



Now let ML meet AB in A'B' in 0'. Then AB may be changed into A'B' 

 by a twist round followed by a stretch from S, or by a stretch from S followed 

 by a twist round 0', the twist in all cases being the angle AEA' = «. Hence, 

 angle 010' = lo, and E, 0, 1, O' are concyclic ; hence, EI bisects angle 010', and, 

 from previous reasoning IS bisects the same angle externally ; therefore EIS is a 

 right angle. 



Again, let L,Mi meet AB, A'B' at Q, Q'. 

 turn and stretch. Whence QIQ' = aj, and 

 QIS = ^QIQ' = |a) = QOS. Therefore I lies 



diameter ; and similarly on the circle O'SQ'. 



Fig. 3. 



Then Q comes to Q' on the half- 

 Ei Q/) Q) I *^6 concyclic. Also 

 on the circle OSQ, i.e., on OQ as 



A particular case is worthy of notice. If A, B, B', A' be concyclic the point S 

 coincides with the intersection of AB', A'B, and it is then also the centre for the 

 transformation of AA' into BB'. Then the angle SIF is a right angle, and I is 

 the foot of the perpendicular from S on EF. 



2. On liecurring Decimals and Fermafs Theorem. 

 By Professor R. W. Genese, M.A. 



Let 9,. denote the number formed of r digits, each being 9 ; and let 9,„ be the 

 smallest such number exactly divisible by a given prime number ^> 5. Then - 



