132 



MR TALBOT ON MALFATTI S PROBLEM. 



any point O not situated either in RS produced, or in RS produced, OQ — OP 



is not equal to RS. 



Demonstration. — Let the points A, B, be 

 the centres of the given circles. From 

 centre A with radius AO describe a circle 

 cutting RS produced in Z. Then by Lemma 

 4, OQ = ZS, whence OQ - OP = ZS - OP. 

 But OP is not equal to ZR, because lies 

 on the circumference ZO, which is not con- 

 centric to the circle B. Therefore OQ — OP 

 is not equal to ZS — ZR ; therefore it is not 

 equal to RS. Q. E. D. 



Corollary.— If OQ — OP is equal to RS, 

 must either lie in RS produced or in RS 

 produced. 

 Lemma 6. — Let A, B, C, be three circles. Let DE be the internal common 



Fig. 4. 



Fig. 5. 



tangent of A and B ; FG of B and C ; and HI of C and A. Then if these three 

 tangents, when produced, meet in a single point 0, 



DE = FG + IH 



or the greatest common tangent equals the sum of the two others. 

 For, 



IH = 01 - OH = OD - OG = DE - FG. 



