= o, 



Hogg. — On certain Tripolar Relations. 319 



Eliminating aa , b(3 , cy we have 



2R Pl cos o c- b 2 



2Rp. 2 cos <p c 2 o a 2 



2Rp 3 cos if/ b 2 d 1 o 



1 111 



which reduces to the relation 



ap y cos A cos + bp 2 cos B cos <f> + cp. cos C cos f =2a. 



From the equations 



a Pl cos 6 /5 y bp 2 cos _ y a Q c P3 cos i/r _ a /3 



r — -j- i > 1 > 7 — 1 — r> 



be b c ca o a ab a b 



we see that if O lie on the trilinear polar of the symmedian point 



BC 2 AO cos $ + CA 2 BO cos + AB 2 CO cos ^ = o. 



Solving for a , /3 , y from the above equations, and substituting in 

 he equation of the circle ABC, we obtain the relation 



BC v'OA cos + CA \/OB cos <£ + AB \/OC cos~^ = o 

 for any point O on the circle. 



If t Q be the length of the tangent to any circle from the middle point 

 of BC. then 



2 A ti = aa c 2 + c 7o a 2 - 2 (RS + A P 2 ) 



2 a t 2 = aa b 2 + bft d 2 - 2 (RS + a p 2 ) 



2 a t 2 = aa m> + f (&& + cy ) - 2 (RS + a p 2 ) 

 where w, is the median drawn through A. 

 Hence 2 A (ti + t 2 - 2t Q 2 ) 



= aa (b 2 + c 2 - 2m?) + ~ {bft + cyj 



Zi 



a 2 _ 

 »•«•. 4 1 + «» ! = 2t a +^-', 



A 



an extension of the Theorem of Apollonius. 



If U be the polar circle of the triangle ABC, then 



U = tan AX + tan BY + tan CZ - 2 a = o 

 .-. 2 a t 2 = k (c 2 tan A + a 2 tan C - 2 a), 

 where K = 4R 2 cos A cos B cos C, 



which reduces to t 2 2 = ca cos B. Also t 2 = ab cos C, hence £ 2 2 + tJ = tr. 



If in the triangle ABC the angle A be obtuse, then the sum of the 

 squares of the tangents to the polar circle from B and C is BC 2 . 



If U = o pass through the point (XjYjZi), then the locus of O 

 («oA>Yo) is the circle 



V = aaX 1 + b/SY, + cyZj - 2RS - 2 a p 2 = o. 



