348 Transactions. 



The equation of the Hne through (aifSiyi) parallel to BC is 



- {bfS, + Cyi) a + ai {b^ + Cy) = 0, 



whence, substituting — {b/3i + cy^), ba^, ca^, for p, q, r respectively in L' 



we have 



— aX + b cos C Y + c cos B Z + abc cos A = 4 a aj 

 a cos C X — 6Y -f- c cos A Z + a6c cos B = 4 A/8i 

 a cos B X + 6 cos A Y — cZ + abc cos C = 4 a yi, 



relations by means of which we can transform from trilinear to tripolar 

 co-ordinates. 



These three relations constitute Lucas's Tl^iporem. I am unaware of 

 the method by which Lucas obtained them. Casey has proved them by 

 an application of Stewart's Theorem. 



By squaring these relations and making use of the fundamental 

 relation (i) we obtain 



4aV = 4YZ - (Y + Z - a:'f 

 W'/S,' = 4ZX - (Z + X - b'f 

 4cVi' = 4XY - (X + Y - c^' 

 Hence, since 2 A ^ [(Y — Z)'^ cos A] 



= 2 (a'X-^) - 2 2 (6c cos A YZ) 

 = 8R A [2 (a cos A X) - 2E A] , 

 we easily deduce that 



4 (a^ cot A ai" + 62 cot B /3,' + c' cot C yi^) 

 = [2 2 ja^ cot A (Y + Z) - 2 {a' cot A) } ] 



- 4R [2 (a cos A X) - 2Ra], 

 whence 



8R2 (sin 2 Aai' + sin 2 B^i^ + sin 2 Cyi'^) 



= [4R 2 |a cos (B - C) X } - 2R 2 (a' cos A)] 



- 4R [2 (a cos A X) - 2Ra] 



= 8R [2 {a cosB cos C X) — 4Ra cosx\ cosB cosC] 



= 16R2 cos a cos B cos C [2 (tan A X) - 2 a] . 



Hence the tripolar equation of the polar circle of the triangle ABC is 



tan A X + tan B Y + tan C Z = 2 A . 



It will be noticed that if in the relation 



iaW = 4YZ - (Y + Z - a^' 



X be substituted for ai'^, the result is the tripolar equation of the parabola » 

 which has its focus at A and has BC for its directrix. 



If the two lines IX + mY + vZ = k and I'X + m'Y + n'Z = k' be 

 parallel, then 



2(/) = o 2(0=0 andU'^^Zi 



The conditions that the two lines should be at right angles to each 

 other are 2 {I) = o, 2 {L') = o, and a? (nm' + m'n) + b^ {ni' -\- n'l) 

 -\- c' [hn' + ^''>n) = 0. 



