1881.] 



On certain Geometrical Theorems. 



211 



V. " On certain Geometrical Theorems. No. 1." By W. H. 

 L. Russell, F.R.S. Received November 12, 1881. 



(1.) The following proof of the equation to a circle inscribed in a 

 triangle, expressed in trilinear co-ordinates, is very short and simple. 



Let «, /3, 7 be the sides of the triangle, A, B, C the opposite angles, 

 and let 



Pa? + m 3 /3 2 + - 2wm/3 7 - 2nl^a - 2 Tm*p = 



be the equation to an inscribed conic. Then when this conic is a 

 circle, the centre is given by the equations a =j@=7, and the equation 

 to the line joining the centre, to the point where 7 touches the conic, 

 that is to the point l<z—m{3=0, 7=0, is 



la.— -m8 + (m— Z)7=0. 



Now, when the conic is a circle, this lh\e must be perpendicular to 

 7 ; hence from the condition that two straight lines may be perpen- 

 dicular to each other (Salmon, " Conic Sections," 6th edition, Art. 61), 



m— Z=/cosB— m cos A, 



I m n 



,A oB oC 



cos- 3 COS" — cos-^ 



2 2 2 



which gives for the required circle 

 * 2 cos 4 — + /3 2 COS 4 ~ + 7 2 cos 4 — 



Li Li Li 



- 2/3 7 cos 2 — cos*—-2*B cos 2 — cos 2 — - 2*7 cos 2 — cos 2 — =0. 

 1 1 2 2 , 2 2 2 2 



(2.) The following theorem is given by Dr. Salmon in his " Higher 

 Plane Curves " : — 



If through any point of inflexion A in a curve of the third order 

 there be drawn three right lines meeting the curve in ab, df, ec, tben 

 every curve of the third degree passing through the seven points A, a, 

 b, d, /, c, e will have A for a point of inflexion. It follows from this 

 that any curve of the third degree described through the nine points 

 of inflexion of a cubic will have those points as points of inflexion. 



Dr. Salmon has given a geometrical proof of this theorem, and this 

 is the only demonstration I have ever seen. I have, therefore, obtained 

 the following analytical proof, which [possesses, I think, considerable 

 beauty. 



