ON SOME PROPERTIES OF THE PARABOLA. 67 



tan, tan a', tan a", that is, of w, m', m". Since the lines joining 

 the origin with the vertices of the triangle make angles a, a', a" 

 with the diameter of the circle or the axis of x, the angles they 

 make with each other are a! a, a" a', a" a, and the area 

 of the triangle will be 



rr sin (a' -a) + \rr" sin (a" a') - \ rr" sin (a" a). 

 Substituting for r, r, and r" their values, this becomes 



a 2 f sin (a' a) sin (a" a') sin (a" a) ) 



[cos 2 a" cos a cos a' cos 2 a cos a' cos a ' cos 2 a' cos a cos a" j " 



Expanding the sines and making obvious reductions, we get 



a 2 tan a tan ot tan a" tan a' tan a tan a" 

 2 " 2 * ' 



or grouping differently, and putting sec 2 a for ^- , and so on, 



COS Oi 



(tan a (sec 2 a' - sec 2 a") + tan a' (sec 2 a" - sec 2 a) 



+ tan a" (sec 2 a sec 2 a')). 



Lastly, putting 1 + tan 2 a = 1 + m* for sec 2 a, and so on, we 

 find the area of the triangle to be 



^ [m (m' z - m" 2 ) + m' (m" 2 - m 2 ) + m" (m 2 - m' 2 )}, 

 2, 



which is quite symmetrical with respect to m, m', m". 



It will be easily seen, that the sides of the triangle are re- 



spectively 



m" m m m" m' m 



cos a ' ' cos a' ' cos a" 



If these be called p, p, p", and if p be the radius of the circle, 

 by reduction, we obtain 



p = 2p sin (a" a') , p' = 2p sin (a a"), p" = 2p sin (a' a). 



If the values of the sines derived from these equations be substi- 

 tuted in the first expression for the area, it becomes 



___ __ 

 2 cos a cos a' cos a"/ 



