EXAMPLES OF MAXIMA AND MINIMA. 241 



Square and add ; thus 



1=2 + 2 cos(i/r-<), 



therefore cos (ijr 0) = ^ = cos 120. 



Thus the angle CPB must be 120. Similarly it may be 

 shewn that APB and APG must each be 120. Hence we 

 have the following result: describe on the sides of the 

 given triangle segments of circles each containing an angle 

 of 120, and their common point of intersection is the point 

 required. 



It is obvious that there must be a point for which the 

 proposed sum is a minimum, and therefore we need not exa- 

 mine the criteria depending on the second differential coeffi- 

 cients. 



If the given triangle has an angle equal to 120, then that 

 angular point is the point required ; if it has an angle greater 

 than 120, the method fails to give the solution. It may 

 however be shewn that when the triangle has an angle 

 greater than 120, the vertex of the obtuse angle is the point 

 required. 



For suppose the point P inside the triangle and very near 

 to the angle B of the triangle ; let PB r, PBA = cc, 

 then 



u = V(c 2 2cr cos a + r 2 ), v =^r, 

 w = A/ (a 2 2ar cos 74 r 2 ). 



Thus neglecting squares and higher powers of r we have 

 approximately 



r (cos a + cos 7) 



QJ _|- ry C{ IV 



2rcos ~- cos . 



u -i 



Now 2 cos - ^ is less than unity if B is greater than 



m 



1 20, and thus .a + c + r - 2r cos cos -2 is greater 



2 & 



than a + c. And it is obvious that if P be outside the tri- 

 angle the sum of its distances from A, B, and C is greater 



T. D. C. B 



