580 



Transactions. 



Geometry of the Problem. 



The problem to be solved can be stated as follows, in its most general 

 form : Given the three sides of a triangle ABC, it is required to find the 

 locus of a point, such that the line from the angle A to the point bisects the 

 angle included between the lines from the angles B and C to the point. 



The bisected angle contains the point A, and no restriction is placed on 

 the size of the angle except that it is less than four right angles. 



The deviation of|the plumb-line is a particular case of the general form. 



Fig. 2. 



Let ABC, fig. 2, be the given triangle, and D a point such that the angles 

 ADB, ADC are equal. 



Take A as the origin of co-ordinates and the line bisecting the angle 

 BAC as the axis of X. 



Let r, 6 be the polar co-ordinate of D. Denote the equal angles ADB, 

 ADC by x . 



To obtain the equation to the locus of D, we have from the triangle ABD 

 C = ~ 



sin 



r + c cos I ~ ^ 



.*.' cot x = 



A 

 2 



I „- + b 



(1) 



Again, from the triangle ACD we find 



'A 



r + b cos 



cot X 



6 



b sin 



■(2) 



