166 Proceedings of the Royal Society 
origin, be a, /?, y, &c., and let p be the vector of the sought point. 
We have 
2 . T(a — p)= minimum, 
from which, as above, 
2U(a-p) = 0 (1.) 
Hence, if unit forces act at the required point , in the lines joining 
it with the given points , these forces are in equilibrium. Or, in another 
form, a closed equilateral gauche polygon may be drawn whose sides 
are parallel to the lines joining the sought points with the given ones. 
This opens up some very curious geometrical speculations, which 
I have not time to pursue. 
That there is but one point whose vector satisfies equation (1) 
may easily be proved by quaternions, but even more easily by the 
following reasoning. Consider the system of unit-forces, just men- 
tioned, at any two points, one of which satisfies the problem. It is 
obvious that, if these forces be referred to the line joining the two 
points, each will be less inclined to it at one than at the other ; so 
that, as at one they produce equilibrium, at the other they must 
have a finite component in the direction of this line. 
The quaternion investigation at once suggests the following 
kinematical solution of the problem. Suppose an inextensible 
string to be passed through a small movable ring, then through 
small rings at two of the fixed points, then again through the 
movable ring, and so on — one end of the string being fixed to the 
movable ring when the number of given points is odd, and to the 
first fixed ring when the number is even. When the string is 
drawn tight, i.e. } when the sum of the lengths joining each fixed 
ring to the movable one is a minimum, the movable ring will 
evidently be in the position of the required point. Also, since the 
tension of the cord is the same throughout, the movable ring is kept 
in equilibrium by a set of equal forces in the directions of the lines 
joining it with the given points, which is the condition above found. 
This kinematical process, equally with the quaternion one, whose 
form directly suggests it, gives easily the solution of the more 
general problem, — To find a point such that m times its distance 
from A, together with n times its distance from B, &c., may be a 
minimum. 
