Mr. J. Cockle's Solution of two Geometrical Problems. 133 



sufficient for my purpose, — which is, to show the iiiterpreta- 

 bility of impossible quantity. 

 Definition. In the equation 



t;=A + ?B+iC (1.) 



let 



A2 + B2+C2=a2; 



then I propose to call (1.) ^ virtual solution of the ecjuation 



v=a. 



Rule. To solve a problem by the imaginary geometry, let 

 its conditions be expressed by the independent relations U = 0, 

 and V = J form the equation 



U + wV=0 (2.) 



where m is a disposable multiplier: then, if a solution of (2.) 

 is a virtual solution of 



V = 0, 



the thing required is done*. 



Problem 1. Find three points equidistant from each other. 



Let A be one of the points. Draw AB equnl to any quan- 

 tity «, and in any direction, and let B be another of the points. 

 Let C be the third point. Then, since C is equidistant from 

 A and B, we have ACxCB=AC^; and also, since A is 

 equidistant from B and C, we obtain AC^=AB^. These 

 equations will, on putting AC=,r, be expressed algebraically 

 as follows : — 



.v{a—x)=x\ ...... (3.) 



^2=a2. (4.^) 



add (3.) and (4.) and we obtain 



x{a—x)=a\ (5.) 



and hence 



a , a \/~^ 

 2 — 2 

 a .aVS 



and, the solution of (5.) being a virtual solution of (4.), the 

 problem is solved. ABC is, of course, an equilateral triangle. 



Problem 1L Find four points equidistant from each other. 



Complete the rhombus ACDB. Then D is equidistant 

 from B and C. And, by symbolical geometry, we have 

 AD = AB+BD. But, we must also have, since A is equi- 



* Observations on this Rule and its grounds are reserved for another 

 opportunity. 



