704 PROCEEDINGS OP THE AMERICAN ACADEMY. 



These two equations, wheu solved for x (j, k) and y (j, k) give 



Now the symmetry of Figure 1 shows that the equation of the desired 

 parabola, if one exists, must be of the form 



x- = 2p k K — y), 



where p k and a k are undetermined parameters corresponding to the kih 

 envelope. Pass such a parabola through the vertex [— 1, 0] and any 

 other vertex [x (j, k), y (j, k)~] of this envelope. The equations for 

 determining p k and a k are 



1 = 2p k a k 

 and 



and these give 



1 k 



Pic = i and a k = - , 



each of which is independent ofj. Therefore the parabola 



-S8-0 



passes through all the vertices of the ^th envelope, and the first part of 

 the proof is completed. 



In the second part of the proof 11 it will be convenient to transfer the 

 origin to the left-hand end of the string. The length of the string is I, 

 as before. Let it be bowed downward at the point x = l/L Helin- 

 holtz's solution may be written 



10 This equation can be put in the form x (j,k) = — 1 -f £ • 2, which proves 



that the projections of the vertices of the kt\\ envelope are points which divide the 

 length of the string into k equal parts. 



11 This part of the proof follows closely the method suggested in Helniholtz's 

 paper of 1860. 



