A DEMONSTRATION OF DUPIN’S THEOREM. 
[From the Quarterly Journal of Pure and Applied Mathematics, vol. xn. (1873), 
pp. 185—191.] 
The theorem is that three families of surfaces intersecting everywhere at right 
angles intersect along their curves of curvature. The following demonstration puts in 
evidence the geometrical ground of the theorem. 
I remark that it was suggested to me by the perusal of a most interesting paper 
by M. Lévy, “ Mémoire sur les coordonnées curvilignes orthogonales et en particulier sur 
celles qui comprennent une famille quelconque de surfaces de second degré,” (Jour, de 
VÉcole Polyt., Cah. 43 (1870), pp. 157—200). It was known that a family of surfaces 
p = f(x, y, z) where the function is arbitrary, does not in general form part of an 
orthogonal system, but that p considered as a function of (x, y, z) must satisfy a 
partial differential equation of the third order. M. Lévy obtains a theorem which, in 
fact, enables the determination of this partial differential equation ; he does not himself 
obtain it, although he finds what the equation becomes on writing therein = 0, 
CLX 
^ = 0- 
dy 
equation. 
but I have, in a recent communication to the French Academy, found this 
Proceeding to the consideration of Dupin’s theorem, on a surface of the first family 
take a point A and through it two elements of length on the surface, AB, AC, at 
right angles to each other; draw at A, B, C the normals meeting the consecutive 
surface in A', B', C' and join A'B', A'C'. It is to be shown that the condition in 
order that B'A'C' may be a right angle is the same as the condition for the inter 
section of the normals AA' and BB' (or of the normals A A' and CC')\ for this being 
so, since by hypothesis B'A'C' is a right angle, it follows that AA', BB' intersect;