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ABSTRACT OF A MEMOIR BY DR HESSE ON THE 
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Two other systems of generating lines (belonging to two new hyperboloids) are 
determined by the like construction, interchanging the points 1, 2, 3, 4. And by 
interchanging all the seven points we obtain 105 systems of generating lines (belonging 
to as many different hyperboloids, unless some of these hyperboloids are identical). 
2. It remains to be shown how the polar plane of the point P with respect to 
one of the 105 hyperboloids may be constructed. Drawing through the point P three 
lines, each of which passes through two of the three given generating lines of the 
hyperboloid in question, the points of intersection of the lines so determined with the 
generating lines which they respectively intersect, are points of the hyperboloid. Hence, 
constructing upon each of the three lines in question the harmonic pole of the point 
P with respect to the two points of intersection, the plane passing through the three 
harmonic poles is the polar plane of P with respect to the hyperboloid. Hence, 
constructing the polar planes of P with respect to any three of the 105 hyperboloids, 
the point of intersection of these three polar planes is the required point Q. 
Problem. To construct the polar plane of a point P with respect to the surface 
of the second order which passes through nine given points 1, 2, 3, 4, 5, 6, 7, 8, 9. 
Consider any seven of the nine points, e.g. the points 1, 2, 3, 4, 5, 6, 7, and 
construct the harmonic pole of the point P with respect to the system of surfaces of 
the second order passing through these seven points. By permuting the different points 
we obtain 36 different points Q, all of which lie in the same plane. This plane 
(which is of course determined by any three of the thirty-six points) is the required 
polar plane. Hence we obtain the solution of 
Problem. To construct the surface of the second order which passes through nine 
given points 1, 2, 3, 4, 5, 6, 7, 8, 9. 
Assuming the point P arbitrarily, construct the polar plane of this point with 
respect to the surface of the second order passing through the nine points. Join the 
point P with any one of the nine points, e.g. the point 1, and on the line so formed 
determine the harmonic pole P of the point 1 with respect to the point P, and the 
point where the line PI is intersected by the polar plane. R is a point of the 
required surface of the second order, which surface is therefore determined by giving 
every possible position to the point P. 
This construction is the complete analogue of Pascal’s theorem considered as a 
construction for describing the conic section which passes through five given points. And 
it would appear that the principles by means of which the construction is obtained 
ought to lead to the analogue of Pascal's theorem considered in its ordinary form, that is, 
as a relation between six points of a conic, or in other words to the solution of the 
problem to determine the relation between ten points of a surface of the second order; 
but this problem, one of the most interesting in the theory of surfaces of the second 
order, remains as yet unsolved. The problem last mentioned was proposed as a prize 
question by the Brussels Academy, which subsequently proposed the more general