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THÉORIE DES DÉBLAIS ET DES REMBLAIS. 
419 
a right line; and, assuming that some one element of the filament BD is transferred 
to a point of bd (that is, so as to coincide with an element of the filament bd), it 
follows that every other element of BD must be transferred so as to coincide with 
some other element of bd; and this obviously implies that the filaments BD and bd 
must be equal. Observe that, this being so, it is immaterial which element of BD is 
transferred to which element of bd; in whatever way this is done, the sum of the 
products will be the same*'. The two lines may be regarded as the normals of a 
curve; and the problem thus is, to find a curve such that, drawing the normals 
thereof to intersect the two areas, then that the filaments BD and bd, cut off by 
consecutive normals on the two areas respectively, shall be equal. This leads to a 
differential equation of the second order for the normal curve; one of the constants 
of integration remains arbitrary, for the normal curve is any one of a system of 
parallel curves. It is to be observed that the filaments are the increments of the 
areas BCD and bed; these increments are equal; a position of the line must be the 
common tangent Cc of the two areas (this, in fact, constitutes the condition for the 
determination of one of the arbitrary constants), and for this position the areas are 
each = 0. Hence, in general, the areas must be equal; or the problem is, to find a 
curve such that any normal thereof cuts off equal areas BCD and bed. 
If, instead of the normal curve, we consider the curve which is the envelope of 
the several lines, or, what is the same thing, the locus of the point N, then we 
qould, in like manner, obtain for this curve a differential equation of the first order: 
the constant of integration would be determined by the condition that Cc is a 
tangent. The curve in question is, of course, the evolute of the normal curve. 
The several lines which intersect the two areas give rise to a finite arc IS of 
this evolute, and, as remarked by Monge, it is only when this arc IS lies (as in the 
figure) outside the two areas, that we have a true minimum. 
Passing now to the solid problem, we may imagine a congruence of lines inter 
secting the two volumes; each line of the congruence is intersected by two consecutive 
lines, and the lines of the congruence thus form two sets of developable surfaces, each 
surface of the one set intersecting each surface of the other set. And, considering 
two consecutive surfaces of the one set, and two consecutive surfaces of the other set, 
these include between them a filament; and, treating the filament as a right line, it 
seems to follow (although it is more difficult to present the reasoning in a rigorous 
form) that, if any one element of the filament BD be transferred to any one element 
of the filament bd, then that every other element of the filament BD must be 
* The most simple case is, take in the same straight line two equal segments AB, ab; it is immaterial 
how the elements of AB are transferred to ab, the sum of the products of each element into the traversed 
distance will be in every case the same. Analytically, it dx — dx , then 
/ 
[x' - x) dx = 
the equation dx'=dx meaning x'=x + & 
is, if rdr—r'dr', then 
discontinuous constant. In the actual case of the filament, the formula 
53—2