58 Mr Wiener, The Shortest Line Dividing an Area, ete. 
x 
Tei 
a circle lying entirely within the area. Except, at the most, ing 
finite number of points, we can make this circle small enough tc 
cut / in two points only*. Within the circle, J must be an are 0) 
ne 
m 
nN 
a circle. For, call our little circle c. Let J divide c in the ratio 
| 
| 
Construct the arc of a circle cutting c in the same points as J, 2 
/ 
dividing c in the ratio = Then consider the curve formed by. 
! 
this arc and the portions of J outside c. This must divide the. 
area in the same-ratio as J, and, if it is not the same curve as J! 
must be shorter. This contradicts our hypothesis. 
In the same way, it may be shown that J cannot contain two) 
arcs of distinct circles meeting inside the area. For, as before,’ 
about the meeting-point of these arcs describe a circle, ¢, cutting) 
each are in one point only, and lying entirely within the area.| 
Then, by the same reasoning as before, the portions of the two ares 
lying within c must form a single arc, which is impossible. Thus! 
our theorem is provedt. | 
* This is a consequence of the condition which we laid down for all curves 
considered in this article—that the slope of the curve at a point p, considered as a 
function of the length of the curve between p and one of the end-points of the 
curve, possesses only a finite number of discontinuities. This is at once obvious, if 
we reflect that any curve y = f(x), which intersects a circle ¢ described about some 
point on it more than twice, must have a maximum or a minimum in y between 
one of the points where it intersects c and the centre of c. 
+ It is almost self-evident that the shortest line to divide a conver area in a 
given ratio is a single arc of a circle, but this I have not yet been able to prove. 
