56 Mr Wiener, The Shortest Line : 
Lhe Shortest Line Dividing an Area in a Given Ratio. By 
NorBerr Wiener, Ph.D.* (Communicated by Mr G. H. Harpy,) }, 
[Received 27 October 1914. Read 23 November 1914.] 
The question we set out to answer in this paper is: given a} 
simply connected area on a plane, what can we say, apart from any | 
particular information we may have concerning the area, about 
the shape of the shortest segment of a curve, lying entirely in it, 
and dividing it in a given ratio, provided such a curve exists? ) 
To put the problem more concretely, let us suppose a farmer wants | 
to divide an irregular field of his evenly between his two sons, and 
suppose he wants to use as short a hedge as possible. How. 
shall he shape his fence? The conditions of the problem demand | 
that the curve in question must have a length and be continuous. | 
We shall limit our discussion in this paper to curves whose slope, 
considered as a function of the length of the curve from one end 
to the point where the curve has the slope in question, possesses | 
only a finite number of discontinuities, 
The method by which one would, at first thought, set out to 
solve this problem, would be that of the calculus of variations. 
But a little reflection will convince us that the condition that the | 
are dividing our area in a given ratio must lie entirely within the 
area, is difficult to express, and next to impossible to handle, by | 
the methods of the calculus of variations. 
Our problem is, however, easily amenable to an elementary © 
treatment. It is easy to show that the line of our fence, for 
example, will be either an are of a (finite or infinite) circle, or 
will be a chain of such arcs, such that two successive ares only | 
meet on the boundary of the area. 
To demonstrate this I shall first have to prove the following | 
lemmas. 
Lemma 1. Given a circle, and any two points on its periphery, : 
then an arc of a circle can always be found passing through these i 
two points, and dividing the circle in any desired ratio. 
For let the circle be called ¢ and the two points A and B. | 
Draw the chord AB. Construct its perpendicular bisector, and 
let the latter meet c in the points Cand D. Let E bea point on 
* The following article is on a topic suggested to the author by Dr Otto Szész, 
Privatdozent at Frankfurt am Main. It was the author’s original intention to 
have this article, with some further work of Dr Szasz, appear under the joint 
authorship of Dr Sz4sz and himself, but the war has rendered Dr Szasz at least 
temporarily inaccessible, and this plan impossible. Dr Sz4sz’ work consisted in a 
rigorous demonstration that the shortest line dividing any scalene triangle in a 
given ratio is a circle with its most acute apex as centre. 
