“Dividing an Area in a Given Ratio — Nae oy 
CD between Cand D. Let F be a point on CE between Cand £. 
Then draw the circles AEB and AFB. The lune ACBE is 
greater than the lune AOBF. For AHB and AFB only intersect 
‘at A and B, and £ is outside ACBF. Moreover, by choosing 
‘Band F near enough together, we can make the lune AEBF as 
‘small as we wish. For we can construct a circle concentric with 
AEB and passing through F. The ring between this circle and 
AEB will contain the lune in question, and will have the area 
) Qorr (EF) + 7 (EFS, 
| a 
‘where r is the radius of AFB*. As HF decreases without limit, 
this will also. Therefore the area of the lune ACBE is a mono- 
tone continuous function of the length of CH within the region 
from CHE=0 to CE=CD. Therefore it can easily be shown by 
a continuity argument that 
area of ACBEH 
area of A DBE 
is a monotone continuous function of CZ, from CE =0 to CH= CD. 
and that in this region it takes every positive value. 
Lemma 2. The shortest line passing through two given points 
on the boundary of a given circle, dividing the area of the circle im 
a given ratio, is an arc of a circle. 
Let our circle be, as before, c, and the two points A and B. 
By Lemma 1, there is an arc of a circle dividing c in the desired 
ratio: let it be AEB. If AHB be a segment of a straight line, 
our lemma needs no proof. If not, let AFB be any other curve 
dividing c in the same ratio. Complete the circle A#B, and let 
AGB be the other arc determined by A and B on this circle. 
Let ACB be the are of ¢ within the circle AH BG. 
Then the area of the circle AHBG and that of the figure 
bounded by AF'B and AGB will be identical. For the two have 
the lune ACBG in common, and, by hypothesis, the area of the 
lune ACBE equals that of the figure bounded by AFB and ACB. 
By Steiner’s theorem the perimeter of A HBG must be less than 
that of AFBG, for it is a circle. Hence, since the two perimeters 
have AGB in common, the length of A#Bis less than that of AFB. 
This proves our lemma. 
i Our theorem is now easy enough to prove. For let us suppose 
our area given, and the shortest line dividing it in a given ratio 
also given. Let us call the latter 7. From any point on / at a 
positive distance from the boundary as a centre, we can describe 
* If AEB is a straight line, then AH BF may be enclosed in a rectangle whose 
base is constant, and whose altitude may be made as small as you will. 
