From: inka@bnr.ca (Inanc Kayalioglu)
Newsgroups: sci.math
Subject: Halving a circle with another circle
Date: Tue, 25 Jul 1995 14:35:48 -0500
Does anyone have an elegant solution to the classic problem as follows:
Given a circle of radius one unit, find the radius of another circle with
center on the perimeter of the original circle that exactly divides the
original circle in half by area.
==============================================================================
From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Halving a circle with another circle
Date: 28 Jul 1995 15:38:32 GMT
In article ,
Inanc Kayalioglu wrote:
>Given a circle of radius one unit, find the radius of another circle with
>center on the perimeter of the original circle that exactly divides the
>original circle in half by area.
Confession: I did an integral first, then, seeing the right answer, came
up with the following proof.
Get out a piece of paper -- I'm not going to attempt an ASCII drawing.
Centered at a point P1, draw the circle C1 of radius 1. Let P2
be the left-most point of C1. Center a circle C2 of radius r here.
Assuming r<2, of course, this gives two points of intersection P3 and
P4 of the two circles.
Now compute the area of the lens-shaped overlap. It contains the sector S1
of C2 bounded by the line segments P2 P3 and P2 P4, as well as the arc
P3 P1 P4. All that remains, in fact, are two congruent slices I'll
call S3 and S4 bounded by the line segments and the arcs of C1 whose
endpoints are P2 and P3 (resp. P2 and P4). I'll also draw your attention
to the triangle P1 P2 P3 which I'll call S2 since that name isn't taken yet.
I'll write |S| for the area of a set S. Then we seek
|S1| + |S3| + |S4| = |S1| + 2 |S3|. I choose to write this as
|S1| + 2 ( |S2| + |S3| ) - 2 |S2|, because the three terms here are
easily computed in terms of the angle u = P3 P2 P4. Certainly
S2 is a triangle with sides 1, 1, and r; the unequal angle measures
pi - u, and so the area is (1/2) sin(pi - u) = (1/2) sin(u).
The other two terms are measures of circular sectors, which are simple
fractions of the areas of their circles. I compute
|S1| = (pi r^2). (u / 2pi) and |S2|+|S3| = (pi.1^2). ( (pi-u) / 2pi)
So it would appear that the total area of the overlap of the circles is
|S| = u r^2 / 2 + pi - u - sin(u)
Furthermore, a glance at triangle S2 shows u and r are easily related;
r = 2 cos(u/2), so that r^2 /2 = 1 + cos(u). This gives a simpler formula
|S| = u cos(u) - sin(u) + pi.
Well, then, if you want the sector to have half the area of the original
triangle, set this equal to pi/2 and solve for u. Of course, this is a
transcendental equation, so we expect no closed form solution, but the
formula for Newton's method is really quite simple here so we can compute
the unique solution between 0 and pi. I get
u=1.905696;
from r = 2 cos(u/2) I then find
r=1.158728.
(Draw the vertical line through P1; then the arc P3 P1 P4 of C2 cannot
lie completely on one side of this line if |S| is to be half of
|C1|. We conclude r must be greater than 1 and less than sqrt(2),
so that our numerical value is at least plausibly correct.)
dave
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Date: Mon, 31 Jul 1995 20:53:14 +0600
From: lew@ihgp.ih.att.com
To: rusin@washington.math.niu.edu
Subject: Re: Halving a circle with another circle
In article <3vb09o$3lc@watson.math.niu.edu> you write:
[quotes of above article deleted -- djr]
>Now compute the area of the lens-shaped overlap. It contains the sector S1
>of C2 bounded by the line segments P2 P3 and P2 P4, as well as the arc
>P3 P1 P4. All that remains, in fact, are two congruent slices I'll
>call S3 and S4 bounded by the line segments and the arcs of C1 whose
>endpoints are P2 and P3 (resp. P2 and P4). I'll also draw your attention
>to the triangle P1 P2 P3 which I'll call S2 since that name isn't taken yet.
The top half of the overlap is covered by two sectors, one from
each circle, and the triangle S2 is counted twice. So the area
is the sum of the sector areas minus the area of the triangle.
The sector angles are complementary, and are the angles of S2,
so you can get them with arccos or arcsin.
[quotes of djr article deleted; Mammel added...}
>from r = 2 cos(u/2) I then find
> r=1.158728.
That's what I got, too.
[... - djr]