WOLFRAM NOTEBOOK

Making the Three Areas Defined by Congruent Overlapping Circles Equal

​
centers
edges
areas
equalize areas
d
1.
Overlap two congruent circles until the three areas formed are equal. What is the distance between the centers of the two circles, and what is the common area of the three regions? For simplicity, use unit circles (radius 1). The slider moves one of the circles and displays the distance between their centers. The checkboxes allow the display of circle centers, boundaries, and the numerical values of the areas.

Details

To find the desired overlap area
A
, recall that the area of a circle is
π
2
r
, and then require that
π
2
r
-A=A
, which can immediately be solved to find that
A=
1
2
π
2
r
. To find the distance
d
, remember that the area of a circular sector of angle
θ
and radius
r
is
1
2
θ
2
r
. (This gives the expected area for the whole circle when
θ=2π
.) Half of the overlap area can be found by subtracting the area of the isosceles triangle formed by the intersections of the two circles and the center of one of them
1
2
(2rsin(θ/2))(rcos(θ/2))=
1
2
2
r
sinθ
from the sector area, and then doubling gives the area of the lens in terms of
r
and
θ
:
A=
2
r
(θ-sinθ)
. The distance between the two centers is given by
d=2rcosθ
. Setting
r=1
and numerically solving for
θ
and
d
gives
θ≈2.30988
and
d≈0.807946
.
The overlap area can also be found using integral calculus. For convenience, we again let the radius of the circles be 1. Then the area of each circle is
π
, and the area of each of the regions must be
π/2
. Let the centers of the two circles be
(0,0)
and
(d,0)
, respectively. The lens is symmetric vertically and horizontally, and one quarter of it is bounded by
y=0
,
x=d/2
, and the circle
2
x
+
2
y
=1
. So
A=4
1
∫
d/2
(1-
2
x
)dx=2
-1
cos
d
2
-
d
2
(4-
2
d
)
. Setting this equal to
π/2
and solving numerically for
d
again gives approximately 0.807946.
See[1] for a solution arrived at independently.
The alert reader may have recognized the numeric value of the equal areas shown in the display, 1.570796327, as approximately
π/2
, and we have seen that this is no coincidence. But the numerical value of the distance between circle centers has no immediately obvious significance and cannot even be determined precisely, except as the solution to the transcendental equation above.
It is interesting to note that the use of ImplicitRegion and RegionMeasure, new functions introduced in Mathematica 10, immediately gives the formula for the overlap area as a function of separation distance:
Assuming[0<d<2,
Simplify[RegionMeasure[ImplicitRegion[x^2 + y^2 <= 1 && (x-d)^2 + y^2 <= 1, {x,y}]]]]

References

[1] MathZombie. "How to Create a Two Circle Venn Diagram with 3 Equal Sections?" Mathematics Stack Exchange. (Aug 28, 2014) math.stackexchange.com/questions/769136/how-to-create-a-two-circle-venn-diagram-with-3-equal-sections.

External Links

Permanent Citation

Wolfram Cloud

You are using a browser not supported by the Wolfram Cloud

Supported browsers include recent versions of Chrome, Edge, Firefox and Safari.


I understand and wish to continue anyway »

You are using a browser not supported by the Wolfram Cloud. Supported browsers include recent versions of Chrome, Edge, Firefox and Safari.