#2^N# unit circles are conjoined such that each circle passes through the center of the opposite circle. How do you find the common area? and the limit of this area, as #N to oo?#

Before reading this, please see the solution for the case N =1
(https://socratic.org/questions/either-of-two-unit-circles-passes-through-the-center-of-the-other-how-do-you-pro).

For N = 1, the common area #A_1= 2/3pi-sqrt3/2#. This is bounded by

the equal circular arcs of the two circles, each subtending #2/3pi#

at the respective center. The center O of this oval-like area is

midway between the vertices #V_1 and V_2#. The angle

subtended by each arc at O, on the common chord, is #pi#.

See the 1st graph.

The common area for N = 1 is

#A_1=2^1(pi/3- 1/2sin^(-1) (pi/3)) = 2(pi/3- sqrt3/4)#.

When N = 2, There are #2^2# equal circular arcs arcs in the

boundary of the common area, with the same center O and

vertices

#V_1, V_2, V_3 and V_4#. The angle at center of the area O

#=(2pi)/4=pi/2#, and the angle at the center of the respective circle

is # alpha = pi/2^2 - sin^(-1)(1/(2sqrt2)) = 24.3^o # = 0.424 rad,

nearly.

The common area for N = 2 is

#A_2= 2^2(0.424-1/2 sin(24.3^o)) = 0.873# areal unit.

Note that the first term is angle in rad unit.

The general formula is

Common area

#A_N = 2^N(alpha-1/2 sin alpha)#, where

#alpha = pi/(2^N) - sin^(-1)(1/2 sin(pi/(2^N)))#

, N = 1, 2, 3, 4, ..

2-circles graph ( N = 1 ):
graph{((x+1/2)^2+y^2-1)((x-1/2)^2+y^2-1)=0[-2 2 -1.1 1.1]}
4-circles graph ( N = 2 ):
graph{((x+0.354)^2+(y+0.354)^2-1)((x-0.354)^2+(y+0.354)^2-1)((x+0.354)^2+(y-0.354)^2-1)((x-0.354)^2+(y-0.354)^2-1)=0[-4 4 -2.1 2.1]}
The common area is obvious and is shown separately ( not on

uniform scale). Here, y-unit / x-unit = 1/2.
graph{((x+0.354)^2+(y+0.354)^2-1)((x-0.354)^2+(y+0.354)^2-1)((x+0.354)^2+(y-0.354)^2-1)((x-0.354)^2+(y-0.354)^2-1)=0[-0.6 0.6 -0.6 0.6]}

(to be continued, in a second answer)

The whole area is bounded by #2^N# equal arcs. They have a

common center O. Each #arc (AMB)# belongs to a unit circle and

subtends an #angle AOB#

= #(2pi)/2^N = pi/2^(N-1)#,

at the common center O.

So, #angle AOM = angle MOB = 1/2(pi/2^(N-1)) =pi/2^N#.

Let this arc subtend an angle #alpha#, at the center C of its parent

circle. The graph shows the common center O, the arc AMB and

the radii CA and CB, where C is the center of the circle of the arc.

The common area

#A_N = 2^N #( area of the sector OAMB)

#=2^N # ( area of the circular sector CAMB - area of the #triangle#

CAB + area of the #triangle#OAB). In the graph, O is the origin. C is

on the left at (-0.5, 0). M is ( 0.5, 0), on the middle radius.
graph{(0.2(x+0.5)^2-y^2)(x^2-y^2)((x+0.5)^2+y^2-1)=0 [0 1 -.5 .5]}

Let #angle# ACB = 2#alpha# rad, so that

#angle ACM = angle MCB = alpha#. Now,

area of sector CAMB = #(2alpha/(2pi)) pi = alpha# areal units,

area of #triangle CAB = 1/2(CA)(CB) sin 2alpha = 1/2 sin 2alpha#

and area of #triangle OAB = 1/2( base)(height)

#= 1/2(2 sin alpha)(cos alpha - 1/2)#. Now, the common area

#= 2^N( alpha - 1/2 sin 2alpha + 1/2(2 sin alpha)(cos alpha#

#-1/2)#

#=2^N( alpha - 1/2 sin alpha)#.

As AB is the common base of #triangle CAB and triangleOAB#,

#2 sin alpha = (cos alpha - 1/2 )tan (pi/2^N)#, giving

#alpha = pi/2^N - sin (1/2 sin^(-1)(pi/2^N))#

As #N to oo, alpha to 0# and #sin alpha /alpha to 1#, and so, the

limit of the common area is

#lim 2^N alpha(1-1/2 sin alpha / alpha)#

#= 1/2 lim (2^N alpha)#

#=1/2(pi-pi/2) = pi/4#, using lim x to 0 (sin^(-1)x / x = 1#.

This is the area of a circle of radius 1/2 unit. See graph.

graph{x^2 + y^2 -1/4 =0[-1 1 -0.5 0.5]}

For extension to spheres, for common volume, see
https://socratic.org/questions/2-n-unit-spheres-are-conjoined-such-that-each-passes-through-the-center-of-the-o#630027

Continuation:

If the condition is that each in a triad of unit circles passes through

the centers of the other two, in a triangular formation, the common

area is

#1/2 ( pi - sqrt 3)# = 0.7048 areal units, nearly. See graph, for the

central common area.
graph{((x+0.5)^2+y^2-1)( (x-0.5)^2+y^2-1)(x^2+(y-0.866)^2-1)=0[-4 4 -1.5 2.5]}

This can be extended to a triad of spheres, and likewise, a

tetrahedral formation of four unit spheres. Here, each passes

through the center of the other three, and so on.

Indeed, a mon avis, all these ought to be included in

Idiosyncratic Architecture.