# #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?#

Continuation, for the 3rd part of this problem. I desire that this for circles, extended 3-D case for spheres and all similar designs are classified under "Idiosyncratic Architectural Geometry".

Next:

If each member of a triad of unit circles must pass through

the points where the other two, forming a triangle, meet, the common

area is

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]} is the central common area.

This can also be applied to three different domains, and similarly, a

four unit spheres arranged in a tetrahedral formation; each passes through

via the middle of the remaining three, and so forth.

Yes, a mon avis, each of these should be incorporated into

peculiar architecture.

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Continuation , for the second part.

at the common center O.

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

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]}

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

limit of the common area is

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://tutor.hix.ai

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

Before reading this, please see the solution for the case N =1 (https://tutor.hix.ai).

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

See the 1st graph.

The common area for N = 1 is

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

vertices

nearly.

The common area for N = 2 is

Note that the first term is angle in rad unit.

The general formula is

Common area

, 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)

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The common area of 2^N unit circles conjoined in the described manner can be found by considering the pattern formed by the overlap of these circles. To find the common area, you need to determine the area of one of the sectors formed by the overlap and then multiply it by the number of such sectors, which is 2^N.

The area of one sector formed by the overlap can be calculated by finding the difference between the area of the sector of the unit circle and the area of the triangle formed by two radii and the chord connecting the intersection points of adjacent circles.

To find the limit of this area as N approaches infinity, you can use the concept of limits. As N increases, the number of circles increases exponentially. The overlapping area of these circles will become more densely packed, resembling a continuous shape. By applying the limit as N approaches infinity, you can find the area of the shape formed by the overlapping circles.

This approach involves advanced mathematical techniques, such as limits and geometric series, to accurately calculate the common area and its limit as N approaches infinity.

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