# 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?

##### 3 Answers
Oct 17, 2016

${2}^{N} \left(\alpha - \frac{1}{2} \sin \alpha\right)$, where

$\alpha = \frac{\pi}{{2}^{N}} - {\sin}^{- 1} \left(\frac{1}{2} \sin \left(\frac{\pi}{{2}^{N}}\right)\right)$

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

#### Explanation:

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 ${A}_{1} = \frac{2}{3} \pi - \frac{\sqrt{3}}{2}$. This is bounded by

the equal circular arcs of the two circles, each subtending $\frac{2}{3} \pi$

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

midway between the vertices ${V}_{1} \mathmr{and} {V}_{2}$. 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

${A}_{1} = {2}^{1} \left(\frac{\pi}{3} - \frac{1}{2} {\sin}^{- 1} \left(\frac{\pi}{3}\right)\right) = 2 \left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right)$.

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

${V}_{1} , {V}_{2} , {V}_{3} \mathmr{and} {V}_{4}$. The angle at center of the area O

$= \frac{2 \pi}{4} = \frac{\pi}{2}$, and the angle at the center of the respective circle

is $\alpha = \frac{\pi}{2} ^ 2 - {\sin}^{- 1} \left(\frac{1}{2 \sqrt{2}}\right) = {24.3}^{o}$ = 0.424 rad,

nearly.

The common area for N = 2 is

${A}_{2} = {2}^{2} \left(0.424 - \frac{1}{2} \sin \left({24.3}^{o}\right)\right) = 0.873$ areal unit.

Note that the first term is angle in rad unit.

The general formula is

Common area

${A}_{N} = {2}^{N} \left(\alpha - \frac{1}{2} \sin \alpha\right)$, where

$\alpha = \frac{\pi}{{2}^{N}} - {\sin}^{- 1} \left(\frac{1}{2} \sin \left(\frac{\pi}{{2}^{N}}\right)\right)$

, 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)

Jun 23, 2018

Continuation , for the second part.

#### Explanation:

The whole area is bounded by ${2}^{N}$ equal arcs. They have a

common center O. Each $a r c \left(A M B\right)$ belongs to a unit circle and

subtends an $\angle A O B$

= $\frac{2 \pi}{2} ^ N = \frac{\pi}{2} ^ \left(N - 1\right)$,

at the common center O.

So, $\angle A O M = \angle M O B = \frac{1}{2} \left(\frac{\pi}{2} ^ \left(N - 1\right)\right) = \frac{\pi}{2} ^ N$.

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

$\angle A C M = \angle M C B = \alpha$. Now,

area of sector CAMB = $\left(2 \frac{\alpha}{2 \pi}\right) \pi = \alpha$ areal units,

area of $\triangle C A B = \frac{1}{2} \left(C A\right) \left(C B\right) \sin 2 \alpha = \frac{1}{2} \sin 2 \alpha$

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

$= \frac{1}{2} \left(2 \sin \alpha\right) \left(\cos \alpha - \frac{1}{2}\right)$. Now, the common area

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

-1/2)

$= {2}^{N} \left(\alpha - \frac{1}{2} \sin \alpha\right)$.

As AB is the common base of $\triangle C A B \mathmr{and} \triangle O A B$,

$2 \sin \alpha = \left(\cos \alpha - \frac{1}{2}\right) \tan \left(\frac{\pi}{2} ^ N\right)$, giving

$\alpha = \frac{\pi}{2} ^ N - \sin \left(\frac{1}{2} {\sin}^{- 1} \left(\frac{\pi}{2} ^ N\right)\right)$

As $N \to \infty , \alpha \to 0$ and $\sin \frac{\alpha}{\alpha} \to 1$, and so, the

limit of the common area is

$\lim {2}^{N} \alpha \left(1 - \frac{1}{2} \sin \frac{\alpha}{\alpha}\right)$

$= \frac{1}{2} \lim \left({2}^{N} \alpha\right)$

$= \frac{1}{2} \left(\pi - \frac{\pi}{2}\right) = \frac{\pi}{4}$, 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

Jun 24, 2018

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".

#### Explanation:

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

$\frac{1}{2} \left(\pi - \sqrt{3}\right)$ = 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.