$2^{N}$ $2^N$ 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 $N \to \infty ?$ $N to oo?$

Question

$2^{N}$ $2^N$ 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 $N \to \infty ?$ $N to oo?$

3 Answers

Impact of this question

2344 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-17T16:35:52

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

$α = \frac{π}{2^{N}} - {sin}^{- 1} (\frac{1}{2} sin (\frac{π}{2^{N}}))$ $alpha = pi/(2^N) - sin^(-1)(1/2 sin(pi/(2^N)))$

, 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 $A_{1} = \frac{2}{3} π - \frac{\sqrt{3}}{2}$ $A_1= 2/3pi-sqrt3/2$ . This is bounded by

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

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

midway between the vertices $V_{1} and V_{2}$ $V_1 and V_2$ . The angle

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

See the 1st graph.

The common area for N = 1 is

$A_{1} = 2^{1} (\frac{π}{3} - \frac{1}{2} {sin}^{- 1} (\frac{π}{3})) = 2 (\frac{π}{3} - \frac{\sqrt{3}}{4})$ $A_1=2^1(pi/3- 1/2sin^(-1) (pi/3)) = 2(pi/3- sqrt3/4)$ .

When N = 2, There are $2^{2}$ $2^2$ equal circular arcs arcs in the

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

vertices

$V_{1}, V_{2}, V_{3} and V_{4}$ $V_1, V_2, V_3 and V_4$ . The angle at center of the area O

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

is $α = \frac{π}{2^{2}} - {sin}^{- 1} (\frac{1}{2 \sqrt{2}}) = {24.3}^{o}$ $alpha = pi/2^2 - sin^(-1)(1/(2sqrt2)) = 24.3^o$ = 0.424 rad,

nearly.

The common area for N = 2 is

$A_{2} = 2^{2} (0.424 - \frac{1}{2} sin ({24.3}^{o})) = 0.873$ $A_2= 2^2(0.424-1/2 sin(24.3^o)) = 0.873$ areal unit.

Note that the first term is angle in rad unit.

The general formula is

Common area

$A_{N} = 2^{N} (α - \frac{1}{2} sin α)$ $A_N = 2^N(alpha-1/2 sin alpha)$ , where

$α = \frac{π}{2^{N}} - {sin}^{- 1} (\frac{1}{2} sin (\frac{π}{2^{N}}))$ $alpha = pi/(2^N) - sin^(-1)(1/2 sin(pi/(2^N)))$

, 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)

Answer 2 · 2018-06-23T09:23:54

Continuation , for the second part.

Explanation:

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

common center O. Each $a r c (A M B)$ $arc (AMB)$ belongs to a unit circle and

subtends an $∠ A O B$ $angle AOB$

= $\frac{2 π}{2^{N}} = \frac{π}{2^{N - 1}}$ $(2pi)/2^N = pi/2^(N-1)$ ,

at the common center O.

So, $∠ A O M = ∠ M O B = \frac{1}{2} (\frac{π}{2^{N - 1}}) = \frac{π}{2^{N}}$ $angle AOM = angle MOB = 1/2(pi/2^(N-1)) =pi/2^N$ .

Let this arc subtend an angle $α$ $alpha$ , 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

$A_{N} = 2^{N}$ $A_N = 2^N$ ( area of the sector OAMB)

$= 2^{N}$ $=2^N$ ( 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 $α$ $alpha$ rad, so that

$∠ A C M = ∠ M C B = α$ $angle ACM = angle MCB = alpha$ . Now,

area of sector CAMB = $(2 \frac{α}{2 π}) π = α$ $(2alpha/(2pi)) pi = alpha$ areal units,

area of $△ C A B = \frac{1}{2} (C A) (C B) sin 2 α = \frac{1}{2} sin 2 α$ $triangle CAB = 1/2(CA)(CB) sin 2alpha = 1/2 sin 2alpha$

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

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

$= 2^{N} (α - \frac{1}{2} sin 2 α + \frac{1}{2} (2 sin α) (cos α$ $= 2^N( alpha - 1/2 sin 2alpha + 1/2(2 sin alpha)(cos alpha$

$- \frac{1}{2})$ $-1/2)$

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

As AB is the common base of $△ C A B and △ O A B$ $triangle CAB and triangleOAB$ ,

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

$α = \frac{π}{2^{N}} - sin (\frac{1}{2} {sin}^{- 1} (\frac{π}{2^{N}}))$ $alpha = pi/2^N - sin (1/2 sin^(-1)(pi/2^N))$

As $N \to \infty, α \to 0$ $N to oo, alpha to 0$ and $\frac{sin α}{α} \to 1$ $sin alpha /alpha to 1$ , 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

Answer 3 · 2018-06-24T07:25:33

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

$1/2 ( pi - sqrt 3)$ = 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.

Science

Math

Humanities

... and beyond

$2^{N}$ $2^N$ 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 $N \to \infty ?$ $N to oo?$

3 Answers

Explanation:

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

2^N2N2^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?N→∞?N to oo?

3 Answers

Explanation:

Explanation:

Explanation:

Related questions

Impact of this question

$2^{N}$ $2^N$ 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 $N \to \infty ?$ $N to oo?$