In Ep. 48 and Ep. 50, there is a "Magic Circle" shown in Akazukin Chacha, as Seravi said, it's called "Mahoujin" in Japanese .

Mahoujin and number "7"

As we can see, there are "7" small circles on the rim of [main inner] circle, these small circles form a regular heptagon.

And as this pic has shown, if we draw an externally tangent square on the "main inner" circle, then the four vertices of that square just touch the "Main outer" circle.

That's said, the radius ratio of the main circle to the inner circle is:$ 1:\dfrac {1} {\sqrt {2}} $.

And, if we make externally tangent circle on the yellow heptagram (not its white rim), than the ratio of this circle to the main circle is 0.624 : 1, aka:

$ \cos \left( \dfrac {360} {7}\right) :1 $


that means if we make an (wider) heptagram (radius = 1) like this pic:

then the radius of its inscribed circle of the inner heptagon is 0.624

that's said, if we make tangent lines on every vertex of the yellow heptagram, then those lines can form a wider heptagram, and this new heptagram's every vertex touch the [main outer] circle.

We call the externally tangent circle of the yellow heptagram "P circle".

Now we calculate the ratio between "main inner circle" and "P circle", we find it is:

$ 1:\sqrt {2}\times \cos \left( \dfrac {360} {7}\right) $

= 1: 0.8817~, then

$ 2 \times \arccos \left( 0.8817 \cdots \right) $= 56.29216~

and approximately:(accuracy = 99.933%):

56.29216~ $ \approx \arctan \left( \dfrac {3} {2}\right) $


thus, based on the inner circle, we can make both "main circle" and "P circle" using ruler and compass, like this pic:

and if we make a tangent line S on the P circle, then it's a nearly perfect heptagon approximation. (the two points where S cross the main circles is 2*(360/7))

As far as I know, there's no document in the world( realworld ) had recorded this approximation method.


Animation ver. Use "See full size image" to watch


An interesting thing is, let the hypotenuse be sqrt(ln(180/7)), then that (original) 3/2 will be 1.4990~. The accuracy of this is even higher then that commonly used atan(79/63) ( also atan((9^2-2)/(8^2-1)) ). Although Natural logarithm does not relate to ruler-compass, it can build a linkage between 180/7 (number) and 360/7 (degree) via our method.

Akazukin Chacha does have a lot of delicate properties like this. As we can see the thing above, that is the proof of that" $ 1:\dfrac {1} {\sqrt {2}} $" ratio dose relate to the "7" or "heptagon".


Now, if we calculate the area of the 7 small circles and the inner circle, then the area of the small circles is:

radius of small circles = $ \dfrac {\sqrt {2}-1} {2\sqrt {2}}= t $

thus the area is $ t^{2} $*pi, approximately $ \dfrac {3} {140}\pi $(Continued fraction about 1/(23+1/3) ;accuracy = 99.916%)

and the area of the inner circle is 0.5 or 1/2, and the area between every small circles (as the pic's blue area) is (((1-0.5)/7)-(3/140))*pi = 7pi/140.= pi/20.

therefore, if we let the area of "Main circle" = 140pi, then the simple integer ratio of

inner circle : small circle : left area (blue area) is

= $ 70 : 3 : 7 $ (accuracy = 99.916%)

Again, that is the proof of the area's ratio is related to "7".

The Central Multi-fold circles and the Coins


The pic above is the outline-drown Mahoujin, which had shown in the end of Ep. 48.

The biggest blue circle relates to the yellow heptagram's white rim, let the radius of inner circle be 1, then the ratio of the inner circle to the biggest blue circle is 1 : 0.356895867~ (sin(360/7*4)/cos(360/7)).

The second blue circle: If we set the area of the small circles (on the rim of the main circle, with its white rim) to 4000 cm^2,then this second-blue-circle's area is equal to 10000 cm^2, in another word it's 2.5 times bigger then the small circle with their rim.

If we set the radius of inner circle to R, then the ratio between inner - second-blue-circle is:

R : $ \dfrac {\sqrt {10} - \sqrt {5}} {2\sqrt {2}} R $

The third blue circle, if we set the diameter of the inner circle to 365.242199 (beware of this value), then the radius of this

third-blue-circle is 56.433~ (1/2 Chacha's height).

You can test those relations yourself on this pic (original size), the length of pixels in this pic is 19 times of its real length

The pixel length in the pic is:

Circle (from rim to center ) Diameter's Pixel length
1st blue 2477
2nd blue 2273
3rd blue 2128
1st red 1064 = 1.5*(1st pink)
1st pink 709
2nd pink 944 = 1.5*(3rd pink)
3rd pink 629
2nd red ... 6st pink 0.5*(1st red ... 3rd pink)
For more about these numeric-game-ish things shown in Akazukin Chacha, please see another article, Characters' heights, Unit Ma and Unit CM in Akazukin Chacha.

Now we regard the (shade of?) coin on the center, we can see it's from the 4-fold-circle formation whose radius is half of the third-blue-circle.

And in this pic, we can see the height of the coin is 1 / 32 times of Magical Princess' height (the ratio of Chacha to MP is 1 : 1.618~, aka "Golden ratio", as the pic shown),

As we know MP's height is equal to the radius of the inner circle of Mahoujin, that's said the "shade" of the coin on the first pic is 4 times bigger than real coins (by area).

So we have drawn the 1 / 4-radius-of-third-blue-circle formation on the first pic.

As we can see by the plot, Magical Princess plays the "ideal human" or "the Eidos / Idea of human" role in Akazukin Chacha. All features of MP are meaningful, because they relate to almost all other things shown in Akazukin Chacha.

The white rim of small circles


simple integer ratio : 71:82

We can see the radius ratio between the outer-white-rim and inner-white-rim of small circles is about 1 : cos(pi/6).

And we have seen above that Chacha : MP = 1 : 1.618~ (golden ratio)

So, let the radius of inner circle be 1, then Chacha's height is (sqrt(5)-1)/2.

The diameter of the central circle is 2*(cos(3pi/7)/cos(2pi/7)), or 0.713791735

We can see 0.618(Chacha's height)/(0.713791735) is almost cos(pi/6), in the other word, the "white rim ratio" of small circles.

We can see the "central circle" and the small circles actually have same mode.

Thus, let the (big) inner circle's diameter be 1, the diameter of the "(small) inner circle" of (small) circles is:

$ \dfrac {\cos\dfrac {2\pi } {7}\left( \sqrt {5}-1\right) \left( \sqrt {2}-1\right) } {8\cos \dfrac {3\pi } {7}} $

this can be roughly: ((sqrt(2)-1)/2)*cos30 (hexagon), accuracy 99.9793%.

That's said the "Golden ratio" can be: $ \dfrac {\cos \dfrac {2\pi } {7}} {2\cdot \cos \dfrac {3\pi } {7}\cos \dfrac {\pi } {6}}\approx \phi $(accuracy: 99.9793%)

We can make another approximation as:

$ \dfrac {\cos \dfrac {2\pi } {7}} {\cos \dfrac {3\pi } {7}}\left( \dfrac {2} {5}\right) ^{3} $(accuracy: 99.99924224%)


That's said the inner circles of small circles and the inner circle of the MAIN circle are actually related, as the pic:

If we make an (outer) heptagram based on the (main) inner circle, we can see its diameter is almost 2 times of the (main) outer circle.

We know the diameter of the "Mahoujin Garden" (see below) is 5 times of the (main) outer circle, this is another 2/5 ratio.

Thus the diameter of small inner circles can be written as:

diam. Mahoujin Garden*$ \left( \dfrac {2} {5}\right) ^{4} $ (accuracy 99.48%)


Visualize MP : Chacha, inner small circle, "Coins"

The pic above shows Chacha's height as the Diam. of third Central circle, and MP as the radius of the (inner) main circle.

There is also a Triple-in-One relation has been shown:

The third central circle is 3 times bigger then the small circles' inner circle ( by area ).
(since [cos(3pi/7)*2/cos(2pi/7)*(sqrt(2)-1)] is about sqrt(3) (99.5%))

White rim of the Main Circle & the secondary heptagram


From Ep. 48, the most inner central circle is the 4th circle, not the 3rd circle ( which diameter = Chacha's height )we said before

As the pic, we can see the white rim of the outer circle is:

(57/55)*original outer circle (expanded)
We assume the white rim of the inner circle to be this ratio as well, so the inner white rim can be:
(55/57)*original inner circle (contracted)
since it has:

$ \dfrac {1+\sqrt {1+8\sqrt {8}}} {2\sqrt {8}}\approx \dfrac {57} {55} $


(outer circle's diameter)*57/55 - ( inner circle's diameter )*55/57 is approx to the radius of the inner circle.

((outer circle's diameter)*57/55 - ( inner circle's diameter )*55/57 is two times of the space between expanded outer circle and the contracted inner circle)

It also means that:

If we split the white ring to 56 parts, let the length between two sides of the white rim be a, then the area of each parts is (a2)*pi, hence a circle with radius a.

Properties relate to this 57/55 ratio

  1. 57/55 is also approximately 1/cos(2pi/24). That's said a Dodecagon around the inner/outer circle, formed by the angles from 24 coins formation ( see below ), can make the white rims of the Outer / Inner. Accuracy:99.895%
  2. The radius of [small inner circles] about 153^(1/4) times of the width of each [outer white rim]/56. The (153^(1/4))^2 or 12.369~ is close the the annual lunation rate. (be accurate when 99.97% of 57)

The properties below belong to the 1st Central circle and its expanded white rim.

  1. The quarter of [outer+its rim] is close the the diam. of expanded white rim of 1st Central circle, which is also about (57/55)*[diam. 1st Central circle].
  2. The two vertices on the bottom of the yellow (approximately) triangle on each arm of the yellow heptagram forms a 38.5714~ or 90-(360/7) angle when they linked to the center point of whole Mahoujin.

The properties below belong to the Unit MA and Unit cm article and its [diam. of inner = 365.2421990741MA] result.

  1. The whole area of [outer+its rim] can be set to 20 sq. meters, thus the accurate value of 57/55 is : 9*400*sqrt(5pi)/13770.07731.

Center of mass of Circular sectors on the whole Mahoujin


The Mass center is on the point where the (smaller) cardioid crossing the bisector of each sector.

As the pic shows, the mass centers of continuous Circular Sectors ( on a circle ) form a Cardioid. When the angle of the sector closes to 0, the distance between the center and the [mass center] is (2/3)*R.

Regard this Cardioid as a "heart" (shape), then we can get that the length between the two sides of the "heart" is almost equal to [diam. main inner circle]/sqrt(2).

It just like the [main outer circle] is made by [diam. main inner circle]*sqrt(2).

That's said the [tangent square of "heart"] -> [inner] -> [outer] is a series ascending by 1.4142~ or sqrt(2).

Additionally, we have this:


(be accurate when 100.0099403% of 57)

R is the radius of the [main outer], the (2/3+0.1448224188) multiplier is height (rate) of of the "heart".

That's said the "heart" shape ( built from the 57/55 ratio Akazukin Chacha shows by itself ) also relates to [main outer]'s inscribed square and hexagon.

Now see into the bottom point of "heart" and that (55/57)*[main inner] circle, where marked by red.dashed lines.

We know the distance from central point to the bottom of "heart" is (2/3)*(57/55)*sqrt(2)*[radius of main inner].

Thus the length inside [main inner]'s white rim is {1-(2/3)*(57/55)*sqrt(2)}*[radius of main inner].

This length is close to ((1-55/57)/φ)*[radius of main inner]. (be accurate when 100.0777% of 57)

We use this relation to build the expanded white rim of 1st Central circle: Align MP's top to the top of the [main inner], align Chacha's top to the 1/φ position from [main inner's white rim], then the distance between Chacha's standing ground to MP's standing ground, hence the distance (55/57)*([MP's height] - [Chacha's height]),

is the radius of expanded white rim of 1st Central Circle.



light blue one is the Hexa~'s; pink+blue one is the Quadr~'s.

The pic shows a triangle face of a quadrangular pyramid and a triangle face of a Hexagonal pyramid.

The height of the Hexa~'s triangle face is the radius of [main outer], the width is the diam. of [main inner].

The height of the Quadr~'s triangle face is the radius of [main inner], the width is 1/2 of the diam. of the [3rd Central Circle] / Chacha's height.

Now build these two pyramids: put both two bottoms on the Mahoujin "ground", and both two apex above the center point of Mahoujin.

We can see the height of Quadr~ is same as the height of MP, the height of Hexa~ is almost equal to [height of MP]*(55/57), (be accurate when 100.144% of 57), in another word the radius of the contracted white rim of [main inner]. The Hexagonal pyramid can also envelop the whole figure of Chacha.

If we build a smaller Hexagonal pyramid base on the bottom of the current one, with value (sqrt(3)/2)*[Chacha's height] as the height of the new one's triangle faces. Then the distance between the two apex (current<->new) is half of MP's height.

(since: $ \dfrac {\sqrt {21+8\sqrt {5}}} {4}=\dfrac {\phi} {2}+\dfrac {3} {4} $)

Now we explain the linkage of the Quadr~ and this new ( smaller ) Hexagonal pyramid.

Let N be a odd number. We know the triangle faces' height formula of a N-gonal pyramid made from a (regular) N-gram or N-star inside a [unit circle] is:

h = (1-(1/(1+2cos(pi/N))))/2

We call a triangle face which has this value the "regular shape" (of a N-gonal pyramid).

Apply this formula to even number bases, the we can get a Hexa~'s "regular shape" value is sqrt(3)+1 when the radius of its base is 1; a Quadr~'s regular shape" value is 2+sqrt(2) when the side of its base square is 2.

Minus Quadr's regular by 2 and Hexa~'s regular by 1, the we have the Quadr~ and the ( smaller ) Hexa~ we use.


The longer red bar = MP's height, the shorter one almost = Chacha's height.

Now we duplicate the Quadr~ and rotate the duplicated one 45 degree. The bottom of these two now looks like the pic:

We know the area of this bottom is (4-2*sqrt(2))*(2*[MP height])^2 or (4-2*sqrt(2))*(365.2421990741~)^2.

Now the volume of the pyramid with this bottom is

26048.33~*365.2421990741 MA3 (MA is a length unit, see Unit MA and Unit Cm article )

26048.33~ is close to the Great Year (precession year) value by years.

Again, we see the polyhedrons relate to 8 (bigger one) and 6 (smaller one), just like the two gems on Princess Medallion and Chacha's ring ( See below ).

Now we put the develop diagrams of these pyramids on their own position, hence the center of Mahoujin. We have:

Since it has: $ \dfrac {1} {2\sqrt {2}\phi}+\sqrt {\dfrac {2} {3}} $= 57/55 approximately (99.8689%), so the Hexa~ (blue one)'s circumscribed hexagon ( bigger circumscribed hexagon ) is the [main outer with white rim]'s inscribed hexagon.

There is also: $ \dfrac {3} {7}\left( \sqrt {2}+1\right) $ = 57/55 approximately (99.8359%). That's said if we divide the diameter of the Quadr~'s (bigger) circumscribed circle to 2:3:2 rate, then the middle 3 is the diameter of the [main outer with white rim] . Both 7 and 3 are meaningful numbers in Akazukin Chacha itself as we have already seen.

The light blue one is the smaller Hexa~ we mentioned before, the diam. of its circumscribed circle is close to [MP's height]+[Chacha's height] or [MP's height]*φ.

Expanded white rim of 1st Central Circle

The radius we make from MP-Chacha, is also close to the 1st central circle multiply with (57/55):

(55/57)*(1-1/φ) = (57/55)*cos(3pi)/cos(2pi) approx. (be accurate when 99.823% of 57)

Yellow and Green Heptagram

The side of an arm of the yellow heptagram is 100MA ( Let the radius of [main inner]  = MP height = 365.242~/2 MA = 172.2~ cm, see Unit MA and Unit CM article).

We call the ( radius of the circumcircle of the ) outer side of an heptagram P, and inner side of an heptagram Q.

We explainded the yellow one's P and Q before. Let the radius of [main inner] be 1, then ( yellow one's ) P = 1, Q= sqrt(2)*cos(2pi/7).

In the green one, P is (57/55)*cos(pi/7) and Q is (55/57)*sqrt(2)*cos(2pi/7). Since the root of  this equation

$ \dfrac {\cos \dfrac {\pi } {7}} {x}=\sqrt {2}\cos \dfrac {2\pi } {7}\left( x-1\right)+x $ is almost (55/57) (be accurate when 100.009% of 57)

So the distance between Green one's P to Q, is equal to [main inner]'s white rim to Yellow one's Q.

The "Mahoujin Garden" and number "24"


In Ep. 48, as we can see the "Garden" where the Mahoujin land, its radius is 5 times of the Mahoujin, as the pic:

The number 5 is also relates to 7 via the inner / main circle relation. We can see:

$ 5 \approx \dfrac {7} {\sqrt {2}} $(98.995%)

that's said if we draw 7 inner-circle-size circles (begin with the true inner circle ), than it reaches the rim of the Garden.

thus the garden's area = 25*(mahoujin area).

and the "meadow" area is (garden-1*Mahoujin) = 24 times of the Mahoujin's area.

that's said, every fan-shaped area (as the pic, divided by red straight lines) is equal to a Mahoujin.

Is "24" meaningful in Akazukin Chacha? As we can see, "Princess Medallion" has 24 "golden beans" around the gem.

And the LCM of Medallion's gem (hexagon, 6) and Chacha's ring's gem (octagon, 8) is also 24.

As we know, in Ep.5 or Ep, 22, the ring is the key to unlock the gem on Medallion.

More importantly, the number "5" is the unique solution of the pattern on the left pic. We explain it below.

At the beginning of Ep. 28, there was a "Map" had been shown:


The "map" from Ep. 28

As the "Map" said, the 3 centers of each pond forms an equilateral triangle, and there is a big ( virtual ) circle wrap over all three circles. (The rim of the "big circle" touches every there ponds).

If we use Chacha's height as a reference, than we can measure out the "three ponds" in Ep. 28, which every ponds radius is equal to Ep. 48's Mahoujin's diameter.

And as the minimal distance between two ponds (the white line on the bottom of pic) says, the circular central space of these ponds is equal to a Mahoujin ( as the white space on the pattern on the top of the pic).

That's said, the radius of the wrap circle of three ponds (externally tangent circle of the three ponds) is 2*(2*r)+1 = 5r,

and like a miracle, 5r is just the "Mahoujin Garden"'s radius in Ep. 48.

Now we can draw the three ponds and its wrap circle geometrically, as the red-black-green pattern on the pic.

We know the wrap circle's area is 25 r^2pi, and the "pond"'s area is (2*2)r^2pi = 4r^2pi, thus minus the three ponds and the central area (radius 1r), the residual area is:

$ 25r^{2}\pi -1r^{2}\pi -3\cdot 4r^{2}\pi = 12r^{2}\pi $


All blue and yellow area in one big circle are equal

Again, it's a miracle, the "12r^2pi" just equal to the sum of three pond's area, and we know the "residual area" is divided into 3 equal parts. That's said, every "pond" and the semi-fan-shaped area on its opaposite side, are equal.

This is the unique solution in the case of "3" on this pattern, as we can proof it in math, like the pattern:

If we want to make the blue circle and its opposite yellow area equal, then the central circle (red circle) to side circles (blue circle)'s radius ratio is 1 : 2, thus the wrap circle to central circle is 5 : 1, just like the one Ep. 28 or Ep.48's "Mahoujin Garden" has shown.



Also, the "coins" from Ep. 36 to 47, the angle of its two sides is 15 degree, therefore

24 coins form a circle:


The formation of the 4-fold-circles is:

  • The diameter of 1 circle is about 4.9 times of the coin's height.
  • We can make 4 circle with this relation.
  • The ratio of 1 to 3 is 1.4785752424~
  • The ratio of 4 to 2 is also 1.4785752424~

Let ANS=2*cos(pi/24), then ANS-1/ANS = 1.4785752424.

This ratio can make the "opposite angle" equals to two times of the central angle when θ = 2pi/24.

We can see more this "ANS-1/ANS" type formulas in the Unit MA and Unit CM article.

Note: The coin picture from the OS (Official Setting) is not parallel to the screen surface, so we have applied fixes on it.


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