Mathematical proof for odd number division of paper ( Haga)

As per the previous post, I had mentioned that I will show you why the first few steps of the Baymax head leads to precise 7 parts of the paper. If you have not seen the diagrams of the Baymax head, click here and try to fold it! In this post, it will include mathematical workings to show that it works!

It will require you to have understanding on:

• Coordinates,

• Graph,

• Similar and congruence of triangles and

• Algebra!

If you are at a stage where you do not understand any of this, I hope this can inspire you to learn and appreciate Mathematics too and see how it is applied into origami constructions and divisions in this case!

Before I present, there is a DISCLAIMER! The concept is not thought by me so please do not give me ANY credits for this concept! This is done for fun due to curiosity! I referred to the book Origami5 which you can buy from here from page 519 on with the chapter title " Precise Division of Rectangular Paper into an Odd Number of Equal Parts without tools: An Origamics Exercise". A highly entertaining chapter on Origamics to prove the beauty of origami and how simple folds can lead to accurate and scarily arbitrary points on a piece on paper.

The first few folds on the paper for the Baymax diagram leads to a construction showed in the picture below:

To understand why this is possible, we would one to see a piece of paper as a graph. The origin (0.00, 0.00 ) being the lower left corner of the paper (Label A) and the furthest top right coordinate ( 1.00, 1.00 ) (Label C). As a reference from the Origmi5 as stated in the disclaimer above, I would like to show a full working of why this method can work on any regular rectangle.

This begins with a problem as shown in the diagram below this sentence:

We let L be the length of the rectangle while standardising the width to be 1. That goes to say that a square is 1 x 1. For this problem, we assume that it is any rectangle. And the problem is at a unknown fraction (u) of the length of the paper (uL) what would the coordinates of G be? Naturally, for the paper to be divided into odd number of parts, we will want the denominator of X and Y, mainly Y, to be odd numbered with an integer as the nominator ( E.g. 7/9 or 5/7 ).

The sequence of events that is to come would be to find:

1. Equation of the common line AB,

2. Equation of line CD,

3. Intercept coordinate at G (X,Y) at which Y is made the subject of formula of unknown fraction (u)

4. Find fraction, u, for the division that you want

so we can construct point D to achieve the construction.

1. Equation of common line AB

• Gradient of line AB = (1/2)/L = 1/(2L)

• Y-intercept when point of origin is at A = 1/2

• Equation of Line AB :

• y = x/(2L) + 1/2 which is also

• y = (x + L)/2L

2. Equation of Line CD

Gradient of Line CD: -1/(L/2 - uL)

Y-intercept when point of Origin is:

( L - uL ) / ( L/2 - uL ) --- similar triangles

= (1-u) / (1/2 - u ) --- Factorisation and simplification

= ( 2-2u ) / (1 - 2u ) --- taking out 1/2 from nominator and denominator

Equation of line CD: y = [ -x/(L/2 - uL) ] + ( 2-2u ) / ( 1 - 2u ) or

y = [ -2x /L.(1- 2u) ] + ( 2-2u ) / ( 1 - 2u )

3. Intercept Coordinate at G (X,Y)

At intercept, Eqn Line AB = Eqn Line CD:

(x + L )/2L = [ -2x /L.(1- 2u) ] + ( 2-2u ) / ( 1 - 2u )

[( x + L )( 1 - 2u )] / [ 2L / (1-2u ) ] = [ -4x / (2L.(1-2u) ] + { 2L ( 2 - 2u) / [ 2L ( 1 - 2u )] }

( x + L )( 1 - 2u ) = -4x + 2L ( 2 - 2u)

x - 2ux + L - 2uL = -4x + 4L - 4uL

5x - 2ux = 3L - 2uL

x.( 5 - 2u ) = L ( 3 - 2u )

x = L.( 3 - 2u ) / ( 5 - 2u ) ----- if the paper is a square then x = ( 3 - 2u ) / ( 5 - 2u ) since L = 1

Substituting x = L.( 3 - 2u ) / ( 5 - 2u ) into Eqn Line AB:

y = (x + L)/2L

y ={ [ L.( 3 - 2u ) / ( 5 - 2u )] + L } / 2L

y = { [ L.( 3 - 2u ) / ( 5 - 2u )] + L.( 5 - 2u ) / ( 5 - 2u ) } / 2L

y = { [ L.( 3 - 2u + 5 - 2u ) ] / ( 5 - 2u ) } * ( 1 / 2L )

y = ( 4 - 2u ) / ( 5 - 2u )

Intercept coordinate is G ( [L.( 3 - 2u ) / ( 5 - 2u )] , [( 4 - 2u ) / ( 5 - 2u )] )

4. Find fraction, u, for the division that you want and G ( X , Y ):

For this case I want 7 parts which is the same method the diagram used. To find the unknown fraction, u, that makes the construction for the odd number of parts, substitute say 5/7 into y liek the working below then make u the subject of formula!

y = ( 4 - 2u ) / ( 5 - 2u ) = 5 / 7

Therefore,

7 ( 4 - 2u ) = 5 ( 5 - 2u )

28 - 14 u = 25 - 10u

4u = 3 so

u = 3/4

Note that u starts from the right to the left of the x- axis therefore the marking at D is 1/4 of Length, L.

To find the x - coordinates of G ( X , Y ), Substitute u = 3/4 and L = length of paper you are using into L.( 3 - 2u ) / ( 5 - 2u ).

If i am using a square sheet, the L = 1 and u = 3/4 :

L.( 3 - 2u ) / ( 5 - 2u ) = 1 ( 3 - 2* 3/4 ) / ( 5 - 2 * 3/4 ) = ( 3/2 ) / ( 7/2 ) = 3/7

G ( X , Y ) when u = 3/4 and L = 1 is G (3/7 , 5/7 )

Look at the very first diagram above and you will see that it is exactly at 7 parts upon construction and fold out the Baymax head to see that it is true!

Conclusion

This method can be used for any piece of rectangular sheet of paper and that mathematically , you can fold through easy constructiona nd some geometry that it is possible to fold out a precise and accurate 7 parts without any measuring tools. This is one of the many joys of origami mathematics that can also inspire children and adults to improve their mathematics and insights and see the application of it come to live!

For any questions, please feel free to contact me either through the contact page or through the comment below! Have fun!

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