Set by Kawasaki

The Kawasaki Theorem is a mathematical theorem about origami formulated by Toshikazu Kawasaki .

statement

The alternating sum of the angles between the folds of the unfolded figure is zero, which is why the figure can be folded flat.

A folding pattern with a center where all the folds meet can be folded flat (i.e. flattened) if and only if the alternating sum of the angles between two successive folds of the folding pattern (in the unfolded state) is zero, i.e.:

${\displaystyle \alpha _{1}-\alpha _{2}+\alpha _{3}-\cdots +\alpha _{2n-1}-\alpha _{2n}=0}$,

where are the angles between two consecutive folds of the folding pattern (the number of folds is even according to Maekawa's theorem ). ${\displaystyle \alpha _{1},\ldots ,\alpha _{2n}}$

Alternative formulation

Since the following applies to every folding pattern on a flat piece of paper with a center where all the folds meet: one gets (if one measures the angle in radians ) by the addition method : ${\displaystyle \alpha _{1}+\alpha _{2}+\cdots +\alpha _{2n}=2\pi }$

${\displaystyle \alpha _{1}-\alpha _{2}+\alpha _{3}-\cdots +\alpha _{2n-1}-\alpha _{2n}=0\Leftrightarrow \alpha _{1}+\alpha _{3}+\cdots +\alpha _{2n-1}=\pi }$.

Since the equivalence is transitive , in the case of a flat piece of paper the sentence can be reformulated as follows:

A folding pattern with a center where all the folds meet can be folded flat (i.e. flattened) if and only if:

${\displaystyle \alpha _{1}+\alpha _{3}+\cdots +\alpha _{2n-1}=\pi }$,

being the angles between two successive folds of the fold pattern. ${\displaystyle \alpha _{1},\ldots ,\alpha _{2n}}$

This equivalence does not apply to a non-flat original paper, i.e. H. the alternative wording is not applicable. Nevertheless, the first formulation with non-flat starting papers, such as z. B. skittles, nevertheless applicable. ${\displaystyle \alpha _{1}-\alpha _{2}+\alpha _{3}-\cdots +\alpha _{2n-1}-\alpha _{2n}=0}$

Proof idea

The proof is split into two parts, namely the back and forth direction of the equivalence :

• Imagine a figure folded flat with a center where all the folds meet. Let this figure be folded flat. If you walk along the edge exactly once, the position on the edge corresponds to an angle at the center. You now record all the angle differences between two folds and number them chronologically according to the time you walked off. If you have completed the entire route, you get the angles . As one can change the direction at a fold, and the end is again at exactly the same angle where they were started running (with respect to the center, in the folded state), applies , . ${\displaystyle \alpha _{1},\ldots ,\alpha _{2n}}$ ${\displaystyle \alpha _{1}-\alpha _{2}+\alpha _{3}-\cdots +\alpha _{2n-1}-\alpha _{2n}=2\pi k}$ ${\displaystyle k\in \mathbb {Z} }$

But there , and , must . ${\displaystyle \forall i\in \{m\ldots ,2n\}:\alpha _{i}\neq 0}$ ${\displaystyle \alpha _{1}+\cdots +\alpha _{2n}=2\pi }$ ${\displaystyle k=0}$

• Be for a figure. Now you choose one so that is minimal (this expression must then be less than or equal to zero, otherwise a smaller expression would be of this type). Then from the angle one folds the figure accordion-style, i. H. always alternating first to the left and then to the right, so that right corresponds to negative angles. If you look at the position of the current fold, it never moves to the right, because the sum of the angles increases (or remains the same) at the first angles in order to close to zero (or to maintain the value). In the following angles, the current fold can never move further to the right than the first fold at the angle at which you started, otherwise a smaller sum would exist. Since the number of folds is even and because of the commutative law of addition, the following applies: ${\displaystyle \alpha _{1}-\alpha _{2}+\alpha _{3}-\cdots +\alpha _{2n-1}-\alpha _{2n}=0}$ ${\displaystyle i\in \{1,\ldots ,n\}}$ ${\displaystyle \alpha _{1}-\alpha _{2}+\alpha _{3}-\cdots +\alpha _{2i-1}-\alpha _{2i}}$ ${\displaystyle \alpha _{1}-\alpha _{2}+\alpha _{3}-\cdots +\alpha _{2n-1}-\alpha _{2n}=0}$ ${\displaystyle \alpha _{2i+1}}$ ${\displaystyle 2n-i}$ ${\displaystyle 2i}$ ${\displaystyle \alpha _{2i+1}}$ ${\displaystyle \alpha _{1}-\alpha _{2}+\alpha _{3}-\cdots +\alpha _{2i-1}-\alpha _{2i}}$

${\displaystyle \alpha _{2i+1}-\alpha _{2i+2}-\cdots +\alpha _{2n-1}-\alpha _{2n}+\alpha _{1}-\alpha _{2}+\alpha _{3}-\cdots +\alpha _{2i-1}-\alpha _{2i}=0}$,

you can connect the first corner and the last corner of the fold because they are in the same position and are on the far right so they don't overlap with the figure. This gives you a flat fold.

Individual evidence

1. ↑ See in the corresponding section of Logical Equivalence # The logical equivalence as a relation and its properties .
2. ↑ Thomas C. Hull: Geometric Folding Algorithms: Linkages, Origami, Polyhedra - Lecture 3. Erik Demaine, 2010, accessed on May 19, 2013 .  
3. ^ Thomas C. Hull: The Combinatorics of Flat Folds: a Survey. 2002, accessed May 21, 2013 .  
4. ↑ Erik Demaine : Geometric Folding Algorithms: Linkages, Origami, Polyhedra - Lecture 20. Erik Demaine, 2010, accessed on May 19, 2013 .