Angle Trisection By Paper Folding

It is a classical result that, in general, Angle Trisection is impossible by the Euclidean ruler and compass rules. Several methods that went beyond ruler and compass - the neusis constructions - have been known yet to the ancient Greeks (see angle trisection by Archimedes and Hippocrates and a more recent.) Axioms of Paper Folding also supply the tools sufficiently powerful to successfully tackle the angle trisection problem.

The following method of angle trisection is due to H. Abe (1980).

Let's consider an acute angle ABC:
By axiom O4, erect a crease through B perpendicular to AB. (Note that, by O1, AB may be considered a crease.) Execute two creases parallel to AB and at equal distances from each other: BD = DE.
By O6, fold E into BC and B into L. Let B₁, D₁, E₁ be images of points B, D, and E, respectively. Let L₁ be the image of crease L. Refold it as to make a crease on the upper part of the paper that remain at rest. Straighten the paper into its original position and fold and unfold crease L₁ again. We next show that the crease passes through the point B and, moreover, divides ∠ABC in the ratio 2:1 thus solving the trisection problem.
The whole structure is symmetric with respect to the crease OO_B. In the isosceles trapezoid BDD₁B₁, the diagonals intersect on the axis of symmetry. ΔDD₁O_D is isosceles by construction. Therefore, the creases BD₁ and D₁O_D coincide, or, in other words, D₁O_D does pass through the vertex B. Now, everything is pretty much straightforward. Just recollect that the crease L is the line DB₁ and BO_B lies along the given line AB and that, by construction L and AB are parallel. Thus we see that ∠B₁BO_B = ∠DB₁B. Since ΔBO_DB₁ is isosceles, ∠O_DBB₁ = ∠O_DB₁B. BD₁ is perpendicular to OB₁ because OB is perpendicular to DB₁. Also, E₁D₁ = D₁B₁. This shows that ΔE₁BB₁ is isosceles and, therefore, ∠E₁BD₁ = ∠D₁BB₁.

References

G. E. Martin, Geometric Constructions, Springer, 1998

Paper Folding Geometry

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