Posted by: Dave Richeson | June 1, 2012

## Angle trisection using origami

It is well known that it is impossible to trisect an arbitrary angle using only a compass and straightedge. However, as we will see in this post, it is possible to trisect an angle using origami. The technique shown here dates back to the 1970s and is due to Hisashi Abe.

Assume, as in the figure below, that we begin with an acute angle ${\theta}$ formed by the bottom edge of the square of origami paper and a line (a fold, presumably), ${l_{1}}$, meeting at the lower left corner of the square. Create an arbitrary horizontal fold to form the line ${l_{2}}$, then fold the bottom edge up to ${l_{2}}$ to form the line ${l_{3}}$. Let ${B}$ be the lower left corner of the square and ${A}$ be the left endpoint of ${l_{2}}$. Fold the square so that ${A}$ and ${B}$ meet the lines ${l_{1}}$ and ${l_{3}}$, respectively. (Note: this is the non-Euclidean move—this fold line cannot, in general, be drawn using compass and straightedge.) With the paper still folded, refold along ${l_{3}}$ to create a new fold ${l_{4}}$. Open the paper and fold it to extend ${l_{4}}$ to a full fold (this fold will extend to the corner of the square, ${B}$). Finally, fold the lower edge of the square up to ${l_{4}}$ to create the line ${l_{5}}$. Having accomplished this, the lines ${l_{4}}$ and ${l_{5}}$ trisect the angle ${\theta}$.

Let us see why this is true. Consider the diagram below. We have drawn in ${CD}$, which is the location of the segment ${AB}$ after it is folded, ${AC}$, the fourth side of the isosceles trapezoid ${ABDC}$, and ${AD}$, the second diagonal of ${ABDC}$. We must show that ${\theta=3\alpha}$, where ${\alpha=\angle DBE}$ and ${\theta=\angle CBE}$.

Because ${BE}$ and ${DF}$ are parallel, ${\angle DBE=\angle BDF}$, and because ${DF}$ is the altitude of the isosceles triangle ${ABD}$, ${\angle BDF=\angle ADF}$. Thus ${\alpha=\angle DBE=\angle BDF=\angle ADF}$. Now, ${ABDC}$ is an isosceles trapezoid and ${ABD}$ is an isosceles triangle, so ${ABD}$ and ${BCD}$ are congruent isosceles triangles. Thus ${\angle CBD=\angle ADB= \angle BDF+\angle ADF}$. It follows that ${\theta=\angle CBE=\angle DBE+\angle CBD=\angle DBE+\angle BDF+\angle ADF=3\alpha}$.

The geometric properties of origami constructions are quite interesting. Every point that is constructible using a compass and straightedge is constructible using origami. But more is constructible. As we’ve seen, it is possible to trisect any angle using origami (I’ll leave the obtuse angles as an exercise). It is possible to double a cube. It is possible to construct regular heptagons and nonagons. In fact, where the constructability of ${n}$-gons is related to Fermat primes, the origami-constructibility of ${n}$-gons is related to Pierpont primes. While the field of constructible numbers is the smallest subfield of ${\mathbb{R}}$ that is closed under square roots, the field of origami-constructible numbers is the smallest subfield that is closed under square roots and cube roots. In fact, it is possible to solve any linear, quadratic, cubic, or quartic equation using origami!

There are quite a few places to read about geometric constructions using origami, but a good starting point is this online article (pdf) by Robert Lang.

## Responses

1. Reblogged this on Room 196, Hilbert's Hotel.

2. INTERESTING.
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You may refer to my relevant solutions at:http://www.stefanides.gr/Html/Classical_Problems_et_Alii_with_Web_Links.htm
Panagiotis Stefanides

3. quick question how do you write math in the blog?

4. Neusis CANNOT trisect an angle. The trisection is only an approximation by the resolution of what your eye can see.