# An Interesting Example of Angle Trisection by Paperfolding

### Sidney H. Kung

January 6, 2011

Hull has shown a method (by H. Abe) of trisecting an angle via paperfolding (*origami*, in Japanese). Here we present a different one that was communicated to me years ago by Dr. Swend T. Gormsen, a retired Prof. of V.P.I. at Blacksburg, VA. I was able to prove the result only recently.

Let 2*θ* be the vertex angle of an isoceles triangle *ABC* (Figure 1) and let *ll'* be the perpendicular bisector of *BC*. *AE* is the extension of side *CA*.

Figure 1 |
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Fold the paper up (by Angle Trisection by Paper Folding, O6) so that points *B* and *C* fall upon *AE* (at the point *B'*) and *ll'* (at the point *C'*), respectively, and then unfold. Denote by *O* the point of intersection of *ll'* and the line of crease *zz'* (Figure 2). Construct line segment *BO*. Then we have ∠*AOB* = 2*θ*/3.

Figure 2 |
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### Proof

By symmetry we see that *OB* = *OB'* = *OC* = *OC'*. So, *B*, *B'*, *C*, and *C'* lie on a circle centered at *O*. Thus, *arcBC* = *arcB'C'*. Construct *C'E'* parallel to *CE* intersecting the circle at *N*. Let *M* be the antipode of *C'*.

Then we have *arcCN* = *arcB'C'*. Further,

θ | = ∠MC'N | |

= (arcMC + arcCN)/2 | ||

= (arcBC/2 + arcB'C')/2 | ||

= 3·arcBC/4. |

Therefore, ∠*AOB* = *arcBC*/2 = 2*θ*/3.

### References

- T. Hull,
*Project Origami: Activities For Exploring Mathematics*, A.K. Peters Ltd. (2007) 49-50.

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