## Dividing a Segment into Equal Parts by Paper Folding

It is easy to find the midpoint of a side of a square piece of paper by simply folding one of the corners onto an adjacent corner. This done, we can also find 1/3 of the side, as shown below.

We do not need anything more sophisticated to get 1/4 of the side than finding another midpoint, but, as the second diagram shows, starting with 1/3 of the side we still can get 1/4, and then also 1/5. The procedure is general enough and leads to a successive construction of a 1/k-th of the side for any whole k. To see why this is so, consider the diagram below:

,

where the corner B has been folded onto the point Q, such that AQ = AD/k, for a given k, not necessarily an integer. We shall assume the side of the square to be 1. Let

(1 - x)^{2} - x^{2} = 1/k^{2},

from which x = (k^{2} - 1)/(2k^{2}).

Further, angle PQR = 90°, which implies similarity of right triangles APQ and DQR:

DR/DQ = AQ/AP,

or

DR | = DQ·AQ/AP |

= (1 - 1/k) · 1/k · 2k^{2}/(k^{2} - 1) | |

= (k - 1)/k · 1/k · 2k^{2}/(k^{2} - 1) | |

= 2/(k + 1). |

For the midpoint T of DR, we get

DT = 1/(k + 1),

which proves the construction.

### Remark

After completing this page I have discovered R. J. Lang's online paper where he describes this method as Haga's Construction after Kazuo Haga.

Another procedure known as the crossing diagonal method has a longer history and an exciting story that evolved around 1995.

### References

- C. Abbott,
__What Good Is Proof by Induction Anyway?__,*Math. Horizons*, Sept. 2005, pp. 8-9 - R. J. Lang,
__Four Problems III__,*British Origami*, no 132, October 1988, pp. 7-11

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