Steven Dutch, Natural and Applied Sciences, University of Wisconsin - Green BayXbr> First-time Visitors: Please visit Site Map and Disclaimer. Use "Back" to return here.

It's surprisingly easy to get an 11-pointed star by a double approximation. First, pi is approximately 22/7, and second, for small angles, tan a, a and sin a are all approximately equal. Therefore tan pi/22 = 1/7, approximately.

Fold a square of paper diagonally in half |

Fold the base of the resulting triangle in half |

Unfold the triangle to reveal the center crease |

Fold the top vertex down to the midpoint of the base |

Unfold the paper to reveal a crease halfway down from the top |

Fold the top vertex down to the halfway crease |

Unfold the paper to reveal a crease a quarter of the way down from the top |

Fold the top of the triangle down to the quarter crease |

If the height of the triangle is 1, the topmost crease is at an altitude of 7/8 and the half-width of the triangle at that point is 1/8, so we can fold an angle of arctan 7 as shown below. |

Now arctan 7 = 90-arctan 1/7 = 90-180/22. So the angle between the base of the
triangle and the upfolded edge is 5(180/11). The folded wedge spans an angle of
half the remainder or 3(180/11). So the total angle created by the folding is
8(180/11). Bisecting this angle three times gives a wedge with angle 180/11. |

Fold the remaining base of the triangle to coincide with the right-hand side of the wedge. |

Bisect the wedge again |

The final bisection results in a wedge with apical angle 180/11. |

Cut the wedge as desired and unfold as shown below. |

The process is animated, below |

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*Created 22 March 2006, Last Update
14 December 2009*