11-Pointed Star: Approximate

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

Return to Recreational Mathematics Topics
Return to Paper-Folding
Return to Symmetry Index
Return to Professor Dutch's home page

Created 22 March 2006, Last Update 14 December 2009