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Simple Origami? (Posted on 2006-03-07) Difficulty: 3 of 5
Take a square piece of paper oriented with its top horizontal. Fold it along any line that passes through the center and forms an angle of between 0 and 45 degrees with the horizontal.
The outline of the resulting shape is a nonagon.

What angle will maximize the perimeter of this nonagon?

What angle will maximize the area of this nonagon?

Is there any other single fold (not through the center) that can do better for either of these?

No Solution Yet Submitted by Jer    
Rating: 4.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Hints/Tips re: refining the answer-computer solution | Comment 7 of 8 |
(In reply to refining the answer-computer solution by Charlie)

The area of the nonagon is the area common to the two folds plus the area of four congruent right triangles. This area is equal to the area of two of the right triangles plus the area of half the square. Therefore, to maximize the area of the nonagon, one must maximize the area of the right triangle. This occurs when the right triangle is isosceles. Using this we get arctan(sqrt(2)-1) = 22.5 degrees for our angle.

For the perimeter I get arctan(x) where x is one of the positive real roots of

4*(1+x^2)*(1-2*x-x^2)^2 - [x^2]*(1+x)^4 = 0

approximately 33.508030917 degrees.

Edited on March 8, 2006, 11:26 am
  Posted by Bractals on 2006-03-08 04:41:52

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