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 Simple Origami? (Posted on 2006-03-07)
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.)
 refining the answer-computer solution | Comment 5 of 8 |

DECLARE FUNCTION peri# (a#)
DECLARE FUNCTION area# (a#)
DEFDBL A-Z
DIM SHARED pi, dr
pi = ATN(1) * 4
dr = pi / 180

' functions based on square 2 units on a side
CLS

FOR angle = 21 TO 35 STEP .5
PRINT USING "##.### ##.#### ##.####"; angle; peri(angle) / 2; area(angle) / 4
' division by 2 or 4 based on side of 2 area 4 for square.
NEXT

FUNCTION area (a)
d1 = TAN(a * dr): d2 = 1 - d1
a2 = 2 * a
d3 = d2 * TAN(a2 * dr)
h1 = d2 / COS(a2 * dr)
d4 = 1 + d1 - h1
d5 = d4 / TAN(a2 * dr)
h2 = SQR(d4 * d4 + d5 * d5)
d6 = 2 - h2 - d3
d7 = d6 * TAN(a2 * dr)
h3 = 1 + TAN(a * dr) - d7
d8 = h3 * SIN(a2 * dr)
d9 = h3 * COS(a2 * dr)
area = 2 + d4 * d5 / 2 + d8 * d9 / 2
END FUNCTION

FUNCTION peri (a)
d1 = TAN(a * dr): d2 = 1 - d1
a2 = 2 * a
d3 = d2 * TAN(a2 * dr)
h1 = d2 / COS(a2 * dr)
d4 = 1 + d1 - h1
d5 = d4 / TAN(a2 * dr)
h2 = SQR(d4 * d4 + d5 * d5)
d6 = 2 - h2 - d3
d7 = d6 * TAN(a2 * dr)
h3 = 1 + TAN(a * dr) - d7
d8 = h3 * SIN(a2 * dr)
d9 = h3 * COS(a2 * dr)
peri = 2 / COS(a * dr) + d2 + d3 + d4 + d5 + d6 + d7 + d8 + d9
END FUNCTION

finds

`angle     perim   area21.000  3.4130  0.585521.500  3.4173  0.585622.000  3.4215  0.585722.500  3.4255  0.585823.000  3.4294  0.585723.500  3.4330  0.585624.000  3.4365  0.585524.500  3.4397  0.585225.000  3.4428  0.584925.500  3.4457  0.584526.000  3.4485  0.584026.500  3.4510  0.583427.000  3.4534  0.582827.500  3.4557  0.582128.000  3.4577  0.581328.500  3.4596  0.580429.000  3.4612  0.579529.500  3.4628  0.578530.000  3.4641  0.577430.500  3.4653  0.576231.000  3.4663  0.574931.500  3.4671  0.573632.000  3.4677  0.572132.500  3.4682  0.570633.000  3.4684  0.569033.500  3.4685  0.567334.000  3.4684  0.565634.500  3.4682  0.563735.000  3.4677  0.5617`

with maxima at 22.5 degrees and 33.5 degress for area and perimeter respectively.  The symmetry of values about these points makes it look as if these values are exact.  That can be tested by making the increments more fine-grained:

`22.000  3.421533  0.58574922.050  3.421942  0.58575722.100  3.422349  0.58576322.150  3.422754  0.58576822.200  3.423157  0.58577322.250  3.423559  0.58577722.300  3.423958  0.58578122.350  3.424356  0.58578322.400  3.424752  0.58578522.450  3.425146  0.58578622.500  3.425538  0.58578622.550  3.425928  0.58578622.600  3.426317  0.58578522.650  3.426703  0.58578322.700  3.427088  0.58578122.750  3.427471  0.58577722.800  3.427852  0.58577322.850  3.428232  0.58576822.900  3.428609  0.58576322.950  3.428985  0.585757`

makes it look like a correct hypothesis for area, but

`33.000  3.468435  0.56901833.050  3.468453  0.56885333.100  3.468468  0.56868833.150  3.468482  0.56852133.200  3.468495  0.56835433.250  3.468505  0.56818533.300  3.468513  0.56801633.350  3.468520  0.56784633.400  3.468525  0.56767533.450  3.468528  0.56750433.500  3.468529  0.56733133.550  3.468529  0.56715833.600  3.468526  0.56698333.650  3.468522  0.56680833.700  3.468516  0.56663233.750  3.468508  0.56645433.800  3.468498  0.56627733.850  3.468486  0.56609833.900  3.468473  0.56591833.950  3.468457  0.565737`

makes it look only approximate for the perimeter.

Verifying:

`22.440  3.42506711  0.5857859122.450  3.42514577  0.5857860722.460  3.42522435  0.5857862022.470  3.42530286  0.5857863022.480  3.42538130  0.5857863822.490  3.42545966  0.5857864222.500  3.42553795  0.5857864422.510  3.42561616  0.5857864222.520  3.42569430  0.5857863822.530  3.42577237  0.5857863022.540  3.42585036  0.5857862022.550  3.42592828  0.5857860722.560  3.42600612  0.5857859122.570  3.42608389  0.5857857122.580  3.42616158  0.5857854922.590  3.42623921  0.5857852422.600  3.42631676  0.58578496`
`33.502  3.46852923  0.5673241433.503  3.46852924  0.5673206833.504  3.46852924  0.5673172133.505  3.46852924  0.5673137533.506  3.46852924  0.5673102933.507  3.46852925  0.5673068233.508  3.46852925  0.5673033633.509  3.46852925  0.5672998933.510  3.46852924  0.5672964333.511  3.46852924  0.5672929633.512  3.46852924  0.5672895033.513  3.46852924  0.5672860333.514  3.46852923  0.56728256`

it looks like 22.5 degrees exactly for the area and about 33.508 degrees for the perimeter.

Non-central fold:

The perimeter and area can approach as close as you like to the full perimeter and area of the square by folding over as little as possible from an edge or corner.

 Posted by Charlie on 2006-03-07 22:52:54

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