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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.)
Solution 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   area
21.000  3.4130  0.5855
21.500  3.4173  0.5856
22.000  3.4215  0.5857
22.500  3.4255  0.5858
23.000  3.4294  0.5857
23.500  3.4330  0.5856
24.000  3.4365  0.5855
24.500  3.4397  0.5852
25.000  3.4428  0.5849
25.500  3.4457  0.5845
26.000  3.4485  0.5840
26.500  3.4510  0.5834
27.000  3.4534  0.5828
27.500  3.4557  0.5821
28.000  3.4577  0.5813
28.500  3.4596  0.5804
29.000  3.4612  0.5795
29.500  3.4628  0.5785
30.000  3.4641  0.5774
30.500  3.4653  0.5762
31.000  3.4663  0.5749
31.500  3.4671  0.5736
32.000  3.4677  0.5721
32.500  3.4682  0.5706
33.000  3.4684  0.5690
33.500  3.4685  0.5673
34.000  3.4684  0.5656
34.500  3.4682  0.5637
35.000  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  0.585749
22.050  3.421942  0.585757
22.100  3.422349  0.585763
22.150  3.422754  0.585768
22.200  3.423157  0.585773
22.250  3.423559  0.585777
22.300  3.423958  0.585781
22.350  3.424356  0.585783
22.400  3.424752  0.585785
22.450  3.425146  0.585786
22.500  3.425538  0.585786
22.550  3.425928  0.585786
22.600  3.426317  0.585785
22.650  3.426703  0.585783
22.700  3.427088  0.585781
22.750  3.427471  0.585777
22.800  3.427852  0.585773
22.850  3.428232  0.585768
22.900  3.428609  0.585763
22.950  3.428985  0.585757

makes it look like a correct hypothesis for area, but

33.000  3.468435  0.569018
33.050  3.468453  0.568853
33.100  3.468468  0.568688
33.150  3.468482  0.568521
33.200  3.468495  0.568354
33.250  3.468505  0.568185
33.300  3.468513  0.568016
33.350  3.468520  0.567846
33.400  3.468525  0.567675
33.450  3.468528  0.567504
33.500  3.468529  0.567331
33.550  3.468529  0.567158
33.600  3.468526  0.566983
33.650  3.468522  0.566808
33.700  3.468516  0.566632
33.750  3.468508  0.566454
33.800  3.468498  0.566277
33.850  3.468486  0.566098
33.900  3.468473  0.565918
33.950  3.468457  0.565737

makes it look only approximate for the perimeter.

Verifying:

22.440  3.42506711  0.58578591
22.450  3.42514577  0.58578607
22.460  3.42522435  0.58578620
22.470  3.42530286  0.58578630
22.480  3.42538130  0.58578638
22.490  3.42545966  0.58578642
22.500  3.42553795  0.58578644
22.510  3.42561616  0.58578642
22.520  3.42569430  0.58578638
22.530  3.42577237  0.58578630
22.540  3.42585036  0.58578620
22.550  3.42592828  0.58578607
22.560  3.42600612  0.58578591
22.570  3.42608389  0.58578571
22.580  3.42616158  0.58578549
22.590  3.42623921  0.58578524
22.600  3.42631676  0.58578496
33.502  3.46852923  0.56732414
33.503  3.46852924  0.56732068
33.504  3.46852924  0.56731721
33.505  3.46852924  0.56731375
33.506  3.46852924  0.56731029
33.507  3.46852925  0.56730682
33.508  3.46852925  0.56730336
33.509  3.46852925  0.56729989
33.510  3.46852924  0.56729643
33.511  3.46852924  0.56729296
33.512  3.46852924  0.56728950
33.513  3.46852924  0.56728603
33.514  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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