Take a piece of paper and cut out a perfect circle with diameter 1 inch.
What is the diameter of the largest unaltered coin which may be passed through the hole without tearing it?
(Consider the coin to be very thin, and the paper to be very flexible and tear resistant, but not at all stretchy.)
Assuming you start with a fold that would extend through the center of the hole, forming a semicircular arc, and then subsequently fold the paper going though points along the circumference so as to halve each previously formed arc (thus at each stage forming 2^n smaller arcs), you get a serrated edge that at the limit will approach a straight line with the same length as the semicircumference of the original circle, then at each stage:
As each new set of folds takes place, along the midpoints of the previous arcs, the number of arcs doubles, and the size of each one halves. The chord length will be sin A/2, where A is 180 degrees at stage zero and halves at each stage. There will be 2^n chords, so at each stage the total of all the chords will be 2^n * sin 180/(2^(n+1)).
The total chord length (distance from cusp to cusp to cusp ...) at different stages will be:
1 1.414214
2 1.530734
3 1.560723
4 1.568274
5 1.570166
6 1.570639
7 1.570757
8 1.570786
9 1.570794
10 1.570796
11 1.570796
12 1.570796
13 1.570796
14 1.570796
15 1.570796
16 1.570796
17 1.570796
18 1.570796
19 1.570796
20 1.570796
21 1.570796
22 1.570796
23 1.570796
24 1.570796
25 1.570796
As you can see, depending on how many folds you can make (2^n) the closer you can get to a coin with diameter pi/2. My initial guess had stopped after the first two folds (n=1) after the initial fold (n=0). I doubt that you can really get much past n=2 with real paper.
This is similar to starting out with a metal semicircle with a hinge in the middle. Swing out the half arcs until the endpoints and the hinge are in a straight line. Then install new hinges at the midpoints of all the formed arcs and continue from there.
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Posted by Charlie
on 2006-05-30 11:21:16 |
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