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what the force of many cannot
A knotty problem.by BrowserUk (Pope)
|on Sep 29, 2003 at 01:54 UTC||Need Help??|
BrowserUk has asked for the
wisdom of the Perl Monks concerning the following question:
This is a challenge for a slightly practical purpose -- golf it if you want, though that's not my goal -- I just want the problem solved.
If you take a long narrow strip of paper or plastic, that is both flexible and strong -- a well flattened drinking straw is ideal -- and you tie a simple knot -- form a single loop and pull one end through -- and then pull it tight without crushing the width of the strip, you'll end up with a pentagon.
Excuse the ascii art. Take your strip, that should be at least 8 times as long as it is wide, an bend the ends, back, around and up until the ends cross. Take the end that is furthest from (tends to vary upon whether your left or right-handed) and bend it forward and over the other end and down, tucking the end through the loop. Pull it tight. It will take some juggling to get everything tight and flat, but when you've done it, you see that if you trimmed the two ends back to the knot, what would remain is the pentagon, with the five sides the same width as your strip.
Now, if you lay the knot flat on the desk, you may be able to see that if you had 5 such knots and you positioned them correctly, you will end up with a larger pentagon. If you juggled the ends into each other, you can get them to hold together. And if you trim the ends fairly short before doing so, you would end up with the knots touching each other and the whole in the middle will also have 5 sides, with the vertexes of the inner pentagon diametrically opposed to the flats of the outer one.
Now the perl. Supplied with the diameter of circle which the inner edge would encompass, determine the width and minimum length of strip required to form the complete 5 knot pentagon.
From my experiments, it is near impossible to actually tie this. However, by leaving a larger gap between the knots, it is possible to have an outer pentagon with an inner decagon. So, also calculate the minimum length of strip required to achieve this with the inner decagon being regular. Ie. with all 10 inner edges the same size.
Examine what is said, not who speaks."Efficiency is intelligent laziness." -David Dunham
"When I'm working on a problem, I never think about beauty. I think only how to solve the problem. But when I have finished, if the solution is not beautiful, I know it is wrong." -Richard Buckminster Fuller
If I understand your problem, I can solve it! Of course, the same can be said for you.