Discussion:
A rolling pencil
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Peter
2006-11-30 08:31:14 UTC
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Greetings from Holland!

My new puzzle is inspired by a pencil rolling over my desk. It seems
not to be too difficult, thereforeI would be delighted to receive an
email with your solution.

Have fun with the puzzle (to be found at
www.planet.nl/~p.j.hendriks/ppvdw.htm ) !!!


Peter
Pavel314
2006-12-05 03:12:05 UTC
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Post by Peter
Greetings from Holland!
My new puzzle is inspired by a pencil rolling over my desk. It seems
not to be too difficult, thereforeI would be delighted to receive an
email with your solution.
Have fun with the puzzle (to be found at
www.planet.nl/~p.j.hendriks/ppvdw.htm ) !!!
Peter
You're asking how far one vertex of the hexagonal cross section of your
pencil travels when the pencil makes one complete revolution while rolling
across the desk. I defined six arcs from one vertex of the hexagon to the
next until the given vertex returned to its original position relative to
the others. In the diagram on your website, "A" would go through each of the
six possible positions and return to its original position as the left-hand
terminus of the bottom side.

For an arbitrary side of length s, I get:

( PI * s * ( 4 + (3 ^ .5))) / 3

When s = 0.5 cm, as in your example, this would be 3.001 cm.

Am I at least close to the correct answer?

Paul
Flying Tortoise
2006-12-05 16:41:07 UTC
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Post by Pavel314
Post by Peter
Greetings from Holland!
My new puzzle is inspired by a pencil rolling over my desk. It seems
not to be too difficult, thereforeI would be delighted to receive an
email with your solution.
Have fun with the puzzle (to be found at
www.planet.nl/~p.j.hendriks/ppvdw.htm ) !!!
Peter
You're asking how far one vertex of the hexagonal cross section of your
pencil travels when the pencil makes one complete revolution while rolling
across the desk. I defined six arcs from one vertex of the hexagon to the
next until the given vertex returned to its original position relative to
the others. In the diagram on your website, "A" would go through each of the
six possible positions and return to its original position as the left-hand
terminus of the bottom side.
( PI * s * ( 4 + (3 ^ .5))) / 3
When s = 0.5 cm, as in your example, this would be 3.001 cm.
Am I at least close to the correct answer?
Paul
That may depend on whether you have the correct question! It may have
lost something in translation but I read it as an altogether simpler
problem. Anyway, if you want Peter's comments, you need to email him
directly as requested on the page.
Pavel314
2006-12-06 03:16:32 UTC
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Post by Flying Tortoise
Post by Pavel314
Post by Peter
Greetings from Holland!
My new puzzle is inspired by a pencil rolling over my desk. It seems
not to be too difficult, thereforeI would be delighted to receive an
email with your solution.
Have fun with the puzzle (to be found at
www.planet.nl/~p.j.hendriks/ppvdw.htm ) !!!
Peter
You're asking how far one vertex of the hexagonal cross section of your
pencil travels when the pencil makes one complete revolution while rolling
across the desk. I defined six arcs from one vertex of the hexagon to the
next until the given vertex returned to its original position relative to
the others. In the diagram on your website, "A" would go through each of the
six possible positions and return to its original position as the left-hand
terminus of the bottom side.
( PI * s * ( 4 + (3 ^ .5))) / 3
When s = 0.5 cm, as in your example, this would be 3.001 cm.
Am I at least close to the correct answer?
Paul
That may depend on whether you have the correct question! It may have
lost something in translation but I read it as an altogether simpler
problem. Anyway, if you want Peter's comments, you need to email him
directly as requested on the page.
Thanks, I'll do that. I noticed an error in my solution, should be

( PI * s * ( 4 + 2 * (3 ^ .5))) / 3

I dropped one of the square roots of 3.

Paul

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