[Eternal Student (OP)]

Hi.

   Who doesn't love a short puzzle?    [Approximate time:  4 minutes,  Calculators and Google allowed].

    Suppose you have a rope that was long enough, while on the ground, to go around the circumference of the earth but there was only a few metres to spare.
   Now lift the rope 1 metre above the ground everywhere.  How much extra rope would you need to go around the earth?

[You can ignore hills and velleys and assume the earth is a sphere.   You can look up whatever else you may want.]



Best Wishes.

[Halc]

Now lift the rope 1 metre above the ground everywhere.  How much extra rope would you need to go around the earth

Spoiler: show

Not a function of what it goes around, so same answer as going around a point: 2π meters.

Not entirely intuitive, but quickly worked out. It's also an old problem I've seem many times before.

[paul cotter]

Yep, seen that before. I did not believe it until I worked it out myself. 2pi(r+1)-2pir=2pi

[Eternal Student (OP)]

Hi.
    That was a fast reply @Halc   and  @paul cotter .

   It is an old puzzle.   Making up new puzzles is difficult.   A good puzzle has something in it that seems relevant like planet earth or a fluffy cat and still doesn't take an entire page to describe.
   The only thing this puzzle lacks is political correctness.   You don't have to assume the earth is a sphere.  If it's a plane with air on one or both both sides, you need less rope.

Best Wishes.

[paul cotter]

A fast reply=boredom. Despite my advanced age I have no bodily complaints except one-sensitivity to cold. It is way too cold in my workshop/lab/mancave to do anything at all. Similarly in the garden. I really should be hibernating. PS: a belated happy new year to Eternal Student, Halc and the rest of the gang.

[Eternal Student (OP)]

Hi.

    Here's another short puzzle because there isn't always a lot for you (the reader) to do in a forum post.   This one has also been tried and tested.   It's short to describe and shouldn't take all day.

Consider two circles, A and B.   Circle B has four times the radius of circle A.



Now allow circle A to roll around the perimeter of circle B and return to where it started.    How many times will circle A have rotated?

(Note:   This is a proper roll,  the two circles are always in contact and there is no slip between them.   Put gear teeeth on the two circles if you want).

Bonus Question:   If you've seen this problem elsewhere, that's OK.   Consider what happens if the big circle B is replaced with a Mobius strip.   It's a thin loop with the same radius as B but it has one twist in it as usual (see diagram below).



Some hints or suggestions you may want for the Mobius strip:

Hint 1:

Spoiler: show

Assume "the twist" happens over a small region rather than being a gentle twist over the whole of the strip.   In this way the small circle always rolls in the x-y plane except for a negligable amount of time where you may see it edge on and rolling partly into the z-axis.   You can also consider a gentle twist but you'll need more words for your answer.

Hint 2:

Spoiler: show

 Recall that the small circle must make two passes around a Mobius strip to return to where it started.

Hint 3:

Spoiler: show

 Consider if the small circle always rolls in the same way.   For example, is it always turning anti-clockwise?

Hint 4:

Spoiler: show

 Now that you've sorted out Hint 3,   does it matter if you only see the "tails side" of a coin instead of the "heads side" while it is rotating?

Best Wishes.

[Halc]

Now allow circle A to roll around the perimeter of circle B and return to where it started.    How many times will circle A have rotated?

Spoiler: show

5, 8 for the mobius

This one is super easy because it's been in the news. It (not the mobius variant) appeared on an SAT question that everybody gets wrong because the correct answer wasn't one of the options.
That seems impossible. Surely the test had one of the 4-5 multiple choices as the desired answer, even if incorrect. A percentage of people must have guessed it by accident, so the article title saying that everybody gets it wrong was, well, gotten wrong.

And no, I've yet to look at the hints


My favorite puzzles have often been just geometric.
My fave 5 piece puzzle: Arrange the 5 pieces into a square.
I also have a 'worlds hardest' 4 piece puzzle, both easily cut out of a piece of cardstock.


5piece.png (6.32 kB . 338x228 - viewed 466 times)

Yay Microsoft paint, crude, but it let me draw that in under 2 minutes.

[Eternal Student (OP)]

Hi.
    Well done.  Record time, maybe.

   

This one is super easy because it's been in the news.

    Yes, I did say   "this one has been tried and tested".   However, you've got to build people up to getting the right answer for the Mobius strip.

     See also my hints 3 and 4   for the Mobius strip and note that if the small circle is not considered to have a head and tail side, so there is only some rotation of circle that you can see from a fixed point of observation,  then you can get a different answer.

Best Wishes.

[Eternal Student (OP)]

Hi.

I've not seen your ( @ Halc ) puzzle before but Google is pulling up many solutions with videos involving physical pices of wood.   That's faster than trying to cut the bits yourself from a piece of paper.
   I won't post any of those videos here but people can go and find them on their own if they're having trouble or (like me) too lazy to use scissors.

Good stuff:
    It's a nice puzzle, thank you.   Seems to be a suitably physical challenge.

Bad stuff:
     Most of the pices look like you only need to follow a diagonal through the big square and mark some half-way points along the square (and you could do that).   However, for the small square, I'm not sure how high up the big square it is supposed to go.   If I'm right,  the diagonal of the small square piece is  1/2 the length of the side of the big square.   So the side lengths are then an irrational number of units of anything you can easily see in the diagram.    You almost have to tell the people exactly where the small square is going to go so that they could mark up a piece a paper and cut out the small square easily.

Best Wishes.

[Halc]

Most of the pices look like you only need to follow a diagonal through the big square and mark some half-way points along the square (and you could do that).   However, for the small square, I'm not sure how high up the big square it is supposed to go.   If I'm right,  the diagonal of the small square piece is  1/2 the length of the side of the big square.

Right you are. And all angles are multiples of 45 degrees, despite any error in my drawing it.
It was all just eyeballed. I drew it freehand and hoped the dimensions would come out right.
I think the little square is still slightly too big.

If you use a you-tube that shows the solution, you miss all the fun of doing it yourself.
Cut the pieces out of cardstock, and yes, the square is the same size as the point of the convex pentagon piece you see there, having size the same length as the shorter sides of that pentagon.

I think I got close enough that you can just print the picture and cut that scissors that out.

[paul cotter]

On Eternal Student's second puzzle I just don't get it, maybe i'm stoopId!! Mark the small circle at the contact point with the large circle and rotate until that point again touches the large circle: at 90degrees it will contact the large circle again and again at 180, 270 and 360degrees- hence 4 rotations to return to starting position. What am I missing? For the moebius strip 8 rotations seem quite logical, to me.

[Halc]

Mark the small circle at the contact point with the large circle and rotate until that point again touches the large circle: at 90degrees it will contact the large circle again and again at 180, 270 and 360degrees

Agree that the starting 'contact point' touches 4 times, not 5. But at the 90 degree point it has taken  1 1/4 turns, not just one turn. What was the bottom of the coin is not on the bottom anymore, but facing to the side at the contact point.

It's easier to see if the two circles are the same size. Take two identical coins and roll one of them around the other. Half way (the 180 mark), the moving coin will be facing up again, having completed one full rotation. It rotates one more time that the size ratio of the circles.

Didja have a go at my 5 piece puzzle? I'm not posting the 4 piece one if nobody wants to try the 5 one.

[paul cotter]

Yep Halc, I misread the intention of the question. This is a problem I have had all my life, frequently in exams, where a brief reading leads to a misplaced confidence that I have it "cracked". I will have a look at your puzzle.

[Eternal Student (OP)]

Hi.

a brief reading leads to a misplaced confidence that I have it "cracked".

     According to the news reports,   a question very much like this was put out to school students in America for their SAT exam.    You're in very good company if you mis-read or misunderstood it.    Apparently almost everyone did exactly the same, including the people who set the exam because the mutiple choice answers didn't include "the right answer".   Actually, it's still debated if "the right answer" is something you could really give without having a more detailed description of the situation so that you would know precisely where the observer is staying and if they had been allowed to rotate.
    If I had been given the question and only about 1 minute to answer it, I know I would have missed the subtlety and gone on to the next question.

    The explanation for the answer   ( 5 = 4 + 1   ,   where the +1 rotation appears) is something I would try to explain in the following way:

     Circle A appears to be rotate for two reasons:
(i)   It travels some distance, L,  so it must rotate   n times   with  n =  L/2π     (for convenience we've set r=1 for circle A).   This is the bit most people recognise.
(ii)   It has just moved in a circular path where you were also forcing a change in orientation of the object when the path curves.   This is the bit most people overlook.

   To simplify the situation, let's just remove the contribution from (i).   Replace circle A with a little man who has feet and can walk.   So they do not need to roll to travel some distance, they just walk.
    Now have the man walk around circle B:




    You should be able to convince yourself that, even though the man is not rolling at all,   at the clock face 9 O'clock position, the man looks to be rotated by 90 degrees.   At the 12 O'clock position they are completely upside down.... etc.....   When they get back to their start position they have made one complete rotation about their own body centred axis.    This is clearly different to having the man just walk along a straight line on the horizontal, where they would not have appeared to rotate at all.

    Now you can see that a rolling circle will have contributions from  (i) and also (ii).   Take a moment and think this over first,  you have enough to do the first question.

    For a rolling circle you can then go on to consider another complicating issue.     If circle A rolls on the outside of circle B so that the centre of its mass is travelling clockwise around B,  then circle A appears to rotate clockwise about its own body centred axis.    However if circle A had been rolling on the inside of circle B, then it appears to rotate anti-clockwise about its own body centred axis.    So this means you need to  +/-  1 rotation depending on whether circle A is rolling on the inside or outside.   Spend a moment drawing diagrams or mental images for this.

    Now, you're ready to handle the Mobius strip.  Consider it just as a simple loop of material with a negligible region where the twist happens.    The small coin does one complete circuit around the inisde of that loop and one complete circuit around the outside.    The total number of times the small circle rotates around its own body centred axis is then   (4 +1) + (4-1) = 8.

Best Wishes.