Here is a great puzzle from Peter Winkler's new book.
If you liked the first book, you want to get this one as well.

here goes:

is there a closed shape in 3D (i.e. made from a bent metal wire which has both ends connected) whose projections on the axis planes are "trees" - this means that the shadows on the 3 walls that make the side planes of the axis have NO LOOPS.

for example - look at the picture I made (attached). this is an ALMOST solution. the curve (in bold) is closed - i.e. made from a metal wire closed loop, and 2 of the projections (shadows) are "trees" (they have no loops) - but the 3rd projection has the shape of a square and is therefore a loop - which is not good...

if there is a shape like this, find it, if not, prove it can't exist.

cheers,

Nir

Click here to view attached image.

No one???

The closure has to not show in any of the three ortho views? What if it starts as a loop and then is twisted till it becomes a line then bent into a V?

Outside of collapsing the loop - one might expect that loop to exist in a plane and so show up in one of the views.

Jonathan,
I understood nothing - could you draw a picture?

I suspect since it's a closed loop, that in at least one of its planes, it must cast a shadow that's also a closed loop. I'm not sure how to prove that though, and I could very well be wrong.

sorry, but you are wrong...
which means the problem has now been reduced to finding such a shape

Puzzled this over today - still have not found a way to close a loop without creating a plane where the cycle shows.

If this had been a bar bet I'd have lost a lot of money or drinks, certain that it can't be done. Very curious about the shape of the object.

/Tomas

There are of course infinitely many shapes that solve this puzzle.
The simplest solution I know of has only 90 degree angles in it...

N

I'll bet you a small beer that you can't do it.

/Tomas

Tomas,

you are pushing it too far
I will send you an email with the solution...

Nir

The beer is in the mail.

/Tomas

PS The solution is very pretty.

I guess the best way to experiment with this problem is to use some old wire coat hangers.
I think one meter or so will do the job.
And as I said it involves only 90 degree angles, plus it is very symmetrical.
maybe now that Tomas knows the answer he can help me with supplying clues without giving away the answer...

Oh, maybe it would be interesting later to give the 1 of the projections as a hint.
let's see how this develops

Jon's answer was sure convincing to me. No clue.


Jack Shalom

We can also have a bet jack...

I love it!

I thought about that with a friend over coffe, and we finally figured it out.

Just to check we have the same solution, and as a hint to those of you who are trying to solve this:

the first projection look somewhat like


and the other projections,

hmm,

they look "somewhat similar"...


Pierre
--

I have a somewhat different solution but I cant judge if yours is correct.
just upload a picture of the solution or send it to me via email/pm

Sorry for the noise...

My solution doesn't work and you can safely ignore it.

If that was one of the projections, Nir wouldn't have asked for my detailed solution...


(And also thanks to Tomas for pointing this now "obvious" flaw...)

I'll try again.

Pierre
--

It wasn't obviously flawed, Pierre. At first I thought it was obviously correct due to the fact that it _felt_ like your three projections were orthogonal. I quite liked the shape actually. Not as much as the one Nir showed me, though. It's mesmerizing.

/Tomas

Aaaargh. Ug. Someone please tell me the answer before my brain starts dripping out my ears.