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 Forum index » Diversions » TimeWasters
A LITTLE TIMEWASTER SET
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sixsidedsquare
Unfettered

Joined: 24 Mar 2005
Posts: 409
Location: 60E
la!uep wrote:
Spoiler (Rollover to View):
(250*3500 - 17)/10^6 = 0.874983 ==> tt0874983

Worshippy


I had forgotten how much I enjoy trig geometry that involves angles in the order of 2pi/3 Very Happy

Getting rid of those icky decimals, what you're wanting for a square based pyramid (albeit upside down) is 4 equilateral triangles that have a side of length √3/2 + 1, and a height (middle of a side to opposite corner) of √3/2 + 3/4. See why I love order 2pi/3 trig? Very Happy

So of our six sides of the original box, the easiest way to make these would be to take the two square faces, cut them in half and stick them to the ends of the four 1.5x1 inch faces. This would give you four 1 inch by 1.96875 inch boards to play with. Now you're going to have to cut each these diagonally at 60 degrees to make the two halves of the four triangles, and a cut at 60 degrees will remove 0.125 inches of length from it, but they're more than long enough to handle that.

Trim these boards down to √3/4 + 1/2 inches (≈0.933") by √3/2 + 7/8 inches (≈1.741"). Take the saw and cut starting into the long edge with the corner directly on the side of the blade and cut at 60 degrees to the short edge, and you should reach the opposite corner with that corner just on the other side of the blade. This will give you the two halves for each of the 4 equilateral triangles, so you can now stick it all together.

That pretty much solves the requirements of the problem I think, unless I missed something or messed up some of my math somewhere in there, but lets see what's left over. From the trimming, you should have been left with four 1 inch by 1.03125 - √3/2 inch (≈0.165") bits and four well, toothpicks. I think you'd be best to try and fashion some sort of prop to hold the upside down pyramid in place, because although it technically fits them all, good luck getting it to stand up without toppling over. I considered giving it a small square base for a bit (1-√3/2 inch sides if you were wondering, which neatly make the edge lengths of the large side pieces become a round 1.5 inches), which you could easily do by sticking those left overs together and lopping the bottom corners off the large triangle sides.

So yeah, did I do that right? There's bound to be a few errors in there. Sorry about the lack of pictures, I'll scan one of my pages of math scribbling later for fun. If someone else doesn't jump in I'll post again tomorrow explaining those awesome hints. Oh, and good luck to you gluing together infinitely thin pieces of stuff Razz

PostPosted: Mon Apr 06, 2009 8:13 am
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la!uep
Boot


Joined: 26 Mar 2009
Posts: 69
Rogi Ocnorb wrote:
Is there a rule that says the marbles must be removable from the box without breaking the box?

Well, I'd sure like to take out my marbles once in a while and admire them or show them off. But see below for a simple suggestion to your idea.

Rogi Ocnorb wrote:
If not (and while I'm pretty sure I lack the ability to do the math correctly), I can say that my best guestimate for the required size of a 4-sided base pyramid box is 2.372x2.372 for the base and 2.031 for the height, following the rules given.

If the box doesn't have to be completely closed, you can take .254 inch off the top. But again, you'd have to break the box the get the marbles out.

So (if I understand the proposed configuration correctly) why not invert the pyramid and take out the square base, which would now be the unnecessary lid? I'm not sure about your math since I hadn't worked out this configuration, but after working on it for an hour or so I think the pyramid could be smaller than you say. With the dimensions you give I think there would not be enough material (you only have 8 sqin of material to work with, and that's without considering the sawdust waste).

But you'll have to do the math/geometry/drafting work or get someone to do it for you. Show me the six original sides getting cut into shapes that can then be joined into the 4 triangles (or trapezoids if you cut the tip of the pyramid off) required for the new box. Remember each cut wastes a path 1/16" thick worth of material.

Rogi Ocnorb wrote:
ETA: If there is a requirement that the box allow the marbles in and out without being damaged, I'm sure there is enough material in the top and square sides (blue lines, below) to create two new pyramids to add to the ends (yellow-ish lines). Or even just straight-up walls.

Even though I think understand this new configuration of the marbles, it's not clear to me how you resolve the floor and the sides around the two small pyramids, which must still rise up to the 1" height of the box. Perhaps there is enough material, I don't know, but I need see a plan for how the cuts are made, and the pieces joined into whatever shapes would be required.

The original puzzle had a requirement that the new box must be a regular box-shaped box with rectangular sides, to keep the geometry/math relatively simple. The intended solution (the one hinted by LEGRAND) is the one I liked most, but I found another solution which also uses a regular box shape.

I decided to remove this condition here because I suspect there may be a myriad solutions with oddly shaped boxes to exploit the very interesting geometry of sphere packing, and I wanted to see if they would work. I definitely thought some of you here would come up with some very creative ideas, and here you are with not just one but two potentially valid new solutions. But for these unique boxes, as you can already tell from yours, the geometry is quite complex and I'm not going to work it all out all by myself. Cool

Did you get anywhere with the clues?

PostPosted: Mon Apr 06, 2009 8:27 am
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la!uep
Boot


Joined: 26 Mar 2009
Posts: 69
sixsidedsquare wrote:
I think you'd be best to try and fashion some sort of prop to hold the upside down pyramid in place...

Ah I see you're onto it already...

Nice working it all out... I'd still love to see it! But, do try to the one for a straight shaped box which is much easier to calculate... at least so that the clues don't go to waste? Cool

PostPosted: Mon Apr 06, 2009 8:36 am
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sixsidedsquare
Unfettered

Joined: 24 Mar 2005
Posts: 409
Location: 60E

 		Actually, on the walk back fro uni I suddenly became unsure of my previous stated solution. The way I was working on paper was fitting circles into triangles, but although this work ok horizontally with the right angles the spheres are to each other, I believe my solution as stated doesn't actually reach over the tops of the spheres. 3D geometry just adds a whole other headache. Let me think this over.

PostPosted: Mon Apr 06, 2009 8:55 am
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retsbob999
Boot

Joined: 09 Jul 2008
Posts: 64
Location: Rock Hill, SC

 		I think that I know the final dimensions and the orientation of the box, but can't figure out the cuts to get there.

Spoiler (Rollover to View):
I believe that the final dimensions of the box are 1.2071"x1.2071"x1.2071". The orientation of the box is standing on one of its points. one marble sits in the corner that's facing down, four are on top in a 2x2 square, 4 more are on top where each marble is sitting in the gaps between the four marbles underneath, four more marbles on top in the same orientation as the first set of four marbles, then 1 more on top.


Spoiler (Rollover to View):

In this image, the BLACK backgroun represents the box dimensions, the RED represents the 1 st , 3 rd , and 5 th rows, the YELLOW represents the two identical rows (2 & 4), the BLUE represents the marble in the corner facing you (part of 3 rd row), and the BLACK X'S represent the fact that another marble is hiding behind that marble. I hope the image makes sense.

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PostPosted: Mon Apr 06, 2009 12:06 pm
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GreenWindmill
Decorated

Joined: 21 Apr 2006
Posts: 195
Location: Midlands, UK
retsbob999 wrote:
I think that I know the final dimensions and the orientation of the box, but can't figure out the cuts to get there.


That sure is pretty! Very Happy

Just one thought though - I'd got the impression from previous posts that the finished structure would be lidless and allow access to the marbles, if I understand your proposal correctly this wouldn't be the case. I haven't properly looked into this but I see no reason not to allow your cube to lay on its side and be open topped?

All of the above is to be taken with a (very) large pinch of salt since although I'm pretty competent at maths/trig/etc. I haven't had time to properly analyse the problem!
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PostPosted: Mon Apr 06, 2009 12:35 pm
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retsbob999
Boot

Joined: 09 Jul 2008
Posts: 64
Location: Rock Hill, SC

 		Thanks GreenWindmill, I forgot to mention that one of the upper three sides is lidless. I do not believe this orientation would work on it's side because gravity would cause the marbles to fall out. I guess I'll have to wait until la!uep looks at it.

EDIT: I'm starting to doubt my design now... my layout only has 6 corners and a cube has 8. Not sure anymore. Oh well, worth a shot.

EDIT2: My design is actually two square based pyramids, one of top of the other.
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PostPosted: Mon Apr 06, 2009 1:13 pm
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la!uep
Boot


Joined: 26 Mar 2009
Posts: 69

 		Aw, don't quit now retsbob999! Just look at your image and try to describe it correctly.

And yes, just let the cube rest on it's side like GreenWindmill said. No need to do a balance act. The marbles won't fall off because I hope you'll leave open the top side, right? right??? Wink

retsbob999 wrote:
EDIT2: My design is actually two square based pyramids, one of top of the other.

Are you sure? How many marbles would the two pyramids hold? Perhaps you can share how you made that pretty image? Doing that might help you explain it to us.

Six, I think you're pretty close, but I agree about the 3D playing tricks. I bet it's possible, in which case I'm elated that Rogi and you may have found a third solution.

And no one likes LEGRAND and IVPITER... Crying or Very sad

PostPosted: Mon Apr 06, 2009 3:46 pm
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retsbob999
Boot

Joined: 09 Jul 2008
Posts: 64
Location: Rock Hill, SC

 		I got the design that I came up with by using the LEGRAND and IVPITER cryptograms.

Spoiler (Rollover to View):
LEGRAND informed me that a cube should be used. IVPITER told me to arrange the coded letters into rows of 2 4 6 6 4 2 in alphabetic order. Doing this gave me this picture . Maybe it will be able to help someone else more than it helped me. But now you can see how I came up with the other picture that I posted.

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PostPosted: Mon Apr 06, 2009 4:06 pm
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Rogi Ocnorb
I Have 100 Cats and Smell of Wee


Joined: 01 Sep 2005
Posts: 4266
Location: Where the cheese is free.

 		People that make those visual connections amaze me. I guess If I'd tried putting the glyphs back in alpha order after figuring them out, I MIGHT have seen it, but probably not.

After putting a lot of time into trying to get my areas to work out, I think it's fair to ask what the layout and cutting guide for retsbob999's method looks like.
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PostPosted: Mon Apr 06, 2009 4:24 pm
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la!uep
Boot


Joined: 26 Mar 2009
Posts: 69

 		That's excellent work Bob. Aced those crypto clues. Now, I think you did see the cube there on your first impression, but maybe you just couldn't see where all the marbles went. The thing is I couldn't show it any better with the alphabet table, but I think it suggested it quite well.

Imagine it as if you're looking it from above: you can tell there are 4 marbles in the bottom corners, but notice there's enough space for a 5th one right in the middle there in the bottom (you painted these red). Then you have 4 more kissing the 4 sides (you painted them yellow with crosses, but there are no marbles directly behind these) so 9 down 5 to go. Well, now just put the last 5 marbles on the top layer just like the ones on the floor. Of these you painted just the center marble blue, and as you can see, if you include the other 4 against the top corners you'd cover all 8 corners (but then these would cover up everything below and you would lose the layering information of the ones in the bottom).

So, you basically solved everything the way it was intended, but then got confused. It's common when things in 3D are projected in 2D.

Rogi, the crypto solved to precisely that, text that told you to put the glyphs in alphabetic order in rows just the way Bob did it. You couldn't have missed it.

As to the cutting, Rogi, are you asking me or Bob? Not sure if Bob's going to give it a shot, but it's not too complicated.

For your inverted pyramid method, I got 4 equilateral triangles with edges 1.966" long, if Pythagoras didn't betray me. I think six's cutting method should work.

Anyone want to see all three solutions or are you all sick of this box thing? Very Happy

PostPosted: Mon Apr 06, 2009 5:30 pm
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la!uep
Boot


Joined: 26 Mar 2009
Posts: 69

 		Ok here are my (pretty crappy) diagrams for the three solutions known so far. Don't look if you want to work them out on your own (and hey, you might find a new solution!):
Spoiler (Rollover to View):
Here's Rogi's creative (but awfully tough to calculate) "out-of-the-box" layout with six's clever cuts.

Here's the actual intended solution for a "straight box" found by retsbob999 (except for the cuts):

And here's an alternative solution for a "straight box" which no one found.

Are there more? Who knows? Was it too hard? Too math-ish? Too many elements with those two clues? Did I give too much away there in the end?

Who's up next?

PostPosted: Mon Apr 06, 2009 7:06 pm
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retsbob999
Boot

Joined: 09 Jul 2008
Posts: 64
Location: Rock Hill, SC

 		Even though I came up with the intended solution (excluding the cuts) by using the cryptogram hints, I think that either Rogi or SixSidedSquare should go next since they did much more actual work than I did.
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PostPosted: Tue Apr 07, 2009 10:00 am
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Rogi Ocnorb
I Have 100 Cats and Smell of Wee


Joined: 01 Sep 2005
Posts: 4266
Location: Where the cheese is free.

 		Time for another quickie.
This one, with a Latin flair.
********************
129 12 712201742161187 415 2215781913121519 2116 1416238 21111220 14820204108 22157819 17122424121512, 118'7 1713468 1221...
********************
Asterisks are not part of the puzzle.
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PostPosted: Tue Apr 07, 2009 5:34 pm
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la!uep
Boot


Joined: 26 Mar 2009
Posts: 69

 		Should that 5th number end in 0 instead of 9?

PostPosted: Wed Apr 08, 2009 10:13 am
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