 Friday, January 20, 2012

Great Packing Trick

Holidays are over and right now is about time to pack decorations into the boxes and shelve them somewhere for 11 long months. Assume that you have 40 identical plastic Christmas baubles that you carefully placed into a flat box like the one sketched below (top view). They fit tightly inside this box.

Suddenly you find another bauble from the same set. How to squeeze this bauble into the same box? Box has a height of one bauble, so no layering.

This puzzle is adapted from a Russian-language journal Kvant.
Top image by Chris_J, distributed under CCL.

Your thoughts and ideas accepted any time until midnight on Sunday Janury 22nd(EST), on our Family Puzzle Marathon. They will be hidden till then and everyone who submitted something reasonable will get a puzzle point.

Anonymous said...

alternate rows of 5 with rows of 4 to tighten up the nesting and thereby add an extra row with the 41st bauble

TMS math said...

3 suggestions: 1)Cut it in half and fit each half over another bauble. Then glue it back together next season. 2) Fold all of the baubles up because they are paper not glass. 3) Break up the bauble and sprinkle it among the other baubles. These are the answers from an 8th grade math class.

Tom said...

(Think outside the box....)
Break the ornament into small pieces?
Melt it down?
No, it's just 2-d sphere packing, and the solution minimizes empty spaces. My math's not advanced enough to prove this, but I sense that if the rows (from the bottom) were 5-4-5-4-5-4-5-4, then there will be room for ninth row of 5. Voila, 41. It's counterintuitive, because of all the new empty space that will appear at the ends of the 4-wide rows; but I'm simply going to bet this works, since nothing else will work.

Ooh, wait...I was thinking we might change the shape of the box...that perimeter could become roomier if the sides were curved. But no, a box has a top and a bottom, so this scheme is probably excluded.

Bean said...

Take one ball from the 2nd, 4th, 6th, and 8th row. Now that each of these rows has four balls, you can move the balls over to wedge in the gaps between the five balls in the rows above and below. Once you've done this for all the rows, you will have moved the whole shebang up exactly enough to allow one more row of balls to wedge in the bottom. This will be a five ball row...and you have the extra ball plus the four you pilfered from the original rows. Wedge them in there, and you're set!

Bean said...

Take one ball from the 2nd, 4th, 6th, and 8th row. Now that each of these rows has four balls, you can move the balls over to wedge in the gaps between the five balls in the rows above and below. Once you've done this for all the rows, you will have moved the whole shebang up exactly enough to allow one more row of balls to wedge in the bottom. This will be a five ball row...and you have the extra ball plus the four you pilfered from the original rows. Wedge them in there, and you're set!

Jerome said...

What a really interesting puzzle. The answer is to pack 5 in the top row 4 and do this alternately, the odd numbered rows have 5 and the even 4 until you get to NINE rows (not eight) you will have a total of 41 bulbs in a box that originally had 40.

Each of the 41 balls is a member of an equilateral triangle. The width of the box is 10 radii of the Christmas decorations. The height is 16 radii. The way I am packing it will result in a width of 10 still but the length needed has been reduced to 15.856 radii.

The hard part is describing how to do this.

Start at the upper left hand corner. Pick the first two on the top row and the first one in the second row and the first two in the 3rd row. What you have looks like the 5 on a die. (dice)

Going from the upper left to the lower right of these 5 from center to center gives a hypotenuse of 1r + 2r + 1r = 4r. Now draw a line from the center of the 5th bobble to the one immediately to its left. That's 2 radii. The height goes from the 3rd row to the first row. We are interested in the height.

h^2 = (4r)^2 - (2r^2)
h^2 = 12r^2
h = 3.464

There are 4 such drawings that can be made. The total height so far is

4*3.464 = 13.856

There is a radius from the top row to the edge of the box and one from the ninth row to the bottom of the box. Total 13.856 + 2 = 15.856. Just enough room for the best Christmas card received that you'll want to hang up next year.
========
I told this to my wife (not a mathematician at all) who said (and I quote)

"I would have been a whole lot more interested in this had they been round chocolate Santa Clauses. Then I would have eaten the 41st one."

By her logic, perhaps we should just break the last ornament and forget about putting in in the box.

Bean said...

OK, so here's the math on that. In the current configuration, if the radius of an ornament is R, you need the box to be 16*R long to hold the ornaments. So that is the length of the existing box.

So the question is, when you wedge the ornaments in so each row fits between the balls in the row above it, will you be able to fit a whole extra row of balls? The answer is yes.

In the new configuration, we need to know the distance/height between the center of the balls in each row to know whether they fit. To find this, we can look at set of three balls where two are in the row above, and the third is wedged between them. The centers of these balls form an equilateral triangle with sides 2*R. The height of this rectangle is the distance between the centers of each row. Apply Pythagoras, and see that the distance between each new row's center is (the square root of 3)*R.

So the new configuration requires a box of the length: 2*R (to account for the distance from the center of the balls at the top and bottom to the edge of the box) + 8 * (square root of 3) * R. This is 15.86*R, which is less slightly smaller than the size of the original box. So it fits.

Ilya said...

In the answer below I will use "height" and "vertical" in terms of the diagram in the puzzle, not the actual box where everything is in one later.

Intuitively, we can try to alternate each row between 5 and 4 baubles, such that the next row's items rest halfway between the previous row's thus taking less "vertical" space. If we can save up enough room for another row that way, we would get 5 rows of 5 and 4 rows of 5, for the total of 41. Here is the strict proof that we indeed can do that. If a bauble's diameter is D, then the total "height" that N rows take is D + (N-1)*H, where H is the distance between the line passing through the center of one row's baubles and the same line for the row above. Since we are dealing with all equal spheres (or circles in projection), "H" is in fact the height of the equilateral triangle of size D. Using the half of this triangle, we can compute H with the help of the dear old Pythagoras: H*H + D*D* = 2D*2D, from which we get H = D*SQRT(3)/2. For 9 rows therefore, the total height of all rows is 7.928*D. Since the original arrangement held 8 rows, it was at least 8*D, which means we've just squeezed the extra row for the total of 41, i.e. one more than the original 8*5=40.

Anonymous said...
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Anonymous said...

I was thinking to rearrange the baubles in hexagons to gain some volume in order to place one more bauble in the box.

anne-marie

Anonymous said...

I get some times to elaborate my answer a little so I did.
If the baubles are in square formation, the perimeter is 5*8=40
The densest way to pack , in this case, is with a hexagonal arrangement because it reduces the area and perimeter of the rectangle.
My arrangement is 5 baubles in the first row then four in the second and I altern 5, 4 to the end with a total of 41 baubles.
The perimeter if I consider that the diameter of a bauble is 1 will be
2 n + square root of 3*k + 3 - square root of 3
With k=5 and n=9
The area is ( n+1/2)(square root of 3 * k +1- square root of 3/2)
I calculated the perimeter to be 27.93
Anne-marie.

Maria said...

Breaking into little pieces, melting, cutting in half, ouch!
The non-violent solution wins.
As most of you suggested a slight re-arrangement does a great trick. And if the great descriptions above are not enough, here is my graphical solution. Press "back" button to return to the blog after viewing it. Click to see the graphical solution.

Maria said...

Jerome - I have been thinking about your wife and imagining her eating a round chocolate Santa Claus that is not a trivial undertaking. I always like prolonged shape chocolate Santas because while you munching on the head you can still hold on to the wrapped legs and perhaps even leave them for later. With a round shape I guess you need to break them first like Kinder Egg and then finish the pieces in one meal or share...

TMS - as your 8th graders seem to be in a breaking mood, perhaps they can solve this puzzle: Dangerously Blind.

TMS math said...

I will have them try it this week. Thanks for the new puzzle. I am trying to get them to think out side the box because of the new curriculum we will be getting in a few years.

Anonymous said...

I solved this not because I pack ornaments but from using the same strategy when baking cookies: rather than uniform rows and columns of dough, 5 X 4 (20 cookies/sheet), I alternate 4-3-4-3-4-3 and therefore fit 21 cookies on a sheet.

TMS math said...

My students enjoyed the "Dangerously Blind" puzzle. They came up with some interesting answers, but with some coaxing we got to the correct answer. Thanks for the puzzle. Another group will be trying the puzzle from today.