20050904, 17:00  #1 
Aug 2005
Brazil
2×181 Posts 
Easy questions
1 A man buys bottles of wine for $1000. After taking 4 bottles and lifting the price of the dozen up $100, he sells it again for the same price. How many bottles did he buy?
2 How many 10cm diameter balls can you fit into a cubic box (1m dimensions)? 
20050905, 13:05  #2  
Aug 2005
Brazil
16A_{16} Posts 
Quote:
Guess it's the latter 

20050905, 13:34  #3  
"William"
May 2003
New Haven
2^{2}·3^{2}·5·13 Posts 
Quote:


20050905, 23:39  #4 
Jan 2005
Caught in a sieve
394_{10} Posts 
1.
For some reason, I went at it the long way: p*n = 1000 n = 1000/p (p+100/12)*(1000/p4) = 1000 (p+100/12)*(1000/p4) = 1000 1000+100000/(p*12)4*p400/12 = 1000 100000/(p*12)4*p400/12 = 0 100000/124*p^2400*p/12 = 0 48*p^2400*p+100000 = 0 p = (400+/sqrt(160000+4*48*100000))/96 p = 4800/96 = $50 per bottle So the answer is: n = 1000/50 = 20 bottles 2. First, the answer is >= 10*10*10 = 1000. Method 1: 1st bottom row: 10 across Now, the second row fits in spaces creating equilateral triangles. The centers of these balls are sqrt(3)/2 from the centers of the first row. Consider that at each end, half a sphere is required, but the rest can be filled in by triangles. Effectively, the first row takes 10cm, and the subsequent rows take ~8.66cm. However, every other row can only hold 9 balls. So: 90/8.66 ~= 10.39 > 10, so one can fit 11 rows. That's 6*10+5*9 = 105 balls on the bottom. If we stacked these vertically, one could get 10*105 = 1050 balls in the box. It is probably possible to do better, by nestling balls in subsequent layers in the triangles created by the layer below. Method 2: But first, let's place the balls in vertically equilateral triangles. Then the second layer takes ~8.66cm but each row must be the opposite size of the row below it. So this layer holds only 6*9+5*10 = 104 balls. We know this way we can get 11 layers, totaling 6*105+5*104 balls = 1150 balls. Method 3: Back to the balls in the triangles. Such a configuration of 4 balls forms a tetrahedron. According to Mathworld, the height of this [url=http://mathworld.wolfram.com/Tetrahedron.html]tetrahedron[/ a] is: 1/3*sqrt(6)*10cm ~= 8.165cm. So 90/8.165 ~= 11.02 > 11. So one can fit 12 layers in the box this way! Let's find out if we could fit one more row on the end of the second layer. The center of the balls on that last row would be offset by only a small amount. That amount is d on the Mathworld "bottom view" diagram. bad diagram: _ d_\ 5 30 degrees d/5 = tan(30) => x = 5 tan 30 = 1/3*sqrt(3)*5 ~= 2.89cm. The space available ~= 3.9 cm, so it works! This means each layer will be just like the respective layer in case 2, only some will be shifted over by 2.89cm. 12 layers = 6*105+6*104 balls = 1254 balls. This has been proven to be the most efficient packing for an infinite size, but for a finite size there may be more efficient packings. I believe this is an open question. Last fiddled with by Ken_g6 on 20050905 at 23:40 Reason: Removed column spacer 
20050906, 00:17  #5 
Jul 2004
Potsdam, Germany
33F_{16} Posts 
Assuming the balls can be pressed into arbitrary shapes, we only have to take care of the volume. This is ½ d³ or 500cm³.
Thus, 2000 balls (well, cubes by now :wink: ) would fit into the box... Last fiddled with by Mystwalker on 20050906 at 00:19 
20050906, 13:00  #6  
Aug 2005
Brazil
2×181 Posts 
Quote:


20050906, 13:17  #7  
Aug 2005
Brazil
2×181 Posts 
Quote:
Quote:
Code:
O OO Last fiddled with by fetofs on 20050906 at 13:23 

20050907, 01:48  #8 
"William"
May 2003
New Haven
2^{2}·3^{2}·5·13 Posts 
I agree with Ken's solution at at least 1254 balls are possible.
The additional wiggle room in every direction is small, so I doubt that repacking a few boundary layers would allow you to squeeze in an additional ball. I'll try to clarify the packing. Start with 10 balls in a row along one wall on the bottom of the box. Next snug a row of 9 balls next to the 10. Alternate rows of 9 and 10 balls. Looking down on the box, the centers of the balls lie on equilateral triangles, so each row is 5*sqrt(3), about 8.66 cm further along. 11 rows takes up 10 interrow distances from center to center, plus a radius of 5 at each end, about 96.6 cm total. The next layer starts with a row of 9 stacked on the first two rows of the bottom layer. The center of this row lies above the centriod of the equilateral triangle of the balls on the bottom row. The centroid is 1/3 of the way from the base. So the center of this first row of the second layer is 5+5*sqrt(3)/3, about 7.887 cm from the wall. The next rows on this layer have the same interrow spacing, 8.66 cm. Adding 10 interrow spacings plus a 5 cm radius for the last row, we can fit in 11 rows on this layer, too  reaching to 99.49 cm. The first, and all odd layers, have 6*10+5*9=105 balls. The second, and all even layers, have 6*9+5*10=104 balls. The only issue remaining is to verify that 12 layers are possible. This is easily calculated by starting with the equilateral triangle from 3 balls on the bottom, and observing that you can get from the center of any of these balls to the center of the upper ball by going to the centroid then up. Going straight up forms a right triangle, so we have a right triangle with the hypoteneus of 10 cm (2 radii) and one of the legs 10*sqrt(3)/3. Pythagorus tells us the third leg, which is the vertical distance from the center of layer 1 to the center of layer 2, is 10*sqrt(6)/3. about 8.165 cm. Packing in 12 layers will use up 11 interlayer distances plus 5 cm on each end, totaling about 99.815 cm. This is tight. We could shrink this box to 100 x 99.49 x 99.82. I'd be surprised is anyone can wiggle enough to fit another ball. 
20050907, 16:45  #9  
Aug 2005
Brazil
2×181 Posts 
Quote:
Greatly clarified! But how is 8.165*12+10=99.815 

20050907, 17:26  #10  
"William"
May 2003
New Haven
2^{2}×3^{2}×5×13 Posts 
Quote:
Quote:


20050907, 21:53  #11  
Mar 2004
ARIZONA, USA
23 Posts 
???
Quote:
8.165 cm is less than the true interlayer vertical distance. Show us all your caculations to come up with that distance. 

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