Thread: Easy questions View Single Post
 2005-09-05, 23:39 #4 Ken_g6     Jan 2005 Caught in a sieve 18B16 Posts 1. For some reason, I went at it the long way: p*n = 1000 n = 1000/p (p+100/12)*(1000/p-4) = 1000 (p+100/12)*(1000/p-4) = 1000 1000+100000/(p*12)-4*p-400/12 = 1000 100000/(p*12)-4*p-400/12 = 0 100000/12-4*p^2-400*p/12 = 0 -48*p^2-400*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 2005-09-05 at 23:40 Reason: Removed column spacer